Battery switching without tears

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.

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.

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.

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

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:Â* Better is the
enemy of good enough.Â* I flew for years with two batteries, two fuses,
and two switches.Â* When battery 1 gets low, flip on battery 2 then flip
off battery 1.Â* 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.

Really?

What is the shorted time when flipping the switch?Â* What's the voltage
difference between the two batteries?Â* What's the total circuit
resistance, including the internal resistance of the batteries?Â*
Theoretical math and practical application do not always agree.Â* It
might be fun to set up such a demonstration and use your o'scope to
measure that current and it's time duration.Â* 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

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

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?Â* What's the voltage
difference between the two batteries?Â* What's the total circuit
resistance, including the internal resistance of the batteries?Â*
Theoretical math and practical application do not always agree.Â* It
might be fun to set up such a demonstration and use your o'scope to
measure that current and it's time duration.Â* 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