View Full Version : Battery switching without tears
On one of my flights last year I had to switch between my avionics battery and engine battery when the avionics battery voltage dropped too low (I had left the master on after the last flight and could only partially charge the avionics battery before launching). The switch over seemed to go okay, but then I noticed that my LX9000 was giving me unbelievably short glide distances. It turns out that the QNH altitude had been reset to the altitude at the time of switching. This was unacceptable, so I resolved to do something about it before this season. The simplest solution was to add a capacitor to the avionics power bus. The capacitor supplies power as the power selector switch is moving, and breaking, from the avionics battery, and connector, or making, to the engine battery (this is called a "break before make" switch. But how big of a capacitor to use? The basic equation involved is:
V = I * t / C or C = I * t / V
where V is voltage, I is current and t is time.
Translation: the bigger the capacitor the smaller the voltage drop. If the requirement is to keep the voltage drop to 1 V, the current is 2 A (my situation) and t is 0.1 s, then C = 0.2 F (200,000 μF). The capacitor would also have to be rated for 16 V, min. That is a pretty big capacitor, so I decided I could tolerate a larger voltage drop (4 V), which cuts the size of the capacitor to 50,000 μF. I ended up finding a suitably sized 39,000 μF capacitor rated for 25 V. A smaller capacitor could by used if the current drain is lower, which is likely for most gliders.
https://www.digikey.com/product-detail/en/kemet/ALS70A393DB025/399-14301-ND/6928303
I installed the capacitor yesterday and monitored the bus voltage during switch-over with an oscilloscope, which was anti-climatic: there was no detectable drop in bus voltage. Apparently the bread-to-make time is very short, perhaps a millisecond. Haven't had a chance to fly with it yet, but should be able to soon. The scope waveforms and capacitor installation can be seen at:
https://flic.kr/s/aHsmMo9rN7
Mike N.
April 5th 20, 03:53 AM
Nice, I have the same issue and have been thinking along these same lines.
Mark Morwood
April 5th 20, 07:45 AM
Thanks for the report. I had wondered about doing the same thing. Just curious, did you include a resistor across the capacitor to make sure it discharges when the master is switched off and no loads are left on?
John Foster
April 5th 20, 08:36 AM
On Saturday, April 4, 2020 at 7:23:44 PM UTC-6, 2G wrote:
> On one of my flights last year I had to switch between my avionics battery and engine battery when the avionics battery voltage dropped too low (I had left the master on after the last flight and could only partially charge the avionics battery before launching). The switch over seemed to go okay, but then I noticed that my LX9000 was giving me unbelievably short glide distances. It turns out that the QNH altitude had been reset to the altitude at the time of switching. This was unacceptable, so I resolved to do something about it before this season. The simplest solution was to add a capacitor to the avionics power bus. The capacitor supplies power as the power selector switch is moving, and breaking, from the avionics battery, and connector, or making, to the engine battery (this is called a "break before make" switch. But how big of a capacitor to use? The basic equation involved is:
>
> V = I * t / C or C = I * t / V
>
> where V is voltage, I is current and t is time.
>
> Translation: the bigger the capacitor the smaller the voltage drop. If the requirement is to keep the voltage drop to 1 V, the current is 2 A (my situation) and t is 0.1 s, then C = 0.2 F (200,000 μF). The capacitor would also have to be rated for 16 V, min. That is a pretty big capacitor, so I decided I could tolerate a larger voltage drop (4 V), which cuts the size of the capacitor to 50,000 μF. I ended up finding a suitably sized 39,000 μF capacitor rated for 25 V. A smaller capacitor could by used if the current drain is lower, which is likely for most gliders.
> https://www.digikey.com/product-detail/en/kemet/ALS70A393DB025/399-14301-ND/6928303
>
> I installed the capacitor yesterday and monitored the bus voltage during switch-over with an oscilloscope, which was anti-climatic: there was no detectable drop in bus voltage. Apparently the bread-to-make time is very short, perhaps a millisecond. Haven't had a chance to fly with it yet, but should be able to soon. The scope waveforms and capacitor installation can be seen at:
> https://flic.kr/s/aHsmMo9rN7
I'm no electrician, but wouldn't it work just fine if you wired both avionics and engine battery in parallel, with a switch (on-off) to each battery. That would mean two switches. Turn the engine battery switch on before turning the avionics battery switch off. Or am I missing something here?
Tim Newport-Peace[_6_]
April 5th 20, 10:17 AM
At 01:23 05 April 2020, 2G wrote:
>On one of my flights last year I had to switch between my avionics
battery
>=
>and engine battery when the avionics battery voltage dropped too low (I
>had=
> left the master on after the last flight and could only partially charge
>t=
>he avionics battery before launching). The switch over seemed to go okay,
>b=
>ut then I noticed that my LX9000 was giving me unbelievably short glide
>dis=
>tances. It turns out that the QNH altitude had been reset to the altitude
>a=
>t the time of switching. This was unacceptable, so I resolved to do
>somethi=
>ng about it before this season. The simplest solution was to add a
>capacito=
>r to the avionics power bus. The capacitor supplies power as the power
>sele=
>ctor switch is moving, and breaking, from the avionics battery, and
>connect=
>or, or making, to the engine battery (this is called a "break before
make"
>=
>switch. But how big of a capacitor to use? The basic equation involved
is:
>
>V =3D I * t / C or C =3D I * t / V
>
>where V is voltage, I is current and t is time.
>
>Translation: the bigger the capacitor the smaller the voltage drop. If
the
>=
>requirement is to keep the voltage drop to 1 V, the current is 2 A (my
>situ=
>ation) and t is 0.1 s, then C =3D 0.2 F (200,000 =CE=BCF). The capacitor
>wo=
>uld also have to be rated for 16 V, min. That is a pretty big capacitor,
>so=
> I decided I could tolerate a larger voltage drop (4 V), which cuts the
>siz=
>e of the capacitor to 50,000 =CE=BCF. I ended up finding a suitably sized
>3=
>9,000 =CE=BCF capacitor rated for 25 V. A smaller capacitor could by used
>i=
>f the current drain is lower, which is likely for most gliders.
>https://www.digikey.com/product-detail/en/kemet/ALS70A393DB025/399-14301-ND=
>/6928303
>
>I installed the capacitor yesterday and monitored the bus voltage during
>sw=
>itch-over with an oscilloscope, which was anti-climatic: there was no
>detec=
>table drop in bus voltage. Apparently the bread-to-make time is very
>short,=
> perhaps a millisecond. Haven't had a chance to fly with it yet, but
>should=
> be able to soon. The scope waveforms and capacitor installation can be
>see=
>n at:
>https://flic.kr/s/aHsmMo9rN7
>
Tim Newport-Peace[_6_]
April 5th 20, 11:02 AM
At 01:23 05 April 2020, 2G wrote:
>On one of my flights last year I had to switch between my avionics
battery
>=
>and engine battery when the avionics battery voltage dropped too low (I
>had=
> left the master on after the last flight and could only partially charge
>t=
>he avionics battery before launching). The switch over seemed to go okay,
>b=
>ut then I noticed that my LX9000 was giving me unbelievably short glide
>dis=
>tances. It turns out that the QNH altitude had been reset to the altitude
>a=
>t the time of switching. This was unacceptable, so I resolved to do
>somethi=
>ng about it before this season. The simplest solution was to add a
>capacito=
>r to the avionics power bus. The capacitor supplies power as the power
>sele=
>ctor switch is moving, and breaking, from the avionics battery, and
>connect=
>or, or making, to the engine battery (this is called a "break before
make"
>=
>switch. But how big of a capacitor to use? The basic equation involved
is:
>
>V =3D I * t / C or C =3D I * t / V
>
>where V is voltage, I is current and t is time.
>
>Translation: the bigger the capacitor the smaller the voltage drop. If
the
>=
>requirement is to keep the voltage drop to 1 V, the current is 2 A (my
>situ=
>ation) and t is 0.1 s, then C =3D 0.2 F (200,000 =CE=BCF). The capacitor
>wo=
>uld also have to be rated for 16 V, min. That is a pretty big capacitor,
>so=
> I decided I could tolerate a larger voltage drop (4 V), which cuts the
>siz=
>e of the capacitor to 50,000 =CE=BCF. I ended up finding a suitably sized
>3=
>9,000 =CE=BCF capacitor rated for 25 V. A smaller capacitor could by used
>i=
>f the current drain is lower, which is likely for most gliders.
>https://www.digikey.com/product-detail/en/kemet/ALS70A393DB025/399-14301-ND=
>/6928303
>
>I installed the capacitor yesterday and monitored the bus voltage during
>sw=
>itch-over with an oscilloscope, which was anti-climatic: there was no
>detec=
>table drop in bus voltage. Apparently the bread-to-make time is very
>short,=
> perhaps a millisecond. Haven't had a chance to fly with it yet, but
>should=
> be able to soon. The scope waveforms and capacitor installation can be
>see=
>n at:
>https://flic.kr/s/aHsmMo9rN7
>
Interesting. Many years ago Volkslogger had a similar issue and it is worth
examining their solution. Only the instrument in question (LX9000) needs
protecting, not the entire Avionics Bus.
They added an Electrolytic Capacitor, same as you did. But they also added
a Schottky Diode in series between the instrument/capacitor and the
supply.
The capacitor now only has to maintain the instrument in question and not
everything on the Supply Bus. In your calculation, the value of I is
greatly reduced (0.6A instead of 2A) and therefore the value of C is also
reduced, making for a much smaller capacitor.
Martin Gregorie[_6_]
April 5th 20, 01:15 PM
On Sat, 04 Apr 2020 18:23:41 -0700, 2G wrote:
> On one of my flights last year I had to switch between my avionics
> battery and engine battery when the avionics battery voltage dropped too
> low (I had left the master on after the last flight and could only
> partially charge the avionics battery before launching).
Some years back I had a momentary disconnection problem with the XLR plug
supplying power to my panel: it carries a common ground plus a positive
line from each of the two batteries I carry - one supplies flight
instruments, the other drives radio and T&B.
I added small capacitors (18mm diameter x 35mm long, 35-50v, so 1000-2000
uF capacity or thereabouts for each capacitor - thats a guestimate since
I can't read their markings without major surgery), one for each battery,
and with a 0.1 ohm resistor in series with it.
This solved the problem completely, but my current draw is lower than the
yours: 500 mA for flight instruments and 200 mA with T&B running and
radio on and receiving.
--
Martin | martin at
Gregorie | gregorie dot org
David S
April 5th 20, 02:22 PM
On Sunday, April 5, 2020 at 3:36:09 AM UTC-4, John Foster wrote:
> I'm no electrician, but wouldn't it work just fine if you wired both avionics and engine battery in parallel, with a switch (on-off) to each battery.. That would mean two switches. Turn the engine battery switch on before turning the avionics battery switch off. Or am I missing something here?
That often works, but the engine battery will be fast-charging the panel battery while both are connected in parallel, so you don't want to leave the switch in that position for very long. If there's a 5 amp fuse between the engine battery and the panel, you don't want to blow it.
John DeRosa's excellent tutorial details several good options for the two-battery issue. Starts at slide #95:
http://derosaweb.net/aviation/presentations/documents/Soaring_Electrical_Wiring_Best_Practices.pdf
Cheers,
...david
David S
April 5th 20, 02:29 PM
On Sunday, April 5, 2020 at 9:22:59 AM UTC-4, David S wrote:
> On Sunday, April 5, 2020 at 3:36:09 AM UTC-4, John Foster wrote:
>
> > I'm no electrician, but wouldn't it work just fine if you wired both avionics and engine battery in parallel, with a switch (on-off) to each battery. That would mean two switches. Turn the engine battery switch on before turning the avionics battery switch off. Or am I missing something here?
>
> That often works, but the engine battery will be fast-charging the panel battery while both are connected in parallel, so you don't want to leave the switch in that position for very long. If there's a 5 amp fuse between the engine battery and the panel, you don't want to blow it.
>
> John DeRosa's excellent tutorial details several good options for the two-battery issue. Starts at slide #95:
> http://derosaweb.net/aviation/presentations/documents/Soaring_Electrical_Wiring_Best_Practices.pdf
>
> Cheers,
> ...david
Actually, slides 119 - 123 provide additional options.
http://derosaweb.net/aviation/presentations/documents/Soaring_Electrical_Wiring_Best_Practices.pdf
>
Page 97 shows the caps.
Page 98 shows why not to switch 2 in parallel for longer than it takes a fuse to blow. (Murphry says it will be the fuse on the good battery;-)
Page 120 shows the diodes. Note that the K2's have more volts to start with, so the diode drop is less of an issue.
I run like page 120, except with only one diode. The theory is that with the flat discharge curve of the K2's, I mostly run on the non-diode batt until is gets near empty and then transition to the second K2. It's automatic and I can watch the bus voltage to see how it's doing. SH wires the front battery in parallel with one of the mid batteries, so if the diode is on one of those, it runs with the vanilla ship wiring. There is a second connector on the diode battery for charging. (The second mid-battery gives me a switchable third, but so far haven't needed it.)
I also built an 'ideal' diode (page 121) but decided it I'd rather keep it simple and see the discharge progress in the bus voltage.
John's pictures make this easy. Aside from the one-diode, two-K2 plan, has anybody got another that is not in his collection?
Dan Marotta
April 5th 20, 04:17 PM
Wouldn't a make-before-break switch solve the problem in a much simpler
fashion?* That's what my Stemme uses and switching between main and tail
batteries is a non event.
On 4/5/2020 4:02 AM, Tim Newport-Peace wrote:
> At 01:23 05 April 2020, 2G wrote:
>> On one of my flights last year I had to switch between my avionics
> battery
>> =
>> and engine battery when the avionics battery voltage dropped too low (I
>> had=
>> left the master on after the last flight and could only partially charge
>> t=
>> he avionics battery before launching). The switch over seemed to go okay,
>> b=
>> ut then I noticed that my LX9000 was giving me unbelievably short glide
>> dis=
>> tances. It turns out that the QNH altitude had been reset to the altitude
>> a=
>> t the time of switching. This was unacceptable, so I resolved to do
>> somethi=
>> ng about it before this season. The simplest solution was to add a
>> capacito=
>> r to the avionics power bus. The capacitor supplies power as the power
>> sele=
>> ctor switch is moving, and breaking, from the avionics battery, and
>> connect=
>> or, or making, to the engine battery (this is called a "break before
> make"
>> =
>> switch. But how big of a capacitor to use? The basic equation involved
> is:
>> V =3D I * t / C or C =3D I * t / V
>>
>> where V is voltage, I is current and t is time.
>>
>> Translation: the bigger the capacitor the smaller the voltage drop. If
> the
>> =
>> requirement is to keep the voltage drop to 1 V, the current is 2 A (my
>> situ=
>> ation) and t is 0.1 s, then C =3D 0.2 F (200,000 =CE=BCF). The capacitor
>> wo=
>> uld also have to be rated for 16 V, min. That is a pretty big capacitor,
>> so=
>> I decided I could tolerate a larger voltage drop (4 V), which cuts the
>> siz=
>> e of the capacitor to 50,000 =CE=BCF. I ended up finding a suitably sized
>> 3=
>> 9,000 =CE=BCF capacitor rated for 25 V. A smaller capacitor could by used
>> i=
>> f the current drain is lower, which is likely for most gliders.
>> https://www.digikey.com/product-detail/en/kemet/ALS70A393DB025/399-14301-ND=
>> /6928303
>>
>> I installed the capacitor yesterday and monitored the bus voltage during
>> sw=
>> itch-over with an oscilloscope, which was anti-climatic: there was no
>> detec=
>> table drop in bus voltage. Apparently the bread-to-make time is very
>> short,=
>> perhaps a millisecond. Haven't had a chance to fly with it yet, but
>> should=
>> be able to soon. The scope waveforms and capacitor installation can be
>> see=
>> n at:
>> https://flic.kr/s/aHsmMo9rN7
>>
> Interesting. Many years ago Volkslogger had a similar issue and it is worth
> examining their solution. Only the instrument in question (LX9000) needs
> protecting, not the entire Avionics Bus.
>
> They added an Electrolytic Capacitor, same as you did. But they also added
> a Schottky Diode in series between the instrument/capacitor and the
> supply.
>
> The capacitor now only has to maintain the instrument in question and not
> everything on the Supply Bus. In your calculation, the value of I is
> greatly reduced (0.6A instead of 2A) and therefore the value of C is also
> reduced, making for a much smaller capacitor.
>
>
>
--
Dan, 5J
jfitch
April 5th 20, 04:45 PM
On Saturday, April 4, 2020 at 6:23:44 PM UTC-7, 2G wrote:
> On one of my flights last year I had to switch between my avionics battery and engine battery when the avionics battery voltage dropped too low (I had left the master on after the last flight and could only partially charge the avionics battery before launching). The switch over seemed to go okay, but then I noticed that my LX9000 was giving me unbelievably short glide distances. It turns out that the QNH altitude had been reset to the altitude at the time of switching. This was unacceptable, so I resolved to do something about it before this season. The simplest solution was to add a capacitor to the avionics power bus. The capacitor supplies power as the power selector switch is moving, and breaking, from the avionics battery, and connector, or making, to the engine battery (this is called a "break before make" switch. But how big of a capacitor to use? The basic equation involved is:
>
> V = I * t / C or C = I * t / V
>
> where V is voltage, I is current and t is time.
>
> Translation: the bigger the capacitor the smaller the voltage drop. If the requirement is to keep the voltage drop to 1 V, the current is 2 A (my situation) and t is 0.1 s, then C = 0.2 F (200,000 μF). The capacitor would also have to be rated for 16 V, min. That is a pretty big capacitor, so I decided I could tolerate a larger voltage drop (4 V), which cuts the size of the capacitor to 50,000 μF. I ended up finding a suitably sized 39,000 μF capacitor rated for 25 V. A smaller capacitor could by used if the current drain is lower, which is likely for most gliders.
> https://www.digikey.com/product-detail/en/kemet/ALS70A393DB025/399-14301-ND/6928303
>
> I installed the capacitor yesterday and monitored the bus voltage during switch-over with an oscilloscope, which was anti-climatic: there was no detectable drop in bus voltage. Apparently the bread-to-make time is very short, perhaps a millisecond. Haven't had a chance to fly with it yet, but should be able to soon. The scope waveforms and capacitor installation can be seen at:
> https://flic.kr/s/aHsmMo9rN7
What is the inrush current when you first switch the power on? Must not be enough to blow the fuse, but that'd be something I'd want to O'scope with a current probe.
On Sunday, April 5, 2020 at 10:18:00 AM UTC-5, Dan Marotta wrote:
> Wouldn't a make-before-break switch solve the problem in a much simpler
> fashion?* That's what my Stemme uses and switching between main and tail
> batteries is a non event.
>
Depends on what happens between make and break.
There will be some amount of surge current from the full battery to the empty one. If the current times the time exceeds the fuse IT, you can blow a battery fuse.
Long wires out to the tail and quick switching could make this less likely.
Martin Gregorie[_6_]
April 5th 20, 05:19 PM
On Sun, 05 Apr 2020 09:10:53 -0700, stu857xx wrote:
> On Sunday, April 5, 2020 at 10:18:00 AM UTC-5, Dan Marotta wrote:
>> Wouldn't a make-before-break switch solve the problem in a much simpler
>> fashion?* That's what my Stemme uses and switching between main and
>> tail batteries is a non event.
>>
>>
> Depends on what happens between make and break.
>
> There will be some amount of surge current from the full battery to the
> empty one. If the current times the time exceeds the fuse IT, you can
> blow a battery fuse.
>
> Long wires out to the tail and quick switching could make this less
> likely.
So would a schottky diode inline with each battery. It will cost you a
whole 0.25v voltage drop and even make the switch unnecessary if you
don't mind drawing from both batteries at once. I power my logger that
way, but split the panel feed so one battery runs flight instruments and
the other does radio and T&B.
Backups? My nav system is a PNA thats good for at least a couple of hours
on internal battery and my backup vario is a Borgelt B.40 thats good for
8 hours plus off the fresh PP3 dry battery it uses for an alternate power
source.
--
Martin | martin at
Gregorie | gregorie dot org
kinsell
April 5th 20, 06:03 PM
On 4/5/20 9:45 AM, jfitch wrote:
> On Saturday, April 4, 2020 at 6:23:44 PM UTC-7, 2G wrote:
>> On one of my flights last year I had to switch between my avionics battery and engine battery when the avionics battery voltage dropped too low (I had left the master on after the last flight and could only partially charge the avionics battery before launching). The switch over seemed to go okay, but then I noticed that my LX9000 was giving me unbelievably short glide distances. It turns out that the QNH altitude had been reset to the altitude at the time of switching. This was unacceptable, so I resolved to do something about it before this season. The simplest solution was to add a capacitor to the avionics power bus. The capacitor supplies power as the power selector switch is moving, and breaking, from the avionics battery, and connector, or making, to the engine battery (this is called a "break before make" switch. But how big of a capacitor to use? The basic equation involved is:
>>
>> V = I * t / C or C = I * t / V
>>
>> where V is voltage, I is current and t is time.
>>
>> Translation: the bigger the capacitor the smaller the voltage drop. If the requirement is to keep the voltage drop to 1 V, the current is 2 A (my situation) and t is 0.1 s, then C = 0.2 F (200,000 μF). The capacitor would also have to be rated for 16 V, min. That is a pretty big capacitor, so I decided I could tolerate a larger voltage drop (4 V), which cuts the size of the capacitor to 50,000 μF. I ended up finding a suitably sized 39,000 μF capacitor rated for 25 V. A smaller capacitor could by used if the current drain is lower, which is likely for most gliders.
>> https://www.digikey.com/product-detail/en/kemet/ALS70A393DB025/399-14301-ND/6928303
>>
>> I installed the capacitor yesterday and monitored the bus voltage during switch-over with an oscilloscope, which was anti-climatic: there was no detectable drop in bus voltage. Apparently the bread-to-make time is very short, perhaps a millisecond. Haven't had a chance to fly with it yet, but should be able to soon. The scope waveforms and capacitor installation can be seen at:
>> https://flic.kr/s/aHsmMo9rN7
>
> What is the inrush current when you first switch the power on? Must not be enough to blow the fuse, but that'd be something I'd want to O'scope with a current probe.
>
Yep. High enough current might eventually erode the switch contacts, or
even damage the capacitor.
Dan Marotta
April 5th 20, 06:35 PM
Well, when I had my LAK-17a, I had a make-before-break and I never blew
a fuse.* The idea is to flip the switch, not see how slowly you can keep
it in the "make" position.* I would imagine the "make" time is of the
order of no more than a couple of thousandths of a second.* Some body
with a scope could measure that time and let us know.
On 4/5/2020 10:10 AM, wrote:
> On Sunday, April 5, 2020 at 10:18:00 AM UTC-5, Dan Marotta wrote:
>> Wouldn't a make-before-break switch solve the problem in a much simpler
>> fashion?* That's what my Stemme uses and switching between main and tail
>> batteries is a non event.
>>
> Depends on what happens between make and break.
>
> There will be some amount of surge current from the full battery to the empty one. If the current times the time exceeds the fuse IT, you can blow a battery fuse.
>
> Long wires out to the tail and quick switching could make this less likely.
--
Dan, 5J
Dan Marotta
April 5th 20, 07:01 PM
You guys must pass the salt like this:
https://www.youtube.com/watch?v=b3PgxWaQMaQ
On 4/5/2020 11:03 AM, kinsell wrote:
> On 4/5/20 9:45 AM, jfitch wrote:
>> On Saturday, April 4, 2020 at 6:23:44 PM UTC-7, 2G wrote:
>>> On one of my flights last year I had to switch between my avionics
>>> battery and engine battery when the avionics battery voltage dropped
>>> too low (I had left the master on after the last flight and could
>>> only partially charge the avionics battery before launching). The
>>> switch over seemed to go okay, but then I noticed that my LX9000 was
>>> giving me unbelievably short glide distances. It turns out that the
>>> QNH altitude had been reset to the altitude at the time of
>>> switching. This was unacceptable, so I resolved to do something
>>> about it before this season. The simplest solution was to add a
>>> capacitor to the avionics power bus. The capacitor supplies power as
>>> the power selector switch is moving, and breaking, from the avionics
>>> battery, and connector, or making, to the engine battery (this is
>>> called a "break before make" switch. But how big of a capacitor to
>>> use? The basic equation involved is:
>>>
>>> V = I * t / C** or** C = I * t / V
>>>
>>> where V is voltage, I is current and t is time.
>>>
>>> Translation: the bigger the capacitor the smaller the voltage drop.
>>> If the requirement is to keep the voltage drop to 1 V, the current
>>> is 2 A (my situation) and t is 0.1 s, then C = 0.2 F (200,000 μF).
>>> The capacitor would also have to be rated for 16 V, min. That is a
>>> pretty big capacitor, so I decided I could tolerate a larger voltage
>>> drop (4 V), which cuts the size of the capacitor to 50,000 μF. I
>>> ended up finding a suitably sized 39,000 μF capacitor rated for 25
>>> V. A smaller capacitor could by used if the current drain is lower,
>>> which is likely for most gliders.
>>> https://www.digikey.com/product-detail/en/kemet/ALS70A393DB025/399-14301-ND/6928303
>>>
>>>
>>> I installed the capacitor yesterday and monitored the bus voltage
>>> during switch-over with an oscilloscope, which was anti-climatic:
>>> there was no detectable drop in bus voltage. Apparently the
>>> bread-to-make time is very short, perhaps a millisecond. Haven't had
>>> a chance to fly with it yet, but should be able to soon. The scope
>>> waveforms and capacitor installation can be seen at:
>>> https://flic.kr/s/aHsmMo9rN7
>>
>> What is the inrush current when you first switch the power on? Must
>> not be enough to blow the fuse, but that'd be something I'd want to
>> O'scope with a current probe.
>>
>
> Yep.* High enough current might eventually erode the switch contacts,
> or even damage the capacitor.
--
Dan, 5J
jfitch
April 5th 20, 07:25 PM
On Sunday, April 5, 2020 at 10:03:31 AM UTC-7, kinsell wrote:
> On 4/5/20 9:45 AM, jfitch wrote:
> > On Saturday, April 4, 2020 at 6:23:44 PM UTC-7, 2G wrote:
> >> On one of my flights last year I had to switch between my avionics battery and engine battery when the avionics battery voltage dropped too low (I had left the master on after the last flight and could only partially charge the avionics battery before launching). The switch over seemed to go okay, but then I noticed that my LX9000 was giving me unbelievably short glide distances. It turns out that the QNH altitude had been reset to the altitude at the time of switching. This was unacceptable, so I resolved to do something about it before this season. The simplest solution was to add a capacitor to the avionics power bus. The capacitor supplies power as the power selector switch is moving, and breaking, from the avionics battery, and connector, or making, to the engine battery (this is called a "break before make" switch. But how big of a capacitor to use? The basic equation involved is:
> >>
> >> V = I * t / C or C = I * t / V
> >>
> >> where V is voltage, I is current and t is time.
> >>
> >> Translation: the bigger the capacitor the smaller the voltage drop. If the requirement is to keep the voltage drop to 1 V, the current is 2 A (my situation) and t is 0.1 s, then C = 0.2 F (200,000 μF). The capacitor would also have to be rated for 16 V, min. That is a pretty big capacitor, so I decided I could tolerate a larger voltage drop (4 V), which cuts the size of the capacitor to 50,000 μF. I ended up finding a suitably sized 39,000 μF capacitor rated for 25 V. A smaller capacitor could by used if the current drain is lower, which is likely for most gliders.
> >> https://www.digikey.com/product-detail/en/kemet/ALS70A393DB025/399-14301-ND/6928303
> >>
> >> I installed the capacitor yesterday and monitored the bus voltage during switch-over with an oscilloscope, which was anti-climatic: there was no detectable drop in bus voltage. Apparently the bread-to-make time is very short, perhaps a millisecond. Haven't had a chance to fly with it yet, but should be able to soon. The scope waveforms and capacitor installation can be seen at:
> >> https://flic.kr/s/aHsmMo9rN7
> >
> > What is the inrush current when you first switch the power on? Must not be enough to blow the fuse, but that'd be something I'd want to O'scope with a current probe.
> >
>
> Yep. High enough current might eventually erode the switch contacts, or
> even damage the capacitor.
The commonly used rotary switches are already being run over spec in modern panels. Schleicher and most others are typically using something like the NKK MRY106 or equivalent, these are called a 2A switch but that is the AC rating, DC rating is 1A. A modern panel uses something like 1.5A, and that is before you key the PTT switch.
When I redid my panel I used an MRT23 which is 3A, and paralleled the contacts for a 6A total. Rotary switches with a higher rating are normally much larger physically.
On Sunday, April 5, 2020 at 1:01:09 PM UTC-5, Dan Marotta wrote:
> You guys must pass the salt like this:
> https://www.youtube.com/watch?v=b3PgxWaQMaQ
>
>
delightful, but not quite the same
A 27 at a recent contest had two separate switches plus an old guy to implement make and eventually break. The result took a Walmart run to get a replacement fuse.
The diode seems a more certain alternative.
Martin Gregorie[_6_]
April 6th 20, 01:28 AM
On Sun, 05 Apr 2020 12:15:53 +0000, Martin Gregorie wrote:
>
> I added small capacitors (18mm diameter x 35mm long, 35-50v, so
> 1000-2000 uF capacity or thereabouts for each capacitor - thats a
> guestimate since I can't read their markings without major surgery), one
> for each battery,
> and with a 0.1 ohm resistor in series with it.
>
Correction: still don't how big the capacitor is but the current limiting
resistor in series with it is 10 ohms (just measured it) so the capacitor
won't charge or discharge more at than 1.2 amps. The resistor is a big
wire-wound resistor in a rectangular ceramic case with the capacitor
taped onto it, so is both mechanically and electrically strong.
But, as I said, since this fixed the problem thats enough to prevent
vario and radio resets if I smack the XLR connector powering my panel. Nav
is fine - got its own internal battery.
The same setup may well sort out a break-before-make switch too, with the
advantage that fitting it across the +12v and ground lines on the panel
side of the switch can most likely be done without disturbing anything
that's already in your panel and is both inexpensive and easy to make and
install.
--
Martin | martin at
Gregorie | gregorie dot org
On Saturday, April 4, 2020 at 11:45:25 PM UTC-7, Mark Morwood wrote:
> Thanks for the report. I had wondered about doing the same thing. Just curious, did you include a resistor across the capacitor to make sure it discharges when the master is switched off and no loads are left on?
No, there is enough parasitic drain to take of that.
Tom
On Sunday, April 5, 2020 at 12:36:09 AM UTC-7, John Foster wrote:
> On Saturday, April 4, 2020 at 7:23:44 PM UTC-6, 2G wrote:
> > On one of my flights last year I had to switch between my avionics battery and engine battery when the avionics battery voltage dropped too low (I had left the master on after the last flight and could only partially charge the avionics battery before launching). The switch over seemed to go okay, but then I noticed that my LX9000 was giving me unbelievably short glide distances. It turns out that the QNH altitude had been reset to the altitude at the time of switching. This was unacceptable, so I resolved to do something about it before this season. The simplest solution was to add a capacitor to the avionics power bus. The capacitor supplies power as the power selector switch is moving, and breaking, from the avionics battery, and connector, or making, to the engine battery (this is called a "break before make" switch. But how big of a capacitor to use? The basic equation involved is:
> >
> > V = I * t / C or C = I * t / V
> >
> > where V is voltage, I is current and t is time.
> >
> > Translation: the bigger the capacitor the smaller the voltage drop. If the requirement is to keep the voltage drop to 1 V, the current is 2 A (my situation) and t is 0.1 s, then C = 0.2 F (200,000 μF). The capacitor would also have to be rated for 16 V, min. That is a pretty big capacitor, so I decided I could tolerate a larger voltage drop (4 V), which cuts the size of the capacitor to 50,000 μF. I ended up finding a suitably sized 39,000 μF capacitor rated for 25 V. A smaller capacitor could by used if the current drain is lower, which is likely for most gliders.
> > https://www.digikey.com/product-detail/en/kemet/ALS70A393DB025/399-14301-ND/6928303
> >
> > I installed the capacitor yesterday and monitored the bus voltage during switch-over with an oscilloscope, which was anti-climatic: there was no detectable drop in bus voltage. Apparently the bread-to-make time is very short, perhaps a millisecond. Haven't had a chance to fly with it yet, but should be able to soon. The scope waveforms and capacitor installation can be seen at:
> > https://flic.kr/s/aHsmMo9rN7
>
> I'm no electrician, but wouldn't it work just fine if you wired both avionics and engine battery in parallel, with a switch (on-off) to each battery.. That would mean two switches. Turn the engine battery switch on before turning the avionics battery switch off. Or am I missing something here?
The glider, an ASH 31 Mi, is already wired with a battery selector switch for the avionics, and is the way to go. You definitely don't want to parallel Pb and LFP battery's accidentally.
Tom
On Sunday, April 5, 2020 at 3:15:06 AM UTC-7, Tim Newport-Peace wrote:
> At 01:23 05 April 2020, 2G wrote:
> >On one of my flights last year I had to switch between my avionics
> battery
> >=
> >and engine battery when the avionics battery voltage dropped too low (I
> >had=
> > left the master on after the last flight and could only partially charge
> >t=
> >he avionics battery before launching). The switch over seemed to go okay,
> >b=
> >ut then I noticed that my LX9000 was giving me unbelievably short glide
> >dis=
> >tances. It turns out that the QNH altitude had been reset to the altitude
> >a=
> >t the time of switching. This was unacceptable, so I resolved to do
> >somethi=
> >ng about it before this season. The simplest solution was to add a
> >capacito=
> >r to the avionics power bus. The capacitor supplies power as the power
> >sele=
> >ctor switch is moving, and breaking, from the avionics battery, and
> >connect=
> >or, or making, to the engine battery (this is called a "break before
> make"
> >=
> >switch. But how big of a capacitor to use? The basic equation involved
> is:
> >
> >V =3D I * t / C or C =3D I * t / V
> >
> >where V is voltage, I is current and t is time.
> >
> >Translation: the bigger the capacitor the smaller the voltage drop. If
> the
> >=
> >requirement is to keep the voltage drop to 1 V, the current is 2 A (my
> >situ=
> >ation) and t is 0.1 s, then C =3D 0.2 F (200,000 =CE=BCF). The capacitor
> >wo=
> >uld also have to be rated for 16 V, min. That is a pretty big capacitor,
> >so=
> > I decided I could tolerate a larger voltage drop (4 V), which cuts the
> >siz=
> >e of the capacitor to 50,000 =CE=BCF. I ended up finding a suitably sized
> >3=
> >9,000 =CE=BCF capacitor rated for 25 V. A smaller capacitor could by used
> >i=
> >f the current drain is lower, which is likely for most gliders.
> >https://www.digikey.com/product-detail/en/kemet/ALS70A393DB025/399-14301-ND=
> >/6928303
> >
> >I installed the capacitor yesterday and monitored the bus voltage during
> >sw=
> >itch-over with an oscilloscope, which was anti-climatic: there was no
> >detec=
> >table drop in bus voltage. Apparently the bread-to-make time is very
> >short,=
> > perhaps a millisecond. Haven't had a chance to fly with it yet, but
> >should=
> > be able to soon. The scope waveforms and capacitor installation can be
> >see=
> >n at:
> >https://flic.kr/s/aHsmMo9rN7
> >
> Interesting. Many years ago Volkslogger had a similar issue and it is worth
> examining their solution. Only the instrument in question (LX9000) needs
> protecting, not the entire Avionics Bus.
>
> They added an Electrolytic Capacitor, same as you did. But they also added
> a Schottky Diode in series between the instrument/capacitor and the
> supply.
>
> The capacitor now only has to maintain the instrument in question and not
> everything on the Supply Bus. In your calculation, the value of I is
> greatly reduced (0.6A instead of 2A) and therefore the value of C is also
> reduced, making for a much smaller capacitor.
As you can see from the installation photo, the physical size of the cap is not an issue.
Tom
On Sunday, April 5, 2020 at 8:18:00 AM UTC-7, Dan Marotta wrote:
> Wouldn't a make-before-break switch solve the problem in a much simpler
> fashion?* That's what my Stemme uses and switching between main and tail
> batteries is a non event.
>
> On 4/5/2020 4:02 AM, Tim Newport-Peace wrote:
> > At 01:23 05 April 2020, 2G wrote:
> >> On one of my flights last year I had to switch between my avionics
> > battery
> >> =
> >> and engine battery when the avionics battery voltage dropped too low (I
> >> had=
> >> left the master on after the last flight and could only partially charge
> >> t=
> >> he avionics battery before launching). The switch over seemed to go okay,
> >> b=
> >> ut then I noticed that my LX9000 was giving me unbelievably short glide
> >> dis=
> >> tances. It turns out that the QNH altitude had been reset to the altitude
> >> a=
> >> t the time of switching. This was unacceptable, so I resolved to do
> >> somethi=
> >> ng about it before this season. The simplest solution was to add a
> >> capacito=
> >> r to the avionics power bus. The capacitor supplies power as the power
> >> sele=
> >> ctor switch is moving, and breaking, from the avionics battery, and
> >> connect=
> >> or, or making, to the engine battery (this is called a "break before
> > make"
> >> =
> >> switch. But how big of a capacitor to use? The basic equation involved
> > is:
> >> V =3D I * t / C or C =3D I * t / V
> >>
> >> where V is voltage, I is current and t is time.
> >>
> >> Translation: the bigger the capacitor the smaller the voltage drop. If
> > the
> >> =
> >> requirement is to keep the voltage drop to 1 V, the current is 2 A (my
> >> situ=
> >> ation) and t is 0.1 s, then C =3D 0.2 F (200,000 =CE=BCF). The capacitor
> >> wo=
> >> uld also have to be rated for 16 V, min. That is a pretty big capacitor,
> >> so=
> >> I decided I could tolerate a larger voltage drop (4 V), which cuts the
> >> siz=
> >> e of the capacitor to 50,000 =CE=BCF. I ended up finding a suitably sized
> >> 3=
> >> 9,000 =CE=BCF capacitor rated for 25 V. A smaller capacitor could by used
> >> i=
> >> f the current drain is lower, which is likely for most gliders.
> >> https://www.digikey.com/product-detail/en/kemet/ALS70A393DB025/399-14301-ND=
> >> /6928303
> >>
> >> I installed the capacitor yesterday and monitored the bus voltage during
> >> sw=
> >> itch-over with an oscilloscope, which was anti-climatic: there was no
> >> detec=
> >> table drop in bus voltage. Apparently the bread-to-make time is very
> >> short,=
> >> perhaps a millisecond. Haven't had a chance to fly with it yet, but
> >> should=
> >> be able to soon. The scope waveforms and capacitor installation can be
> >> see=
> >> n at:
> >> https://flic.kr/s/aHsmMo9rN7
> >>
> > Interesting. Many years ago Volkslogger had a similar issue and it is worth
> > examining their solution. Only the instrument in question (LX9000) needs
> > protecting, not the entire Avionics Bus.
> >
> > They added an Electrolytic Capacitor, same as you did. But they also added
> > a Schottky Diode in series between the instrument/capacitor and the
> > supply.
> >
> > The capacitor now only has to maintain the instrument in question and not
> > everything on the Supply Bus. In your calculation, the value of I is
> > greatly reduced (0.6A instead of 2A) and therefore the value of C is also
> > reduced, making for a much smaller capacitor.
> >
> >
> >
>
> --
> Dan, 5J
That would require changing the switch on the panel. The course of least resistance (hence the title "Battery switch without tears") was to simply add a capacitor. It also avoids shorting of two dissimilar batteries.
Tom
On Sunday, April 5, 2020 at 8:45:30 AM UTC-7, jfitch wrote:
> On Saturday, April 4, 2020 at 6:23:44 PM UTC-7, 2G wrote:
> > On one of my flights last year I had to switch between my avionics battery and engine battery when the avionics battery voltage dropped too low (I had left the master on after the last flight and could only partially charge the avionics battery before launching). The switch over seemed to go okay, but then I noticed that my LX9000 was giving me unbelievably short glide distances. It turns out that the QNH altitude had been reset to the altitude at the time of switching. This was unacceptable, so I resolved to do something about it before this season. The simplest solution was to add a capacitor to the avionics power bus. The capacitor supplies power as the power selector switch is moving, and breaking, from the avionics battery, and connector, or making, to the engine battery (this is called a "break before make" switch. But how big of a capacitor to use? The basic equation involved is:
> >
> > V = I * t / C or C = I * t / V
> >
> > where V is voltage, I is current and t is time.
> >
> > Translation: the bigger the capacitor the smaller the voltage drop. If the requirement is to keep the voltage drop to 1 V, the current is 2 A (my situation) and t is 0.1 s, then C = 0.2 F (200,000 μF). The capacitor would also have to be rated for 16 V, min. That is a pretty big capacitor, so I decided I could tolerate a larger voltage drop (4 V), which cuts the size of the capacitor to 50,000 μF. I ended up finding a suitably sized 39,000 μF capacitor rated for 25 V. A smaller capacitor could by used if the current drain is lower, which is likely for most gliders.
> > https://www.digikey.com/product-detail/en/kemet/ALS70A393DB025/399-14301-ND/6928303
> >
> > I installed the capacitor yesterday and monitored the bus voltage during switch-over with an oscilloscope, which was anti-climatic: there was no detectable drop in bus voltage. Apparently the bread-to-make time is very short, perhaps a millisecond. Haven't had a chance to fly with it yet, but should be able to soon. The scope waveforms and capacitor installation can be seen at:
> > https://flic.kr/s/aHsmMo9rN7
>
> What is the inrush current when you first switch the power on? Must not be enough to blow the fuse, but that'd be something I'd want to O'scope with a current probe.
I have no idea what the inrush current is. The time period of this current is so short that it doesn't trip my Klixon circuit breaker.
Tom
On Sunday, April 5, 2020 at 9:19:23 AM UTC-7, Martin Gregorie wrote:
> On Sun, 05 Apr 2020 09:10:53 -0700, stu857xx wrote:
>
> > On Sunday, April 5, 2020 at 10:18:00 AM UTC-5, Dan Marotta wrote:
> >> Wouldn't a make-before-break switch solve the problem in a much simpler
> >> fashion?* That's what my Stemme uses and switching between main and
> >> tail batteries is a non event.
> >>
> >>
> > Depends on what happens between make and break.
> >
> > There will be some amount of surge current from the full battery to the
> > empty one. If the current times the time exceeds the fuse IT, you can
> > blow a battery fuse.
> >
> > Long wires out to the tail and quick switching could make this less
> > likely.
>
> So would a schottky diode inline with each battery. It will cost you a
> whole 0.25v voltage drop and even make the switch unnecessary if you
> don't mind drawing from both batteries at once. I power my logger that
> way, but split the panel feed so one battery runs flight instruments and
> the other does radio and T&B.
>
> Backups? My nav system is a PNA thats good for at least a couple of hours
> on internal battery and my backup vario is a Borgelt B.40 thats good for
> 8 hours plus off the fresh PP3 dry battery it uses for an alternate power
> source.
>
>
>
> --
> Martin | martin at
> Gregorie | gregorie dot org
Schottky diodes are unnecessary and I would not recommend them do to their voltage drop. The cap works just fine and isn't invasive (it is wired in parallel to the existing voltage bus).
Tom
On Sunday, April 5, 2020 at 10:03:31 AM UTC-7, kinsell wrote:
> On 4/5/20 9:45 AM, jfitch wrote:
> > On Saturday, April 4, 2020 at 6:23:44 PM UTC-7, 2G wrote:
> >> On one of my flights last year I had to switch between my avionics battery and engine battery when the avionics battery voltage dropped too low (I had left the master on after the last flight and could only partially charge the avionics battery before launching). The switch over seemed to go okay, but then I noticed that my LX9000 was giving me unbelievably short glide distances. It turns out that the QNH altitude had been reset to the altitude at the time of switching. This was unacceptable, so I resolved to do something about it before this season. The simplest solution was to add a capacitor to the avionics power bus. The capacitor supplies power as the power selector switch is moving, and breaking, from the avionics battery, and connector, or making, to the engine battery (this is called a "break before make" switch. But how big of a capacitor to use? The basic equation involved is:
> >>
> >> V = I * t / C or C = I * t / V
> >>
> >> where V is voltage, I is current and t is time.
> >>
> >> Translation: the bigger the capacitor the smaller the voltage drop. If the requirement is to keep the voltage drop to 1 V, the current is 2 A (my situation) and t is 0.1 s, then C = 0.2 F (200,000 μF). The capacitor would also have to be rated for 16 V, min. That is a pretty big capacitor, so I decided I could tolerate a larger voltage drop (4 V), which cuts the size of the capacitor to 50,000 μF. I ended up finding a suitably sized 39,000 μF capacitor rated for 25 V. A smaller capacitor could by used if the current drain is lower, which is likely for most gliders.
> >> https://www.digikey.com/product-detail/en/kemet/ALS70A393DB025/399-14301-ND/6928303
> >>
> >> I installed the capacitor yesterday and monitored the bus voltage during switch-over with an oscilloscope, which was anti-climatic: there was no detectable drop in bus voltage. Apparently the bread-to-make time is very short, perhaps a millisecond. Haven't had a chance to fly with it yet, but should be able to soon. The scope waveforms and capacitor installation can be seen at:
> >> https://flic.kr/s/aHsmMo9rN7
> >
> > What is the inrush current when you first switch the power on? Must not be enough to blow the fuse, but that'd be something I'd want to O'scope with a current probe.
> >
>
> Yep. High enough current might eventually erode the switch contacts, or
> even damage the capacitor.
These capacitors are intended for power supply applications and can handle high currents (note the size of the connector posts), although there aren't high currents in my panel.
Tom
On Sunday, April 5, 2020 at 11:25:37 AM UTC-7, jfitch wrote:
> On Sunday, April 5, 2020 at 10:03:31 AM UTC-7, kinsell wrote:
> > On 4/5/20 9:45 AM, jfitch wrote:
> > > On Saturday, April 4, 2020 at 6:23:44 PM UTC-7, 2G wrote:
> > >> On one of my flights last year I had to switch between my avionics battery and engine battery when the avionics battery voltage dropped too low (I had left the master on after the last flight and could only partially charge the avionics battery before launching). The switch over seemed to go okay, but then I noticed that my LX9000 was giving me unbelievably short glide distances. It turns out that the QNH altitude had been reset to the altitude at the time of switching. This was unacceptable, so I resolved to do something about it before this season. The simplest solution was to add a capacitor to the avionics power bus. The capacitor supplies power as the power selector switch is moving, and breaking, from the avionics battery, and connector, or making, to the engine battery (this is called a "break before make" switch. But how big of a capacitor to use? The basic equation involved is:
> > >>
> > >> V = I * t / C or C = I * t / V
> > >>
> > >> where V is voltage, I is current and t is time.
> > >>
> > >> Translation: the bigger the capacitor the smaller the voltage drop. If the requirement is to keep the voltage drop to 1 V, the current is 2 A (my situation) and t is 0.1 s, then C = 0.2 F (200,000 μF). The capacitor would also have to be rated for 16 V, min. That is a pretty big capacitor, so I decided I could tolerate a larger voltage drop (4 V), which cuts the size of the capacitor to 50,000 μF. I ended up finding a suitably sized 39,000 μF capacitor rated for 25 V. A smaller capacitor could by used if the current drain is lower, which is likely for most gliders.
> > >> https://www.digikey.com/product-detail/en/kemet/ALS70A393DB025/399-14301-ND/6928303
> > >>
> > >> I installed the capacitor yesterday and monitored the bus voltage during switch-over with an oscilloscope, which was anti-climatic: there was no detectable drop in bus voltage. Apparently the bread-to-make time is very short, perhaps a millisecond. Haven't had a chance to fly with it yet, but should be able to soon. The scope waveforms and capacitor installation can be seen at:
> > >> https://flic.kr/s/aHsmMo9rN7
> > >
> > > What is the inrush current when you first switch the power on? Must not be enough to blow the fuse, but that'd be something I'd want to O'scope with a current probe.
> > >
> >
> > Yep. High enough current might eventually erode the switch contacts, or
> > even damage the capacitor.
>
> The commonly used rotary switches are already being run over spec in modern panels. Schleicher and most others are typically using something like the NKK MRY106 or equivalent, these are called a 2A switch but that is the AC rating, DC rating is 1A. A modern panel uses something like 1.5A, and that is before you key the PTT switch.
>
> When I redid my panel I used an MRT23 which is 3A, and paralleled the contacts for a 6A total. Rotary switches with a higher rating are normally much larger physically.
Schleicher parallels two sets of contacts of that 4-throw, dual pole switch..
Tom
krasw
April 6th 20, 07:26 AM
On Sunday, 5 April 2020 13:15:06 UTC+3, Tim Newport-Peace wrote:
> Interesting. Many years ago Volkslogger had a similar issue and it is worth
> examining their solution. Only the instrument in question (LX9000) needs
> protecting, not the entire Avionics Bus.
>
Wait, don't all other instrument "see" the capacitor in LX9000 as well? It does not matter where the capacitor is physically, it still supplies current to the whole circuit (which is closed, apart for batteries for a short while)?
Tim Newport-Peace[_6_]
April 6th 20, 10:29 AM
At 06:26 06 April 2020, krasw wrote:
>On Sunday, 5 April 2020 13:15:06 UTC+3, Tim Newport-Peace wrote:
>> Interesting. Many years ago Volkslogger had a similar issue and it is
>worth
>> examining their solution. Only the instrument in question (LX9000)
needs
>> protecting, not the entire Avionics Bus.
>>
>
>Wait, don't all other instrument "see" the capacitor in LX9000 as well?
It
>does not matter where the capacitor is physically, it still supplies
>current to the whole circuit (which is closed, apart for batteries for a
>short while)?
>
The Schottky Diode mentioned is there to prevent this. It isolates the
instrument and it's capacitor from the rest of the instruments.
12v---->diode----->instrument/capacitor.
jfitch
April 6th 20, 04:02 PM
On Sunday, April 5, 2020 at 7:02:13 PM UTC-7, 2G wrote:
> On Sunday, April 5, 2020 at 11:25:37 AM UTC-7, jfitch wrote:
> > On Sunday, April 5, 2020 at 10:03:31 AM UTC-7, kinsell wrote:
> > > On 4/5/20 9:45 AM, jfitch wrote:
> > > > On Saturday, April 4, 2020 at 6:23:44 PM UTC-7, 2G wrote:
> > > >> On one of my flights last year I had to switch between my avionics battery and engine battery when the avionics battery voltage dropped too low (I had left the master on after the last flight and could only partially charge the avionics battery before launching). The switch over seemed to go okay, but then I noticed that my LX9000 was giving me unbelievably short glide distances. It turns out that the QNH altitude had been reset to the altitude at the time of switching. This was unacceptable, so I resolved to do something about it before this season. The simplest solution was to add a capacitor to the avionics power bus. The capacitor supplies power as the power selector switch is moving, and breaking, from the avionics battery, and connector, or making, to the engine battery (this is called a "break before make" switch. But how big of a capacitor to use? The basic equation involved is:
> > > >>
> > > >> V = I * t / C or C = I * t / V
> > > >>
> > > >> where V is voltage, I is current and t is time.
> > > >>
> > > >> Translation: the bigger the capacitor the smaller the voltage drop.. If the requirement is to keep the voltage drop to 1 V, the current is 2 A (my situation) and t is 0.1 s, then C = 0.2 F (200,000 μF). The capacitor would also have to be rated for 16 V, min. That is a pretty big capacitor, so I decided I could tolerate a larger voltage drop (4 V), which cuts the size of the capacitor to 50,000 μF. I ended up finding a suitably sized 39,000 μF capacitor rated for 25 V. A smaller capacitor could by used if the current drain is lower, which is likely for most gliders.
> > > >> https://www.digikey.com/product-detail/en/kemet/ALS70A393DB025/399-14301-ND/6928303
> > > >>
> > > >> I installed the capacitor yesterday and monitored the bus voltage during switch-over with an oscilloscope, which was anti-climatic: there was no detectable drop in bus voltage. Apparently the bread-to-make time is very short, perhaps a millisecond. Haven't had a chance to fly with it yet, but should be able to soon. The scope waveforms and capacitor installation can be seen at:
> > > >> https://flic.kr/s/aHsmMo9rN7
> > > >
> > > > What is the inrush current when you first switch the power on? Must not be enough to blow the fuse, but that'd be something I'd want to O'scope with a current probe.
> > > >
> > >
> > > Yep. High enough current might eventually erode the switch contacts, or
> > > even damage the capacitor.
> >
> > The commonly used rotary switches are already being run over spec in modern panels. Schleicher and most others are typically using something like the NKK MRY106 or equivalent, these are called a 2A switch but that is the AC rating, DC rating is 1A. A modern panel uses something like 1.5A, and that is before you key the PTT switch.
> >
> > When I redid my panel I used an MRT23 which is 3A, and paralleled the contacts for a 6A total. Rotary switches with a higher rating are normally much larger physically.
>
> Schleicher parallels two sets of contacts of that 4-throw, dual pole switch.
>
> Tom
That gets you to 2A - still less than the panel draws with the PTT pressed I'd bet. The time constant of the Klixon is quite long compared to the inrush current (if the impedance is low). Do you have a high bandwidth current probe for your O'scope? The peak current may depend mainly on the arcing characteristics of the contacts in the switch.
It's too bad they can't figure out how to make an LX9000 that can take a momentary interruption in power - many others have figured out how to do this..
kinsell
April 6th 20, 04:05 PM
On 4/5/20 8:00 PM, 2G wrote:
> On Sunday, April 5, 2020 at 10:03:31 AM UTC-7, kinsell wrote:
>> On 4/5/20 9:45 AM, jfitch wrote:
>>> On Saturday, April 4, 2020 at 6:23:44 PM UTC-7, 2G wrote:
>>>> On one of my flights last year I had to switch between my avionics battery and engine battery when the avionics battery voltage dropped too low (I had left the master on after the last flight and could only partially charge the avionics battery before launching). The switch over seemed to go okay, but then I noticed that my LX9000 was giving me unbelievably short glide distances. It turns out that the QNH altitude had been reset to the altitude at the time of switching. This was unacceptable, so I resolved to do something about it before this season. The simplest solution was to add a capacitor to the avionics power bus. The capacitor supplies power as the power selector switch is moving, and breaking, from the avionics battery, and connector, or making, to the engine battery (this is called a "break before make" switch. But how big of a capacitor to use? The basic equation involved is:
>>>>
>>>> V = I * t / C or C = I * t / V
>>>>
>>>> where V is voltage, I is current and t is time.
>>>>
>>>> Translation: the bigger the capacitor the smaller the voltage drop. If the requirement is to keep the voltage drop to 1 V, the current is 2 A (my situation) and t is 0.1 s, then C = 0.2 F (200,000 μF). The capacitor would also have to be rated for 16 V, min. That is a pretty big capacitor, so I decided I could tolerate a larger voltage drop (4 V), which cuts the size of the capacitor to 50,000 μF. I ended up finding a suitably sized 39,000 μF capacitor rated for 25 V. A smaller capacitor could by used if the current drain is lower, which is likely for most gliders.
>>>> https://www.digikey.com/product-detail/en/kemet/ALS70A393DB025/399-14301-ND/6928303
>>>>
>>>> I installed the capacitor yesterday and monitored the bus voltage during switch-over with an oscilloscope, which was anti-climatic: there was no detectable drop in bus voltage. Apparently the bread-to-make time is very short, perhaps a millisecond. Haven't had a chance to fly with it yet, but should be able to soon. The scope waveforms and capacitor installation can be seen at:
>>>> https://flic.kr/s/aHsmMo9rN7
>>>
>>> What is the inrush current when you first switch the power on? Must not be enough to blow the fuse, but that'd be something I'd want to O'scope with a current probe.
>>>
>>
>> Yep. High enough current might eventually erode the switch contacts, or
>> even damage the capacitor.
>
> These capacitors are intended for power supply applications and can handle high currents (note the size of the connector posts), although there aren't high currents in my panel.
>
> Tom
>
There sure are high currents when you power up the panel. Current into
the cap is capacitance times dv/dt, where capacitance is large and dv/dt
is near infinite. Do the math. The switch may survive for a while then
fail due to the repeated surges. The fact the Klixon didn't trip
doesn't mean anything, fuses and breakers can take huge overloads for
brief periods without tripping.
There's better ways of doing this with diodes than just putting on a
monster cap.
Dave
Tango Eight
April 6th 20, 04:33 PM
Switches are generally rated according to interrupt capacity, and that's why the AC and DC ratings are different. In the immortal words of a friend of mine in the business, "DC just doesn't switch worth a ****". As the contacts open, an arc is struck. With AC current, the arc self extinguishes (for small switches) when the current goes to zero. The arc persists longer with DC, causes more contact wear, hence a reduced rating.
A "1A DC" switch is not going to fail or in any way be stressed by keying the mic and momentarily running 2, or even three or four amps. Try not to turn off the master while you are talking on the radio :-).
My panel has a CN2, CNv, radio, flarm. I have the standard issue Schleicher rotary switch, no modifications. Switching between batteries is simple: turn the knob. No drama, no instrument problems.
Evan Ludeman
Dan Marotta
April 6th 20, 07:20 PM
MAKE before BREAK, not the reverse, which will momentarily interrupt
power to your panel.* Of course, if you're really, really fast, you
might get by...
On 4/5/2020 6:28 PM, Martin Gregorie wrote:
> The same setup may well sort out a break-before-make switch too, with the
> advantage that fitting it across the +12v and ground lines on the panel
> side of the switch can most likely be done without disturbing anything
> that's already in your panel and is both inexpensive and easy to make and
> install.
--
Dan, 5J
Dan Marotta
April 6th 20, 07:23 PM
Tom,
Isn't the rotary panel switch in a Schleicher glider a make before break?
OK, I'll stop posting to this thread now.* My solution, which I've
relied on for years without ever losing a connection or blowing a fuse
seems to be too simple to be acceptable.
On 4/5/2020 7:45 PM, 2G wrote:
> The glider, an ASH 31 Mi, is already wired with a battery selector switch for the avionics, and is the way to go. You definitely don't want to parallel Pb and LFP battery's accidentally.
>
> Tom
--
Dan, 5J
Dan Marotta
April 6th 20, 07:24 PM
I'll grant you that!
On 4/5/2020 7:52 PM, 2G wrote:
> That would require changing the switch on the panel. The course of least resistance (hence the title "Battery switch without tears") was to simply add a capacitor. It also avoids shorting of two dissimilar batteries.
>
> Tom
--
Dan, 5J
jfitch
April 6th 20, 07:54 PM
On Monday, April 6, 2020 at 8:33:37 AM UTC-7, Tango Eight wrote:
> Switches are generally rated according to interrupt capacity, and that's why the AC and DC ratings are different. In the immortal words of a friend of mine in the business, "DC just doesn't switch worth a ****". As the contacts open, an arc is struck. With AC current, the arc self extinguishes (for small switches) when the current goes to zero. The arc persists longer with DC, causes more contact wear, hence a reduced rating.
>
> A "1A DC" switch is not going to fail or in any way be stressed by keying the mic and momentarily running 2, or even three or four amps. Try not to turn off the master while you are talking on the radio :-).
>
> My panel has a CN2, CNv, radio, flarm. I have the standard issue Schleicher rotary switch, no modifications. Switching between batteries is simple: turn the knob. No drama, no instrument problems.
>
> Evan Ludeman
No, a 1A DC switch will probably not instantly fail at 2A. An engine redlined at 7000 will probably not instantly fail at 8000. A glider rated at 5.5G will probably not instantly fail at 7G. Still, designing something to routinely operate out of spec isn't considered good practice in my shop. And the switch might fail with an inrush current of 20A, while switching, after awhile.
CN, like most other manufacturers designed to the reality of switching power sources in operation. LX, apparently, did not. I have no problem with switching my break-before-make rotary switches on a full panel of (non LX) electronics - nothing resets.
The continuation of this thread is proof that we are quarantined. ;)
On Monday, April 6, 2020 at 11:33:37 AM UTC-4, Tango Eight wrote:
> Switches are generally rated according to interrupt capacity, and that's why the AC and DC ratings are different. In the immortal words of a friend of mine in the business, "DC just doesn't switch worth a ****". As the contacts open, an arc is struck. With AC current, the arc self extinguishes (for small switches) when the current goes to zero. The arc persists longer with DC, causes more contact wear, hence a reduced rating.
>
> A "1A DC" switch is not going to fail or in any way be stressed by keying the mic and momentarily running 2, or even three or four amps. Try not to turn off the master while you are talking on the radio :-).
>
> My panel has a CN2, CNv, radio, flarm. I have the standard issue Schleicher rotary switch, no modifications. Switching between batteries is simple: turn the knob. No drama, no instrument problems.
>
> Evan Ludeman
Case in point to illustrate Evan's statement: I've tried to use a small water heating tank to use excess solar power to heat some water. Its heating element is rated 120VAC 15A. It has a thermostat with an internal switch to match. I fed it about 50VDC 6A from solar panels. That switch died in short order. Also note that we're warned not to disconnect solar panel connectors while the sun is shining and the power is being used, because the arc will damage the contacts in those connectors.
On Monday, April 6, 2020 at 8:02:53 AM UTC-7, jfitch wrote:
> On Sunday, April 5, 2020 at 7:02:13 PM UTC-7, 2G wrote:
> > On Sunday, April 5, 2020 at 11:25:37 AM UTC-7, jfitch wrote:
> > > On Sunday, April 5, 2020 at 10:03:31 AM UTC-7, kinsell wrote:
> > > > On 4/5/20 9:45 AM, jfitch wrote:
> > > > > On Saturday, April 4, 2020 at 6:23:44 PM UTC-7, 2G wrote:
> > > > >> On one of my flights last year I had to switch between my avionics battery and engine battery when the avionics battery voltage dropped too low (I had left the master on after the last flight and could only partially charge the avionics battery before launching). The switch over seemed to go okay, but then I noticed that my LX9000 was giving me unbelievably short glide distances. It turns out that the QNH altitude had been reset to the altitude at the time of switching. This was unacceptable, so I resolved to do something about it before this season. The simplest solution was to add a capacitor to the avionics power bus. The capacitor supplies power as the power selector switch is moving, and breaking, from the avionics battery, and connector, or making, to the engine battery (this is called a "break before make" switch. But how big of a capacitor to use? The basic equation involved is:
> > > > >>
> > > > >> V = I * t / C or C = I * t / V
> > > > >>
> > > > >> where V is voltage, I is current and t is time.
> > > > >>
> > > > >> Translation: the bigger the capacitor the smaller the voltage drop. If the requirement is to keep the voltage drop to 1 V, the current is 2 A (my situation) and t is 0.1 s, then C = 0.2 F (200,000 μF). The capacitor would also have to be rated for 16 V, min. That is a pretty big capacitor, so I decided I could tolerate a larger voltage drop (4 V), which cuts the size of the capacitor to 50,000 μF. I ended up finding a suitably sized 39,000 μF capacitor rated for 25 V. A smaller capacitor could by used if the current drain is lower, which is likely for most gliders.
> > > > >> https://www.digikey.com/product-detail/en/kemet/ALS70A393DB025/399-14301-ND/6928303
> > > > >>
> > > > >> I installed the capacitor yesterday and monitored the bus voltage during switch-over with an oscilloscope, which was anti-climatic: there was no detectable drop in bus voltage. Apparently the bread-to-make time is very short, perhaps a millisecond. Haven't had a chance to fly with it yet, but should be able to soon. The scope waveforms and capacitor installation can be seen at:
> > > > >> https://flic.kr/s/aHsmMo9rN7
> > > > >
> > > > > What is the inrush current when you first switch the power on? Must not be enough to blow the fuse, but that'd be something I'd want to O'scope with a current probe.
> > > > >
> > > >
> > > > Yep. High enough current might eventually erode the switch contacts, or
> > > > even damage the capacitor.
> > >
> > > The commonly used rotary switches are already being run over spec in modern panels. Schleicher and most others are typically using something like the NKK MRY106 or equivalent, these are called a 2A switch but that is the AC rating, DC rating is 1A. A modern panel uses something like 1.5A, and that is before you key the PTT switch.
> > >
> > > When I redid my panel I used an MRT23 which is 3A, and paralleled the contacts for a 6A total. Rotary switches with a higher rating are normally much larger physically.
> >
> > Schleicher parallels two sets of contacts of that 4-throw, dual pole switch.
> >
> > Tom
>
> That gets you to 2A - still less than the panel draws with the PTT pressed I'd bet. The time constant of the Klixon is quite long compared to the inrush current (if the impedance is low). Do you have a high bandwidth current probe for your O'scope? The peak current may depend mainly on the arcing characteristics of the contacts in the switch.
>
> It's too bad they can't figure out how to make an LX9000 that can take a momentary interruption in power - many others have figured out how to do this.
The scope and probe are 100 MHz. I don't have a current probe (would have to use a series resistor which would change it from "without tears" to "many tears." Remember, the capacitor is providing the energy to the instruments when the switch is between break and make, so no current is required from the battery when it makes.
On Monday, April 6, 2020 at 8:05:20 AM UTC-7, kinsell wrote:
> On 4/5/20 8:00 PM, 2G wrote:
> > On Sunday, April 5, 2020 at 10:03:31 AM UTC-7, kinsell wrote:
> >> On 4/5/20 9:45 AM, jfitch wrote:
> >>> On Saturday, April 4, 2020 at 6:23:44 PM UTC-7, 2G wrote:
> >>>> On one of my flights last year I had to switch between my avionics battery and engine battery when the avionics battery voltage dropped too low (I had left the master on after the last flight and could only partially charge the avionics battery before launching). The switch over seemed to go okay, but then I noticed that my LX9000 was giving me unbelievably short glide distances. It turns out that the QNH altitude had been reset to the altitude at the time of switching. This was unacceptable, so I resolved to do something about it before this season. The simplest solution was to add a capacitor to the avionics power bus. The capacitor supplies power as the power selector switch is moving, and breaking, from the avionics battery, and connector, or making, to the engine battery (this is called a "break before make" switch. But how big of a capacitor to use? The basic equation involved is:
> >>>>
> >>>> V = I * t / C or C = I * t / V
> >>>>
> >>>> where V is voltage, I is current and t is time.
> >>>>
> >>>> Translation: the bigger the capacitor the smaller the voltage drop. If the requirement is to keep the voltage drop to 1 V, the current is 2 A (my situation) and t is 0.1 s, then C = 0.2 F (200,000 μF). The capacitor would also have to be rated for 16 V, min. That is a pretty big capacitor, so I decided I could tolerate a larger voltage drop (4 V), which cuts the size of the capacitor to 50,000 μF. I ended up finding a suitably sized 39,000 μF capacitor rated for 25 V. A smaller capacitor could by used if the current drain is lower, which is likely for most gliders.
> >>>> https://www.digikey.com/product-detail/en/kemet/ALS70A393DB025/399-14301-ND/6928303
> >>>>
> >>>> I installed the capacitor yesterday and monitored the bus voltage during switch-over with an oscilloscope, which was anti-climatic: there was no detectable drop in bus voltage. Apparently the bread-to-make time is very short, perhaps a millisecond. Haven't had a chance to fly with it yet, but should be able to soon. The scope waveforms and capacitor installation can be seen at:
> >>>> https://flic.kr/s/aHsmMo9rN7
> >>>
> >>> What is the inrush current when you first switch the power on? Must not be enough to blow the fuse, but that'd be something I'd want to O'scope with a current probe.
> >>>
> >>
> >> Yep. High enough current might eventually erode the switch contacts, or
> >> even damage the capacitor.
> >
> > These capacitors are intended for power supply applications and can handle high currents (note the size of the connector posts), although there aren't high currents in my panel.
> >
> > Tom
> >
>
> There sure are high currents when you power up the panel. Current into
> the cap is capacitance times dv/dt, where capacitance is large and dv/dt
> is near infinite. Do the math. The switch may survive for a while then
> fail due to the repeated surges. The fact the Klixon didn't trip
> doesn't mean anything, fuses and breakers can take huge overloads for
> brief periods without tripping.
>
> There's better ways of doing this with diodes than just putting on a
> monster cap.
>
> Dave
Remember how they come up with these current ratings. It is based on heating of the contact - a very short current pulse (μsec) that results in virtually no heating.
On Monday, April 6, 2020 at 11:23:21 AM UTC-7, Dan Marotta wrote:
> Tom,
>
> Isn't the rotary panel switch in a Schleicher glider a make before break?
>
> OK, I'll stop posting to this thread now.* My solution, which I've
> relied on for years without ever losing a connection or blowing a fuse
> seems to be too simple to be acceptable.
>
> On 4/5/2020 7:45 PM, 2G wrote:
> > The glider, an ASH 31 Mi, is already wired with a battery selector switch for the avionics, and is the way to go. You definitely don't want to parallel Pb and LFP battery's accidentally.
> >
> > Tom
>
> --
> Dan, 5J
Dan,
Think about it: if it were MAKE before BREAK I wouldn't have had this problem and sure as hell wouldn't post a solution to a non-existent problem.
On Monday, April 6, 2020 at 9:40:36 PM UTC-7, 2G wrote:
> On Monday, April 6, 2020 at 8:05:20 AM UTC-7, kinsell wrote:
> > On 4/5/20 8:00 PM, 2G wrote:
> > > On Sunday, April 5, 2020 at 10:03:31 AM UTC-7, kinsell wrote:
> > >> On 4/5/20 9:45 AM, jfitch wrote:
> > >>> On Saturday, April 4, 2020 at 6:23:44 PM UTC-7, 2G wrote:
> > >>>> On one of my flights last year I had to switch between my avionics battery and engine battery when the avionics battery voltage dropped too low (I had left the master on after the last flight and could only partially charge the avionics battery before launching). The switch over seemed to go okay, but then I noticed that my LX9000 was giving me unbelievably short glide distances. It turns out that the QNH altitude had been reset to the altitude at the time of switching. This was unacceptable, so I resolved to do something about it before this season. The simplest solution was to add a capacitor to the avionics power bus. The capacitor supplies power as the power selector switch is moving, and breaking, from the avionics battery, and connector, or making, to the engine battery (this is called a "break before make" switch. But how big of a capacitor to use? The basic equation involved is:
> > >>>>
> > >>>> V = I * t / C or C = I * t / V
> > >>>>
> > >>>> where V is voltage, I is current and t is time.
> > >>>>
> > >>>> Translation: the bigger the capacitor the smaller the voltage drop.. If the requirement is to keep the voltage drop to 1 V, the current is 2 A (my situation) and t is 0.1 s, then C = 0.2 F (200,000 μF). The capacitor would also have to be rated for 16 V, min. That is a pretty big capacitor, so I decided I could tolerate a larger voltage drop (4 V), which cuts the size of the capacitor to 50,000 μF. I ended up finding a suitably sized 39,000 μF capacitor rated for 25 V. A smaller capacitor could by used if the current drain is lower, which is likely for most gliders.
> > >>>> https://www.digikey.com/product-detail/en/kemet/ALS70A393DB025/399-14301-ND/6928303
> > >>>>
> > >>>> I installed the capacitor yesterday and monitored the bus voltage during switch-over with an oscilloscope, which was anti-climatic: there was no detectable drop in bus voltage. Apparently the bread-to-make time is very short, perhaps a millisecond. Haven't had a chance to fly with it yet, but should be able to soon. The scope waveforms and capacitor installation can be seen at:
> > >>>> https://flic.kr/s/aHsmMo9rN7
> > >>>
> > >>> What is the inrush current when you first switch the power on? Must not be enough to blow the fuse, but that'd be something I'd want to O'scope with a current probe.
> > >>>
> > >>
> > >> Yep. High enough current might eventually erode the switch contacts, or
> > >> even damage the capacitor.
> > >
> > > These capacitors are intended for power supply applications and can handle high currents (note the size of the connector posts), although there aren't high currents in my panel.
> > >
> > > Tom
> > >
> >
> > There sure are high currents when you power up the panel. Current into
> > the cap is capacitance times dv/dt, where capacitance is large and dv/dt
> > is near infinite. Do the math. The switch may survive for a while then
> > fail due to the repeated surges. The fact the Klixon didn't trip
> > doesn't mean anything, fuses and breakers can take huge overloads for
> > brief periods without tripping.
> >
> > There's better ways of doing this with diodes than just putting on a
> > monster cap.
> >
> > Dave
>
> Remember how they come up with these current ratings. It is based on heating of the contact - a very short current pulse (μsec) that results in virtually no heating.
I flew with this mod today and it worked perfectly - no glitch in the LX9000 QNH.
Tom
Dan Marotta
April 7th 20, 05:32 PM
Well then, I must be really confused.* If the switch was a BREAK before
MAKE, the equipment would be momentarily without power.* Some devices
can survive that, perhaps with internal capacitance, but others will
lose power momentarily.* This may or may not be an issue.
Inrush current seems to be a big bugaboo to some people and, in some
cases it is, but what is the time constant associated with the inrush?*
How much current are we talking about and for how long? People keep
talking about the high battery "charging" the low battery during the
milliseconds that they are in parallel. Theoretically, yes, practically,
hogwash.* The heat required to blow a fuse or burn a wire does not rise
instantaneously; there's counter EMF to reduce the current...* So many
details that meant a bunch when you were taking a test back in school
but, practically speaking, don't mean squat in this case.
So, switch your batteries and protect your circuits however you wish.*
I'll stick with my make before break switches.* I'll even report back
when something fails due to their use, but don't hold your breath.
On 4/6/2020 10:43 PM, 2G wrote:
> On Monday, April 6, 2020 at 11:23:21 AM UTC-7, Dan Marotta wrote:
>> Tom,
>>
>> Isn't the rotary panel switch in a Schleicher glider a make before break?
>>
>> OK, I'll stop posting to this thread now.* My solution, which I've
>> relied on for years without ever losing a connection or blowing a fuse
>> seems to be too simple to be acceptable.
>>
>> On 4/5/2020 7:45 PM, 2G wrote:
>>> The glider, an ASH 31 Mi, is already wired with a battery selector switch for the avionics, and is the way to go. You definitely don't want to parallel Pb and LFP battery's accidentally.
>>>
>>> Tom
>> --
>> Dan, 5J
> Dan,
>
> Think about it: if it were MAKE before BREAK I wouldn't have had this problem and sure as hell wouldn't post a solution to a non-existent problem.
--
Dan, 5J
jfitch
April 7th 20, 05:44 PM
On Monday, April 6, 2020 at 9:36:21 PM UTC-7, 2G wrote:
> On Monday, April 6, 2020 at 8:02:53 AM UTC-7, jfitch wrote:
> > On Sunday, April 5, 2020 at 7:02:13 PM UTC-7, 2G wrote:
> > > On Sunday, April 5, 2020 at 11:25:37 AM UTC-7, jfitch wrote:
> > > > On Sunday, April 5, 2020 at 10:03:31 AM UTC-7, kinsell wrote:
> > > > > On 4/5/20 9:45 AM, jfitch wrote:
> > > > > > On Saturday, April 4, 2020 at 6:23:44 PM UTC-7, 2G wrote:
> > > > > >> On one of my flights last year I had to switch between my avionics battery and engine battery when the avionics battery voltage dropped too low (I had left the master on after the last flight and could only partially charge the avionics battery before launching). The switch over seemed to go okay, but then I noticed that my LX9000 was giving me unbelievably short glide distances. It turns out that the QNH altitude had been reset to the altitude at the time of switching. This was unacceptable, so I resolved to do something about it before this season. The simplest solution was to add a capacitor to the avionics power bus. The capacitor supplies power as the power selector switch is moving, and breaking, from the avionics battery, and connector, or making, to the engine battery (this is called a "break before make" switch. But how big of a capacitor to use? The basic equation involved is:
> > > > > >>
> > > > > >> V = I * t / C or C = I * t / V
> > > > > >>
> > > > > >> where V is voltage, I is current and t is time.
> > > > > >>
> > > > > >> Translation: the bigger the capacitor the smaller the voltage drop. If the requirement is to keep the voltage drop to 1 V, the current is 2 A (my situation) and t is 0.1 s, then C = 0.2 F (200,000 μF). The capacitor would also have to be rated for 16 V, min. That is a pretty big capacitor, so I decided I could tolerate a larger voltage drop (4 V), which cuts the size of the capacitor to 50,000 μF. I ended up finding a suitably sized 39,000 μF capacitor rated for 25 V. A smaller capacitor could by used if the current drain is lower, which is likely for most gliders..
> > > > > >> https://www.digikey.com/product-detail/en/kemet/ALS70A393DB025/399-14301-ND/6928303
> > > > > >>
> > > > > >> I installed the capacitor yesterday and monitored the bus voltage during switch-over with an oscilloscope, which was anti-climatic: there was no detectable drop in bus voltage. Apparently the bread-to-make time is very short, perhaps a millisecond. Haven't had a chance to fly with it yet, but should be able to soon. The scope waveforms and capacitor installation can be seen at:
> > > > > >> https://flic.kr/s/aHsmMo9rN7
> > > > > >
> > > > > > What is the inrush current when you first switch the power on? Must not be enough to blow the fuse, but that'd be something I'd want to O'scope with a current probe.
> > > > > >
> > > > >
> > > > > Yep. High enough current might eventually erode the switch contacts, or
> > > > > even damage the capacitor.
> > > >
> > > > The commonly used rotary switches are already being run over spec in modern panels. Schleicher and most others are typically using something like the NKK MRY106 or equivalent, these are called a 2A switch but that is the AC rating, DC rating is 1A. A modern panel uses something like 1.5A, and that is before you key the PTT switch.
> > > >
> > > > When I redid my panel I used an MRT23 which is 3A, and paralleled the contacts for a 6A total. Rotary switches with a higher rating are normally much larger physically.
> > >
> > > Schleicher parallels two sets of contacts of that 4-throw, dual pole switch.
> > >
> > > Tom
> >
> > That gets you to 2A - still less than the panel draws with the PTT pressed I'd bet. The time constant of the Klixon is quite long compared to the inrush current (if the impedance is low). Do you have a high bandwidth current probe for your O'scope? The peak current may depend mainly on the arcing characteristics of the contacts in the switch.
> >
> > It's too bad they can't figure out how to make an LX9000 that can take a momentary interruption in power - many others have figured out how to do this.
>
> The scope and probe are 100 MHz. I don't have a current probe (would have to use a series resistor which would change it from "without tears" to "many tears." Remember, the capacitor is providing the energy to the instruments when the switch is between break and make, so no current is required from the battery when it makes.
The issue with the switch will not be switching between two batteries. It is switching on the first battery, charging the capacitor. DC contact rating for a switch is determined from two things: heating of the contacts and arcing resistance during switching. The latter is generally more critical, which is why usually the AC rating is higher. A rotary switch of this type is particularly susceptible as it is not "snap acting".
> Inrush current seems to be a big bugaboo
Seem to have time to kill. Lets try for some numbers.
Here's the fuse data sheet
https://www.littelfuse.com/~/media/automotive/datasheets/fuses/passenger-car-and-commercial-vehicle/blade-fuses/littelfuse_atof_datasheet.pdf
I use a 7.5 amp fuse on each battery.
Assuming the battery full and empty voltages differ by 1 volt.
The current with both batteries on the bus also depends on the wiring resistance. If there is 0.1 ohms, then the fuse current is 10 amps.
at 25C, that could blow it in a few seconds.
The I2T rating for the fuse is 60. That is measured over 8 milliseconds. So I think that says the fuse will blow quickly with 86 amps.
Seems like unless you have really good wiring or a big charge difference, quickly switching make before break should be ok.
That matches your experience with the Stemme and also the Walmart run at the recent contest where the switch was left with both batteries on the bus for a while.
Dan Marotta
April 7th 20, 11:31 PM
Good practice (checklist discipline) holds that all switches be off
before turning the battery on.* At the end of the day, all equipment
should be switched off before turning off the battery.* The switch can't
suffer much arcing under those conditions.* As said elsewhere, the
voltage difference between two batteries being switched between, is not
much more than 1 volt and the time that they are in parallel is probably
less than 100 mSec using a make before break switch.
I should have made the above clear up front.
On 4/7/2020 10:44 AM, jfitch wrote:
> On Monday, April 6, 2020 at 9:36:21 PM UTC-7, 2G wrote:
>> On Monday, April 6, 2020 at 8:02:53 AM UTC-7, jfitch wrote:
>>> On Sunday, April 5, 2020 at 7:02:13 PM UTC-7, 2G wrote:
>>>> On Sunday, April 5, 2020 at 11:25:37 AM UTC-7, jfitch wrote:
>>>>> On Sunday, April 5, 2020 at 10:03:31 AM UTC-7, kinsell wrote:
>>>>>> On 4/5/20 9:45 AM, jfitch wrote:
>>>>>>> On Saturday, April 4, 2020 at 6:23:44 PM UTC-7, 2G wrote:
>>>>>>>> On one of my flights last year I had to switch between my avionics battery and engine battery when the avionics battery voltage dropped too low (I had left the master on after the last flight and could only partially charge the avionics battery before launching). The switch over seemed to go okay, but then I noticed that my LX9000 was giving me unbelievably short glide distances. It turns out that the QNH altitude had been reset to the altitude at the time of switching. This was unacceptable, so I resolved to do something about it before this season. The simplest solution was to add a capacitor to the avionics power bus. The capacitor supplies power as the power selector switch is moving, and breaking, from the avionics battery, and connector, or making, to the engine battery (this is called a "break before make" switch. But how big of a capacitor to use? The basic equation involved is:
>>>>>>>>
>>>>>>>> V = I * t / C or C = I * t / V
>>>>>>>>
>>>>>>>> where V is voltage, I is current and t is time.
>>>>>>>>
>>>>>>>> Translation: the bigger the capacitor the smaller the voltage drop. If the requirement is to keep the voltage drop to 1 V, the current is 2 A (my situation) and t is 0.1 s, then C = 0.2 F (200,000 μF). The capacitor would also have to be rated for 16 V, min. That is a pretty big capacitor, so I decided I could tolerate a larger voltage drop (4 V), which cuts the size of the capacitor to 50,000 μF. I ended up finding a suitably sized 39,000 μF capacitor rated for 25 V. A smaller capacitor could by used if the current drain is lower, which is likely for most gliders.
>>>>>>>> https://www.digikey.com/product-detail/en/kemet/ALS70A393DB025/399-14301-ND/6928303
>>>>>>>>
>>>>>>>> I installed the capacitor yesterday and monitored the bus voltage during switch-over with an oscilloscope, which was anti-climatic: there was no detectable drop in bus voltage. Apparently the bread-to-make time is very short, perhaps a millisecond. Haven't had a chance to fly with it yet, but should be able to soon. The scope waveforms and capacitor installation can be seen at:
>>>>>>>> https://flic.kr/s/aHsmMo9rN7
>>>>>>> What is the inrush current when you first switch the power on? Must not be enough to blow the fuse, but that'd be something I'd want to O'scope with a current probe.
>>>>>>>
>>>>>> Yep. High enough current might eventually erode the switch contacts, or
>>>>>> even damage the capacitor.
>>>>> The commonly used rotary switches are already being run over spec in modern panels. Schleicher and most others are typically using something like the NKK MRY106 or equivalent, these are called a 2A switch but that is the AC rating, DC rating is 1A. A modern panel uses something like 1.5A, and that is before you key the PTT switch.
>>>>>
>>>>> When I redid my panel I used an MRT23 which is 3A, and paralleled the contacts for a 6A total. Rotary switches with a higher rating are normally much larger physically.
>>>> Schleicher parallels two sets of contacts of that 4-throw, dual pole switch.
>>>>
>>>> Tom
>>> That gets you to 2A - still less than the panel draws with the PTT pressed I'd bet. The time constant of the Klixon is quite long compared to the inrush current (if the impedance is low). Do you have a high bandwidth current probe for your O'scope? The peak current may depend mainly on the arcing characteristics of the contacts in the switch.
>>>
>>> It's too bad they can't figure out how to make an LX9000 that can take a momentary interruption in power - many others have figured out how to do this.
>> The scope and probe are 100 MHz. I don't have a current probe (would have to use a series resistor which would change it from "without tears" to "many tears." Remember, the capacitor is providing the energy to the instruments when the switch is between break and make, so no current is required from the battery when it makes.
> The issue with the switch will not be switching between two batteries. It is switching on the first battery, charging the capacitor. DC contact rating for a switch is determined from two things: heating of the contacts and arcing resistance during switching. The latter is generally more critical, which is why usually the AC rating is higher. A rotary switch of this type is particularly susceptible as it is not "snap acting".
--
Dan, 5J
On Tuesday, April 7, 2020 at 9:32:07 AM UTC-7, Dan Marotta wrote:
> Well then, I must be really confused.* If the switch was a BREAK before
> MAKE, the equipment would be momentarily without power.* Some devices
> can survive that, perhaps with internal capacitance, but others will
> lose power momentarily.* This may or may not be an issue.
>
> Inrush current seems to be a big bugaboo to some people and, in some
> cases it is, but what is the time constant associated with the inrush?*
> How much current are we talking about and for how long? People keep
> talking about the high battery "charging" the low battery during the
> milliseconds that they are in parallel. Theoretically, yes, practically,
> hogwash.* The heat required to blow a fuse or burn a wire does not rise
> instantaneously; there's counter EMF to reduce the current...* So many
> details that meant a bunch when you were taking a test back in school
> but, practically speaking, don't mean squat in this case.
>
> So, switch your batteries and protect your circuits however you wish.*
> I'll stick with my make before break switches.* I'll even report back
> when something fails due to their use, but don't hold your breath.
>
> On 4/6/2020 10:43 PM, 2G wrote:
> > On Monday, April 6, 2020 at 11:23:21 AM UTC-7, Dan Marotta wrote:
> >> Tom,
> >>
> >> Isn't the rotary panel switch in a Schleicher glider a make before break?
> >>
> >> OK, I'll stop posting to this thread now.* My solution, which I've
> >> relied on for years without ever losing a connection or blowing a fuse
> >> seems to be too simple to be acceptable.
> >>
> >> On 4/5/2020 7:45 PM, 2G wrote:
> >>> The glider, an ASH 31 Mi, is already wired with a battery selector switch for the avionics, and is the way to go. You definitely don't want to parallel Pb and LFP battery's accidentally.
> >>>
> >>> Tom
> >> --
> >> Dan, 5J
> > Dan,
> >
> > Think about it: if it were MAKE before BREAK I wouldn't have had this problem and sure as hell wouldn't post a solution to a non-existent problem.
>
> --
> Dan, 5J
If my power switch was a break-before-make (also called a shorting switch) I would not have had a problem and certainly wouldn't have posted a fix to a non-existent problem. But it isn't. It was easier for me to put in a capacitor than replace the switch. It also protects against any switch bounce while switching.
People seem to be overly concerned with inrush current. High inrush current would only happen if there were a substantial amount of power supply capacitance in the LX9000 that needed charging. If that were the case it would handle a millisecond battery switch over without any problem, but it doesn't.. Also, there is no inrush current during battery switching as the caps in the LX9000 are already charged.
Tom
Andy Blackburn[_3_]
April 8th 20, 04:06 AM
I put a power resistor in the circuit to keep the current surge down. I undersized the capacitor so if I mess up on the with rotation I can lose power.. Typically I'll shut off some non-essential equipment to lower the draw if I really don't want a computer reset.
Andy
On Sunday, April 5, 2020 at 8:45:30 AM UTC-7, jfitch wrote:
> What is the inrush current when you first switch the power on? Must not be enough to blow the fuse, but that'd be something I'd want to O'scope with a current probe.
On Tuesday, April 7, 2020 at 8:06:46 PM UTC-7, Andy Blackburn wrote:
> I put a power resistor in the circuit to keep the current surge down. I undersized the capacitor so if I mess up on the with rotation I can lose power. Typically I'll shut off some non-essential equipment to lower the draw if I really don't want a computer reset.
>
> Andy
>
> On Sunday, April 5, 2020 at 8:45:30 AM UTC-7, jfitch wrote:
>
> > What is the inrush current when you first switch the power on? Must not be enough to blow the fuse, but that'd be something I'd want to O'scope with a current probe.
You didn't say what the resistor size was. There is already a resistor in the circuit - it is called the internal resistance of the battery,wiring resistance, switch contact resistance and the equivalent series resistance (ESR) of the capacitor.
Andy Blackburn[_3_]
April 8th 20, 05:07 AM
On Tuesday, April 7, 2020 at 8:58:53 PM UTC-7, 2G wrote:
> On Tuesday, April 7, 2020 at 8:06:46 PM UTC-7, Andy Blackburn wrote:
> > I put a power resistor in the circuit to keep the current surge down. I undersized the capacitor so if I mess up on the with rotation I can lose power. Typically I'll shut off some non-essential equipment to lower the draw if I really don't want a computer reset.
> >
> > Andy
> >
> > On Sunday, April 5, 2020 at 8:45:30 AM UTC-7, jfitch wrote:
> >
> > > What is the inrush current when you first switch the power on? Must not be enough to blow the fuse, but that'd be something I'd want to O'scope with a current probe.
>
> You didn't say what the resistor size was. There is already a resistor in the circuit - it is called the internal resistance of the battery,wiring resistance, switch contact resistance and the equivalent series resistance (ESR) of the capacitor.
I think it was only a few Ohms, but it was a big sucker so it could take the current. I figured bet to play it safe so you know where you are dissipating the energy. Probably unnecessary, but I am a belt + suspenders kind of guy.
Andy
On Tuesday, April 7, 2020 at 8:58:53 PM UTC-7, 2G wrote:
> On Tuesday, April 7, 2020 at 8:06:46 PM UTC-7, Andy Blackburn wrote:
> > I put a power resistor in the circuit to keep the current surge down. I undersized the capacitor so if I mess up on the with rotation I can lose power. Typically I'll shut off some non-essential equipment to lower the draw if I really don't want a computer reset.
> >
> > Andy
> >
> > On Sunday, April 5, 2020 at 8:45:30 AM UTC-7, jfitch wrote:
> >
> > > What is the inrush current when you first switch the power on? Must not be enough to blow the fuse, but that'd be something I'd want to O'scope with a current probe.
>
> You didn't say what the resistor size was. There is already a resistor in the circuit - it is called the internal resistance of the battery,wiring resistance, switch contact resistance and the equivalent series resistance (ESR) of the capacitor.
I forgot to mention that aluminum electrolytic capacitors have very significant inductance at frequencies above 10 KHz. Inductance (also called a choke because they choke off current) limits inrush current. The joules of energy being transferred is pretty low (about 0.6 J), but I have no problem with adding a small series resistor (1 ohm - don't worry about the wattage as very little power is supplied by the cap).
Tom
kinsell
April 8th 20, 05:55 AM
On 4/7/20 10:07 PM, Andy Blackburn wrote:
> On Tuesday, April 7, 2020 at 8:58:53 PM UTC-7, 2G wrote:
>> On Tuesday, April 7, 2020 at 8:06:46 PM UTC-7, Andy Blackburn wrote:
>>> I put a power resistor in the circuit to keep the current surge down. I undersized the capacitor so if I mess up on the with rotation I can lose power. Typically I'll shut off some non-essential equipment to lower the draw if I really don't want a computer reset.
>>>
>>> Andy
>>>
>>> On Sunday, April 5, 2020 at 8:45:30 AM UTC-7, jfitch wrote:
>>>
>>>> What is the inrush current when you first switch the power on? Must not be enough to blow the fuse, but that'd be something I'd want to O'scope with a current probe.
>>
>> You didn't say what the resistor size was. There is already a resistor in the circuit - it is called the internal resistance of the battery,wiring resistance, switch contact resistance and the equivalent series resistance (ESR) of the capacitor.
>
> I think it was only a few Ohms, but it was a big sucker so it could take the current. I figured bet to play it safe so you know where you are dissipating the energy. Probably unnecessary, but I am a belt + suspenders kind of guy.
>
> Andy
>
For people who didn't sleep through their EE101 class, they can figure
out how to parallel a diode across the resistor to allow very slow
charging of the capacitor, yet rapid discharge to power the panel
temporarily. Charging the cap in seconds instead of milliseconds would
give oh about a factor of 1000 reduction in the inrush current. There's
better ways of doing it, but this would be an adequate solution.
The batteries Tom uses are both capable of delivering several hundreds
of amps at a reasonable voltage to a starter motor, so that should give
an idea of how much current limiting they provide.
The really funny thing is, most people reading this have no need at all
to ever switch batteries. Instead of two small batteries, one big one
is so much easier to manage. If they have to be broken into multiple
units, then just wire them in parallel and let them all provide power
until depleted. If you switch them, then you risk switching too soon
and wasting capacity in the first one, or switching too late and ruining
a flight log or messing up a flight computer when you key the mike and
don't realize how weak the battery was.
-Dave
Tim Newport-Peace[_6_]
April 8th 20, 09:41 AM
At 04:20 08 April 2020, 2G wrote:
>On Tuesday, April 7, 2020 at 8:58:53 PM UTC-7, 2G wrote:
>> On Tuesday, April 7, 2020 at 8:06:46 PM UTC-7, Andy Blackburn wrote:
>> > I put a power resistor in the circuit to keep the current surge down.
>I=
> undersized the capacitor so if I mess up on the with rotation I can lose
>p=
>ower. Typically I'll shut off some non-essential equipment to lower the
>dra=
>w if I really don't want a computer reset.
>> >=20
>> > Andy
>> >=20
>> > On Sunday, April 5, 2020 at 8:45:30 AM UTC-7, jfitch wrote:
>> >=20
>> > > What is the inrush current when you first switch the power on? Must
>n=
>ot be enough to blow the fuse, but that'd be something I'd want to
O'scope
>=
>with a current probe.
>>=20
>> You didn't say what the resistor size was. There is already a resistor
>in=
> the circuit - it is called the internal resistance of the battery,wiring
>r=
>esistance, switch contact resistance and the equivalent series resistance
>(=
>ESR) of the capacitor.
>
>I forgot to mention that aluminum electrolytic capacitors have very
>signifi=
>cant inductance at frequencies above 10 KHz. Inductance (also called a
>chok=
>e because they choke off current) limits inrush current. The joules of
>ener=
>gy being transferred is pretty low (about 0.6 J), but I have no problem
>wit=
>h adding a small series resistor (1 ohm - don't worry about the wattage
as
>=
>very little power is supplied by the cap).
>
>Tom
>
Has anyone reported this problem to LXV, and if so what was their response
please? There seems to be an opportunity for a firmware fix to this
problem, not requiring any additional hardware.
In the specification for Flight Recorders, a new IGC file is not started if
power has been interrupted for less than one minute. This is to allow a
change of battery, breaker reset or change a fuse. Exactly the problem we
are discussing.
If the LX firmware were enhanced to inhibit re-calculating QNH where power
has been lost for less than n seconds, this problem goes away.
Discuss.
> >
> Has anyone reported this problem to LXV, and if so what was their response
> please? There seems to be an opportunity for a firmware fix to this
> problem, not requiring any additional hardware.
>
> In the specification for Flight Recorders, a new IGC file is not started if
> power has been interrupted for less than one minute. This is to allow a
> change of battery, breaker reset or change a fuse. Exactly the problem we
> are discussing.
>
> If the LX firmware were enhanced to inhibit re-calculating QNH where power
> has been lost for less than n seconds, this problem goes away.
>
> Discuss.
The LXNav manual says that the internal logger will continue to operate 'for a short time' if power is lost. The manual also says that the proper shutdown procedure should always be used, eg when changing batteries, since if power is suddenly cut the operating system may be scrambled.
Martin Gregorie[_6_]
April 8th 20, 02:54 PM
On Wed, 08 Apr 2020 06:43:22 -0700, towsked wrote:
>> Has anyone reported this problem to LXV, and if so what was their
>> response please? There seems to be an opportunity for a firmware fix to
>> this problem, not requiring any additional hardware.
>>
>> In the specification for Flight Recorders, a new IGC file is not
>> started if power has been interrupted for less than one minute. This is
>> to allow a change of battery, breaker reset or change a fuse. Exactly
>> the problem we are discussing.
>>
>> If the LX firmware were enhanced to inhibit re-calculating QNH where
>> power has been lost for less than n seconds, this problem goes away.
>>
>> Discuss.
>
>
> The LXNav manual says that the internal logger will continue to operate
> 'for a short time' if power is lost. The manual also says that the
> proper shutdown procedure should always be used, eg when changing
> batteries, since if power is suddenly cut the operating system may be
> scrambled.
What OS does LXNav use these days?
I'm asking because this sounds remarkably like similar problems with
WinCE, where it was well-known that powering off without shutting down
properly was almost as certain to corrupt the SD card as yanking the
card out of a running PDA.
--
Martin | martin at
Gregorie | gregorie dot org
kinsell
April 8th 20, 03:57 PM
On 4/8/20 7:54 AM, Martin Gregorie wrote:
> On Wed, 08 Apr 2020 06:43:22 -0700, towsked wrote:
>
>
>>> Has anyone reported this problem to LXV, and if so what was their
>>> response please? There seems to be an opportunity for a firmware fix to
>>> this problem, not requiring any additional hardware.
>>>
>>> In the specification for Flight Recorders, a new IGC file is not
>>> started if power has been interrupted for less than one minute. This is
>>> to allow a change of battery, breaker reset or change a fuse. Exactly
>>> the problem we are discussing.
>>>
>>> If the LX firmware were enhanced to inhibit re-calculating QNH where
>>> power has been lost for less than n seconds, this problem goes away.
>>>
>>> Discuss.
>>
>>
>> The LXNav manual says that the internal logger will continue to operate
>> 'for a short time' if power is lost. The manual also says that the
>> proper shutdown procedure should always be used, eg when changing
>> batteries, since if power is suddenly cut the operating system may be
>> scrambled.
>
> What OS does LXNav use these days?
>
> I'm asking because this sounds remarkably like similar problems with
> WinCE, where it was well-known that powering off without shutting down
> properly was almost as certain to corrupt the SD card as yanking the
> card out of a running PDA.
>
>
The Stratux ads-b receiver has the same issue, high probability of
corrupting the memory if you don't shut it down manually before removing
power. Developers are aware of the problem, but haven't come up with a
proper fix.
Runs an embedded linux, they're not storing any data during normal
operation, so baffling why that's so hard to fix.
-Dave
I havent had to switch batteries in flight lately but my power flarm will reset and start a new log it interupted in flight. Very annoying as I use it almost exclusively for logging as its download is very fast and simple. My backup logger is the venerable SN10B, it does not reset and continues right where it left off, Thanks YO! I have thought about adding a cap to the circuit for the powerflarm. Lots of good ideas here so far but I am trying to keep it as simple as possible.
jfitch
April 8th 20, 05:03 PM
On Tuesday, April 7, 2020 at 6:19:36 PM UTC-7, 2G wrote:
> On Tuesday, April 7, 2020 at 9:32:07 AM UTC-7, Dan Marotta wrote:
> > Well then, I must be really confused.* If the switch was a BREAK before
> > MAKE, the equipment would be momentarily without power.* Some devices
> > can survive that, perhaps with internal capacitance, but others will
> > lose power momentarily.* This may or may not be an issue.
> >
> > Inrush current seems to be a big bugaboo to some people and, in some
> > cases it is, but what is the time constant associated with the inrush?*
> > How much current are we talking about and for how long? People keep
> > talking about the high battery "charging" the low battery during the
> > milliseconds that they are in parallel. Theoretically, yes, practically,
> > hogwash.* The heat required to blow a fuse or burn a wire does not rise
> > instantaneously; there's counter EMF to reduce the current...* So many
> > details that meant a bunch when you were taking a test back in school
> > but, practically speaking, don't mean squat in this case.
> >
> > So, switch your batteries and protect your circuits however you wish.*
> > I'll stick with my make before break switches.* I'll even report back
> > when something fails due to their use, but don't hold your breath.
> >
> > On 4/6/2020 10:43 PM, 2G wrote:
> > > On Monday, April 6, 2020 at 11:23:21 AM UTC-7, Dan Marotta wrote:
> > >> Tom,
> > >>
> > >> Isn't the rotary panel switch in a Schleicher glider a make before break?
> > >>
> > >> OK, I'll stop posting to this thread now.* My solution, which I've
> > >> relied on for years without ever losing a connection or blowing a fuse
> > >> seems to be too simple to be acceptable.
> > >>
> > >> On 4/5/2020 7:45 PM, 2G wrote:
> > >>> The glider, an ASH 31 Mi, is already wired with a battery selector switch for the avionics, and is the way to go. You definitely don't want to parallel Pb and LFP battery's accidentally.
> > >>>
> > >>> Tom
> > >> --
> > >> Dan, 5J
> > > Dan,
> > >
> > > Think about it: if it were MAKE before BREAK I wouldn't have had this problem and sure as hell wouldn't post a solution to a non-existent problem.
> >
> > --
> > Dan, 5J
>
> If my power switch was a break-before-make (also called a shorting switch) I would not have had a problem and certainly wouldn't have posted a fix to a non-existent problem. But it isn't. It was easier for me to put in a capacitor than replace the switch. It also protects against any switch bounce while switching.
>
> People seem to be overly concerned with inrush current. High inrush current would only happen if there were a substantial amount of power supply capacitance in the LX9000 that needed charging. If that were the case it would handle a millisecond battery switch over without any problem, but it doesn't. Also, there is no inrush current during battery switching as the caps in the LX9000 are already charged.
>
> Tom
The inrush current to the LX9000 is not the issue, nor is switching between batteries while the cap is charged. The issue (if there is one) is the inrush to the discharged cap when you first turn on the mains. It is a pretty complex and loosely speced system. Caps are very loosely spec'd, as are battery internal resistance, switch resistance, and of course the wiring in each glider is unique. That's why I suggested measuring it. When he switch is turned on, it will arc and bounce because that is what switches do. How much is the question. Quite easy to measure, hard to calculate accurately. It is quite possible that the step impedance in the system limits the inrush to an acceptable value - but you don't know what you don't know.
It would have been easy for LX to spec a switching power supply with enough residual energy to cover, say 20 ms switching interval. Apparently they did not. Or perhaps they are using linear supplies, in which case the garbage can is the appropriate resting place. I'll second Dave's suggestion of using one large battery bank instead of seperate small ones. It is better for battery life too. (In the 31 the second battery may be the starting battery so that isn't practical). Or better still, use a higher capacity LFP - my small instrument battery will power the panel for around 12 hours.
Dave Nadler
April 8th 20, 05:52 PM
On Wednesday, April 8, 2020 at 12:00:15 PM UTC-4, wrote:
> ... the venerable SN10B, it does not reset and continues right where it
> left off, Thanks YO!
Yes, I designed SN10 this way decades ago, because, Big Surprise!!
....glider instruments see power interruptions !!!
The SN10 has an internal short-term backup supply.
When the power is interrupted, it does a smooth shut-down (OS I wrote).
SN10 memory has a long-term backup battery, and next time power is applied,
the SN10 resumes wherever it left off. SN10s never do a hard reboot unless
new software is installed.
Of course, we have had SN10s damaged when customers added a capacitor
to their logger (powered thru SN10), and the inrush current eventually
blew up the SN10 power switch. Also some loggers had giant internal caps
to try prevent corruption, however...
Sensible design dictates avionics have internal short-term supplies
with power-interrupt sensing and implement smooth shutdown, especially
where they write to flash for logging, saving configuration, etc.
Otherwise things like SD cards or internal flash get corrupted with
data loss when power is interrupted during a write operation.
Of course, lots of design out there is not sensible...
Hope that helps clear some of the confusion,
Best Regards, Dave
Martin Gregorie[_6_]
April 8th 20, 05:58 PM
On Wed, 08 Apr 2020 08:57:43 -0600, kinsell wrote:
> Runs an embedded linux, they're not storing any data during normal
> operation, so baffling why that's so hard to fix.
>
When does it commit data to non-volatile memory, i.e. does in
intentionally buffer it and only commit at intervals or as part of
shutting down?
FWIW The same problem occurs with Raspberry Pis.
As you may or may not know, these run a Debian Linux clone as their OS
and use SD cards for non-volatile memory by default. Much of the time
people can get away with simply pulling the plug when the Pi appears to
be idle, but if you do that while the Pi is flushing its caches to its SD
card or, more rarely, the card is in the middle of a wear-levelling
process, then the SD card will become corrupted and possibly permanently
damaged if it was wear-levelling when the power vanished. Which is why
everybody soon learns to shut the Pi down with a 'sudo stop' command
before powering it off. I think its worse with SD cards simply because
their internal controllers and cheap, rather basic and have no power
buffering. Use an SSD instead or, even better, a hard drive and the
problem largely goes away because ext4 is a journalling filesystem, so
has built-in recovery.
I agree this is a tricky problem, and maybe best solved with some sort of
cheap'n cheerful UPS. Here's a suggestion along those lines:
Pimoroni sell the PowerBoost 1000 Charger, a small and fairly cheap UPS
circuit ($US 19.39 from Amazon), which you connect between a 5v power
supply and the device you want to power via a UPS socket. You also
connect a suitable sized 1S (3.7 volt) Lithium-ion battery to it - the
sort used to power small RC models would be fine - and there's a power
buffer for any UPS-powered device that doesn't have an internal battery.
Its good to supply up to 1000mA provided that the battery is rated for
that current. Some soldering is needed.
It comes with a selection of sockets, but they're all sat loosely in
place on the board so you can solder the ones you want on, sling the
others and solder any permanent connections you need.
I have one but haven't used it yet - I'm planning to make a PDA from a 4"
touch screen and a Pi Zero WL and use this Pimoroni plus an RC model 1S
LiPO battery to power it. Add a case made from epoxyboard and it should
be ready to rock'n roll.
--
Martin | martin at
Gregorie | gregorie dot org
Dave Nadler
April 8th 20, 07:32 PM
On Wednesday, April 8, 2020 at 12:58:39 PM UTC-4, Martin Gregorie wrote:
> On Wed, 08 Apr 2020 08:57:43 -0600, kinsell wrote:
>
> > Runs an embedded linux, they're not storing any data during normal
> > operation, so baffling why that's so hard to fix.
> >
> When does it commit data to non-volatile memory, i.e. does in
> intentionally buffer it and only commit at intervals or as part of
> shutting down?
A common problem is forgetting to put /tmp in RAM.
Leaving it on a uSD will eventually end badly.
> FWIW The same problem occurs with Raspberry Pis.
>
> As you may or may not know, these run a Debian Linux clone as their OS
> and use SD cards for non-volatile memory by default. Much of the time
> people can get away with simply pulling the plug when the Pi appears to
> be idle, but if you do that while the Pi is flushing its caches to its SD
> card or, more rarely, the card is in the middle of a wear-levelling
> process, then the SD card will become corrupted and possibly permanently
> damaged if it was wear-levelling when the power vanished. Which is why
> everybody soon learns to shut the Pi down with a 'sudo stop' command
> before powering it off. I think its worse with SD cards simply because
> their internal controllers and cheap, rather basic and have no power
> buffering. Use an SSD instead or, even better, a hard drive and the
> problem largely goes away because ext4 is a journalling filesystem, so
> has built-in recovery.
>
> I agree this is a tricky problem, and maybe best solved with some sort of
> cheap'n cheerful UPS. Here's a suggestion along those lines:
>
> Pimoroni sell the PowerBoost 1000 Charger, a small and fairly cheap UPS
> circuit ($US 19.39 from Amazon), which you connect between a 5v power
> supply and the device you want to power via a UPS socket. You also
> connect a suitable sized 1S (3.7 volt) Lithium-ion battery to it - the
> sort used to power small RC models would be fine - and there's a power
> buffer for any UPS-powered device that doesn't have an internal battery.
> Its good to supply up to 1000mA provided that the battery is rated for
> that current. Some soldering is needed.
>
> It comes with a selection of sockets, but they're all sat loosely in
> place on the board so you can solder the ones you want on, sling the
> others and solder any permanent connections you need.
>
> I have one but haven't used it yet - I'm planning to make a PDA from a 4"
> touch screen and a Pi Zero WL and use this Pimoroni plus an RC model 1S
> LiPO battery to power it. Add a case made from epoxyboard and it should
> be ready to rock'n roll.
Another Pi solution is this one with superCap: https://juice4halt.com/
Beware the window before supercap completely charged (should finish
shortly after boot)...
Enjoy,
Best Regards, Dave
On Wednesday, April 8, 2020 at 9:04:00 AM UTC-7, jfitch wrote:
> On Tuesday, April 7, 2020 at 6:19:36 PM UTC-7, 2G wrote:
> > On Tuesday, April 7, 2020 at 9:32:07 AM UTC-7, Dan Marotta wrote:
> > > Well then, I must be really confused.* If the switch was a BREAK before
> > > MAKE, the equipment would be momentarily without power.* Some devices
> > > can survive that, perhaps with internal capacitance, but others will
> > > lose power momentarily.* This may or may not be an issue.
> > >
> > > Inrush current seems to be a big bugaboo to some people and, in some
> > > cases it is, but what is the time constant associated with the inrush?*
> > > How much current are we talking about and for how long? People keep
> > > talking about the high battery "charging" the low battery during the
> > > milliseconds that they are in parallel. Theoretically, yes, practically,
> > > hogwash.* The heat required to blow a fuse or burn a wire does not rise
> > > instantaneously; there's counter EMF to reduce the current...* So many
> > > details that meant a bunch when you were taking a test back in school
> > > but, practically speaking, don't mean squat in this case.
> > >
> > > So, switch your batteries and protect your circuits however you wish.*
> > > I'll stick with my make before break switches.* I'll even report back
> > > when something fails due to their use, but don't hold your breath.
> > >
> > > On 4/6/2020 10:43 PM, 2G wrote:
> > > > On Monday, April 6, 2020 at 11:23:21 AM UTC-7, Dan Marotta wrote:
> > > >> Tom,
> > > >>
> > > >> Isn't the rotary panel switch in a Schleicher glider a make before break?
> > > >>
> > > >> OK, I'll stop posting to this thread now.* My solution, which I've
> > > >> relied on for years without ever losing a connection or blowing a fuse
> > > >> seems to be too simple to be acceptable.
> > > >>
> > > >> On 4/5/2020 7:45 PM, 2G wrote:
> > > >>> The glider, an ASH 31 Mi, is already wired with a battery selector switch for the avionics, and is the way to go. You definitely don't want to parallel Pb and LFP battery's accidentally.
> > > >>>
> > > >>> Tom
> > > >> --
> > > >> Dan, 5J
> > > > Dan,
> > > >
> > > > Think about it: if it were MAKE before BREAK I wouldn't have had this problem and sure as hell wouldn't post a solution to a non-existent problem.
> > >
> > > --
> > > Dan, 5J
> >
> > If my power switch was a break-before-make (also called a shorting switch) I would not have had a problem and certainly wouldn't have posted a fix to a non-existent problem. But it isn't. It was easier for me to put in a capacitor than replace the switch. It also protects against any switch bounce while switching.
> >
> > People seem to be overly concerned with inrush current. High inrush current would only happen if there were a substantial amount of power supply capacitance in the LX9000 that needed charging. If that were the case it would handle a millisecond battery switch over without any problem, but it doesn't. Also, there is no inrush current during battery switching as the caps in the LX9000 are already charged.
> >
> > Tom
>
> The inrush current to the LX9000 is not the issue, nor is switching between batteries while the cap is charged. The issue (if there is one) is the inrush to the discharged cap when you first turn on the mains. It is a pretty complex and loosely speced system. Caps are very loosely spec'd, as are battery internal resistance, switch resistance, and of course the wiring in each glider is unique. That's why I suggested measuring it. When he switch is turned on, it will arc and bounce because that is what switches do. How much is the question. Quite easy to measure, hard to calculate accurately. It is quite possible that the step impedance in the system limits the inrush to an acceptable value - but you don't know what you don't know.
>
> It would have been easy for LX to spec a switching power supply with enough residual energy to cover, say 20 ms switching interval. Apparently they did not. Or perhaps they are using linear supplies, in which case the garbage can is the appropriate resting place. I'll second Dave's suggestion of using one large battery bank instead of seperate small ones. It is better for battery life too. (In the 31 the second battery may be the starting battery so that isn't practical). Or better still, use a higher capacity LFP - my small instrument battery will power the panel for around 12 hours.
All decent suggestions. I can deduce the series resistance from the time constant of the waveforms, which is about 200 mohm. This includes the battery internal resistance, wiring resistance, and capacitor ESR. So forget about "hundreds of amps" flowing. Any more measurements will have to wait until I get out to the airport again, but including a series resistor of a few ohms, depending upon you total panel current drain, won't hurt. I did look up the NKK switch you mentioned and that is definitely not the one used in my panel.
Tom
kinsell
April 10th 20, 10:19 PM
On 4/8/20 10:58 AM, Martin Gregorie wrote:
> On Wed, 08 Apr 2020 08:57:43 -0600, kinsell wrote:
>
>> Runs an embedded linux, they're not storing any data during normal
>> operation, so baffling why that's so hard to fix.
>>
> When does it commit data to non-volatile memory, i.e. does in
> intentionally buffer it and only commit at intervals or as part of
> shutting down?
>
> FWIW The same problem occurs with Raspberry Pis.
>
> As you may or may not know, these run a Debian Linux clone as their OS
> and use SD cards for non-volatile memory by default. Much of the time
> people can get away with simply pulling the plug when the Pi appears to
> be idle, but if you do that while the Pi is flushing its caches to its SD
> card or, more rarely, the card is in the middle of a wear-levelling
> process, then the SD card will become corrupted and possibly permanently
> damaged if it was wear-levelling when the power vanished. Which is why
> everybody soon learns to shut the Pi down with a 'sudo stop' command
> before powering it off. I think its worse with SD cards simply because
> their internal controllers and cheap, rather basic and have no power
> buffering. Use an SSD instead or, even better, a hard drive and the
> problem largely goes away because ext4 is a journalling filesystem, so
> has built-in recovery.
>
> I agree this is a tricky problem, and maybe best solved with some sort of
> cheap'n cheerful UPS. Here's a suggestion along those lines:
>
> Pimoroni sell the PowerBoost 1000 Charger, a small and fairly cheap UPS
> circuit ($US 19.39 from Amazon), which you connect between a 5v power
> supply and the device you want to power via a UPS socket. You also
> connect a suitable sized 1S (3.7 volt) Lithium-ion battery to it - the
> sort used to power small RC models would be fine - and there's a power
> buffer for any UPS-powered device that doesn't have an internal battery.
> Its good to supply up to 1000mA provided that the battery is rated for
> that current. Some soldering is needed.
>
> It comes with a selection of sockets, but they're all sat loosely in
> place on the board so you can solder the ones you want on, sling the
> others and solder any permanent connections you need.
>
> I have one but haven't used it yet - I'm planning to make a PDA from a 4"
> touch screen and a Pi Zero WL and use this Pimoroni plus an RC model 1S
> LiPO battery to power it. Add a case made from epoxyboard and it should
> be ready to rock'n roll.
>
>
The Stratux project does run on a Rspberry Pi using the micro-SD for
storage. Yes, running other software on that hardware can lead to the
same file system corruption problem. What's interesting is I can load
FlightAware software on the same hardware and never see the problem. In
both applications, there is no user data that needs to be stored to
flash memory, other than a tiny bit of configuration data when they are
first set up. It is possible to run on a read-only file system, if some
provision is made for the configuration data. People have experimented
with this, but the images distributed for Stratux have never had a
solution that I'm aware of.
So many devices these days have a computer with flash memory. If your
smart TV becomes unbootable after a power fail, would that be considered
acceptable? Of course not. But for some reason Stratux doesn't seem to
get fixed. Putting a UPS on a microcomputer seems like an ugly Band-Aid
(rtm) that shouldn't be necessary. Just a simple file system check on
power up might be adequate. There's a Sentry Mini receiver available
now for just $300, I'll bet it doesn't lose its mind if you just pull power.
-Dave
jfitch
April 11th 20, 12:55 AM
On Wednesday, April 8, 2020 at 2:51:22 PM UTC-7, 2G wrote:
> On Wednesday, April 8, 2020 at 9:04:00 AM UTC-7, jfitch wrote:
> > On Tuesday, April 7, 2020 at 6:19:36 PM UTC-7, 2G wrote:
> > > On Tuesday, April 7, 2020 at 9:32:07 AM UTC-7, Dan Marotta wrote:
> > > > Well then, I must be really confused.* If the switch was a BREAK before
> > > > MAKE, the equipment would be momentarily without power.* Some devices
> > > > can survive that, perhaps with internal capacitance, but others will
> > > > lose power momentarily.* This may or may not be an issue.
> > > >
> > > > Inrush current seems to be a big bugaboo to some people and, in some
> > > > cases it is, but what is the time constant associated with the inrush?*
> > > > How much current are we talking about and for how long? People keep
> > > > talking about the high battery "charging" the low battery during the
> > > > milliseconds that they are in parallel. Theoretically, yes, practically,
> > > > hogwash.* The heat required to blow a fuse or burn a wire does not rise
> > > > instantaneously; there's counter EMF to reduce the current...* So many
> > > > details that meant a bunch when you were taking a test back in school
> > > > but, practically speaking, don't mean squat in this case.
> > > >
> > > > So, switch your batteries and protect your circuits however you wish.*
> > > > I'll stick with my make before break switches.* I'll even report back
> > > > when something fails due to their use, but don't hold your breath.
> > > >
> > > > On 4/6/2020 10:43 PM, 2G wrote:
> > > > > On Monday, April 6, 2020 at 11:23:21 AM UTC-7, Dan Marotta wrote:
> > > > >> Tom,
> > > > >>
> > > > >> Isn't the rotary panel switch in a Schleicher glider a make before break?
> > > > >>
> > > > >> OK, I'll stop posting to this thread now.* My solution, which I've
> > > > >> relied on for years without ever losing a connection or blowing a fuse
> > > > >> seems to be too simple to be acceptable.
> > > > >>
> > > > >> On 4/5/2020 7:45 PM, 2G wrote:
> > > > >>> The glider, an ASH 31 Mi, is already wired with a battery selector switch for the avionics, and is the way to go. You definitely don't want to parallel Pb and LFP battery's accidentally.
> > > > >>>
> > > > >>> Tom
> > > > >> --
> > > > >> Dan, 5J
> > > > > Dan,
> > > > >
> > > > > Think about it: if it were MAKE before BREAK I wouldn't have had this problem and sure as hell wouldn't post a solution to a non-existent problem.
> > > >
> > > > --
> > > > Dan, 5J
> > >
> > > If my power switch was a break-before-make (also called a shorting switch) I would not have had a problem and certainly wouldn't have posted a fix to a non-existent problem. But it isn't. It was easier for me to put in a capacitor than replace the switch. It also protects against any switch bounce while switching.
> > >
> > > People seem to be overly concerned with inrush current. High inrush current would only happen if there were a substantial amount of power supply capacitance in the LX9000 that needed charging. If that were the case it would handle a millisecond battery switch over without any problem, but it doesn't. Also, there is no inrush current during battery switching as the caps in the LX9000 are already charged.
> > >
> > > Tom
> >
> > The inrush current to the LX9000 is not the issue, nor is switching between batteries while the cap is charged. The issue (if there is one) is the inrush to the discharged cap when you first turn on the mains. It is a pretty complex and loosely speced system. Caps are very loosely spec'd, as are battery internal resistance, switch resistance, and of course the wiring in each glider is unique. That's why I suggested measuring it. When he switch is turned on, it will arc and bounce because that is what switches do. How much is the question. Quite easy to measure, hard to calculate accurately.. It is quite possible that the step impedance in the system limits the inrush to an acceptable value - but you don't know what you don't know.
> >
> > It would have been easy for LX to spec a switching power supply with enough residual energy to cover, say 20 ms switching interval. Apparently they did not. Or perhaps they are using linear supplies, in which case the garbage can is the appropriate resting place. I'll second Dave's suggestion of using one large battery bank instead of seperate small ones. It is better for battery life too. (In the 31 the second battery may be the starting battery so that isn't practical). Or better still, use a higher capacity LFP - my small instrument battery will power the panel for around 12 hours.
>
> All decent suggestions. I can deduce the series resistance from the time constant of the waveforms, which is about 200 mohm. This includes the battery internal resistance, wiring resistance, and capacitor ESR. So forget about "hundreds of amps" flowing. Any more measurements will have to wait until I get out to the airport again, but including a series resistor of a few ohms, depending upon you total panel current drain, won't hurt. I did look up the NKK switch you mentioned and that is definitely not the one used in my panel.
>
> Tom
What switch did Schleicher use on your 31? Mine originally had that NKK switch, searching for a replacement I could not find any with better specs than the NKK MRT. There are many different manufacturers of an equivalent MRY106, but all seemed to mirror the spec.
jfitch
April 11th 20, 12:58 AM
On Wednesday, April 8, 2020 at 2:51:22 PM UTC-7, 2G wrote:
> On Wednesday, April 8, 2020 at 9:04:00 AM UTC-7, jfitch wrote:
> > On Tuesday, April 7, 2020 at 6:19:36 PM UTC-7, 2G wrote:
> > > On Tuesday, April 7, 2020 at 9:32:07 AM UTC-7, Dan Marotta wrote:
> > > > Well then, I must be really confused.* If the switch was a BREAK before
> > > > MAKE, the equipment would be momentarily without power.* Some devices
> > > > can survive that, perhaps with internal capacitance, but others will
> > > > lose power momentarily.* This may or may not be an issue.
> > > >
> > > > Inrush current seems to be a big bugaboo to some people and, in some
> > > > cases it is, but what is the time constant associated with the inrush?*
> > > > How much current are we talking about and for how long? People keep
> > > > talking about the high battery "charging" the low battery during the
> > > > milliseconds that they are in parallel. Theoretically, yes, practically,
> > > > hogwash.* The heat required to blow a fuse or burn a wire does not rise
> > > > instantaneously; there's counter EMF to reduce the current...* So many
> > > > details that meant a bunch when you were taking a test back in school
> > > > but, practically speaking, don't mean squat in this case.
> > > >
> > > > So, switch your batteries and protect your circuits however you wish.*
> > > > I'll stick with my make before break switches.* I'll even report back
> > > > when something fails due to their use, but don't hold your breath.
> > > >
> > > > On 4/6/2020 10:43 PM, 2G wrote:
> > > > > On Monday, April 6, 2020 at 11:23:21 AM UTC-7, Dan Marotta wrote:
> > > > >> Tom,
> > > > >>
> > > > >> Isn't the rotary panel switch in a Schleicher glider a make before break?
> > > > >>
> > > > >> OK, I'll stop posting to this thread now.* My solution, which I've
> > > > >> relied on for years without ever losing a connection or blowing a fuse
> > > > >> seems to be too simple to be acceptable.
> > > > >>
> > > > >> On 4/5/2020 7:45 PM, 2G wrote:
> > > > >>> The glider, an ASH 31 Mi, is already wired with a battery selector switch for the avionics, and is the way to go. You definitely don't want to parallel Pb and LFP battery's accidentally.
> > > > >>>
> > > > >>> Tom
> > > > >> --
> > > > >> Dan, 5J
> > > > > Dan,
> > > > >
> > > > > Think about it: if it were MAKE before BREAK I wouldn't have had this problem and sure as hell wouldn't post a solution to a non-existent problem.
> > > >
> > > > --
> > > > Dan, 5J
> > >
> > > If my power switch was a break-before-make (also called a shorting switch) I would not have had a problem and certainly wouldn't have posted a fix to a non-existent problem. But it isn't. It was easier for me to put in a capacitor than replace the switch. It also protects against any switch bounce while switching.
> > >
> > > People seem to be overly concerned with inrush current. High inrush current would only happen if there were a substantial amount of power supply capacitance in the LX9000 that needed charging. If that were the case it would handle a millisecond battery switch over without any problem, but it doesn't. Also, there is no inrush current during battery switching as the caps in the LX9000 are already charged.
> > >
> > > Tom
> >
> > The inrush current to the LX9000 is not the issue, nor is switching between batteries while the cap is charged. The issue (if there is one) is the inrush to the discharged cap when you first turn on the mains. It is a pretty complex and loosely speced system. Caps are very loosely spec'd, as are battery internal resistance, switch resistance, and of course the wiring in each glider is unique. That's why I suggested measuring it. When he switch is turned on, it will arc and bounce because that is what switches do. How much is the question. Quite easy to measure, hard to calculate accurately.. It is quite possible that the step impedance in the system limits the inrush to an acceptable value - but you don't know what you don't know.
> >
> > It would have been easy for LX to spec a switching power supply with enough residual energy to cover, say 20 ms switching interval. Apparently they did not. Or perhaps they are using linear supplies, in which case the garbage can is the appropriate resting place. I'll second Dave's suggestion of using one large battery bank instead of seperate small ones. It is better for battery life too. (In the 31 the second battery may be the starting battery so that isn't practical). Or better still, use a higher capacity LFP - my small instrument battery will power the panel for around 12 hours.
>
> All decent suggestions. I can deduce the series resistance from the time constant of the waveforms, which is about 200 mohm. This includes the battery internal resistance, wiring resistance, and capacitor ESR. So forget about "hundreds of amps" flowing. Any more measurements will have to wait until I get out to the airport again, but including a series resistor of a few ohms, depending upon you total panel current drain, won't hurt. I did look up the NKK switch you mentioned and that is definitely not the one used in my panel.
>
> Tom
200 mohms seems like it would still send a lot of current through the switch, at least briefly. 70A?
2G
April 11th 20, 02:03 AM
On Friday, April 10, 2020 at 4:55:12 PM UTC-7, jfitch wrote:
> On Wednesday, April 8, 2020 at 2:51:22 PM UTC-7, 2G wrote:
> > On Wednesday, April 8, 2020 at 9:04:00 AM UTC-7, jfitch wrote:
> > > On Tuesday, April 7, 2020 at 6:19:36 PM UTC-7, 2G wrote:
> > > > On Tuesday, April 7, 2020 at 9:32:07 AM UTC-7, Dan Marotta wrote:
> > > > > Well then, I must be really confused.* If the switch was a BREAK before
> > > > > MAKE, the equipment would be momentarily without power.* Some devices
> > > > > can survive that, perhaps with internal capacitance, but others will
> > > > > lose power momentarily.* This may or may not be an issue.
> > > > >
> > > > > Inrush current seems to be a big bugaboo to some people and, in some
> > > > > cases it is, but what is the time constant associated with the inrush?*
> > > > > How much current are we talking about and for how long? People keep
> > > > > talking about the high battery "charging" the low battery during the
> > > > > milliseconds that they are in parallel. Theoretically, yes, practically,
> > > > > hogwash.* The heat required to blow a fuse or burn a wire does not rise
> > > > > instantaneously; there's counter EMF to reduce the current...* So many
> > > > > details that meant a bunch when you were taking a test back in school
> > > > > but, practically speaking, don't mean squat in this case.
> > > > >
> > > > > So, switch your batteries and protect your circuits however you wish.*
> > > > > I'll stick with my make before break switches.* I'll even report back
> > > > > when something fails due to their use, but don't hold your breath..
> > > > >
> > > > > On 4/6/2020 10:43 PM, 2G wrote:
> > > > > > On Monday, April 6, 2020 at 11:23:21 AM UTC-7, Dan Marotta wrote:
> > > > > >> Tom,
> > > > > >>
> > > > > >> Isn't the rotary panel switch in a Schleicher glider a make before break?
> > > > > >>
> > > > > >> OK, I'll stop posting to this thread now.* My solution, which I've
> > > > > >> relied on for years without ever losing a connection or blowing a fuse
> > > > > >> seems to be too simple to be acceptable.
> > > > > >>
> > > > > >> On 4/5/2020 7:45 PM, 2G wrote:
> > > > > >>> The glider, an ASH 31 Mi, is already wired with a battery selector switch for the avionics, and is the way to go. You definitely don't want to parallel Pb and LFP battery's accidentally.
> > > > > >>>
> > > > > >>> Tom
> > > > > >> --
> > > > > >> Dan, 5J
> > > > > > Dan,
> > > > > >
> > > > > > Think about it: if it were MAKE before BREAK I wouldn't have had this problem and sure as hell wouldn't post a solution to a non-existent problem.
> > > > >
> > > > > --
> > > > > Dan, 5J
> > > >
> > > > If my power switch was a break-before-make (also called a shorting switch) I would not have had a problem and certainly wouldn't have posted a fix to a non-existent problem. But it isn't. It was easier for me to put in a capacitor than replace the switch. It also protects against any switch bounce while switching.
> > > >
> > > > People seem to be overly concerned with inrush current. High inrush current would only happen if there were a substantial amount of power supply capacitance in the LX9000 that needed charging. If that were the case it would handle a millisecond battery switch over without any problem, but it doesn't. Also, there is no inrush current during battery switching as the caps in the LX9000 are already charged.
> > > >
> > > > Tom
> > >
> > > The inrush current to the LX9000 is not the issue, nor is switching between batteries while the cap is charged. The issue (if there is one) is the inrush to the discharged cap when you first turn on the mains. It is a pretty complex and loosely speced system. Caps are very loosely spec'd, as are battery internal resistance, switch resistance, and of course the wiring in each glider is unique. That's why I suggested measuring it. When he switch is turned on, it will arc and bounce because that is what switches do. How much is the question. Quite easy to measure, hard to calculate accurately. It is quite possible that the step impedance in the system limits the inrush to an acceptable value - but you don't know what you don't know.
> > >
> > > It would have been easy for LX to spec a switching power supply with enough residual energy to cover, say 20 ms switching interval. Apparently they did not. Or perhaps they are using linear supplies, in which case the garbage can is the appropriate resting place. I'll second Dave's suggestion of using one large battery bank instead of seperate small ones. It is better for battery life too. (In the 31 the second battery may be the starting battery so that isn't practical). Or better still, use a higher capacity LFP - my small instrument battery will power the panel for around 12 hours.
> >
> > All decent suggestions. I can deduce the series resistance from the time constant of the waveforms, which is about 200 mohm. This includes the battery internal resistance, wiring resistance, and capacitor ESR. So forget about "hundreds of amps" flowing. Any more measurements will have to wait until I get out to the airport again, but including a series resistor of a few ohms, depending upon you total panel current drain, won't hurt. I did look up the NKK switch you mentioned and that is definitely not the one used in my panel.
> >
> > Tom
>
> What switch did Schleicher use on your 31? Mine originally had that NKK switch, searching for a replacement I could not find any with better specs than the NKK MRT. There are many different manufacturers of an equivalent MRY106, but all seemed to mirror the spec.
I don't know what they used, but it is definitely not the NKK model you posted.
Tom
2G
April 11th 20, 02:07 AM
On Friday, April 10, 2020 at 4:58:04 PM UTC-7, jfitch wrote:
> On Wednesday, April 8, 2020 at 2:51:22 PM UTC-7, 2G wrote:
> > On Wednesday, April 8, 2020 at 9:04:00 AM UTC-7, jfitch wrote:
> > > On Tuesday, April 7, 2020 at 6:19:36 PM UTC-7, 2G wrote:
> > > > On Tuesday, April 7, 2020 at 9:32:07 AM UTC-7, Dan Marotta wrote:
> > > > > Well then, I must be really confused.* If the switch was a BREAK before
> > > > > MAKE, the equipment would be momentarily without power.* Some devices
> > > > > can survive that, perhaps with internal capacitance, but others will
> > > > > lose power momentarily.* This may or may not be an issue.
> > > > >
> > > > > Inrush current seems to be a big bugaboo to some people and, in some
> > > > > cases it is, but what is the time constant associated with the inrush?*
> > > > > How much current are we talking about and for how long? People keep
> > > > > talking about the high battery "charging" the low battery during the
> > > > > milliseconds that they are in parallel. Theoretically, yes, practically,
> > > > > hogwash.* The heat required to blow a fuse or burn a wire does not rise
> > > > > instantaneously; there's counter EMF to reduce the current...* So many
> > > > > details that meant a bunch when you were taking a test back in school
> > > > > but, practically speaking, don't mean squat in this case.
> > > > >
> > > > > So, switch your batteries and protect your circuits however you wish.*
> > > > > I'll stick with my make before break switches.* I'll even report back
> > > > > when something fails due to their use, but don't hold your breath..
> > > > >
> > > > > On 4/6/2020 10:43 PM, 2G wrote:
> > > > > > On Monday, April 6, 2020 at 11:23:21 AM UTC-7, Dan Marotta wrote:
> > > > > >> Tom,
> > > > > >>
> > > > > >> Isn't the rotary panel switch in a Schleicher glider a make before break?
> > > > > >>
> > > > > >> OK, I'll stop posting to this thread now.* My solution, which I've
> > > > > >> relied on for years without ever losing a connection or blowing a fuse
> > > > > >> seems to be too simple to be acceptable.
> > > > > >>
> > > > > >> On 4/5/2020 7:45 PM, 2G wrote:
> > > > > >>> The glider, an ASH 31 Mi, is already wired with a battery selector switch for the avionics, and is the way to go. You definitely don't want to parallel Pb and LFP battery's accidentally.
> > > > > >>>
> > > > > >>> Tom
> > > > > >> --
> > > > > >> Dan, 5J
> > > > > > Dan,
> > > > > >
> > > > > > Think about it: if it were MAKE before BREAK I wouldn't have had this problem and sure as hell wouldn't post a solution to a non-existent problem.
> > > > >
> > > > > --
> > > > > Dan, 5J
> > > >
> > > > If my power switch was a break-before-make (also called a shorting switch) I would not have had a problem and certainly wouldn't have posted a fix to a non-existent problem. But it isn't. It was easier for me to put in a capacitor than replace the switch. It also protects against any switch bounce while switching.
> > > >
> > > > People seem to be overly concerned with inrush current. High inrush current would only happen if there were a substantial amount of power supply capacitance in the LX9000 that needed charging. If that were the case it would handle a millisecond battery switch over without any problem, but it doesn't. Also, there is no inrush current during battery switching as the caps in the LX9000 are already charged.
> > > >
> > > > Tom
> > >
> > > The inrush current to the LX9000 is not the issue, nor is switching between batteries while the cap is charged. The issue (if there is one) is the inrush to the discharged cap when you first turn on the mains. It is a pretty complex and loosely speced system. Caps are very loosely spec'd, as are battery internal resistance, switch resistance, and of course the wiring in each glider is unique. That's why I suggested measuring it. When he switch is turned on, it will arc and bounce because that is what switches do. How much is the question. Quite easy to measure, hard to calculate accurately. It is quite possible that the step impedance in the system limits the inrush to an acceptable value - but you don't know what you don't know.
> > >
> > > It would have been easy for LX to spec a switching power supply with enough residual energy to cover, say 20 ms switching interval. Apparently they did not. Or perhaps they are using linear supplies, in which case the garbage can is the appropriate resting place. I'll second Dave's suggestion of using one large battery bank instead of seperate small ones. It is better for battery life too. (In the 31 the second battery may be the starting battery so that isn't practical). Or better still, use a higher capacity LFP - my small instrument battery will power the panel for around 12 hours.
> >
> > All decent suggestions. I can deduce the series resistance from the time constant of the waveforms, which is about 200 mohm. This includes the battery internal resistance, wiring resistance, and capacitor ESR. So forget about "hundreds of amps" flowing. Any more measurements will have to wait until I get out to the airport again, but including a series resistor of a few ohms, depending upon you total panel current drain, won't hurt. I did look up the NKK switch you mentioned and that is definitely not the one used in my panel.
> >
> > Tom
>
> 200 mohms seems like it would still send a lot of current through the switch, at least briefly. 70A?
I modeled it in Spice, included series inductance, and saw peak currents in the range of 10-20A for a few msec.
Tom
On Wednesday, April 8, 2020 at 12:55:49 AM UTC-4, kinsell wrote:
>
> The really funny thing is, most people reading this have no need at all
> to ever switch batteries. Instead of two small batteries, one big one
> is so much easier to manage. If they have to be broken into multiple
> units, then just wire them in parallel and let them all provide power
> until depleted. If you switch them, then you risk switching too soon
> and wasting capacity in the first one, or switching too late and ruining
> a flight log or messing up a flight computer when you key the mike and
> don't realize how weak the battery was.
>
> -Dave
Funny that nobody responded to this comment from Dave. I think he's right on. Other than motorgliders that need an engine-starting battery separate from the avionics, why do we need a 2-battery setup? I've flown with a single battery for 25 years now, and have NEVER had a problem with that. Had plenty of other glitches in flight recorders etc, but not the battery's fault. Having two batteries (perhaps one in the tail) with separate wires to the panel can add capacity and redundancy even if they are simply paralleled within the panel.
Can one battery go bad (shorted cell) and load down the other one? Theoretically yes. Not likely, if you test your batteries once a season and stop using any that show a decline in capacity. But you can have a separate on/off switch for each battery, and normally have both turned on for the whole flight. If you have a voltmeter in the panel (some radios have it built-in) you can turn off one battery at a time just to check the condition of the other one.
2G
April 13th 20, 04:51 AM
On Saturday, April 11, 2020 at 1:39:09 PM UTC-7, wrote:
> On Wednesday, April 8, 2020 at 12:55:49 AM UTC-4, kinsell wrote:
> >
> > The really funny thing is, most people reading this have no need at all
> > to ever switch batteries. Instead of two small batteries, one big one
> > is so much easier to manage. If they have to be broken into multiple
> > units, then just wire them in parallel and let them all provide power
> > until depleted. If you switch them, then you risk switching too soon
> > and wasting capacity in the first one, or switching too late and ruining
> > a flight log or messing up a flight computer when you key the mike and
> > don't realize how weak the battery was.
> >
> > -Dave
>
> Funny that nobody responded to this comment from Dave. I think he's right on. Other than motorgliders that need an engine-starting battery separate from the avionics, why do we need a 2-battery setup? I've flown with a single battery for 25 years now, and have NEVER had a problem with that. Had plenty of other glitches in flight recorders etc, but not the battery's fault. Having two batteries (perhaps one in the tail) with separate wires to the panel can add capacity and redundancy even if they are simply paralleled within the panel.
>
> Can one battery go bad (shorted cell) and load down the other one? Theoretically yes. Not likely, if you test your batteries once a season and stop using any that show a decline in capacity. But you can have a separate on/off switch for each battery, and normally have both turned on for the whole flight. If you have a voltmeter in the panel (some radios have it built-in) you can turn off one battery at a time just to check the condition of the other one.
That's a reasonable question and there is, of course, a reasonable answer. You need a backup battery if your main battery fails in flight.
2G
April 13th 20, 04:55 AM
On Friday, April 10, 2020 at 6:07:46 PM UTC-7, 2G wrote:
> On Friday, April 10, 2020 at 4:58:04 PM UTC-7, jfitch wrote:
> > On Wednesday, April 8, 2020 at 2:51:22 PM UTC-7, 2G wrote:
> > > On Wednesday, April 8, 2020 at 9:04:00 AM UTC-7, jfitch wrote:
> > > > On Tuesday, April 7, 2020 at 6:19:36 PM UTC-7, 2G wrote:
> > > > > On Tuesday, April 7, 2020 at 9:32:07 AM UTC-7, Dan Marotta wrote:
> > > > > > Well then, I must be really confused.* If the switch was a BREAK before
> > > > > > MAKE, the equipment would be momentarily without power.* Some devices
> > > > > > can survive that, perhaps with internal capacitance, but others will
> > > > > > lose power momentarily.* This may or may not be an issue.
> > > > > >
> > > > > > Inrush current seems to be a big bugaboo to some people and, in some
> > > > > > cases it is, but what is the time constant associated with the inrush?*
> > > > > > How much current are we talking about and for how long? People keep
> > > > > > talking about the high battery "charging" the low battery during the
> > > > > > milliseconds that they are in parallel. Theoretically, yes, practically,
> > > > > > hogwash.* The heat required to blow a fuse or burn a wire does not rise
> > > > > > instantaneously; there's counter EMF to reduce the current...* So many
> > > > > > details that meant a bunch when you were taking a test back in school
> > > > > > but, practically speaking, don't mean squat in this case.
> > > > > >
> > > > > > So, switch your batteries and protect your circuits however you wish.*
> > > > > > I'll stick with my make before break switches.* I'll even report back
> > > > > > when something fails due to their use, but don't hold your breath.
> > > > > >
> > > > > > On 4/6/2020 10:43 PM, 2G wrote:
> > > > > > > On Monday, April 6, 2020 at 11:23:21 AM UTC-7, Dan Marotta wrote:
> > > > > > >> Tom,
> > > > > > >>
> > > > > > >> Isn't the rotary panel switch in a Schleicher glider a make before break?
> > > > > > >>
> > > > > > >> OK, I'll stop posting to this thread now.* My solution, which I've
> > > > > > >> relied on for years without ever losing a connection or blowing a fuse
> > > > > > >> seems to be too simple to be acceptable.
> > > > > > >>
> > > > > > >> On 4/5/2020 7:45 PM, 2G wrote:
> > > > > > >>> The glider, an ASH 31 Mi, is already wired with a battery selector switch for the avionics, and is the way to go. You definitely don't want to parallel Pb and LFP battery's accidentally.
> > > > > > >>>
> > > > > > >>> Tom
> > > > > > >> --
> > > > > > >> Dan, 5J
> > > > > > > Dan,
> > > > > > >
> > > > > > > Think about it: if it were MAKE before BREAK I wouldn't have had this problem and sure as hell wouldn't post a solution to a non-existent problem.
> > > > > >
> > > > > > --
> > > > > > Dan, 5J
> > > > >
> > > > > If my power switch was a break-before-make (also called a shorting switch) I would not have had a problem and certainly wouldn't have posted a fix to a non-existent problem. But it isn't. It was easier for me to put in a capacitor than replace the switch. It also protects against any switch bounce while switching.
> > > > >
> > > > > People seem to be overly concerned with inrush current. High inrush current would only happen if there were a substantial amount of power supply capacitance in the LX9000 that needed charging. If that were the case it would handle a millisecond battery switch over without any problem, but it doesn't. Also, there is no inrush current during battery switching as the caps in the LX9000 are already charged.
> > > > >
> > > > > Tom
> > > >
> > > > The inrush current to the LX9000 is not the issue, nor is switching between batteries while the cap is charged. The issue (if there is one) is the inrush to the discharged cap when you first turn on the mains. It is a pretty complex and loosely speced system. Caps are very loosely spec'd, as are battery internal resistance, switch resistance, and of course the wiring in each glider is unique. That's why I suggested measuring it. When he switch is turned on, it will arc and bounce because that is what switches do.. How much is the question. Quite easy to measure, hard to calculate accurately. It is quite possible that the step impedance in the system limits the inrush to an acceptable value - but you don't know what you don't know.
> > > >
> > > > It would have been easy for LX to spec a switching power supply with enough residual energy to cover, say 20 ms switching interval. Apparently they did not. Or perhaps they are using linear supplies, in which case the garbage can is the appropriate resting place. I'll second Dave's suggestion of using one large battery bank instead of seperate small ones. It is better for battery life too. (In the 31 the second battery may be the starting battery so that isn't practical). Or better still, use a higher capacity LFP - my small instrument battery will power the panel for around 12 hours.
> > >
> > > All decent suggestions. I can deduce the series resistance from the time constant of the waveforms, which is about 200 mohm. This includes the battery internal resistance, wiring resistance, and capacitor ESR. So forget about "hundreds of amps" flowing. Any more measurements will have to wait until I get out to the airport again, but including a series resistor of a few ohms, depending upon you total panel current drain, won't hurt. I did look up the NKK switch you mentioned and that is definitely not the one used in my panel.
> > >
> > > Tom
> >
> > 200 mohms seems like it would still send a lot of current through the switch, at least briefly. 70A?
>
> I modeled it in Spice, included series inductance, and saw peak currents in the range of 10-20A for a few msec.
>
> Tom
Well, today I did measure the inrush current: the peak current was 9A, very close to what I had simulated with Spice. This current is very brief and totally within the capability of the switch to handle, but a few ohms of series resistance will cut it down to a couple of amps if you are anal about it.
Tim Newport-Peace[_6_]
April 13th 20, 11:36 AM
At 03:55 13 April 2020, 2G wrote:
<snip>
>
>Well, today I did measure the inrush current: the peak current was 9A,
>very=
> close to what I had simulated with Spice. This current is very brief and
>t=
>otally within the capability of the switch to handle, but a few ohms of
>ser=
>ies resistance will cut it down to a couple of amps if you are anal about
>i=
>t.
>
>Tom
>
A while ago, I asked if anyone else has experienced your problem.
Apparently not.
The LX9000 has (presumably) a capacitance of its own which sustains it over
changing batteries in most cases.
If there is a high load from other devices on the Bus then there may not be
sufficient capacitance in the LX9000 and it need some help.
If we isolate the LX9000 from the other devices on the Bus with a schottky
diode, then other devices on the Bus will not drain the LX9000’s
capacitance during switch-over as the diode will be back-biased. If it
still needs some help, a small capacitance can be added on the LX9000 side
of the diode.
This will also lead to a reduced inrush current as the connected
capacitance will be less.
Mike N.
April 13th 20, 12:21 PM
I see this issue with my S80 in my Ventus Ct with two batteries.
The S80 powers my Flarm as designed.
When switching between my main battery and second battery I often, but not always have the Flarm power cycle and sometimes the S80 as well.
I too was thinking about sourcing a make before break switch but I am going to put in a capacitor.
Bruce
April 13th 20, 02:48 PM
On Monday, April 13, 2020 at 4:21:35 AM UTC-7, Mike N. wrote:
> I see this issue with my S80 in my Ventus Ct with two batteries.
>
> The S80 powers my Flarm as designed.
>
> When switching between my main battery and second battery I often, but not always have the Flarm power cycle and sometimes the S80 as well.
>
> I too was thinking about sourcing a make before break switch but I am going to put in a capacitor.
I have an S10 with an internal battery that is no problem, But I also use XCSoar on a OpenVario computer that I built from a kit. That has a big problem, as I switch batteries a couple of times a flight to keep the main battery as charged as possible for starts. So a year ago, I got with an electronics guy at our local supplier and he recommended a properly sized Capacitor and it has worked great. I don't remember the size or model, but it is hidden in the panel.
I researched true make before break switchs and they tended to be large and expensive. Not all switches that say the make before break really do what they advertise.
Bruce
Bruce
MNLou
April 13th 20, 04:07 PM
Interesting Mike -
I have a Flarm / S8 / Oudie set up. I never have any issues while switching batteries.
I wonder of the Oudie battery somehow supplies the "make" current while the switch is being thrown?
Lou
2G
April 13th 20, 05:41 PM
On Monday, April 13, 2020 at 3:45:04 AM UTC-7, Tim Newport-Peace wrote:
> At 03:55 13 April 2020, 2G wrote:
> <snip>
> >
> >Well, today I did measure the inrush current: the peak current was 9A,
> >very=
> > close to what I had simulated with Spice. This current is very brief and
> >t=
> >otally within the capability of the switch to handle, but a few ohms of
> >ser=
> >ies resistance will cut it down to a couple of amps if you are anal about
> >i=
> >t.
> >
> >Tom
> >
> A while ago, I asked if anyone else has experienced your problem.
> Apparently not.
> The LX9000 has (presumably) a capacitance of its own which sustains it over
> changing batteries in most cases.
> If there is a high load from other devices on the Bus then there may not be
> sufficient capacitance in the LX9000 and it need some help.
>
> If we isolate the LX9000 from the other devices on the Bus with a schottky
> diode, then other devices on the Bus will not drain the LX9000’s
> capacitance during switch-over as the diode will be back-biased. If it
> still needs some help, a small capacitance can be added on the LX9000 side
> of the diode.
> This will also lead to a reduced inrush current as the connected
> capacitance will be less.
Hell of a lot easier just to add a capacitor to the bus line, and it can be significantly smaller than the 39000μ that I used (10000μ would probably be fine).
Andy Blackburn[_3_]
April 13th 20, 08:11 PM
I did because I am.
Big old power resistors. I think they were 1-2 ohms. However, after considering battery performance implications of deep discharge I normally fly with the switch set to both batteries.
Andy Blackburn
9B
>
.... a few ohms of series resistance will cut it down to a couple of amps if you are anal about it.
>
> Tom
kinsell
April 13th 20, 09:08 PM
That could be a significant voltage drop across the resistors. I'd use
a couple power Schottky diodes instead.
On 4/13/20 1:11 PM, Andy Blackburn wrote:
> I did because I am.
>
> Big old power resistors. I think they were 1-2 ohms. However, after considering battery performance implications of deep discharge I normally fly with the switch set to both batteries.
>
> Andy Blackburn
> 9B
>
>>
> ... a few ohms of series resistance will cut it down to a couple of amps if you are anal about it.
>>
>> Tom
>
2G
April 14th 20, 01:46 AM
On Monday, April 13, 2020 at 1:08:16 PM UTC-7, kinsell wrote:
> That could be a significant voltage drop across the resistors. I'd use
> a couple power Schottky diodes instead.
>
>
>
> On 4/13/20 1:11 PM, Andy Blackburn wrote:
> > I did because I am.
> >
> > Big old power resistors. I think they were 1-2 ohms. However, after considering battery performance implications of deep discharge I normally fly with the switch set to both batteries.
> >
> > Andy Blackburn
> > 9B
> >
> >>
> > ... a few ohms of series resistance will cut it down to a couple of amps if you are anal about it.
> >>
> >> Tom
> >
Anybody who took EE101 (and most everybody else) can tell you that:
E = I * R
If
I = 1A
R = 1Ω
then
E = 1V
Andy Blackburn[_3_]
April 14th 20, 06:25 AM
I used Shottky diodes plus power resistors plus capacitors. I'm no EE but I took enough circuits courses to handle this problem. The Shottky diodes keep the batteries from cross-discharging each other, the capacitors keep the instruments powered when the switch is disconnected from battery 1 and before it is connected to battery 2 and the resistors keep the capacitors from drawing too much current when you power them up since they make the circuit (even with the diodes) look like a direct short initially.
Andy
On Monday, April 13, 2020 at 1:08:16 PM UTC-7, kinsell wrote:
> That could be a significant voltage drop across the resistors. I'd use
> a couple power Schottky diodes instead.
kinsell
April 14th 20, 01:49 PM
On 4/10/20 3:19 PM, kinsell wrote:
> On 4/8/20 10:58 AM, Martin Gregorie wrote:
>> On Wed, 08 Apr 2020 08:57:43 -0600, kinsell wrote:
>>
>>> Runs an embedded linux, they're not storing any data during normal
>>> operation, so baffling why that's so hard to fix.
>>>
>> When does it commit data to non-volatile memory, i.e. does in
>> intentionally buffer it and only commit at intervals or as part of
>> shutting down?
>> FWIW The same problem occurs with Raspberry Pis.
>>
>> As you may or may not know, these run a Debian Linux clone as their OS
>> and use SD cards for non-volatile memory by default. Much of the time
>> people can get away with simply pulling the plug when the Pi appears to
>> be idle, but if you do that while the Pi is flushing its caches to its SD
>> card or, more rarely, the card is in the middle of a wear-levelling
>> process, then the SD card will become corrupted and possibly permanently
>> damaged if it was wear-levelling when the power vanished. Which is why
>> everybody soon learns to shut the Pi down with a 'sudo stop' command
>> before powering it off. I think its worse with SD cards simply because
>> their internal controllers and cheap, rather basic and have no power
>> buffering. Use an SSD instead or, even better, a hard drive and the
>> problem largely goes away because ext4 is a journalling filesystem, so
>> has built-in recovery.
>>
>> I agree this is a tricky problem, and maybe best solved with some sort of
>> cheap'n cheerful UPS. Here's a suggestion along those lines:
>>
>> Pimoroni sell the PowerBoost 1000 Charger, a small and fairly cheap UPS
>> circuit ($US 19.39 from Amazon), which you connect between a 5v power
>> supply and the device you want to power via a UPS socket. You also
>> connect a suitable sized 1S (3.7 volt) Lithium-ion battery to it - the
>> sort used to power small RC models would be fine - and there's a power
>> buffer for any UPS-powered device that doesn't have an internal battery.
>> Its good to supply up to 1000mA provided that the battery is rated for
>> that current. Some soldering is needed.
>>
>> It comes with a selection of sockets, but they're all sat loosely in
>> place on the board so you can solder the ones you want on, sling the
>> others and solder any permanent connections you need.
>>
>> I have one but haven't used it yet - I'm planning to make a PDA from a 4"
>> touch screen and a Pi Zero WL and use this Pimoroni plus an RC model 1S
>> LiPO battery to power it. Add a case made from epoxyboard* and it should
>> be ready to rock'n roll.
>>
>>
>
> The Stratux project does run on a Rspberry Pi using the micro-SD for
> storage.* Yes, running other software on that hardware can lead to the
> same file system corruption problem.* What's interesting is I can load
> FlightAware software on the same hardware and never see the problem.* In
> both applications, there is no user data that needs to be stored to
> flash memory, other than a tiny bit of configuration data when they are
> first set up.* It is possible to run on a read-only file system, if some
> provision is made for the configuration data.* People have experimented
> with this, but the images distributed for Stratux have never had a
> solution that I'm aware of.
>
> So many devices these days have a computer with flash memory.* If your
> smart TV becomes unbootable after a power fail, would that be considered
> acceptable?* Of course not.* But for some reason Stratux doesn't seem to
> get fixed.* Putting a UPS on a microcomputer seems like an ugly Band-Aid
> (rtm) that shouldn't be necessary.* Just a simple file system check on
> power up might be adequate.* There's a Sentry Mini receiver available
> now for just $300, I'll bet it doesn't lose its mind if you just pull
> power.
>
> -Dave
>
Looking at a couple lightly-used Stratux boxes, one of then had a
completely full file system, the other was almost full. Seems to be
more of a science fair project than a serious instrument to be used for
navigation.
kinsell
April 14th 20, 01:54 PM
On 4/13/20 11:25 PM, Andy Blackburn wrote:
> I used Shottky diodes plus power resistors plus capacitors. I'm no EE but I took enough circuits courses to handle this problem. The Shottky diodes keep the batteries from cross-discharging each other, the capacitors keep the instruments powered when the switch is disconnected from battery 1 and before it is connected to battery 2 and the resistors keep the capacitors from drawing too much current when you power them up since they make the circuit (even with the diodes) look like a direct short initially.
>
> Andy
>
> On Monday, April 13, 2020 at 1:08:16 PM UTC-7, kinsell wrote:
>> That could be a significant voltage drop across the resistors. I'd use
>> a couple power Schottky diodes instead.
You might want to take a serious look at the IR drop across the
resistors. You have a complex solution to a simple problem that could
actually be causing more problems than it solves.
On Tuesday, April 14, 2020 at 1:25:55 AM UTC-4, Andy Blackburn wrote:
> I used Shottky diodes plus power resistors plus capacitors. I'm no EE but I took enough circuits courses to handle this problem. The Shottky diodes keep the batteries from cross-discharging each other, the capacitors keep the instruments powered when the switch is disconnected from battery 1 and before it is connected to battery 2 and the resistors keep the capacitors from drawing too much current when you power them up since they make the circuit (even with the diodes) look like a direct short initially.
>
> Andy
KISS. Just the two diodes (and no switch) should be enough. Whichever battery is stronger (higher voltage) would take the load. Automatically. No switching needed. With the higher voltage of LiFePO4 batteries (relative to lead-acid) the voltage drop in the diode is acceptable, especially if it's the Schottky type.
Or, if you really want to remove the voltage drop in the diodes, add an SPDT switch (perhaps one with also a center-off position) IN PARALLEL to the diodes. No matter which position that switch is in, both batteries will still be connected. But the battery the switch leads to will feed the avionics with no voltage drop since the switch bypasses the diode on that side. The other diode will meanwhile prevent current from going INTO the other battery. The middle-off position (or no switch at all) is the safest though, since if either battery develops a shorted cell (or shorted or loose wiring, blown battery fuse, etc) without your knowledge, it won't affect the other battery, thanks to the diodes.
On Tuesday, April 14, 2020 at 9:31:13 AM UTC-4, wrote:
> On Tuesday, April 14, 2020 at 1:25:55 AM UTC-4, Andy Blackburn wrote:
> > I used Shottky diodes plus power resistors plus capacitors. I'm no EE but I took enough circuits courses to handle this problem. The Shottky diodes keep the batteries from cross-discharging each other, the capacitors keep the instruments powered when the switch is disconnected from battery 1 and before it is connected to battery 2 and the resistors keep the capacitors from drawing too much current when you power them up since they make the circuit (even with the diodes) look like a direct short initially.
> >
> > Andy
>
> KISS. Just the two diodes (and no switch) should be enough. Whichever battery is stronger (higher voltage) would take the load. Automatically. No switching needed. With the higher voltage of LiFePO4 batteries (relative to lead-acid) the voltage drop in the diode is acceptable, especially if it's the Schottky type.
>
> Or, if you really want to remove the voltage drop in the diodes, add an SPDT switch (perhaps one with also a center-off position) IN PARALLEL to the diodes. No matter which position that switch is in, both batteries will still be connected. But the battery the switch leads to will feed the avionics with no voltage drop since the switch bypasses the diode on that side. The other diode will meanwhile prevent current from going INTO the other battery. The middle-off position (or no switch at all) is the safest though, since if either battery develops a shorted cell (or shorted or loose wiring, blown battery fuse, etc) without your knowledge, it won't affect the other battery and the avionics, thanks to the two diodes.
- Clarification: I meant a diode between each battery and the avionics bus as a whole. Not separately for a specific instrument.
2G
April 15th 20, 02:28 AM
On Tuesday, April 14, 2020 at 6:38:00 AM UTC-7, wrote:
> On Tuesday, April 14, 2020 at 9:31:13 AM UTC-4, wrote:
> > On Tuesday, April 14, 2020 at 1:25:55 AM UTC-4, Andy Blackburn wrote:
> > > I used Shottky diodes plus power resistors plus capacitors. I'm no EE but I took enough circuits courses to handle this problem. The Shottky diodes keep the batteries from cross-discharging each other, the capacitors keep the instruments powered when the switch is disconnected from battery 1 and before it is connected to battery 2 and the resistors keep the capacitors from drawing too much current when you power them up since they make the circuit (even with the diodes) look like a direct short initially.
> > >
> > > Andy
> >
> > KISS. Just the two diodes (and no switch) should be enough. Whichever battery is stronger (higher voltage) would take the load. Automatically. No switching needed. With the higher voltage of LiFePO4 batteries (relative to lead-acid) the voltage drop in the diode is acceptable, especially if it's the Schottky type.
> >
> > Or, if you really want to remove the voltage drop in the diodes, add an SPDT switch (perhaps one with also a center-off position) IN PARALLEL to the diodes. No matter which position that switch is in, both batteries will still be connected. But the battery the switch leads to will feed the avionics with no voltage drop since the switch bypasses the diode on that side.. The other diode will meanwhile prevent current from going INTO the other battery. The middle-off position (or no switch at all) is the safest though, since if either battery develops a shorted cell (or shorted or loose wiring, blown battery fuse, etc) without your knowledge, it won't affect the other battery and the avionics, thanks to the two diodes.
>
> - Clarification: I meant a diode between each battery and the avionics bus as a whole. Not separately for a specific instrument.
I measured the inrush current once again and found that the vertical of the scope was set for a 1X probe instead of the 10X actually being used. This meant that the peak current was 90A instead of 9A, which is a bit high. I added a 1.1 ohm resistor and the peak current dropped to 6A. A simulation shows that a 2 ohm resistor drops it to 3A. This is a good value to use if you have a 1A current drain as the voltage drop will be 2V. The wattage of resistor is unimportant because so little energy is being dissipated by the resistor. The energy transferred remains constant regardless of the resistor value as it is the energy required to charge the capacitor (the current pulse lengthens for larger resistor values).
Tom
>
> I measured the inrush current once again and found that the vertical of the scope was set for a 1X probe instead of the 10X actually being used. This meant that the peak current was 90A instead of 9A, which is a bit high. I added a 1.1 ohm resistor and the peak current dropped to 6A.
My understanding of that test and the numbers don't add up. I'm thinking you are charging a cap with a 12 V battery and measuring the current with and without and added 1.1 ohm series resistor?
If I combine the two measurements to find the open circuit voltage of the battery and overall wiring resistance I get 7.07 volts and 0.078 ohms.
Probably I don't understand what you are testing. We seem to have lots of time to kill. Could you send a picture of the circuits with the probe attached?
While I am reading these posts with interest, I confess to being an electrical illiterate. I just use two batteries, each with a fuse, and two switches. When switching, I turn on #2 before turning off #1.
If these circuits with diodes, resistors, make-before-break switches and so on are superior, please explain why, and if the case is compelling, a circuit diagram would be appreciated so that I might take advantage of the information.
After all, in aviation "R & D" actually stands for "Ripoff and Duplicate."
David S
April 15th 20, 03:38 PM
Your procedure generally works, but while both batteries are connected, battery #2 will be recharging battery #1 and powering your panel, so there's a chance you could blow the fuse on battery #2. That leaves you to finish your flight with only one nearly depleted battery.
Cheers,
...david
On Wednesday, April 15, 2020 at 9:18:08 AM UTC-4, wrote:
> While I am reading these posts with interest, I confess to being an electrical illiterate. I just use two batteries, each with a fuse, and two switches. When switching, I turn on #2 before turning off #1.
>
> If these circuits with diodes, resistors, make-before-break switches and so on are superior, please explain why, and if the case is compelling, a circuit diagram would be appreciated so that I might take advantage of the information.
>
> After all, in aviation "R & D" actually stands for "Ripoff and Duplicate."
David- Thanks for the advice. However, I rarely have both battery switches ON for more than a second. Is this still a potential problem? I can add diodes if necessary.
kinsell
April 15th 20, 04:03 PM
No, two similar batteries have virtually no ability to cross charge like
that. People take introductory EE classes, learn about ideal voltage
sources, and assume batteries are like that. They're not.
Years ago, in a previous incantation of this same discussion, I
suggested someone take a charged and discharged battery, connect them
through an ammeter, and report the results. They did, and couldn't even
see a flicker of the needle. They concluded that ampere hours of charge
had instantly transferred from one battery to the other, before the
needle had a chance to twitch. I pointed out the ammeter would be
nothing more than a smoking hole in the table if that were true.
Mark can use his procedure if he wants, or better yet just keep both
switches on an forget about flipping switches. Unless he's a former 747
captain, who likes to fiddle with lots of switches.
Dave
P.S. Assumes batteries of the same chemistry
On 4/15/20 8:38 AM, David S wrote:
> Your procedure generally works, but while both batteries are connected, battery #2 will be recharging battery #1 and powering your panel, so there's a chance you could blow the fuse on battery #2. That leaves you to finish your flight with only one nearly depleted battery.
>
> Cheers,
> ...david
>
>
> On Wednesday, April 15, 2020 at 9:18:08 AM UTC-4, wrote:
>> While I am reading these posts with interest, I confess to being an electrical illiterate. I just use two batteries, each with a fuse, and two switches. When switching, I turn on #2 before turning off #1.
>>
>> If these circuits with diodes, resistors, make-before-break switches and so on are superior, please explain why, and if the case is compelling, a circuit diagram would be appreciated so that I might take advantage of the information.
>>
>> After all, in aviation "R & D" actually stands for "Ripoff and Duplicate."
>
Unless he's a former 747 captain, who likes to fiddle with lots of switches.
Thanks for that advice, also. It's pretty much what I suspected. But, while I don't mind flipping switches, some people tell me I am better at pushing their buttons.
David S
April 15th 20, 04:54 PM
If you run with both batteries connected, a failure that should have blown one battery fuse with either blow both fuses or none at all. Both results are bad.
Cheers,
...david
On Wednesday, April 15, 2020 at 11:04:01 AM UTC-4, kinsell wrote:
> Mark can use his procedure if he wants, or better yet just keep both
> switches on an forget about flipping switches. Unless he's a former 747
> captain, who likes to fiddle with lots of switches.
NM
April 15th 20, 05:12 PM
On Tuesday, April 14, 2020 at 9:28:12 PM UTC-4, 2G wrote:
> On Tuesday, April 14, 2020 at 6:38:00 AM UTC-7, wrote:
> > On Tuesday, April 14, 2020 at 9:31:13 AM UTC-4, wrote:
> > > On Tuesday, April 14, 2020 at 1:25:55 AM UTC-4, Andy Blackburn wrote:
> > > > I used Shottky diodes plus power resistors plus capacitors. I'm no EE but I took enough circuits courses to handle this problem. The Shottky diodes keep the batteries from cross-discharging each other, the capacitors keep the instruments powered when the switch is disconnected from battery 1 and before it is connected to battery 2 and the resistors keep the capacitors from drawing too much current when you power them up since they make the circuit (even with the diodes) look like a direct short initially.
> > > >
> > > > Andy
> > >
> > > KISS. Just the two diodes (and no switch) should be enough. Whichever battery is stronger (higher voltage) would take the load. Automatically.. No switching needed. With the higher voltage of LiFePO4 batteries (relative to lead-acid) the voltage drop in the diode is acceptable, especially if it's the Schottky type.
> > >
> > > Or, if you really want to remove the voltage drop in the diodes, add an SPDT switch (perhaps one with also a center-off position) IN PARALLEL to the diodes. No matter which position that switch is in, both batteries will still be connected. But the battery the switch leads to will feed the avionics with no voltage drop since the switch bypasses the diode on that side. The other diode will meanwhile prevent current from going INTO the other battery. The middle-off position (or no switch at all) is the safest though, since if either battery develops a shorted cell (or shorted or loose wiring, blown battery fuse, etc) without your knowledge, it won't affect the other battery and the avionics, thanks to the two diodes.
> >
> > - Clarification: I meant a diode between each battery and the avionics bus as a whole. Not separately for a specific instrument.
>
> I measured the inrush current once again and found that the vertical of the scope was set for a 1X probe instead of the 10X actually being used. This meant that the peak current was 90A instead of 9A, which is a bit high. I added a 1.1 ohm resistor and the peak current dropped to 6A. A simulation shows that a 2 ohm resistor drops it to 3A. This is a good value to use if you have a 1A current drain as the voltage drop will be 2V. The wattage of resistor is unimportant because so little energy is being dissipated by the resistor. The energy transferred remains constant regardless of the resistor value as it is the energy required to charge the capacitor (the current pulse lengthens for larger resistor values).
>
> Tom
Tom,good plan to increase the value of the resistor to limit the current inrush. One could even make the resistor 100 ohms and have a reversed diode in parallel with it so that when the capacitor is needed to sustain the instrument during switching, the current would flow in the reverse direction through the diode, bypassing the resistor. You now have the best of both worlds - slow charge of the capacitor when the power is turned on, and a fast discharge to support the instrument without IR drop across the resistor, instead the drop would just be the bias voltage of the diode.
I've seen a lot of recommendations to keep both batteries connected and not bother with switching. One reason I don't want to do that is that I have a couple of slim 2.3 Ah SLA batteries in the tail of my ASW 24 as a backup (yes, they fit--barely) and the usual LiFePO4 battery up front for the primary source. Ignoring the issues of batteries with different capacities and chemistries running in parallel, pulling out the tail battery pack every night and charging it is just another chore. At a contest, I never do unless I've had to go to it (only a few times in many decades). I'm lazy.
Chip Bearden
JB
Chip how much loss do you experience getting that juice from the tail all the way up to your panel?
On Wednesday, April 15, 2020 at 1:25:50 PM UTC-4, wrote:
> I've seen a lot of recommendations to keep both batteries connected and not bother with switching. One reason I don't want to do that is that I have a couple of slim 2.3 Ah SLA batteries in the tail of my ASW 24 as a backup (yes, they fit--barely) and the usual LiFePO4 battery up front for the primary source. Ignoring the issues of batteries with different capacities and chemistries running in parallel, pulling out the tail battery pack every night and charging it is just another chore. At a contest, I never do unless I've had to go to it (only a few times in many decades). I'm lazy.
>
> Chip Bearden
> JB
Chip, your situation sure does call for a switch. Or two switches. Or two diodes. But certainly not simply in parallel. Note that with two diodes the "switching" would be automatic: as long as your LiFePO4 battery up front is not almost completely discharged, there would be no draw on the tail batteries since the diode between them and the "bus" would have a voltage on it in the "wrong" (non-conducting) direction.
> Chip, your situation sure does call for a switch. Or two switches. Or two diodes. But certainly not simply in parallel. Note that with two diodes the "switching" would be automatic: as long as your LiFePO4 battery up front is not almost completely discharged, there would be no draw on the tail batteries since the diode between them and the "bus" would have a voltage on it in the "wrong" (non-conducting) direction.
I use two switches. Switch on the tail battery before switching off the front battery. Since I hardly ever need it and the front battery has never been terribly discharged, I haven't experienced big inrush currents and fuses blown. I'm looking at adding a capacitor since that is apparently what my LNAV had for 25 years when I had a single DPDT switch to go back and forth if needed and never had anything powercycle.
I'm also a little less vulnerable than some--maybe. My PowerFLARM is a portable and I keep batteries in it that would be good for the whole day if necessary. Likewise my glide computer is TopHat running on a Kobo (with an internal battery) that gets its GPS data from the FLARM with a backup of a Dell Streak that is fully self contained and can be connected to a big USB battery if I need to run it all day. So in primary/backup battery failure, what I would lose would be the radio (of course) and the ClearNav vario (a big loss because I have the LCD screen with thermal assistant). But I also have a Winter running on a triple probe. If I lose the triple probe, the ClearNav uses electronic TE.
I try to think about redundancy because over the years I've lost (at contests) primary battery power (loose fuse holder), pitot pressure (hole in the ASI diaphragm, which also affected the LNAV), the LNAV as well as the ClearNav (forgot to connect tubing I disconnected during troubleshooting--operator error), and a bad battery (didn't check capacity in the spring--laziness).
There is some voltage drop from the tail battery due to the wiring but the current draw isn't as much as some guys have (e.g., no transponder). I haven't flown this year to note how much and I can't recall what I saw a few years ago when I stuck a voltmeter on it uninstalled and then again at the buss in the panel with and without keying the mike.
There are so many things that CAN go wrong these days that I try to protect myself against as many as possible. As I mentioned in a another thread a day or so ago, the most dangerous words in technology are "It SHOULD work" followed by "THAT shouldn't happen" (or "Haven't seen THAT before") followed by "Works on/in MY machine/glider". All true. Throw in interface problems and software bugs and it's wonder we can fly a contest without a major failure.
Chip Bearden
JB
Dan Marotta
April 16th 20, 12:05 AM
This whole discussion is a living example of the old saw:* Better is the
enemy of good enough.* I flew for years with two batteries, two fuses,
and two switches.* When battery 1 gets low, flip on battery 2 then flip
off battery 1.* Never had an issue.
On 4/15/2020 10:12 AM, NM wrote:
> On Tuesday, April 14, 2020 at 9:28:12 PM UTC-4, 2G wrote:
>> On Tuesday, April 14, 2020 at 6:38:00 AM UTC-7, wrote:
>>> On Tuesday, April 14, 2020 at 9:31:13 AM UTC-4, wrote:
>>>> On Tuesday, April 14, 2020 at 1:25:55 AM UTC-4, Andy Blackburn wrote:
>>>>> I used Shottky diodes plus power resistors plus capacitors. I'm no EE but I took enough circuits courses to handle this problem. The Shottky diodes keep the batteries from cross-discharging each other, the capacitors keep the instruments powered when the switch is disconnected from battery 1 and before it is connected to battery 2 and the resistors keep the capacitors from drawing too much current when you power them up since they make the circuit (even with the diodes) look like a direct short initially.
>>>>>
>>>>> Andy
>>>> KISS. Just the two diodes (and no switch) should be enough. Whichever battery is stronger (higher voltage) would take the load. Automatically. No switching needed. With the higher voltage of LiFePO4 batteries (relative to lead-acid) the voltage drop in the diode is acceptable, especially if it's the Schottky type.
>>>>
>>>> Or, if you really want to remove the voltage drop in the diodes, add an SPDT switch (perhaps one with also a center-off position) IN PARALLEL to the diodes. No matter which position that switch is in, both batteries will still be connected. But the battery the switch leads to will feed the avionics with no voltage drop since the switch bypasses the diode on that side. The other diode will meanwhile prevent current from going INTO the other battery. The middle-off position (or no switch at all) is the safest though, since if either battery develops a shorted cell (or shorted or loose wiring, blown battery fuse, etc) without your knowledge, it won't affect the other battery and the avionics, thanks to the two diodes.
>>> - Clarification: I meant a diode between each battery and the avionics bus as a whole. Not separately for a specific instrument.
>> I measured the inrush current once again and found that the vertical of the scope was set for a 1X probe instead of the 10X actually being used. This meant that the peak current was 90A instead of 9A, which is a bit high. I added a 1.1 ohm resistor and the peak current dropped to 6A. A simulation shows that a 2 ohm resistor drops it to 3A. This is a good value to use if you have a 1A current drain as the voltage drop will be 2V. The wattage of resistor is unimportant because so little energy is being dissipated by the resistor. The energy transferred remains constant regardless of the resistor value as it is the energy required to charge the capacitor (the current pulse lengthens for larger resistor values).
>>
>> Tom
> Tom,good plan to increase the value of the resistor to limit the current inrush. One could even make the resistor 100 ohms and have a reversed diode in parallel with it so that when the capacitor is needed to sustain the instrument during switching, the current would flow in the reverse direction through the diode, bypassing the resistor. You now have the best of both worlds - slow charge of the capacitor when the power is turned on, and a fast discharge to support the instrument without IR drop across the resistor, instead the drop would just be the bias voltage of the diode.
--
Dan, 5J
Dan Marotta
April 16th 20, 12:08 AM
Chip,
I solved that problem in my LAK-17a by tapping a charging pigtail into
the wire from the tail battery.* The charging connector (power pole) was
in the baggage compartment.* It was simple to charge it without removing
it.* It was a pain to access the wires to do that, but I only had to do
it once.
On 4/15/2020 11:25 AM, wrote:
> I've seen a lot of recommendations to keep both batteries connected and not bother with switching. One reason I don't want to do that is that I have a couple of slim 2.3 Ah SLA batteries in the tail of my ASW 24 as a backup (yes, they fit--barely) and the usual LiFePO4 battery up front for the primary source. Ignoring the issues of batteries with different capacities and chemistries running in parallel, pulling out the tail battery pack every night and charging it is just another chore. At a contest, I never do unless I've had to go to it (only a few times in many decades). I'm lazy.
>
> Chip Bearden
> JB
--
Dan, 5J
Dan,
Thanks to so many of us showing up crewless and to CDs who call tasks with the intent of getting most or all pilots home, it's relatively rare these days that I have to hook up the trailer to my van very often during a contest: once in the past five nationals, IIRC (not counting several aero retrieves at Elmira).
And at sites with favorable weather and my new polyurethane finish, I'm tying out at night much of the time at national contest sites unless storms threaten. As I said, removing the tail battery is no big deal, although if I'm tied out, then I have to remove the elevator to get access. I'm just lazy. It's easier not to put that battery on the line unless I have to--then it's good for the whole contest.
Hence, my two switch (or my DPDT switch before it got shot up at TSA) approach makes the most sense for me. For those using batteries with wildly different capacities (e.g., motorgliders) or avionics drawing a lot more amps than mine do, a different approach might be in order.
Chip Bearden
JB
krasw
April 16th 20, 05:48 AM
On Thursday, 16 April 2020 02:05:45 UTC+3, Dan Marotta wrote:
> This whole discussion is a living example of the old saw:* Better is the
> enemy of good enough.* I flew for years with two batteries, two fuses,
> and two switches.* When battery 1 gets low, flip on battery 2 then flip
> off battery 1.* Never had an issue.
>
Less components means less problems with bad components, less solder joints that eventually break with vibration, all of this is better than good. I run same setup, on-off switch for both batteries, usually only one of them online, except when getting my engine (turbo) out in case battery runs out of juice right then. Zero problems ever. Glider came out of factory with both batteries wired parallel btw.
2G
April 16th 20, 07:17 AM
On Wednesday, April 15, 2020 at 5:57:07 AM UTC-7, wrote:
> >
> > I measured the inrush current once again and found that the vertical of the scope was set for a 1X probe instead of the 10X actually being used. This meant that the peak current was 90A instead of 9A, which is a bit high. I added a 1.1 ohm resistor and the peak current dropped to 6A.
>
> My understanding of that test and the numbers don't add up. I'm thinking you are charging a cap with a 12 V battery and measuring the current with and without and added 1.1 ohm series resistor?
>
> If I combine the two measurements to find the open circuit voltage of the battery and overall wiring resistance I get 7.07 volts and 0.078 ohms.
>
> Probably I don't understand what you are testing. We seem to have lots of time to kill. Could you send a picture of the circuits with the probe attached?
The circuit is very simply: a capacitor and a switch to a battery. In the second case, a 1.1 ohm resistor is added in series between the capacitor and the battery. You are mistaking a d.c. circuit and an a.c. circuit: when the switch closes current will flow from the battery to the capacitor. This current flow is modeled by a differential equation:
i = C dv/dt
or
dv/dt = i / C
Integrated wrt time gives you:
V = 1/C Int[i dt]
In other words, the voltage on the capacitor increases as current flows into it. If you look at it in the steady state, or d.c., you will miss this entirely.
I measured the current with a 0.0015 ohm current shunt and an oscilloscope.
2G
April 16th 20, 07:21 AM
On Wednesday, April 15, 2020 at 6:18:08 AM UTC-7, wrote:
> While I am reading these posts with interest, I confess to being an electrical illiterate. I just use two batteries, each with a fuse, and two switches. When switching, I turn on #2 before turning off #1.
>
> If these circuits with diodes, resistors, make-before-break switches and so on are superior, please explain why, and if the case is compelling, a circuit diagram would be appreciated so that I might take advantage of the information.
>
> After all, in aviation "R & D" actually stands for "Ripoff and Duplicate."
A make-before-break is also called a "shorting" switch. If you use such a switch you WILL short the two batteries together, which could result in a large current flow from the battery with the higher voltage to the battery with the lower voltage. This large current could blow your protection fuse(s).. This is especially the case if you have two separate switches.
2G
April 16th 20, 07:25 AM
On Wednesday, April 15, 2020 at 8:04:01 AM UTC-7, kinsell wrote:
> No, two similar batteries have virtually no ability to cross charge like
> that. People take introductory EE classes, learn about ideal voltage
> sources, and assume batteries are like that. They're not.
>
> Years ago, in a previous incantation of this same discussion, I
> suggested someone take a charged and discharged battery, connect them
> through an ammeter, and report the results. They did, and couldn't even
> see a flicker of the needle. They concluded that ampere hours of charge
> had instantly transferred from one battery to the other, before the
> needle had a chance to twitch. I pointed out the ammeter would be
> nothing more than a smoking hole in the table if that were true.
>
> Mark can use his procedure if he wants, or better yet just keep both
> switches on an forget about flipping switches. Unless he's a former 747
> captain, who likes to fiddle with lots of switches.
>
> Dave
>
> P.S. Assumes batteries of the same chemistry
>
>
>
> On 4/15/20 8:38 AM, David S wrote:
> > Your procedure generally works, but while both batteries are connected, battery #2 will be recharging battery #1 and powering your panel, so there's a chance you could blow the fuse on battery #2. That leaves you to finish your flight with only one nearly depleted battery.
> >
> > Cheers,
> > ...david
> >
> >
> > On Wednesday, April 15, 2020 at 9:18:08 AM UTC-4, wrote:
> >> While I am reading these posts with interest, I confess to being an electrical illiterate. I just use two batteries, each with a fuse, and two switches. When switching, I turn on #2 before turning off #1.
> >>
> >> If these circuits with diodes, resistors, make-before-break switches and so on are superior, please explain why, and if the case is compelling, a circuit diagram would be appreciated so that I might take advantage of the information.
> >>
> >> After all, in aviation "R & D" actually stands for "Ripoff and Duplicate."
> >
You need to retake EE101 - if you EVER took it. Your ammeter's bandwidth (do you even know what "bandwidth" means?) WILL NOT remotely see the current flow - it just can't mechanically respond to the very short current pulse.
2G
April 16th 20, 07:36 AM
On Wednesday, April 15, 2020 at 9:12:33 AM UTC-7, NM wrote:
> On Tuesday, April 14, 2020 at 9:28:12 PM UTC-4, 2G wrote:
> > On Tuesday, April 14, 2020 at 6:38:00 AM UTC-7, wrote:
> > > On Tuesday, April 14, 2020 at 9:31:13 AM UTC-4, wrote:
> > > > On Tuesday, April 14, 2020 at 1:25:55 AM UTC-4, Andy Blackburn wrote:
> > > > > I used Shottky diodes plus power resistors plus capacitors. I'm no EE but I took enough circuits courses to handle this problem. The Shottky diodes keep the batteries from cross-discharging each other, the capacitors keep the instruments powered when the switch is disconnected from battery 1 and before it is connected to battery 2 and the resistors keep the capacitors from drawing too much current when you power them up since they make the circuit (even with the diodes) look like a direct short initially.
> > > > >
> > > > > Andy
> > > >
> > > > KISS. Just the two diodes (and no switch) should be enough. Whichever battery is stronger (higher voltage) would take the load. Automatically. No switching needed. With the higher voltage of LiFePO4 batteries (relative to lead-acid) the voltage drop in the diode is acceptable, especially if it's the Schottky type.
> > > >
> > > > Or, if you really want to remove the voltage drop in the diodes, add an SPDT switch (perhaps one with also a center-off position) IN PARALLEL to the diodes. No matter which position that switch is in, both batteries will still be connected. But the battery the switch leads to will feed the avionics with no voltage drop since the switch bypasses the diode on that side. The other diode will meanwhile prevent current from going INTO the other battery. The middle-off position (or no switch at all) is the safest though, since if either battery develops a shorted cell (or shorted or loose wiring, blown battery fuse, etc) without your knowledge, it won't affect the other battery and the avionics, thanks to the two diodes.
> > >
> > > - Clarification: I meant a diode between each battery and the avionics bus as a whole. Not separately for a specific instrument.
> >
> > I measured the inrush current once again and found that the vertical of the scope was set for a 1X probe instead of the 10X actually being used. This meant that the peak current was 90A instead of 9A, which is a bit high. I added a 1.1 ohm resistor and the peak current dropped to 6A. A simulation shows that a 2 ohm resistor drops it to 3A. This is a good value to use if you have a 1A current drain as the voltage drop will be 2V. The wattage of resistor is unimportant because so little energy is being dissipated by the resistor. The energy transferred remains constant regardless of the resistor value as it is the energy required to charge the capacitor (the current pulse lengthens for larger resistor values).
> >
> > Tom
>
> Tom,good plan to increase the value of the resistor to limit the current inrush. One could even make the resistor 100 ohms and have a reversed diode in parallel with it so that when the capacitor is needed to sustain the instrument during switching, the current would flow in the reverse direction through the diode, bypassing the resistor. You now have the best of both worlds - slow charge of the capacitor when the power is turned on, and a fast discharge to support the instrument without IR drop across the resistor, instead the drop would just be the bias voltage of the diode.
I don't really think that the resistor is necessary, but offer it to those that are overly concerned about the inrush current. The bottom line is the energy that is transferred from the battery to the capacitor heating the switch contacts. Switches have current ratings to limit the temperature rise to tolerable levels when the current is flowing continuously; this short current pulse will not raise the switch contact temperatures to any significant level. Remember, the SAME amount of energy will be transferred between the battery and the capacitor REGARDLESS of the resistor value (joules = C * V). This translates to the SAME amount of switch contact heating. If you make the resistor insanely large so the time constant is on the order of minutes, the heat will dissipate and lower the maximum temperature. A better approach is to use a smaller capacitor that still maintains voltage during switching.
Tom
2G
April 16th 20, 07:38 AM
On Wednesday, April 15, 2020 at 4:05:45 PM UTC-7, Dan Marotta wrote:
> This whole discussion is a living example of the old saw:* Better is the
> enemy of good enough.* I flew for years with two batteries, two fuses,
> and two switches.* When battery 1 gets low, flip on battery 2 then flip
> off battery 1.* Never had an issue.
>
> On 4/15/2020 10:12 AM, NM wrote:
> > On Tuesday, April 14, 2020 at 9:28:12 PM UTC-4, 2G wrote:
> >> On Tuesday, April 14, 2020 at 6:38:00 AM UTC-7, wrote:
> >>> On Tuesday, April 14, 2020 at 9:31:13 AM UTC-4, wrote:
> >>>> On Tuesday, April 14, 2020 at 1:25:55 AM UTC-4, Andy Blackburn wrote:
> >>>>> I used Shottky diodes plus power resistors plus capacitors. I'm no EE but I took enough circuits courses to handle this problem. The Shottky diodes keep the batteries from cross-discharging each other, the capacitors keep the instruments powered when the switch is disconnected from battery 1 and before it is connected to battery 2 and the resistors keep the capacitors from drawing too much current when you power them up since they make the circuit (even with the diodes) look like a direct short initially.
> >>>>>
> >>>>> Andy
> >>>> KISS. Just the two diodes (and no switch) should be enough. Whichever battery is stronger (higher voltage) would take the load. Automatically. No switching needed. With the higher voltage of LiFePO4 batteries (relative to lead-acid) the voltage drop in the diode is acceptable, especially if it's the Schottky type.
> >>>>
> >>>> Or, if you really want to remove the voltage drop in the diodes, add an SPDT switch (perhaps one with also a center-off position) IN PARALLEL to the diodes. No matter which position that switch is in, both batteries will still be connected. But the battery the switch leads to will feed the avionics with no voltage drop since the switch bypasses the diode on that side. The other diode will meanwhile prevent current from going INTO the other battery. The middle-off position (or no switch at all) is the safest though, since if either battery develops a shorted cell (or shorted or loose wiring, blown battery fuse, etc) without your knowledge, it won't affect the other battery and the avionics, thanks to the two diodes.
> >>> - Clarification: I meant a diode between each battery and the avionics bus as a whole. Not separately for a specific instrument.
> >> I measured the inrush current once again and found that the vertical of the scope was set for a 1X probe instead of the 10X actually being used. This meant that the peak current was 90A instead of 9A, which is a bit high.. I added a 1.1 ohm resistor and the peak current dropped to 6A. A simulation shows that a 2 ohm resistor drops it to 3A. This is a good value to use if you have a 1A current drain as the voltage drop will be 2V. The wattage of resistor is unimportant because so little energy is being dissipated by the resistor. The energy transferred remains constant regardless of the resistor value as it is the energy required to charge the capacitor (the current pulse lengthens for larger resistor values).
> >>
> >> Tom
> > Tom,good plan to increase the value of the resistor to limit the current inrush. One could even make the resistor 100 ohms and have a reversed diode in parallel with it so that when the capacitor is needed to sustain the instrument during switching, the current would flow in the reverse direction through the diode, bypassing the resistor. You now have the best of both worlds - slow charge of the capacitor when the power is turned on, and a fast discharge to support the instrument without IR drop across the resistor, instead the drop would just be the bias voltage of the diode.
>
> --
> Dan, 5J
Yeah, until you get the sequence backwards.
Dan Marotta
April 16th 20, 04:56 PM
Really?
What is the shorted time when flipping the switch?* What's the voltage
difference between the two batteries?* What's the total circuit
resistance, including the internal resistance of the batteries?*
Theoretical math and practical application do not always agree.* It
might be fun to set up such a demonstration and use your o'scope to
measure that current and it's time duration.* Compare that to the "blow
time" of any fuses.
Seriously, I've done it for years without any problems, but I recognize
that past performance is no guarantee of future results. I'd be curious
about the results and you have the equipment to do it.
On 4/16/2020 12:21 AM, 2G wrote:
> On Wednesday, April 15, 2020 at 6:18:08 AM UTC-7, wrote:
>> While I am reading these posts with interest, I confess to being an electrical illiterate. I just use two batteries, each with a fuse, and two switches. When switching, I turn on #2 before turning off #1.
>>
>> If these circuits with diodes, resistors, make-before-break switches and so on are superior, please explain why, and if the case is compelling, a circuit diagram would be appreciated so that I might take advantage of the information.
>>
>> After all, in aviation "R & D" actually stands for "Ripoff and Duplicate."
> A make-before-break is also called a "shorting" switch. If you use such a switch you WILL short the two batteries together, which could result in a large current flow from the battery with the higher voltage to the battery with the lower voltage. This large current could blow your protection fuse(s). This is especially the case if you have two separate switches.
--
Dan, 5J
jfitch
April 16th 20, 05:29 PM
On Wednesday, April 15, 2020 at 11:36:35 PM UTC-7, 2G wrote:
> On Wednesday, April 15, 2020 at 9:12:33 AM UTC-7, NM wrote:
> > On Tuesday, April 14, 2020 at 9:28:12 PM UTC-4, 2G wrote:
> > > On Tuesday, April 14, 2020 at 6:38:00 AM UTC-7, wrote:
> > > > On Tuesday, April 14, 2020 at 9:31:13 AM UTC-4, wrote:
> > > > > On Tuesday, April 14, 2020 at 1:25:55 AM UTC-4, Andy Blackburn wrote:
> > > > > > I used Shottky diodes plus power resistors plus capacitors. I'm no EE but I took enough circuits courses to handle this problem. The Shottky diodes keep the batteries from cross-discharging each other, the capacitors keep the instruments powered when the switch is disconnected from battery 1 and before it is connected to battery 2 and the resistors keep the capacitors from drawing too much current when you power them up since they make the circuit (even with the diodes) look like a direct short initially.
> > > > > >
> > > > > > Andy
> > > > >
> > > > > KISS. Just the two diodes (and no switch) should be enough. Whichever battery is stronger (higher voltage) would take the load. Automatically. No switching needed. With the higher voltage of LiFePO4 batteries (relative to lead-acid) the voltage drop in the diode is acceptable, especially if it's the Schottky type.
> > > > >
> > > > > Or, if you really want to remove the voltage drop in the diodes, add an SPDT switch (perhaps one with also a center-off position) IN PARALLEL to the diodes. No matter which position that switch is in, both batteries will still be connected. But the battery the switch leads to will feed the avionics with no voltage drop since the switch bypasses the diode on that side. The other diode will meanwhile prevent current from going INTO the other battery. The middle-off position (or no switch at all) is the safest though, since if either battery develops a shorted cell (or shorted or loose wiring, blown battery fuse, etc) without your knowledge, it won't affect the other battery and the avionics, thanks to the two diodes.
> > > >
> > > > - Clarification: I meant a diode between each battery and the avionics bus as a whole. Not separately for a specific instrument.
> > >
> > > I measured the inrush current once again and found that the vertical of the scope was set for a 1X probe instead of the 10X actually being used. This meant that the peak current was 90A instead of 9A, which is a bit high. I added a 1.1 ohm resistor and the peak current dropped to 6A. A simulation shows that a 2 ohm resistor drops it to 3A. This is a good value to use if you have a 1A current drain as the voltage drop will be 2V. The wattage of resistor is unimportant because so little energy is being dissipated by the resistor. The energy transferred remains constant regardless of the resistor value as it is the energy required to charge the capacitor (the current pulse lengthens for larger resistor values).
> > >
> > > Tom
> >
> > Tom,good plan to increase the value of the resistor to limit the current inrush. One could even make the resistor 100 ohms and have a reversed diode in parallel with it so that when the capacitor is needed to sustain the instrument during switching, the current would flow in the reverse direction through the diode, bypassing the resistor. You now have the best of both worlds - slow charge of the capacitor when the power is turned on, and a fast discharge to support the instrument without IR drop across the resistor, instead the drop would just be the bias voltage of the diode.
>
> I don't really think that the resistor is necessary, but offer it to those that are overly concerned about the inrush current. The bottom line is the energy that is transferred from the battery to the capacitor heating the switch contacts. Switches have current ratings to limit the temperature rise to tolerable levels when the current is flowing continuously; this short current pulse will not raise the switch contact temperatures to any significant level. Remember, the SAME amount of energy will be transferred between the battery and the capacitor REGARDLESS of the resistor value (joules = C * V). This translates to the SAME amount of switch contact heating. If you make the resistor insanely large so the time constant is on the order of minutes, the heat will dissipate and lower the maximum temperature. A better approach is to use a smaller capacitor that still maintains voltage during switching.
>
> Tom
The concern isn't heating in the switch contacts due to their specified resistance. It is high current arcing during switching. This can cause erosion of the contacts, in more extreme cases welding them together. A 12V battery is quite capable of generating a vey high energy arc, one can be used to arc weld steel. Arcs are peculiar phenomena with negative resistance, not easy to measure their presence or characteristics. All DC switched arc on make, the current of the arc is limited by the impedance of the circuit - in this case very low.
There is little mortal danger, the switch isn't going to catch fire or explode. It probably will have a markedly shorter life. In the worst case it may weld itself on one day. The capacitor may be the easiest solution, but not the most elegant, and it may not be without tears. The best solution is to parallel the batteries always so that routine switching is unnecessary. This has higher reliability, will result in longer battery life, and requires no operator action. If circumstances make that impossible then as suggested above a select switch shunted with diodes, followed by an on-off switch is safe, zero energy loss, and keeps all components in spec.
I'm wondering if the lack of actual problems with, for example, the two switch approach many of us use compared with all the bad things that COULD happen has to do with the circumstances in which we actually use the switches.
I've only been spurred to switch batteries a few times over the years (except to check the voltage). My radio display starts blinking when the voltage gets low. That's not so useful because the radio is mounted low and partially behind the control stick, but I did see it once. My ClearNav vario has a low-voltage warning setting that I think is set at 11.5 v. So between those two devices, I'm fairly sure I would see the need to switch away from an unexpectedly discharging battery before the voltage dropped precipitously (recognizing you don't get much warning for LiFePO4 types). And my backup battery, a gel cell pack all the way back in the tail, usually reads 12 point something anyway. So maybe there would be a 1 volt delta between the two batteries in a most likely failure mode--which hardly ever occurs? Would that be likely to cause a big surge of current from one battery into another?
I don't have a master switch--I use the two battery switches in lieu of that. But normally switching on a battery is the first thing I do and switching it off is the last thing; i.e., there's almost never much of a load across the switch contacts when they separate or meet.
I'm not saying bad things couldn't happen. But the fact that few if any have reported such things might be because they seldom occur in our standard operating mode.
Waiting anxiously for the experts to weigh in. This is more entertaining than arguing about the Coronavirus.
Chip Bearden
JB
On Wednesday, April 15, 2020 at 10:38:25 AM UTC-4, David S wrote:
> Your procedure generally works, but while both batteries are connected, battery #2 will be recharging battery #1 and powering your panel, so there's a chance you could blow the fuse on battery #2. That leaves you to finish your flight with only one nearly depleted battery.
>
> Cheers,
> ...david
>
>
> On Wednesday, April 15, 2020 at 9:18:08 AM UTC-4, wrote:
> > While I am reading these posts with interest, I confess to being an electrical illiterate. I just use two batteries, each with a fuse, and two switches. When switching, I turn on #2 before turning off #1.
> >
> > If these circuits with diodes, resistors, make-before-break switches and so on are superior, please explain why, and if the case is compelling, a circuit diagram would be appreciated so that I might take advantage of the information.
> >
> > After all, in aviation "R & D" actually stands for "Ripoff and Duplicate."
I have been doing this for more than 30 years and have never blown a fuse or had any adverse effect.
UH
2G
April 16th 20, 06:23 PM
On Thursday, April 16, 2020 at 8:56:54 AM UTC-7, Dan Marotta wrote:
> Really?
>
> What is the shorted time when flipping the switch?* What's the voltage
> difference between the two batteries?* What's the total circuit
> resistance, including the internal resistance of the batteries?*
> Theoretical math and practical application do not always agree.* It
> might be fun to set up such a demonstration and use your o'scope to
> measure that current and it's time duration.* Compare that to the "blow
> time" of any fuses.
>
> Seriously, I've done it for years without any problems, but I recognize
> that past performance is no guarantee of future results. I'd be curious
> about the results and you have the equipment to do it.
>
> On 4/16/2020 12:21 AM, 2G wrote:
> > On Wednesday, April 15, 2020 at 6:18:08 AM UTC-7, wrote:
> >> While I am reading these posts with interest, I confess to being an electrical illiterate. I just use two batteries, each with a fuse, and two switches. When switching, I turn on #2 before turning off #1.
> >>
> >> If these circuits with diodes, resistors, make-before-break switches and so on are superior, please explain why, and if the case is compelling, a circuit diagram would be appreciated so that I might take advantage of the information.
> >>
> >> After all, in aviation "R & D" actually stands for "Ripoff and Duplicate."
> > A make-before-break is also called a "shorting" switch. If you use such a switch you WILL short the two batteries together, which could result in a large current flow from the battery with the higher voltage to the battery with the lower voltage. This large current could blow your protection fuse(s). This is especially the case if you have two separate switches.
>
> --
> Dan, 5J
Dan,
The I-35W bridge in Minneapolis worked fine for 40 years before it collapsed. The problem was a design error that was there since Day 1.
How much current flows between your two batteries? A lot! The internal resistance of the batteries is probably 0.01 ohm apiece, 18 AWG wire is 0.06 ohm/ft and your switch contact is typically 0.01 ohm. The big unknown is the location of your batteries and the length of the wire. However, the longer the run the more likely they used a smaller gauge wire, so let's start with 10 ft. This totals 0.1 ohm. Looking at the worst-case scenario, you may have a 5 V difference which results in 50 A of current. Since you are manually flipping switches, this current could last a few seconds. The largest factor here is the length and gauge of the wire. If I were you I would measure the actual current the way I did: with a scope and current probe or shunt.
Tom
2G
April 16th 20, 06:27 PM
On Thursday, April 16, 2020 at 9:29:50 AM UTC-7, jfitch wrote:
> On Wednesday, April 15, 2020 at 11:36:35 PM UTC-7, 2G wrote:
> > On Wednesday, April 15, 2020 at 9:12:33 AM UTC-7, NM wrote:
> > > On Tuesday, April 14, 2020 at 9:28:12 PM UTC-4, 2G wrote:
> > > > On Tuesday, April 14, 2020 at 6:38:00 AM UTC-7, wrote:
> > > > > On Tuesday, April 14, 2020 at 9:31:13 AM UTC-4, wrote:
> > > > > > On Tuesday, April 14, 2020 at 1:25:55 AM UTC-4, Andy Blackburn wrote:
> > > > > > > I used Shottky diodes plus power resistors plus capacitors. I'm no EE but I took enough circuits courses to handle this problem. The Shottky diodes keep the batteries from cross-discharging each other, the capacitors keep the instruments powered when the switch is disconnected from battery 1 and before it is connected to battery 2 and the resistors keep the capacitors from drawing too much current when you power them up since they make the circuit (even with the diodes) look like a direct short initially.
> > > > > > >
> > > > > > > Andy
> > > > > >
> > > > > > KISS. Just the two diodes (and no switch) should be enough. Whichever battery is stronger (higher voltage) would take the load. Automatically. No switching needed. With the higher voltage of LiFePO4 batteries (relative to lead-acid) the voltage drop in the diode is acceptable, especially if it's the Schottky type.
> > > > > >
> > > > > > Or, if you really want to remove the voltage drop in the diodes, add an SPDT switch (perhaps one with also a center-off position) IN PARALLEL to the diodes. No matter which position that switch is in, both batteries will still be connected. But the battery the switch leads to will feed the avionics with no voltage drop since the switch bypasses the diode on that side. The other diode will meanwhile prevent current from going INTO the other battery. The middle-off position (or no switch at all) is the safest though, since if either battery develops a shorted cell (or shorted or loose wiring, blown battery fuse, etc) without your knowledge, it won't affect the other battery and the avionics, thanks to the two diodes.
> > > > >
> > > > > - Clarification: I meant a diode between each battery and the avionics bus as a whole. Not separately for a specific instrument.
> > > >
> > > > I measured the inrush current once again and found that the vertical of the scope was set for a 1X probe instead of the 10X actually being used. This meant that the peak current was 90A instead of 9A, which is a bit high. I added a 1.1 ohm resistor and the peak current dropped to 6A. A simulation shows that a 2 ohm resistor drops it to 3A. This is a good value to use if you have a 1A current drain as the voltage drop will be 2V. The wattage of resistor is unimportant because so little energy is being dissipated by the resistor. The energy transferred remains constant regardless of the resistor value as it is the energy required to charge the capacitor (the current pulse lengthens for larger resistor values).
> > > >
> > > > Tom
> > >
> > > Tom,good plan to increase the value of the resistor to limit the current inrush. One could even make the resistor 100 ohms and have a reversed diode in parallel with it so that when the capacitor is needed to sustain the instrument during switching, the current would flow in the reverse direction through the diode, bypassing the resistor. You now have the best of both worlds - slow charge of the capacitor when the power is turned on, and a fast discharge to support the instrument without IR drop across the resistor, instead the drop would just be the bias voltage of the diode.
> >
> > I don't really think that the resistor is necessary, but offer it to those that are overly concerned about the inrush current. The bottom line is the energy that is transferred from the battery to the capacitor heating the switch contacts. Switches have current ratings to limit the temperature rise to tolerable levels when the current is flowing continuously; this short current pulse will not raise the switch contact temperatures to any significant level. Remember, the SAME amount of energy will be transferred between the battery and the capacitor REGARDLESS of the resistor value (joules = C * V). This translates to the SAME amount of switch contact heating. If you make the resistor insanely large so the time constant is on the order of minutes, the heat will dissipate and lower the maximum temperature. A better approach is to use a smaller capacitor that still maintains voltage during switching.
> >
> > Tom
>
> The concern isn't heating in the switch contacts due to their specified resistance. It is high current arcing during switching. This can cause erosion of the contacts, in more extreme cases welding them together. A 12V battery is quite capable of generating a vey high energy arc, one can be used to arc weld steel. Arcs are peculiar phenomena with negative resistance, not easy to measure their presence or characteristics. All DC switched arc on make, the current of the arc is limited by the impedance of the circuit - in this case very low.
>
> There is little mortal danger, the switch isn't going to catch fire or explode. It probably will have a markedly shorter life. In the worst case it may weld itself on one day. The capacitor may be the easiest solution, but not the most elegant, and it may not be without tears. The best solution is to parallel the batteries always so that routine switching is unnecessary. This has higher reliability, will result in longer battery life, and requires no operator action. If circumstances make that impossible then as suggested above a select switch shunted with diodes, followed by an on-off switch is safe, zero energy loss, and keeps all components in spec.
As I have mentioned several times now, a small series resistor will reduce the current down to acceptable levels. How many times must I repeat this?
Tom
I got so mad the last time we fought this battery switching battle, I went out and hooked a fresh 12v battery to a low 10v battery with the parallel circuit completed with an amp-meter! I saw no spike and only 300m/a current flow as the fresh battery tried to charge the low battery.........the switches didn’t melt, the amp-meter didn’t explode and the fuses didn’t blow!
I have recently solved the issue by only using one battery............15a/h lithium iron battery............have t seen voltage below 12.2 v, even after 4 hours at 2a average current flow!
Please don’t stop arguing this issue, it’s so entertaining to watch the EE’s chase imaginary amps around!
JJ
Dan Marotta
April 16th 20, 08:01 PM
Observation of real life events makes this seem to be as frightening as
a comet hitting the earth.* I have a degree in electrical engineering
and I don't see the problem.* What I do see is people quoting theory.*
They are, of course, absolutely correct and their elegant and sometimes
complicated solutions will also work.* But it is my firm belief from
practical experience and observation that this is really just ****ing in
a rain storm.
You're right, Chip.* Nothing bad is going to happen with your setup.*
And yes, this is a lot more entertaining than that virus thing.
On 4/16/2020 11:03 AM, wrote:
> I'm wondering if the lack of actual problems with, for example, the two switch approach many of us use compared with all the bad things that COULD happen has to do with the circumstances in which we actually use the switches.
>
> I've only been spurred to switch batteries a few times over the years (except to check the voltage). My radio display starts blinking when the voltage gets low. That's not so useful because the radio is mounted low and partially behind the control stick, but I did see it once. My ClearNav vario has a low-voltage warning setting that I think is set at 11.5 v. So between those two devices, I'm fairly sure I would see the need to switch away from an unexpectedly discharging battery before the voltage dropped precipitously (recognizing you don't get much warning for LiFePO4 types). And my backup battery, a gel cell pack all the way back in the tail, usually reads 12 point something anyway. So maybe there would be a 1 volt delta between the two batteries in a most likely failure mode--which hardly ever occurs? Would that be likely to cause a big surge of current from one battery into another?
>
> I don't have a master switch--I use the two battery switches in lieu of that. But normally switching on a battery is the first thing I do and switching it off is the last thing; i.e., there's almost never much of a load across the switch contacts when they separate or meet.
>
> I'm not saying bad things couldn't happen. But the fact that few if any have reported such things might be because they seldom occur in our standard operating mode.
>
> Waiting anxiously for the experts to weigh in. This is more entertaining than arguing about the Coronavirus.
>
> Chip Bearden
> JB
--
Dan, 5J
OK JJ or 9B so if I only want to keep the powerflarm running when switching to bat 2 what size cap and diode do I put in the +12vdc line to the powerflarm and does the cap go in series in the +12 line? or across the +12 -12 or ground?
CH
On Thursday, April 16, 2020 at 1:35:35 PM UTC-4, wrote:
> I got so mad the last time we fought this battery switching battle, I went out and hooked a fresh 12v battery to a low 10v battery with the parallel circuit completed with an amp-meter! I saw no spike and only 300m/a current flow as the fresh battery tried to charge the low battery.........the switches didn’t melt, the amp-meter didn’t explode and the fuses didn’t blow!
> I have recently solved the issue by only using one battery............15a/h lithium iron battery............have t seen voltage below 12.2 v, even after 4 hours at 2a average current flow!
> Please don’t stop arguing this issue, it’s so entertaining to watch the EE’s chase imaginary amps around!
> JJ
Batteries don't keep the same voltage once you start charging them - the voltage jumps up very quickly. The remaining voltage difference driving that current is small (*** assuming the two batteries are of similar chemistry ***). Divided by the internal resistances of both batteries, the resulting current is reasonable. That is why nothing bad happens, usually, during the short period that both batteries (the strong one and the weak one) are connected.
Meanwhile my single 12AH lithium iron phosphate battery seems to have infinite capacity, in the sense that any flight I've done with it, even 6 hours, didn't discharge it too deeply. Unlike lead-acid batteries, these newfangled lithium batteries retain most of their capacity for quite a few seasons of use, output a voltage well over 12V until almost fully discharged, and can be discharged deeply without damaging future capacity or longevity. Much superior tech, and now quite affordable (probably lower cost per year than lead-acid).
2G
April 16th 20, 08:07 PM
On Thursday, April 16, 2020 at 10:35:35 AM UTC-7, wrote:
> I got so mad the last time we fought this battery switching battle, I went out and hooked a fresh 12v battery to a low 10v battery with the parallel circuit completed with an amp-meter! I saw no spike and only 300m/a current flow as the fresh battery tried to charge the low battery.........the switches didn’t melt, the amp-meter didn’t explode and the fuses didn’t blow!
> I have recently solved the issue by only using one battery............15a/h lithium iron battery............have t seen voltage below 12.2 v, even after 4 hours at 2a average current flow!
> Please don’t stop arguing this issue, it’s so entertaining to watch the EE’s chase imaginary amps around!
> JJ
I will add that Jon's concern about arcing is misplaced; arcing is associated with inductive loads when the current is interrupted. What happens in an inductor is that there is a magnetic field that is built up that has to have a place to go when the current is suddenly interrupted. The voltage in the circuit goes to very high levels as a result and will cause an arc in a mechanical switch. This is usually dealt with by a fly-back diode that allows the current to continue to flow and die off gradually.
If arcing had been taking place I would have seen it on the scope waveforms as a series of spikes in the current waveform. This did not occur in any of the many times I tested it, and it should not occur as capacitors are effective in minimizing or eliminating spikes.
Tom
Dan Marotta
April 16th 20, 08:09 PM
Well, Tom,* I don't have a scope and you do, so I ask again:* Why don't
you measure it?
But I will say that it's absolutely impossible to have a 5 volt
difference between the batteries.* I've never seen a 12 volt SLA higher
than 13.6 volts and that's fresh off the charger.* By the time you plug
it into your system, it's closer to 12.2 or 12.4 volts.* An 11.4 volt
battery will run a variometer, but likely won't transmit over your
radio.* A DPST switch will switch over in mili seconds, not seconds.
We're not talking about bridges that carried heavy trucks for 40 years
with a design defect.* If your wiring is 40 years old, I'd suggest
changing it.
They also said that a jet fuel fire couldn't weaken a steel beam
sufficiently to cause a building to collapse.* Come to* Moriarty and
have a look at the steel post and beam hangar that slumped to the ground
last week after a fire.
On 4/16/2020 11:23 AM, 2G wrote:
> On Thursday, April 16, 2020 at 8:56:54 AM UTC-7, Dan Marotta wrote:
>> Really?
>>
>> What is the shorted time when flipping the switch?* What's the voltage
>> difference between the two batteries?* What's the total circuit
>> resistance, including the internal resistance of the batteries?
>> Theoretical math and practical application do not always agree.* It
>> might be fun to set up such a demonstration and use your o'scope to
>> measure that current and it's time duration.* Compare that to the "blow
>> time" of any fuses.
>>
>> Seriously, I've done it for years without any problems, but I recognize
>> that past performance is no guarantee of future results. I'd be curious
>> about the results and you have the equipment to do it.
>>
>> On 4/16/2020 12:21 AM, 2G wrote:
>>> On Wednesday, April 15, 2020 at 6:18:08 AM UTC-7, wrote:
>>>> While I am reading these posts with interest, I confess to being an electrical illiterate. I just use two batteries, each with a fuse, and two switches. When switching, I turn on #2 before turning off #1.
>>>>
>>>> If these circuits with diodes, resistors, make-before-break switches and so on are superior, please explain why, and if the case is compelling, a circuit diagram would be appreciated so that I might take advantage of the information.
>>>>
>>>> After all, in aviation "R & D" actually stands for "Ripoff and Duplicate."
>>> A make-before-break is also called a "shorting" switch. If you use such a switch you WILL short the two batteries together, which could result in a large current flow from the battery with the higher voltage to the battery with the lower voltage. This large current could blow your protection fuse(s). This is especially the case if you have two separate switches.
>> --
>> Dan, 5J
> Dan,
>
> The I-35W bridge in Minneapolis worked fine for 40 years before it collapsed. The problem was a design error that was there since Day 1.
>
> How much current flows between your two batteries? A lot! The internal resistance of the batteries is probably 0.01 ohm apiece, 18 AWG wire is 0.06 ohm/ft and your switch contact is typically 0.01 ohm. The big unknown is the location of your batteries and the length of the wire. However, the longer the run the more likely they used a smaller gauge wire, so let's start with 10 ft. This totals 0.1 ohm. Looking at the worst-case scenario, you may have a 5 V difference which results in 50 A of current. Since you are manually flipping switches, this current could last a few seconds. The largest factor here is the length and gauge of the wire. If I were you I would measure the actual current the way I did: with a scope and current probe or shunt.
>
> Tom
--
Dan, 5J
Dan Marotta
April 16th 20, 08:12 PM
I'm not disagreeing with you, Tom.* You are absolutely correct about the
resistor.* All that I am saying is that, while you are theoretically
correct, in actual practice, I've never seen a problem in a glider
electrical system that did not have the resistor.
Do you own a resistor manufacturing company? :-D
On 4/16/2020 11:27 AM, 2G wrote:
> On Thursday, April 16, 2020 at 9:29:50 AM UTC-7, jfitch wrote:
>> On Wednesday, April 15, 2020 at 11:36:35 PM UTC-7, 2G wrote:
>>> On Wednesday, April 15, 2020 at 9:12:33 AM UTC-7, NM wrote:
>>>> On Tuesday, April 14, 2020 at 9:28:12 PM UTC-4, 2G wrote:
>>>>> On Tuesday, April 14, 2020 at 6:38:00 AM UTC-7, wrote:
>>>>>> On Tuesday, April 14, 2020 at 9:31:13 AM UTC-4, wrote:
>>>>>>> On Tuesday, April 14, 2020 at 1:25:55 AM UTC-4, Andy Blackburn wrote:
>>>>>>>> I used Shottky diodes plus power resistors plus capacitors. I'm no EE but I took enough circuits courses to handle this problem. The Shottky diodes keep the batteries from cross-discharging each other, the capacitors keep the instruments powered when the switch is disconnected from battery 1 and before it is connected to battery 2 and the resistors keep the capacitors from drawing too much current when you power them up since they make the circuit (even with the diodes) look like a direct short initially.
>>>>>>>>
>>>>>>>> Andy
>>>>>>> KISS. Just the two diodes (and no switch) should be enough. Whichever battery is stronger (higher voltage) would take the load. Automatically. No switching needed. With the higher voltage of LiFePO4 batteries (relative to lead-acid) the voltage drop in the diode is acceptable, especially if it's the Schottky type.
>>>>>>>
>>>>>>> Or, if you really want to remove the voltage drop in the diodes, add an SPDT switch (perhaps one with also a center-off position) IN PARALLEL to the diodes. No matter which position that switch is in, both batteries will still be connected. But the battery the switch leads to will feed the avionics with no voltage drop since the switch bypasses the diode on that side. The other diode will meanwhile prevent current from going INTO the other battery. The middle-off position (or no switch at all) is the safest though, since if either battery develops a shorted cell (or shorted or loose wiring, blown battery fuse, etc) without your knowledge, it won't affect the other battery and the avionics, thanks to the two diodes.
>>>>>> - Clarification: I meant a diode between each battery and the avionics bus as a whole. Not separately for a specific instrument.
>>>>> I measured the inrush current once again and found that the vertical of the scope was set for a 1X probe instead of the 10X actually being used. This meant that the peak current was 90A instead of 9A, which is a bit high. I added a 1.1 ohm resistor and the peak current dropped to 6A. A simulation shows that a 2 ohm resistor drops it to 3A. This is a good value to use if you have a 1A current drain as the voltage drop will be 2V. The wattage of resistor is unimportant because so little energy is being dissipated by the resistor. The energy transferred remains constant regardless of the resistor value as it is the energy required to charge the capacitor (the current pulse lengthens for larger resistor values).
>>>>>
>>>>> Tom
>>>> Tom,good plan to increase the value of the resistor to limit the current inrush. One could even make the resistor 100 ohms and have a reversed diode in parallel with it so that when the capacitor is needed to sustain the instrument during switching, the current would flow in the reverse direction through the diode, bypassing the resistor. You now have the best of both worlds - slow charge of the capacitor when the power is turned on, and a fast discharge to support the instrument without IR drop across the resistor, instead the drop would just be the bias voltage of the diode.
>>> I don't really think that the resistor is necessary, but offer it to those that are overly concerned about the inrush current. The bottom line is the energy that is transferred from the battery to the capacitor heating the switch contacts. Switches have current ratings to limit the temperature rise to tolerable levels when the current is flowing continuously; this short current pulse will not raise the switch contact temperatures to any significant level. Remember, the SAME amount of energy will be transferred between the battery and the capacitor REGARDLESS of the resistor value (joules = C * V). This translates to the SAME amount of switch contact heating. If you make the resistor insanely large so the time constant is on the order of minutes, the heat will dissipate and lower the maximum temperature. A better approach is to use a smaller capacitor that still maintains voltage during switching.
>>>
>>> Tom
>> The concern isn't heating in the switch contacts due to their specified resistance. It is high current arcing during switching. This can cause erosion of the contacts, in more extreme cases welding them together. A 12V battery is quite capable of generating a vey high energy arc, one can be used to arc weld steel. Arcs are peculiar phenomena with negative resistance, not easy to measure their presence or characteristics. All DC switched arc on make, the current of the arc is limited by the impedance of the circuit - in this case very low.
>>
>> There is little mortal danger, the switch isn't going to catch fire or explode. It probably will have a markedly shorter life. In the worst case it may weld itself on one day. The capacitor may be the easiest solution, but not the most elegant, and it may not be without tears. The best solution is to parallel the batteries always so that routine switching is unnecessary. This has higher reliability, will result in longer battery life, and requires no operator action. If circumstances make that impossible then as suggested above a select switch shunted with diodes, followed by an on-off switch is safe, zero energy loss, and keeps all components in spec.
> As I have mentioned several times now, a small series resistor will reduce the current down to acceptable levels. How many times must I repeat this?
>
> Tom
--
Dan, 5J
Dan Marotta
April 16th 20, 08:14 PM
Give 'em hell, John!
On 4/16/2020 11:35 AM, wrote:
> I got so mad the last time we fought this battery switching battle, I went out and hooked a fresh 12v battery to a low 10v battery with the parallel circuit completed with an amp-meter! I saw no spike and only 300m/a current flow as the fresh battery tried to charge the low battery.........the switches didn’t melt, the amp-meter didn’t explode and the fuses didn’t blow!
> I have recently solved the issue by only using one battery............15a/h lithium iron battery............have t seen voltage below 12.2 v, even after 4 hours at 2a average current flow!
> Please don’t stop arguing this issue, it’s so entertaining to watch the EE’s chase imaginary amps around!
> JJ
--
Dan, 5J
2G
April 16th 20, 08:16 PM
On Thursday, April 16, 2020 at 12:09:32 PM UTC-7, Dan Marotta wrote:
> Well, Tom,* I don't have a scope and you do, so I ask again:* Why don't
> you measure it?
>
> But I will say that it's absolutely impossible to have a 5 volt
> difference between the batteries.* I've never seen a 12 volt SLA higher
> than 13.6 volts and that's fresh off the charger.* By the time you plug
> it into your system, it's closer to 12.2 or 12.4 volts.* An 11.4 volt
> battery will run a variometer, but likely won't transmit over your
> radio.* A DPST switch will switch over in mili seconds, not seconds.
>
> We're not talking about bridges that carried heavy trucks for 40 years
> with a design defect.* If your wiring is 40 years old, I'd suggest
> changing it.
>
> They also said that a jet fuel fire couldn't weaken a steel beam
> sufficiently to cause a building to collapse.* Come to* Moriarty and
> have a look at the steel post and beam hangar that slumped to the ground
> last week after a fire.
>
> On 4/16/2020 11:23 AM, 2G wrote:
> > On Thursday, April 16, 2020 at 8:56:54 AM UTC-7, Dan Marotta wrote:
> >> Really?
> >>
> >> What is the shorted time when flipping the switch?* What's the voltage
> >> difference between the two batteries?* What's the total circuit
> >> resistance, including the internal resistance of the batteries?
> >> Theoretical math and practical application do not always agree.* It
> >> might be fun to set up such a demonstration and use your o'scope to
> >> measure that current and it's time duration.* Compare that to the "blow
> >> time" of any fuses.
> >>
> >> Seriously, I've done it for years without any problems, but I recognize
> >> that past performance is no guarantee of future results. I'd be curious
> >> about the results and you have the equipment to do it.
> >>
> >> On 4/16/2020 12:21 AM, 2G wrote:
> >>> On Wednesday, April 15, 2020 at 6:18:08 AM UTC-7, wrote:
> >>>> While I am reading these posts with interest, I confess to being an electrical illiterate. I just use two batteries, each with a fuse, and two switches. When switching, I turn on #2 before turning off #1.
> >>>>
> >>>> If these circuits with diodes, resistors, make-before-break switches and so on are superior, please explain why, and if the case is compelling, a circuit diagram would be appreciated so that I might take advantage of the information.
> >>>>
> >>>> After all, in aviation "R & D" actually stands for "Ripoff and Duplicate."
> >>> A make-before-break is also called a "shorting" switch. If you use such a switch you WILL short the two batteries together, which could result in a large current flow from the battery with the higher voltage to the battery with the lower voltage. This large current could blow your protection fuse(s). This is especially the case if you have two separate switches.
> >> --
> >> Dan, 5J
> > Dan,
> >
> > The I-35W bridge in Minneapolis worked fine for 40 years before it collapsed. The problem was a design error that was there since Day 1.
> >
> > How much current flows between your two batteries? A lot! The internal resistance of the batteries is probably 0.01 ohm apiece, 18 AWG wire is 0.06 ohm/ft and your switch contact is typically 0.01 ohm. The big unknown is the location of your batteries and the length of the wire. However, the longer the run the more likely they used a smaller gauge wire, so let's start with 10 ft. This totals 0.1 ohm. Looking at the worst-case scenario, you may have a 5 V difference which results in 50 A of current. Since you are manually flipping switches, this current could last a few seconds. The largest factor here is the length and gauge of the wire. If I were you I would measure the actual current the way I did: with a scope and current probe or shunt.
> >
> > Tom
>
> --
> Dan, 5J
I will measure it. The 5V, which I said was worst case, comes from a 9V battery being connected to a 14V battery. These voltages will equalize very quickly and impossible to see with an ammeter or a DVM because they don't have the bandwidth to observe millisecond events.
Tom
Dan Marotta
April 16th 20, 08:24 PM
Agreed.
On 4/16/2020 1:16 PM, 2G wrote:
> On Thursday, April 16, 2020 at 12:09:32 PM UTC-7, Dan Marotta wrote:
>> Well, Tom,* I don't have a scope and you do, so I ask again:* Why don't
>> you measure it?
>>
>> But I will say that it's absolutely impossible to have a 5 volt
>> difference between the batteries.* I've never seen a 12 volt SLA higher
>> than 13.6 volts and that's fresh off the charger.* By the time you plug
>> it into your system, it's closer to 12.2 or 12.4 volts.* An 11.4 volt
>> battery will run a variometer, but likely won't transmit over your
>> radio.* A DPST switch will switch over in mili seconds, not seconds.
>>
>> We're not talking about bridges that carried heavy trucks for 40 years
>> with a design defect.* If your wiring is 40 years old, I'd suggest
>> changing it.
>>
>> They also said that a jet fuel fire couldn't weaken a steel beam
>> sufficiently to cause a building to collapse.* Come to* Moriarty and
>> have a look at the steel post and beam hangar that slumped to the ground
>> last week after a fire.
>>
>> On 4/16/2020 11:23 AM, 2G wrote:
>>> On Thursday, April 16, 2020 at 8:56:54 AM UTC-7, Dan Marotta wrote:
>>>> Really?
>>>>
>>>> What is the shorted time when flipping the switch?* What's the voltage
>>>> difference between the two batteries?* What's the total circuit
>>>> resistance, including the internal resistance of the batteries?
>>>> Theoretical math and practical application do not always agree.* It
>>>> might be fun to set up such a demonstration and use your o'scope to
>>>> measure that current and it's time duration.* Compare that to the "blow
>>>> time" of any fuses.
>>>>
>>>> Seriously, I've done it for years without any problems, but I recognize
>>>> that past performance is no guarantee of future results. I'd be curious
>>>> about the results and you have the equipment to do it.
>>>>
>>>> On 4/16/2020 12:21 AM, 2G wrote:
>>>>> On Wednesday, April 15, 2020 at 6:18:08 AM UTC-7, wrote:
>>>>>> While I am reading these posts with interest, I confess to being an electrical illiterate. I just use two batteries, each with a fuse, and two switches. When switching, I turn on #2 before turning off #1.
>>>>>>
>>>>>> If these circuits with diodes, resistors, make-before-break switches and so on are superior, please explain why, and if the case is compelling, a circuit diagram would be appreciated so that I might take advantage of the information.
>>>>>>
>>>>>> After all, in aviation "R & D" actually stands for "Ripoff and Duplicate."
>>>>> A make-before-break is also called a "shorting" switch. If you use such a switch you WILL short the two batteries together, which could result in a large current flow from the battery with the higher voltage to the battery with the lower voltage. This large current could blow your protection fuse(s). This is especially the case if you have two separate switches.
>>>> --
>>>> Dan, 5J
>>> Dan,
>>>
>>> The I-35W bridge in Minneapolis worked fine for 40 years before it collapsed. The problem was a design error that was there since Day 1.
>>>
>>> How much current flows between your two batteries? A lot! The internal resistance of the batteries is probably 0.01 ohm apiece, 18 AWG wire is 0.06 ohm/ft and your switch contact is typically 0.01 ohm. The big unknown is the location of your batteries and the length of the wire. However, the longer the run the more likely they used a smaller gauge wire, so let's start with 10 ft. This totals 0.1 ohm. Looking at the worst-case scenario, you may have a 5 V difference which results in 50 A of current. Since you are manually flipping switches, this current could last a few seconds. The largest factor here is the length and gauge of the wire. If I were you I would measure the actual current the way I did: with a scope and current probe or shunt.
>>>
>>> Tom
>> --
>> Dan, 5J
> I will measure it. The 5V, which I said was worst case, comes from a 9V battery being connected to a 14V battery. These voltages will equalize very quickly and impossible to see with an ammeter or a DVM because they don't have the bandwidth to observe millisecond events.
>
> Tom
--
Dan, 5J
CH.........I have switched on the fresh battery just before switching off the low battery for over 40 years and counting. No resisters, no capacitors, no nothing!
I documented this with my little test! Now, I’m using the same batteries, same capacity and same type batteries!
Hope this helps,
JJ
JJ, unfortunately, the previous owner installed a 3 position switch in bat1/off/bat2 so the switiching takes just a fraction of a second longer than if It was just a 2 position. And nothing else in my panel is affected other than the powerflarm and then only once in a while ( last time was in a nationals :(
jfitch
April 17th 20, 06:46 AM
On Thursday, April 16, 2020 at 12:07:53 PM UTC-7, 2G wrote:
> On Thursday, April 16, 2020 at 10:35:35 AM UTC-7, wrote:
> > I got so mad the last time we fought this battery switching battle, I went out and hooked a fresh 12v battery to a low 10v battery with the parallel circuit completed with an amp-meter! I saw no spike and only 300m/a current flow as the fresh battery tried to charge the low battery.........the switches didn’t melt, the amp-meter didn’t explode and the fuses didn’t blow!
> > I have recently solved the issue by only using one battery............15a/h lithium iron battery............have t seen voltage below 12.2 v, even after 4 hours at 2a average current flow!
> > Please don’t stop arguing this issue, it’s so entertaining to watch the EE’s chase imaginary amps around!
> > JJ
>
> I will add that Jon's concern about arcing is misplaced; arcing is associated with inductive loads when the current is interrupted. What happens in an inductor is that there is a magnetic field that is built up that has to have a place to go when the current is suddenly interrupted. The voltage in the circuit goes to very high levels as a result and will cause an arc in a mechanical switch. This is usually dealt with by a fly-back diode that allows the current to continue to flow and die off gradually.
>
> If arcing had been taking place I would have seen it on the scope waveforms as a series of spikes in the current waveform. This did not occur in any of the many times I tested it, and it should not occur as capacitors are effective in minimizing or eliminating spikes.
>
> Tom
Arcing will occur on break in an inductive circuit and on make in a capacitive circuit. It is quite easy to strike an arc with a 12V battery. It will happen without the capacitor - the increased capacitance will increase the energy and/or duration. You are very unlikely to see that on an oscilloscope - it will not create the voltage spike that an inductive load will. Open the switch body up and switch it on in a dark room - you'll see it. I'll be convinced when you show me the manufacturers spec that says 90A is ok on a 2A switch "as long as it doesn't last too long" or recommending it to switch large capacitive loads.
By far the simplest solution is to get a 12AH LFP and fly all day with the voltage above 13 the whole time. (By the way, you do get a warning near the end of the LFP's charge, on my 12AH it takes about 1.5 hrs to go from 12.5 to 11.5V at about the 9 hour mark).
On boats we use make-before-break switches to switch between battery banks (to preserve alternator diodes). The banks are hundreds of AH, with many thousands of Amps available. There are no problems with current when switching, even if a 10V bank is continuously connected to a 14V bank by selecting "Both" on the switch. The reason is the charge acceptance rate on LA batteries is quite low, even from a constant voltage, unlimited current source. Typically less than 1/2C, so in a glider with 12AH batteries, you might get around 6 amps for a few minutes between a fresh and completely discharged battery worst case. At about 1/2C charge rate, the battery terminals will be at max charge voltage even starting from completely flat. The IV characteristics of an LA battery are very different than capacitors. Switching LFP batteries this way may create problems - it depends on the particular batteries. Switching a charged LFP battery onto a discharged LA will again hit the same 1/2C current limit. Switching a charged LA battery onto a discharged LFP could create high current, depending on the batteries.
So it isn't a mystery why this has been done for 40 years with no problems. But those gliders don't have huge capacitors on the bus.
jfitch
April 17th 20, 07:00 AM
On Thursday, April 16, 2020 at 10:27:48 AM UTC-7, 2G wrote:
> On Thursday, April 16, 2020 at 9:29:50 AM UTC-7, jfitch wrote:
> > On Wednesday, April 15, 2020 at 11:36:35 PM UTC-7, 2G wrote:
> > > On Wednesday, April 15, 2020 at 9:12:33 AM UTC-7, NM wrote:
> > > > On Tuesday, April 14, 2020 at 9:28:12 PM UTC-4, 2G wrote:
> > > > > On Tuesday, April 14, 2020 at 6:38:00 AM UTC-7, wrote:
> > > > > > On Tuesday, April 14, 2020 at 9:31:13 AM UTC-4, wrote:
> > > > > > > On Tuesday, April 14, 2020 at 1:25:55 AM UTC-4, Andy Blackburn wrote:
> > > > > > > > I used Shottky diodes plus power resistors plus capacitors. I'm no EE but I took enough circuits courses to handle this problem. The Shottky diodes keep the batteries from cross-discharging each other, the capacitors keep the instruments powered when the switch is disconnected from battery 1 and before it is connected to battery 2 and the resistors keep the capacitors from drawing too much current when you power them up since they make the circuit (even with the diodes) look like a direct short initially..
> > > > > > > >
> > > > > > > > Andy
> > > > > > >
> > > > > > > KISS. Just the two diodes (and no switch) should be enough. Whichever battery is stronger (higher voltage) would take the load. Automatically. No switching needed. With the higher voltage of LiFePO4 batteries (relative to lead-acid) the voltage drop in the diode is acceptable, especially if it's the Schottky type.
> > > > > > >
> > > > > > > Or, if you really want to remove the voltage drop in the diodes, add an SPDT switch (perhaps one with also a center-off position) IN PARALLEL to the diodes. No matter which position that switch is in, both batteries will still be connected. But the battery the switch leads to will feed the avionics with no voltage drop since the switch bypasses the diode on that side. The other diode will meanwhile prevent current from going INTO the other battery. The middle-off position (or no switch at all) is the safest though, since if either battery develops a shorted cell (or shorted or loose wiring, blown battery fuse, etc) without your knowledge, it won't affect the other battery and the avionics, thanks to the two diodes.
> > > > > >
> > > > > > - Clarification: I meant a diode between each battery and the avionics bus as a whole. Not separately for a specific instrument.
> > > > >
> > > > > I measured the inrush current once again and found that the vertical of the scope was set for a 1X probe instead of the 10X actually being used. This meant that the peak current was 90A instead of 9A, which is a bit high. I added a 1.1 ohm resistor and the peak current dropped to 6A. A simulation shows that a 2 ohm resistor drops it to 3A. This is a good value to use if you have a 1A current drain as the voltage drop will be 2V. The wattage of resistor is unimportant because so little energy is being dissipated by the resistor. The energy transferred remains constant regardless of the resistor value as it is the energy required to charge the capacitor (the current pulse lengthens for larger resistor values).
> > > > >
> > > > > Tom
> > > >
> > > > Tom,good plan to increase the value of the resistor to limit the current inrush. One could even make the resistor 100 ohms and have a reversed diode in parallel with it so that when the capacitor is needed to sustain the instrument during switching, the current would flow in the reverse direction through the diode, bypassing the resistor. You now have the best of both worlds - slow charge of the capacitor when the power is turned on, and a fast discharge to support the instrument without IR drop across the resistor, instead the drop would just be the bias voltage of the diode.
> > >
> > > I don't really think that the resistor is necessary, but offer it to those that are overly concerned about the inrush current. The bottom line is the energy that is transferred from the battery to the capacitor heating the switch contacts. Switches have current ratings to limit the temperature rise to tolerable levels when the current is flowing continuously; this short current pulse will not raise the switch contact temperatures to any significant level. Remember, the SAME amount of energy will be transferred between the battery and the capacitor REGARDLESS of the resistor value (joules = C * V). This translates to the SAME amount of switch contact heating. If you make the resistor insanely large so the time constant is on the order of minutes, the heat will dissipate and lower the maximum temperature. A better approach is to use a smaller capacitor that still maintains voltage during switching.
> > >
> > > Tom
> >
> > The concern isn't heating in the switch contacts due to their specified resistance. It is high current arcing during switching. This can cause erosion of the contacts, in more extreme cases welding them together. A 12V battery is quite capable of generating a vey high energy arc, one can be used to arc weld steel. Arcs are peculiar phenomena with negative resistance, not easy to measure their presence or characteristics. All DC switched arc on make, the current of the arc is limited by the impedance of the circuit - in this case very low.
> >
> > There is little mortal danger, the switch isn't going to catch fire or explode. It probably will have a markedly shorter life. In the worst case it may weld itself on one day. The capacitor may be the easiest solution, but not the most elegant, and it may not be without tears. The best solution is to parallel the batteries always so that routine switching is unnecessary. This has higher reliability, will result in longer battery life, and requires no operator action. If circumstances make that impossible then as suggested above a select switch shunted with diodes, followed by an on-off switch is safe, zero energy loss, and keeps all components in spec.
>
> As I have mentioned several times now, a small series resistor will reduce the current down to acceptable levels. How many times must I repeat this?
>
> Tom
A small resistor will reduce the inrush current to the capacitor, but it will also use power all day. A 1 ohm will reduce your effective battery capacity by 8% - making the need for switching all the more likely!
Martin Gregorie[_6_]
April 17th 20, 01:36 PM
On Thu, 16 Apr 2020 23:00:28 -0700, jfitch wrote:
> On Thursday, April 16, 2020 at 10:27:48 AM UTC-7, 2G wrote:
>> On Thursday, April 16, 2020 at 9:29:50 AM UTC-7, jfitch wrote:
>> > On Wednesday, April 15, 2020 at 11:36:35 PM UTC-7, 2G wrote:
>> > > On Wednesday, April 15, 2020 at 9:12:33 AM UTC-7, NM wrote:
>> > > > On Tuesday, April 14, 2020 at 9:28:12 PM UTC-4, 2G wrote:
>> > > > > On Tuesday, April 14, 2020 at 6:38:00 AM UTC-7,
>> > > > > wrote:
>> > > > > > On Tuesday, April 14, 2020 at 9:31:13 AM UTC-4,
>> > > > > > wrote:
>> > > > > > > On Tuesday, April 14, 2020 at 1:25:55 AM UTC-4, Andy
>> > > > > > > Blackburn wrote:
>> > > > > > > > I used Shottky diodes plus power resistors plus
>> > > > > > > > capacitors. I'm no EE but I took enough circuits courses
>> > > > > > > > to handle this problem. The Shottky diodes keep the
>> > > > > > > > batteries from cross-discharging each other, the
>> > > > > > > > capacitors keep the instruments powered when the switch
>> > > > > > > > is disconnected from battery 1 and before it is connected
>> > > > > > > > to battery 2 and the resistors keep the capacitors from
>> > > > > > > > drawing too much current when you power them up since
>> > > > > > > > they make the circuit (even with the diodes) look like a
>> > > > > > > > direct short initially.
>> > > > > > > >
>> > > > > > > > Andy
>> > > > > > >
>> > > > > > > KISS. Just the two diodes (and no switch) should be
>> > > > > > > enough. Whichever battery is stronger (higher voltage)
>> > > > > > > would take the load. Automatically. No switching needed.
>> > > > > > > With the higher voltage of LiFePO4 batteries (relative to
>> > > > > > > lead-acid) the voltage drop in the diode is acceptable,
>> > > > > > > especially if it's the Schottky type.
>> > > > > > >
>> > > > > > > Or, if you really want to remove the voltage drop in the
>> > > > > > > diodes, add an SPDT switch (perhaps one with also a
>> > > > > > > center-off position) IN PARALLEL to the diodes. No matter
>> > > > > > > which position that switch is in, both batteries will still
>> > > > > > > be connected. But the battery the switch leads to will
>> > > > > > > feed the avionics with no voltage drop since the switch
>> > > > > > > bypasses the diode on that side. The other diode will
>> > > > > > > meanwhile prevent current from going INTO the other
>> > > > > > > battery. The middle-off position (or no switch at all) is
>> > > > > > > the safest though, since if either battery develops a
>> > > > > > > shorted cell (or shorted or loose wiring, blown battery
>> > > > > > > fuse, etc) without your knowledge, it won't affect the
>> > > > > > > other battery and the avionics, thanks to the two diodes.
>> > > > > >
>> > > > > > - Clarification: I meant a diode between each battery and the
>> > > > > > avionics bus as a whole. Not separately for a specific
>> > > > > > instrument.
>> > > > >
>> > > > > I measured the inrush current once again and found that the
>> > > > > vertical of the scope was set for a 1X probe instead of the 10X
>> > > > > actually being used. This meant that the peak current was 90A
>> > > > > instead of 9A, which is a bit high. I added a 1.1 ohm resistor
>> > > > > and the peak current dropped to 6A. A simulation shows that a 2
>> > > > > ohm resistor drops it to 3A. This is a good value to use if you
>> > > > > have a 1A current drain as the voltage drop will be 2V. The
>> > > > > wattage of resistor is unimportant because so little energy is
>> > > > > being dissipated by the resistor. The energy transferred
>> > > > > remains constant regardless of the resistor value as it is the
>> > > > > energy required to charge the capacitor (the current pulse
>> > > > > lengthens for larger resistor values).
>> > > > >
>> > > > > Tom
>> > > >
>> > > > Tom,good plan to increase the value of the resistor to limit the
>> > > > current inrush. One could even make the resistor 100 ohms and
>> > > > have a reversed diode in parallel with it so that when the
>> > > > capacitor is needed to sustain the instrument during switching,
>> > > > the current would flow in the reverse direction through the
>> > > > diode, bypassing the resistor. You now have the best of both
>> > > > worlds - slow charge of the capacitor when the power is turned
>> > > > on, and a fast discharge to support the instrument without IR
>> > > > drop across the resistor, instead the drop would just be the bias
>> > > > voltage of the diode.
>> > >
>> > > I don't really think that the resistor is necessary, but offer it
>> > > to those that are overly concerned about the inrush current. The
>> > > bottom line is the energy that is transferred from the battery to
>> > > the capacitor heating the switch contacts. Switches have current
>> > > ratings to limit the temperature rise to tolerable levels when the
>> > > current is flowing continuously; this short current pulse will not
>> > > raise the switch contact temperatures to any significant level.
>> > > Remember, the SAME amount of energy will be transferred between the
>> > > battery and the capacitor REGARDLESS of the resistor value (joules
>> > > = C * V). This translates to the SAME amount of switch contact
>> > > heating. If you make the resistor insanely large so the time
>> > > constant is on the order of minutes, the heat will dissipate and
>> > > lower the maximum temperature. A better approach is to use a
>> > > smaller capacitor that still maintains voltage during switching.
>> > >
>> > > Tom
>> >
>> > The concern isn't heating in the switch contacts due to their
>> > specified resistance. It is high current arcing during switching.
>> > This can cause erosion of the contacts, in more extreme cases welding
>> > them together. A 12V battery is quite capable of generating a vey
>> > high energy arc, one can be used to arc weld steel. Arcs are peculiar
>> > phenomena with negative resistance, not easy to measure their
>> > presence or characteristics. All DC switched arc on make, the current
>> > of the arc is limited by the impedance of the circuit - in this case
>> > very low.
>> >
>> > There is little mortal danger, the switch isn't going to catch fire
>> > or explode. It probably will have a markedly shorter life. In the
>> > worst case it may weld itself on one day. The capacitor may be the
>> > easiest solution, but not the most elegant, and it may not be without
>> > tears. The best solution is to parallel the batteries always so that
>> > routine switching is unnecessary. This has higher reliability, will
>> > result in longer battery life, and requires no operator action. If
>> > circumstances make that impossible then as suggested above a select
>> > switch shunted with diodes, followed by an on-off switch is safe,
>> > zero energy loss, and keeps all components in spec.
>>
>> As I have mentioned several times now, a small series resistor will
>> reduce the current down to acceptable levels. How many times must I
>> repeat this?
>>
>> Tom
>
> A small resistor will reduce the inrush current to the capacitor, but it
> will also use power all day. A 1 ohm will reduce your effective battery
> capacity by 8% - making the need for switching all the more likely!
How do you work that out?
There is no inflow to the capacitor once it is charged to the same
voltage as the prime battery. This happens when you first connect the
battery and set the switch to connect the prime battery to the panel.
Don't forget that the capacitor is on the PANEL side of the switch.
When you switch power over from prime to backup battery the capacitor
will discharge for a millisec or two when no battery is connected and the
capacitor is running the panel, followed by an equally quick inflow as
the backup battery tops up the capacitor to match its voltage.
If you want to be really picky, there will also be a very small outflow
from the capacitor: as the battery voltage slowly drops under load, the
capacitor will discharge slowly to match the battery voltage.
--
Martin | martin at
Gregorie | gregorie dot org
2G
April 18th 20, 04:57 AM
On Thursday, April 16, 2020 at 10:46:57 PM UTC-7, jfitch wrote:
> On Thursday, April 16, 2020 at 12:07:53 PM UTC-7, 2G wrote:
> > On Thursday, April 16, 2020 at 10:35:35 AM UTC-7, wrote:
> > > I got so mad the last time we fought this battery switching battle, I went out and hooked a fresh 12v battery to a low 10v battery with the parallel circuit completed with an amp-meter! I saw no spike and only 300m/a current flow as the fresh battery tried to charge the low battery.........the switches didn’t melt, the amp-meter didn’t explode and the fuses didn’t blow!
> > > I have recently solved the issue by only using one battery.............15a/h lithium iron battery............have t seen voltage below 12.2 v, even after 4 hours at 2a average current flow!
> > > Please don’t stop arguing this issue, it’s so entertaining to watch the EE’s chase imaginary amps around!
> > > JJ
> >
> > I will add that Jon's concern about arcing is misplaced; arcing is associated with inductive loads when the current is interrupted. What happens in an inductor is that there is a magnetic field that is built up that has to have a place to go when the current is suddenly interrupted. The voltage in the circuit goes to very high levels as a result and will cause an arc in a mechanical switch. This is usually dealt with by a fly-back diode that allows the current to continue to flow and die off gradually.
> >
> > If arcing had been taking place I would have seen it on the scope waveforms as a series of spikes in the current waveform. This did not occur in any of the many times I tested it, and it should not occur as capacitors are effective in minimizing or eliminating spikes.
> >
> > Tom
>
> Arcing will occur on break in an inductive circuit and on make in a capacitive circuit. It is quite easy to strike an arc with a 12V battery. It will happen without the capacitor - the increased capacitance will increase the energy and/or duration. You are very unlikely to see that on an oscilloscope - it will not create the voltage spike that an inductive load will. Open the switch body up and switch it on in a dark room - you'll see it. I'll be convinced when you show me the manufacturers spec that says 90A is ok on a 2A switch "as long as it doesn't last too long" or recommending it to switch large capacitive loads.
>
> By far the simplest solution is to get a 12AH LFP and fly all day with the voltage above 13 the whole time. (By the way, you do get a warning near the end of the LFP's charge, on my 12AH it takes about 1.5 hrs to go from 12..5 to 11.5V at about the 9 hour mark).
>
> On boats we use make-before-break switches to switch between battery banks (to preserve alternator diodes). The banks are hundreds of AH, with many thousands of Amps available. There are no problems with current when switching, even if a 10V bank is continuously connected to a 14V bank by selecting "Both" on the switch. The reason is the charge acceptance rate on LA batteries is quite low, even from a constant voltage, unlimited current source. Typically less than 1/2C, so in a glider with 12AH batteries, you might get around 6 amps for a few minutes between a fresh and completely discharged battery worst case. At about 1/2C charge rate, the battery terminals will be at max charge voltage even starting from completely flat. The IV characteristics of an LA battery are very different than capacitors. Switching LFP batteries this way may create problems - it depends on the particular batteries. Switching a charged LFP battery onto a discharged LA will again hit the same 1/2C current limit. Switching a charged LA battery onto a discharged LFP could create high current, depending on the batteries.
>
> So it isn't a mystery why this has been done for 40 years with no problems. But those gliders don't have huge capacitors on the bus.
Arcing is a chaotic process that produces erratic current spikes - go do your research and you will see. I saw none of that on the waveforms I recorded. Again, for what must be the FIFTH TIME add a series resistor to reduce the current level if you wish. I REPEAT, add a series resistor to reduce the current level if you wish. Do you got it this time, Jon?????
Continuing the diversion...
Not a problem here and maybe optimal given the times, but
what is is about the field of EE?
We seem ready, willing and able to have a discussion unencumbered by reality.
Other sorts of E's seem more grounded.
I wonder if it is because with SW and IC's we can make things with so many frilly details that we forget the real?
Or perhaps making really new stuff is about imagining the unreal and then making it happen. That might explain being wired differently?
I think we’re arguing over two different events. In a typical SH bird, that has one switch for Bat-1 or Bat-2, all power will be shut off for a brief moment as batteries are switched.........Yes, this will cause a voltage spike as electronic fields collapse across coils, etc in the radio and other electronic equipment, that was drawing about 2 amps before the power was removed. Several ways to prevent the momentary power loss have been recommended and I believe they will all work!
The procedure that Hank, Chip and I use does not remove all power, it only adds about 2 volts to the power supply! We turn on the fresh battery with its own SPST switch, just before we switch off the low voltage battery with another SPST switch. I have tested this and observed no voltage spike, and read 300 m/a (1/3 amp) as the fresh battery tried to recharge the low battery. This event is little more that what the old “voltage regulator” did when cars had generators. The voltage regulator would shut off generator voltage when the battery was fully charged, then turn on generator voltage as it sensed the battery voltage was getting low. The same event occurs when your buddy offers to “give you a jump” when your car won’t start. Using jumper cables, he adds about 2 volts to your car battery.
Does this help to put to rest this never ending argument?
JJ
> Does this help to put to rest this never ending argument?
> JJ
Of course not! This is RAS! I expect this argument to go on until the Sun burns out. (Because of a lack of diodes and resistors.)
jfitch
April 18th 20, 04:49 PM
On Friday, April 17, 2020 at 5:36:50 AM UTC-7, Martin Gregorie wrote:
> On Thu, 16 Apr 2020 23:00:28 -0700, jfitch wrote:
>
> > On Thursday, April 16, 2020 at 10:27:48 AM UTC-7, 2G wrote:
> >> On Thursday, April 16, 2020 at 9:29:50 AM UTC-7, jfitch wrote:
> >> > On Wednesday, April 15, 2020 at 11:36:35 PM UTC-7, 2G wrote:
> >> > > On Wednesday, April 15, 2020 at 9:12:33 AM UTC-7, NM wrote:
> >> > > > On Tuesday, April 14, 2020 at 9:28:12 PM UTC-4, 2G wrote:
> >> > > > > On Tuesday, April 14, 2020 at 6:38:00 AM UTC-7,
> >> > > > > wrote:
> >> > > > > > On Tuesday, April 14, 2020 at 9:31:13 AM UTC-4,
> >> > > > > > wrote:
> >> > > > > > > On Tuesday, April 14, 2020 at 1:25:55 AM UTC-4, Andy
> >> > > > > > > Blackburn wrote:
> >> > > > > > > > I used Shottky diodes plus power resistors plus
> >> > > > > > > > capacitors. I'm no EE but I took enough circuits courses
> >> > > > > > > > to handle this problem. The Shottky diodes keep the
> >> > > > > > > > batteries from cross-discharging each other, the
> >> > > > > > > > capacitors keep the instruments powered when the switch
> >> > > > > > > > is disconnected from battery 1 and before it is connected
> >> > > > > > > > to battery 2 and the resistors keep the capacitors from
> >> > > > > > > > drawing too much current when you power them up since
> >> > > > > > > > they make the circuit (even with the diodes) look like a
> >> > > > > > > > direct short initially.
> >> > > > > > > >
> >> > > > > > > > Andy
> >> > > > > > >
> >> > > > > > > KISS. Just the two diodes (and no switch) should be
> >> > > > > > > enough. Whichever battery is stronger (higher voltage)
> >> > > > > > > would take the load. Automatically. No switching needed.
> >> > > > > > > With the higher voltage of LiFePO4 batteries (relative to
> >> > > > > > > lead-acid) the voltage drop in the diode is acceptable,
> >> > > > > > > especially if it's the Schottky type.
> >> > > > > > >
> >> > > > > > > Or, if you really want to remove the voltage drop in the
> >> > > > > > > diodes, add an SPDT switch (perhaps one with also a
> >> > > > > > > center-off position) IN PARALLEL to the diodes. No matter
> >> > > > > > > which position that switch is in, both batteries will still
> >> > > > > > > be connected. But the battery the switch leads to will
> >> > > > > > > feed the avionics with no voltage drop since the switch
> >> > > > > > > bypasses the diode on that side. The other diode will
> >> > > > > > > meanwhile prevent current from going INTO the other
> >> > > > > > > battery. The middle-off position (or no switch at all) is
> >> > > > > > > the safest though, since if either battery develops a
> >> > > > > > > shorted cell (or shorted or loose wiring, blown battery
> >> > > > > > > fuse, etc) without your knowledge, it won't affect the
> >> > > > > > > other battery and the avionics, thanks to the two diodes.
> >> > > > > >
> >> > > > > > - Clarification: I meant a diode between each battery and the
> >> > > > > > avionics bus as a whole. Not separately for a specific
> >> > > > > > instrument.
> >> > > > >
> >> > > > > I measured the inrush current once again and found that the
> >> > > > > vertical of the scope was set for a 1X probe instead of the 10X
> >> > > > > actually being used. This meant that the peak current was 90A
> >> > > > > instead of 9A, which is a bit high. I added a 1.1 ohm resistor
> >> > > > > and the peak current dropped to 6A. A simulation shows that a 2
> >> > > > > ohm resistor drops it to 3A. This is a good value to use if you
> >> > > > > have a 1A current drain as the voltage drop will be 2V. The
> >> > > > > wattage of resistor is unimportant because so little energy is
> >> > > > > being dissipated by the resistor. The energy transferred
> >> > > > > remains constant regardless of the resistor value as it is the
> >> > > > > energy required to charge the capacitor (the current pulse
> >> > > > > lengthens for larger resistor values).
> >> > > > >
> >> > > > > Tom
> >> > > >
> >> > > > Tom,good plan to increase the value of the resistor to limit the
> >> > > > current inrush. One could even make the resistor 100 ohms and
> >> > > > have a reversed diode in parallel with it so that when the
> >> > > > capacitor is needed to sustain the instrument during switching,
> >> > > > the current would flow in the reverse direction through the
> >> > > > diode, bypassing the resistor. You now have the best of both
> >> > > > worlds - slow charge of the capacitor when the power is turned
> >> > > > on, and a fast discharge to support the instrument without IR
> >> > > > drop across the resistor, instead the drop would just be the bias
> >> > > > voltage of the diode.
> >> > >
> >> > > I don't really think that the resistor is necessary, but offer it
> >> > > to those that are overly concerned about the inrush current. The
> >> > > bottom line is the energy that is transferred from the battery to
> >> > > the capacitor heating the switch contacts. Switches have current
> >> > > ratings to limit the temperature rise to tolerable levels when the
> >> > > current is flowing continuously; this short current pulse will not
> >> > > raise the switch contact temperatures to any significant level.
> >> > > Remember, the SAME amount of energy will be transferred between the
> >> > > battery and the capacitor REGARDLESS of the resistor value (joules
> >> > > = C * V). This translates to the SAME amount of switch contact
> >> > > heating. If you make the resistor insanely large so the time
> >> > > constant is on the order of minutes, the heat will dissipate and
> >> > > lower the maximum temperature. A better approach is to use a
> >> > > smaller capacitor that still maintains voltage during switching.
> >> > >
> >> > > Tom
> >> >
> >> > The concern isn't heating in the switch contacts due to their
> >> > specified resistance. It is high current arcing during switching.
> >> > This can cause erosion of the contacts, in more extreme cases welding
> >> > them together. A 12V battery is quite capable of generating a vey
> >> > high energy arc, one can be used to arc weld steel. Arcs are peculiar
> >> > phenomena with negative resistance, not easy to measure their
> >> > presence or characteristics. All DC switched arc on make, the current
> >> > of the arc is limited by the impedance of the circuit - in this case
> >> > very low.
> >> >
> >> > There is little mortal danger, the switch isn't going to catch fire
> >> > or explode. It probably will have a markedly shorter life. In the
> >> > worst case it may weld itself on one day. The capacitor may be the
> >> > easiest solution, but not the most elegant, and it may not be without
> >> > tears. The best solution is to parallel the batteries always so that
> >> > routine switching is unnecessary. This has higher reliability, will
> >> > result in longer battery life, and requires no operator action. If
> >> > circumstances make that impossible then as suggested above a select
> >> > switch shunted with diodes, followed by an on-off switch is safe,
> >> > zero energy loss, and keeps all components in spec.
> >>
> >> As I have mentioned several times now, a small series resistor will
> >> reduce the current down to acceptable levels. How many times must I
> >> repeat this?
> >>
> >> Tom
> >
> > A small resistor will reduce the inrush current to the capacitor, but it
> > will also use power all day. A 1 ohm will reduce your effective battery
> > capacity by 8% - making the need for switching all the more likely!
>
> How do you work that out?
>
> There is no inflow to the capacitor once it is charged to the same
> voltage as the prime battery. This happens when you first connect the
> battery and set the switch to connect the prime battery to the panel.
> Don't forget that the capacitor is on the PANEL side of the switch.
>
> When you switch power over from prime to backup battery the capacitor
> will discharge for a millisec or two when no battery is connected and the
> capacitor is running the panel, followed by an equally quick inflow as
> the backup battery tops up the capacitor to match its voltage.
>
> If you want to be really picky, there will also be a very small outflow
> from the capacitor: as the battery voltage slowly drops under load, the
> capacitor will discharge slowly to match the battery voltage.
>
>
> --
> Martin | martin at
> Gregorie | gregorie dot org
Martin, yes it was late at night and I was thinking (or not...) that he had put the resistor in the battery lead side of the circuit - which would of course be stupid. If in the capacitor lead, there is no ongoing energy loss.. Which is why I deleted the post within 15 minutes when rational thought returned.
The comment about arcing is generally true, whether you limit the inrush current to 90A or 9A. Switches arc even with resistive loads. It is a dominant failure mode of switches. Even at 2A, the life will be determined by contact erosion due to arcing. If you are switching 9A instead of the switch rating of 2A, you will reduce the spec'd life of that switch. Take an old one apart, take microphotographs of the contacts, and report back. The contact erosion from arcing will be obvious. Whether you see it on an oscilloscope will depend on many details of the setup. Again, the switch isn't going to burst into flames due to the capacitor - it just isn't the most elegant solution to the problem, or the most simple. The make before break scheme that JJ expounds is simpler and has less high current switching, but wiring all the batteries as one large bank is simpler, more reliable, and results in longer battery life.
2G
April 18th 20, 08:07 PM
On Saturday, April 18, 2020 at 8:49:45 AM UTC-7, jfitch wrote:
> On Friday, April 17, 2020 at 5:36:50 AM UTC-7, Martin Gregorie wrote:
> > On Thu, 16 Apr 2020 23:00:28 -0700, jfitch wrote:
> >
> > > On Thursday, April 16, 2020 at 10:27:48 AM UTC-7, 2G wrote:
> > >> On Thursday, April 16, 2020 at 9:29:50 AM UTC-7, jfitch wrote:
> > >> > On Wednesday, April 15, 2020 at 11:36:35 PM UTC-7, 2G wrote:
> > >> > > On Wednesday, April 15, 2020 at 9:12:33 AM UTC-7, NM wrote:
> > >> > > > On Tuesday, April 14, 2020 at 9:28:12 PM UTC-4, 2G wrote:
> > >> > > > > On Tuesday, April 14, 2020 at 6:38:00 AM UTC-7,
> > >> > > > > wrote:
> > >> > > > > > On Tuesday, April 14, 2020 at 9:31:13 AM UTC-4,
> > >> > > > > > wrote:
> > >> > > > > > > On Tuesday, April 14, 2020 at 1:25:55 AM UTC-4, Andy
> > >> > > > > > > Blackburn wrote:
> > >> > > > > > > > I used Shottky diodes plus power resistors plus
> > >> > > > > > > > capacitors. I'm no EE but I took enough circuits courses
> > >> > > > > > > > to handle this problem. The Shottky diodes keep the
> > >> > > > > > > > batteries from cross-discharging each other, the
> > >> > > > > > > > capacitors keep the instruments powered when the switch
> > >> > > > > > > > is disconnected from battery 1 and before it is connected
> > >> > > > > > > > to battery 2 and the resistors keep the capacitors from
> > >> > > > > > > > drawing too much current when you power them up since
> > >> > > > > > > > they make the circuit (even with the diodes) look like a
> > >> > > > > > > > direct short initially.
> > >> > > > > > > >
> > >> > > > > > > > Andy
> > >> > > > > > >
> > >> > > > > > > KISS. Just the two diodes (and no switch) should be
> > >> > > > > > > enough. Whichever battery is stronger (higher voltage)
> > >> > > > > > > would take the load. Automatically. No switching needed.
> > >> > > > > > > With the higher voltage of LiFePO4 batteries (relative to
> > >> > > > > > > lead-acid) the voltage drop in the diode is acceptable,
> > >> > > > > > > especially if it's the Schottky type.
> > >> > > > > > >
> > >> > > > > > > Or, if you really want to remove the voltage drop in the
> > >> > > > > > > diodes, add an SPDT switch (perhaps one with also a
> > >> > > > > > > center-off position) IN PARALLEL to the diodes. No matter
> > >> > > > > > > which position that switch is in, both batteries will still
> > >> > > > > > > be connected. But the battery the switch leads to will
> > >> > > > > > > feed the avionics with no voltage drop since the switch
> > >> > > > > > > bypasses the diode on that side. The other diode will
> > >> > > > > > > meanwhile prevent current from going INTO the other
> > >> > > > > > > battery. The middle-off position (or no switch at all) is
> > >> > > > > > > the safest though, since if either battery develops a
> > >> > > > > > > shorted cell (or shorted or loose wiring, blown battery
> > >> > > > > > > fuse, etc) without your knowledge, it won't affect the
> > >> > > > > > > other battery and the avionics, thanks to the two diodes..
> > >> > > > > >
> > >> > > > > > - Clarification: I meant a diode between each battery and the
> > >> > > > > > avionics bus as a whole. Not separately for a specific
> > >> > > > > > instrument.
> > >> > > > >
> > >> > > > > I measured the inrush current once again and found that the
> > >> > > > > vertical of the scope was set for a 1X probe instead of the 10X
> > >> > > > > actually being used. This meant that the peak current was 90A
> > >> > > > > instead of 9A, which is a bit high. I added a 1.1 ohm resistor
> > >> > > > > and the peak current dropped to 6A. A simulation shows that a 2
> > >> > > > > ohm resistor drops it to 3A. This is a good value to use if you
> > >> > > > > have a 1A current drain as the voltage drop will be 2V. The
> > >> > > > > wattage of resistor is unimportant because so little energy is
> > >> > > > > being dissipated by the resistor. The energy transferred
> > >> > > > > remains constant regardless of the resistor value as it is the
> > >> > > > > energy required to charge the capacitor (the current pulse
> > >> > > > > lengthens for larger resistor values).
> > >> > > > >
> > >> > > > > Tom
> > >> > > >
> > >> > > > Tom,good plan to increase the value of the resistor to limit the
> > >> > > > current inrush. One could even make the resistor 100 ohms and
> > >> > > > have a reversed diode in parallel with it so that when the
> > >> > > > capacitor is needed to sustain the instrument during switching,
> > >> > > > the current would flow in the reverse direction through the
> > >> > > > diode, bypassing the resistor. You now have the best of both
> > >> > > > worlds - slow charge of the capacitor when the power is turned
> > >> > > > on, and a fast discharge to support the instrument without IR
> > >> > > > drop across the resistor, instead the drop would just be the bias
> > >> > > > voltage of the diode.
> > >> > >
> > >> > > I don't really think that the resistor is necessary, but offer it
> > >> > > to those that are overly concerned about the inrush current. The
> > >> > > bottom line is the energy that is transferred from the battery to
> > >> > > the capacitor heating the switch contacts. Switches have current
> > >> > > ratings to limit the temperature rise to tolerable levels when the
> > >> > > current is flowing continuously; this short current pulse will not
> > >> > > raise the switch contact temperatures to any significant level.
> > >> > > Remember, the SAME amount of energy will be transferred between the
> > >> > > battery and the capacitor REGARDLESS of the resistor value (joules
> > >> > > = C * V). This translates to the SAME amount of switch contact
> > >> > > heating. If you make the resistor insanely large so the time
> > >> > > constant is on the order of minutes, the heat will dissipate and
> > >> > > lower the maximum temperature. A better approach is to use a
> > >> > > smaller capacitor that still maintains voltage during switching.
> > >> > >
> > >> > > Tom
> > >> >
> > >> > The concern isn't heating in the switch contacts due to their
> > >> > specified resistance. It is high current arcing during switching.
> > >> > This can cause erosion of the contacts, in more extreme cases welding
> > >> > them together. A 12V battery is quite capable of generating a vey
> > >> > high energy arc, one can be used to arc weld steel. Arcs are peculiar
> > >> > phenomena with negative resistance, not easy to measure their
> > >> > presence or characteristics. All DC switched arc on make, the current
> > >> > of the arc is limited by the impedance of the circuit - in this case
> > >> > very low.
> > >> >
> > >> > There is little mortal danger, the switch isn't going to catch fire
> > >> > or explode. It probably will have a markedly shorter life. In the
> > >> > worst case it may weld itself on one day. The capacitor may be the
> > >> > easiest solution, but not the most elegant, and it may not be without
> > >> > tears. The best solution is to parallel the batteries always so that
> > >> > routine switching is unnecessary. This has higher reliability, will
> > >> > result in longer battery life, and requires no operator action. If
> > >> > circumstances make that impossible then as suggested above a select
> > >> > switch shunted with diodes, followed by an on-off switch is safe,
> > >> > zero energy loss, and keeps all components in spec.
> > >>
> > >> As I have mentioned several times now, a small series resistor will
> > >> reduce the current down to acceptable levels. How many times must I
> > >> repeat this?
> > >>
> > >> Tom
> > >
> > > A small resistor will reduce the inrush current to the capacitor, but it
> > > will also use power all day. A 1 ohm will reduce your effective battery
> > > capacity by 8% - making the need for switching all the more likely!
> >
> > How do you work that out?
> >
> > There is no inflow to the capacitor once it is charged to the same
> > voltage as the prime battery. This happens when you first connect the
> > battery and set the switch to connect the prime battery to the panel.
> > Don't forget that the capacitor is on the PANEL side of the switch.
> >
> > When you switch power over from prime to backup battery the capacitor
> > will discharge for a millisec or two when no battery is connected and the
> > capacitor is running the panel, followed by an equally quick inflow as
> > the backup battery tops up the capacitor to match its voltage.
> >
> > If you want to be really picky, there will also be a very small outflow
> > from the capacitor: as the battery voltage slowly drops under load, the
> > capacitor will discharge slowly to match the battery voltage.
> >
> >
> > --
> > Martin | martin at
> > Gregorie | gregorie dot org
>
> Martin, yes it was late at night and I was thinking (or not...) that he had put the resistor in the battery lead side of the circuit - which would of course be stupid. If in the capacitor lead, there is no ongoing energy loss. Which is why I deleted the post within 15 minutes when rational thought returned.
>
> The comment about arcing is generally true, whether you limit the inrush current to 90A or 9A. Switches arc even with resistive loads. It is a dominant failure mode of switches. Even at 2A, the life will be determined by contact erosion due to arcing. If you are switching 9A instead of the switch rating of 2A, you will reduce the spec'd life of that switch. Take an old one apart, take microphotographs of the contacts, and report back. The contact erosion from arcing will be obvious. Whether you see it on an oscilloscope will depend on many details of the setup. Again, the switch isn't going to burst into flames due to the capacitor - it just isn't the most elegant solution to the problem, or the most simple. The make before break scheme that JJ expounds is simpler and has less high current switching, but wiring all the batteries as one large bank is simpler, more reliable, and results in longer battery life.
Of course the resistor is in series with the capacitor, forming what is called a snubber circuit. "Snubber" means it snubs transients, which is what arcing produces. But there is no arcing as evidenced by the current waveform.. No wonder, the gap for a 12V circuit to arc is about 160 μinches. Switch manufacturers don't specify instantaneous current limits, only continuous currents. So the datasheet simply doesn't apply to instantaneous currents.
No, YOU should take apart an old part and take the microphotographs of switches you claim to have.
Here are high-speed videos of actual arcing:
https://www.arcsuppressiontechnologies.com/
Here is a demonstration of using a capacitor to suppress arcs:
https://www.youtube.com/watch?v=Xr5_gUrUZxY
Dan Marotta
April 18th 20, 11:28 PM
I hear the arc from a bolt of lightening on my aircraft radio. Nothing
from flipping a battery switch...* In fact, there's not a switch in my
cockpit that causes the slightest crackle in my radio. Sure there's and
arc, but it is (to coin a phrase) a Tempest in a Teapot.
On 4/18/2020 1:07 PM, 2G wrote:
> On Saturday, April 18, 2020 at 8:49:45 AM UTC-7, jfitch wrote:
>> On Friday, April 17, 2020 at 5:36:50 AM UTC-7, Martin Gregorie wrote:
>>> On Thu, 16 Apr 2020 23:00:28 -0700, jfitch wrote:
>>>
>>>> On Thursday, April 16, 2020 at 10:27:48 AM UTC-7, 2G wrote:
>>>>> On Thursday, April 16, 2020 at 9:29:50 AM UTC-7, jfitch wrote:
>>>>>> On Wednesday, April 15, 2020 at 11:36:35 PM UTC-7, 2G wrote:
>>>>>>> On Wednesday, April 15, 2020 at 9:12:33 AM UTC-7, NM wrote:
>>>>>>>> On Tuesday, April 14, 2020 at 9:28:12 PM UTC-4, 2G wrote:
>>>>>>>>> On Tuesday, April 14, 2020 at 6:38:00 AM UTC-7,
>>>>>>>>> wrote:
>>>>>>>>>> On Tuesday, April 14, 2020 at 9:31:13 AM UTC-4,
>>>>>>>>>> wrote:
>>>>>>>>>>> On Tuesday, April 14, 2020 at 1:25:55 AM UTC-4, Andy
>>>>>>>>>>> Blackburn wrote:
>>>>>>>>>>>> I used Shottky diodes plus power resistors plus
>>>>>>>>>>>> capacitors. I'm no EE but I took enough circuits courses
>>>>>>>>>>>> to handle this problem. The Shottky diodes keep the
>>>>>>>>>>>> batteries from cross-discharging each other, the
>>>>>>>>>>>> capacitors keep the instruments powered when the switch
>>>>>>>>>>>> is disconnected from battery 1 and before it is connected
>>>>>>>>>>>> to battery 2 and the resistors keep the capacitors from
>>>>>>>>>>>> drawing too much current when you power them up since
>>>>>>>>>>>> they make the circuit (even with the diodes) look like a
>>>>>>>>>>>> direct short initially.
>>>>>>>>>>>>
>>>>>>>>>>>> Andy
>>>>>>>>>>> KISS. Just the two diodes (and no switch) should be
>>>>>>>>>>> enough. Whichever battery is stronger (higher voltage)
>>>>>>>>>>> would take the load. Automatically. No switching needed.
>>>>>>>>>>> With the higher voltage of LiFePO4 batteries (relative to
>>>>>>>>>>> lead-acid) the voltage drop in the diode is acceptable,
>>>>>>>>>>> especially if it's the Schottky type.
>>>>>>>>>>>
>>>>>>>>>>> Or, if you really want to remove the voltage drop in the
>>>>>>>>>>> diodes, add an SPDT switch (perhaps one with also a
>>>>>>>>>>> center-off position) IN PARALLEL to the diodes. No matter
>>>>>>>>>>> which position that switch is in, both batteries will still
>>>>>>>>>>> be connected. But the battery the switch leads to will
>>>>>>>>>>> feed the avionics with no voltage drop since the switch
>>>>>>>>>>> bypasses the diode on that side. The other diode will
>>>>>>>>>>> meanwhile prevent current from going INTO the other
>>>>>>>>>>> battery. The middle-off position (or no switch at all) is
>>>>>>>>>>> the safest though, since if either battery develops a
>>>>>>>>>>> shorted cell (or shorted or loose wiring, blown battery
>>>>>>>>>>> fuse, etc) without your knowledge, it won't affect the
>>>>>>>>>>> other battery and the avionics, thanks to the two diodes.
>>>>>>>>>> - Clarification: I meant a diode between each battery and the
>>>>>>>>>> avionics bus as a whole. Not separately for a specific
>>>>>>>>>> instrument.
>>>>>>>>> I measured the inrush current once again and found that the
>>>>>>>>> vertical of the scope was set for a 1X probe instead of the 10X
>>>>>>>>> actually being used. This meant that the peak current was 90A
>>>>>>>>> instead of 9A, which is a bit high. I added a 1.1 ohm resistor
>>>>>>>>> and the peak current dropped to 6A. A simulation shows that a 2
>>>>>>>>> ohm resistor drops it to 3A. This is a good value to use if you
>>>>>>>>> have a 1A current drain as the voltage drop will be 2V. The
>>>>>>>>> wattage of resistor is unimportant because so little energy is
>>>>>>>>> being dissipated by the resistor. The energy transferred
>>>>>>>>> remains constant regardless of the resistor value as it is the
>>>>>>>>> energy required to charge the capacitor (the current pulse
>>>>>>>>> lengthens for larger resistor values).
>>>>>>>>>
>>>>>>>>> Tom
>>>>>>>> Tom,good plan to increase the value of the resistor to limit the
>>>>>>>> current inrush. One could even make the resistor 100 ohms and
>>>>>>>> have a reversed diode in parallel with it so that when the
>>>>>>>> capacitor is needed to sustain the instrument during switching,
>>>>>>>> the current would flow in the reverse direction through the
>>>>>>>> diode, bypassing the resistor. You now have the best of both
>>>>>>>> worlds - slow charge of the capacitor when the power is turned
>>>>>>>> on, and a fast discharge to support the instrument without IR
>>>>>>>> drop across the resistor, instead the drop would just be the bias
>>>>>>>> voltage of the diode.
>>>>>>> I don't really think that the resistor is necessary, but offer it
>>>>>>> to those that are overly concerned about the inrush current. The
>>>>>>> bottom line is the energy that is transferred from the battery to
>>>>>>> the capacitor heating the switch contacts. Switches have current
>>>>>>> ratings to limit the temperature rise to tolerable levels when the
>>>>>>> current is flowing continuously; this short current pulse will not
>>>>>>> raise the switch contact temperatures to any significant level.
>>>>>>> Remember, the SAME amount of energy will be transferred between the
>>>>>>> battery and the capacitor REGARDLESS of the resistor value (joules
>>>>>>> = C * V). This translates to the SAME amount of switch contact
>>>>>>> heating. If you make the resistor insanely large so the time
>>>>>>> constant is on the order of minutes, the heat will dissipate and
>>>>>>> lower the maximum temperature. A better approach is to use a
>>>>>>> smaller capacitor that still maintains voltage during switching.
>>>>>>>
>>>>>>> Tom
>>>>>> The concern isn't heating in the switch contacts due to their
>>>>>> specified resistance. It is high current arcing during switching.
>>>>>> This can cause erosion of the contacts, in more extreme cases welding
>>>>>> them together. A 12V battery is quite capable of generating a vey
>>>>>> high energy arc, one can be used to arc weld steel. Arcs are peculiar
>>>>>> phenomena with negative resistance, not easy to measure their
>>>>>> presence or characteristics. All DC switched arc on make, the current
>>>>>> of the arc is limited by the impedance of the circuit - in this case
>>>>>> very low.
>>>>>>
>>>>>> There is little mortal danger, the switch isn't going to catch fire
>>>>>> or explode. It probably will have a markedly shorter life. In the
>>>>>> worst case it may weld itself on one day. The capacitor may be the
>>>>>> easiest solution, but not the most elegant, and it may not be without
>>>>>> tears. The best solution is to parallel the batteries always so that
>>>>>> routine switching is unnecessary. This has higher reliability, will
>>>>>> result in longer battery life, and requires no operator action. If
>>>>>> circumstances make that impossible then as suggested above a select
>>>>>> switch shunted with diodes, followed by an on-off switch is safe,
>>>>>> zero energy loss, and keeps all components in spec.
>>>>> As I have mentioned several times now, a small series resistor will
>>>>> reduce the current down to acceptable levels. How many times must I
>>>>> repeat this?
>>>>>
>>>>> Tom
>>>> A small resistor will reduce the inrush current to the capacitor, but it
>>>> will also use power all day. A 1 ohm will reduce your effective battery
>>>> capacity by 8% - making the need for switching all the more likely!
>>> How do you work that out?
>>>
>>> There is no inflow to the capacitor once it is charged to the same
>>> voltage as the prime battery. This happens when you first connect the
>>> battery and set the switch to connect the prime battery to the panel.
>>> Don't forget that the capacitor is on the PANEL side of the switch.
>>>
>>> When you switch power over from prime to backup battery the capacitor
>>> will discharge for a millisec or two when no battery is connected and the
>>> capacitor is running the panel, followed by an equally quick inflow as
>>> the backup battery tops up the capacitor to match its voltage.
>>>
>>> If you want to be really picky, there will also be a very small outflow
>>> from the capacitor: as the battery voltage slowly drops under load, the
>>> capacitor will discharge slowly to match the battery voltage.
>>>
>>>
>>> --
>>> Martin | martin at
>>> Gregorie | gregorie dot org
>> Martin, yes it was late at night and I was thinking (or not...) that he had put the resistor in the battery lead side of the circuit - which would of course be stupid. If in the capacitor lead, there is no ongoing energy loss. Which is why I deleted the post within 15 minutes when rational thought returned.
>>
>> The comment about arcing is generally true, whether you limit the inrush current to 90A or 9A. Switches arc even with resistive loads. It is a dominant failure mode of switches. Even at 2A, the life will be determined by contact erosion due to arcing. If you are switching 9A instead of the switch rating of 2A, you will reduce the spec'd life of that switch. Take an old one apart, take microphotographs of the contacts, and report back. The contact erosion from arcing will be obvious. Whether you see it on an oscilloscope will depend on many details of the setup. Again, the switch isn't going to burst into flames due to the capacitor - it just isn't the most elegant solution to the problem, or the most simple. The make before break scheme that JJ expounds is simpler and has less high current switching, but wiring all the batteries as one large bank is simpler, more reliable, and results in longer battery life.
> Of course the resistor is in series with the capacitor, forming what is called a snubber circuit. "Snubber" means it snubs transients, which is what arcing produces. But there is no arcing as evidenced by the current waveform. No wonder, the gap for a 12V circuit to arc is about 160 μinches. Switch manufacturers don't specify instantaneous current limits, only continuous currents. So the datasheet simply doesn't apply to instantaneous currents.
>
> No, YOU should take apart an old part and take the microphotographs of switches you claim to have.
>
> Here are high-speed videos of actual arcing:
>
> https://www.arcsuppressiontechnologies.com/
>
> Here is a demonstration of using a capacitor to suppress arcs:
>
> https://www.youtube.com/watch?v=Xr5_gUrUZxY
--
Dan, 5J
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