Master Switches

On Tuesday, April 23, 2013 7:39:54 AM UTC-4, wrote:
I'm setting up a way to switch between each battery independently versus two or three batteries slaved together via the use of a simple circuit of diodes (rectifiers) and capacitors. The switches we all use are break before make so you don't want to lose power to instruments when switching between batteries. I found a suitable rotary mini-switch with up to 6 positions so you can do any combination of batteries you like (individually or slaved together). I realize the diodes cost about 0.15 volts on the output side so being able to get a little extra juice might be better but I also discovered that batteries last longer if you don't regularly deep discharge them. Since I have two main batteries in the baggage area plus one in the tail I figure my main scenario will be to slave the two mains together via diodes so theres no cross-charging and each draws down at half the rate. On a typical day they only get halfway drained and keep the tail battery as a fallback. I don't want to take the tail off to recharge the tail battery every day.. This means I need a switch to go between Batt 1, Batt 2, Batt 1+2 and Batt 3. A position for Batt 1+2+3 is an option, but not super attractive if you wan an emergency reserve.



I've been using around 2500 micro farads of capacitor with a 1 ohm power resistor across the leads for several years. Works great.In v2 I'm upping the capacitance.



9B



On Tuesday, April 23, 2013 1:06:58 AM UTC-7, Jim White wrote:

If your batteries are dissimilarly discharged during flight (which is



possibly the reason you want to switch batteries) then the instantaneous



current flowing through your circuit when you join them could be quite high



and cause you more problems.







There are two better ways to solve this problem:







1) Insert a capacitor across the power as has been suggested. A further



enhancement would be a choke across the switch.







2) Power your loggers etc off a separate bus which is connected through a



rectifier bridge to both batteries. The logger will draw power from the



higher voltage battery and neither battery can charge the other. There will



be a voltage drop of between 0.5 and 1 volt but most loggers will work down



to 9v so no problem.







Jim







ps: avoid thermal trips if voltage loss is an issue.

Using a single rotary switch to control all your power sources creates a single point of failure.
FWIW

I guess I'm not as worried about switching batteries as you guys. I have 2 batteries powering 2 separate busses with the instruments split onto each bus about equally, with some consideration for redundancy: Bus 1 has SN10 and PFB/Oudie. Bus 2 has radio, backup vario, and Themi backup logger.

Each bus can be powered by either battery 1 or battery 2 (or turned off). With 8AH batteries I rarely run low on either battery (I recharge both each night, rotating 3 batteries through the system.

If one battery gets low, I can power all the panel off the other battery if needed. In that situation I would probably shut down unneeded systems.

Works fine so far ( 12 years since I put it in the glider).

Kirk
66

On Apr 23, 6:39*am, wrote:

via the use of a simple circuit of diodes (rectifiers)...I realize the diodes cost about 0.15 volts on the output side

A couple of points on the use of diodes.

1) For those that don't understand their use, diodes only allow
current to flow one direction (i.e. from the battery to your
electronics) sort of like your sump pump's check valve. If you
connect two batteries in parallel without diodes then one battery
could charge the other (I don't think this is as bad a thing as it is
made out to be). However, if you have a diode on each battery, then
each battery can still power your electronics but cannot charge each
other.

2) 9B mentions that the "diodes cost about 0.15 volts" (called the
forward voltage drop). But this all depends on the type of diode.
The most common silicon diodes have significantly higher voltage
drops. Take for example the Radio Shack 1N4001 "rectifier" diode
(276-1101) which is good for 1 amp. According to the Motorola spec
sheet (http://www.futurlec.com/Diodes/1N4001.shtml) the voltage drop
is between 0.93-1.1 volts! That is a lot of voltage to loose in our
glider's closed power systems and could cause older electronics to
drop out due to low voltage on long flights.

There are other technologies that can handle the current flow we need
and prevent cross-charging of the batteries. One is a pair of
Schottky diodes which have very low forward voltage (http://
en.wikipedia.org/wiki/Schottky_diode). Another interesting item is
the MOSFET "Ideal Diode" which basically has zero forward voltage drop
- see http://www.linear.com/product/LTC4358 - which is made for just
our situation and good to 5 amps!

FWIW - I run my batteries in parallel and have Bat A and Bat B
switches (both on all the time) and no diodes. No issues. YMMV.

My $0.02.

- John

On Tuesday, April 23, 2013 1:06:58 AM UTC-7, Jim White wrote:
If your batteries are dissimilarly discharged during flight (which is

possibly the reason you want to switch batteries) then the instantaneous

current flowing through your circuit when you join them could be quite high

and cause you more problems.


Jim


Jim,

I have often connected dissimilarly charged batteries with no ill effects. My ASH26E has a miniature 5 amp rated "shorting switch" to connect the avionic bat to the engine bat (both 18AH) for charging. No problems.

This is maybe due to the more deeply the battery is discharged, the higher it's initial internal resistance. That, and the fact that the voltage delta between evan a 50% discharged and a fully charged battery is not that great.

bumper

Snipped from another thread...

via the use of a simple circuit of diodes (rectifiers)...I realize the
diodes cost about 0.15 volts on the output side

A couple of points on the use of diodes.

1) For those that don't understand their use, diodes only allow current to
flow one direction (i.e. from the battery to your electronics) sort of like
your sump pump's check valve. If you connect two batteries in parallel
without diodes then one battery could charge the other (I don't think this
is as bad a thing as it is made out to be).

The engineer in me - N.B. I am not an electrical engineer, nor have I ever
played one on TV - has long wondered about the validity of, but never bothered
to research, the widely held caution I'll state as: "Connect two batteries
together in parallel at your own risk because THE STRONGER WILL CHARGE THE
WEAKER!!!"

Being curious, but still too lazy to do real research on the matter, I'll ask
RAS readers instead. :-)

So what? If the starting place is nominally identically sized/charged dual
batteries (e.g. in a sailplane), why should the weaker battery care whether it
gets charging current from another battery or a regulated power supply? Why
should the stronger one care where it's sending its current?

As to the last question, I realize internal resistance of the "being charged"
battery matters, at least theoretically, but in the real world are the risks
(given the underlying starting assumptions) genuinely problematical...or
purely theoretical...or somewhere in-between?

Wouldn't two "nearly identical batteries" directly wired "otherwise
electrically sensibly" together in parallel pretty quickly self-equalize their
charges?

Hoping to receive more factually-based replies than "theoretically-based
handwaving" I am...just curious.

Bob W.

On Tuesday, April 23, 2013 4:45:50 AM UTC-7, John Godfrey (QT) wrote:

Using a single rotary switch to control all your power sources creates a single point of failure.

FWIW

Oh snap! Now I need two independent switches?!?

9B

Bob,
I did the 'test' the last time this came up on RAS. I hooked a fully charged battery (13v) to discharged battery (10v), in parallel and my multi-meter read
120 m/a as the fully charged battery started 'charging' the dead battery. 120m/a is nothing ,,,,,,,,,,, just over a tenth of an amp and that is only for the brief second that both batteries are connected to the buss as I switch from battery one to battery 2.
Can't we finally put this to rest?
JJ

On Apr 24, 9:56*am, BobW wrote:
The engineer in me - N.B. I am not an electrical engineer, nor have I ever
played one on TV - has long wondered about the validity of, but never bothered
to research, the widely held caution I'll state as: "Connect two batteries
together in parallel at your own risk because THE STRONGER WILL CHARGE THE
WEAKER!!!"

Two (or more) batteries wired in parallel will basically act as a
single battery. Basically, the voltages will track, during both
charging and discharging. Moreover, types and capacities do NOT have
to be identical.

Now, if we have a common bus and separate switch for each battery, the
situation changes. The "the stronger will charge the weaker" scenario
occurs when you parallel two batteries at different states of
discharge. There are two things to worry about he (a) wasting the
"stronger" battery charge and (b) current spike, potentially blowing
fuses, damaging wiring and switches.

Let's start with (b). I tested this scenario a year or so ago so my
memory is a bit fuzzy, but here is what I can recall. I had two 9 Ah
batteries, one fully charged and one discharged. I put a current clamp
on a wire, connected the clamp to a scope and flipped the switch.
Result - peak current of two amps or so, decaying *very* quickly. One
could not possibly see those two amps using a standard meter, which
probably explains JJ's results. Anyway, those two amps should not
damage anything.

Now, let's have a look at (a). For the sake of the discussion I am
going to assume that the "charging" current does not decay, but stays
at 2 A - basically, a worst-case scenario. Our pilot notices that his
first battery is dying and flips a switch, bringing a fresh battery
online. He then looks up, scans for traffic, maybe answers a radio
call. Ten seconds later he flips another switch, disconnecting the
first battery. How much of the "good" battery charge did he loose? Two
amps times ten seconds, or 0.002777 hours equals 0.00555 AH or 5.55
mAh. Really, nothing to worry about.

As I wrote, I tested it a while ago and do not remember all the
details. Should someone express interest, I can redo the experiment
and post the waveforms.

I do not understand where would the "voltage spike," mentioned earlier
in this thread, come from. Maybe if one had really long wires with a
bunch of distributed inductance...

Bart

Bart wrote:
On Apr 24, 9:56 am, BobW wrote:
The engineer in me - N.B. I am not an electrical engineer, nor have I ever
played one on TV - has long wondered about the validity of, but never bothered
to research, the widely held caution I'll state as: "Connect two batteries
together in parallel at your own risk because THE STRONGER WILL CHARGE THE
WEAKER!!!"

Two (or more) batteries wired in parallel will basically act as a
single battery.

Maybe, maybe not, depending on what you mean by "act as a
single battery".


Basically, the voltages will track, during both
charging and discharging. Moreover, types and capacities do NOT have
to be identical.

The voltages will indeed track, but that does not mean the currents,
temperatures, pressures, and chemical reactions will track. And
it is those aspects that cause concern.

No two cells in a battery are identical, and they each form a
link in a chain. And we know which link breaks.

Two batteries in parallel are like two chains in parallel.
I think you can see where the analogy is heading.


Now, if we have a common bus and separate switch for each battery, the
situation changes. The "the stronger will charge the weaker" scenario
occurs when you parallel two batteries at different states of
discharge.

Which will /always/ be the case, even with nominally identical
batteries.


There are two things to worry about he (a) wasting the
"stronger" battery charge and (b) current spike, potentially blowing
fuses, damaging wiring and switches.

"Spike" implies short duration; that may not be the case,
particularly when you consider "short compared with what".


Let's start with (b). I tested this scenario a year or so ago so my
memory is a bit fuzzy, but here is what I can recall. I had two 9 Ah
batteries, one fully charged and one discharged. I put a current clamp
on a wire, connected the clamp to a scope and flipped the switch.
Result - peak current of two amps or so, decaying *very* quickly. One
could not possibly see those two amps using a standard meter, which
probably explains JJ's results. Anyway, those two amps should not
damage anything.

It is unwise to base "good practice" on a "single anecdote"


Now, let's have a look at (a). For the sake of the discussion I am
going to assume that the "charging" current does not decay, but stays
at 2 A - basically, a worst-case scenario.

Why do you presume 2A is worst case? Why not 4A or 8A etc? It
would be unwise to base a recommendation on a single test
under one set of conditions.


Our pilot notices that his
first battery is dying and flips a switch, bringing a fresh battery
online. He then looks up, scans for traffic, maybe answers a radio
call. Ten seconds later he flips another switch, disconnecting the
first battery. How much of the "good" battery charge did he loose? Two
amps times ten seconds, or 0.002777 hours equals 0.00555 AH or 5.55
mAh. Really, nothing to worry about.

I agree such loss of charge is not a serious concern.
But there are other causes for concern.


As I wrote, I tested it a while ago and do not remember all the
details. Should someone express interest, I can redo the experiment
and post the waveforms.

If you do, please state *all* the conditions of *all* the
equipment in the test -- which is not a trivial exercise
and will clarify the limitations of the test.


I do not understand where would the "voltage spike," mentioned earlier
in this thread, come from. Maybe if one had really long wires with a
bunch of distributed inductance...

I'm not sure what that "voltage spike" was, but you can
indeed get "interesting" voltages associated with large/fast
current changes, due to a variety of mechanisms. Whether or
not those spikes are significant will depend on the protection
and margins built into the other equipment attached to the power
supply.

If you go against the battery manufacturers recommendations
and/or generally recognised "good practice", then if push
comes to shove, what would be the attitude of insurance
companies and courts of law?

On Wednesday, April 24, 2013 4:34:01 PM UTC-4, Bart wrote:
On Apr 24, 9:56*am, BobW wrote:

The engineer in me - N.B. I am not an electrical engineer, nor have I ever

played one on TV - has long wondered about the validity of, but never bothered

to research, the widely held caution I'll state as: "Connect two batteries

together in parallel at your own risk because THE STRONGER WILL CHARGE THE

WEAKER!!!"



Two (or more) batteries wired in parallel will basically act as a

single battery. Basically, the voltages will track, during both

charging and discharging. Moreover, types and capacities do NOT have

to be identical.



Now, if we have a common bus and separate switch for each battery, the

situation changes. The "the stronger will charge the weaker" scenario

occurs when you parallel two batteries at different states of

discharge. There are two things to worry about he (a) wasting the

"stronger" battery charge and (b) current spike, potentially blowing

fuses, damaging wiring and switches.



Let's start with (b). I tested this scenario a year or so ago so my

memory is a bit fuzzy, but here is what I can recall. I had two 9 Ah

batteries, one fully charged and one discharged. I put a current clamp

on a wire, connected the clamp to a scope and flipped the switch.

Result - peak current of two amps or so, decaying *very* quickly. One

could not possibly see those two amps using a standard meter, which

probably explains JJ's results. Anyway, those two amps should not

damage anything.



Now, let's have a look at (a). For the sake of the discussion I am

going to assume that the "charging" current does not decay, but stays

at 2 A - basically, a worst-case scenario. Our pilot notices that his

first battery is dying and flips a switch, bringing a fresh battery

online. He then looks up, scans for traffic, maybe answers a radio

call. Ten seconds later he flips another switch, disconnecting the

first battery. How much of the "good" battery charge did he loose? Two

amps times ten seconds, or 0.002777 hours equals 0.00555 AH or 5.55

mAh. Really, nothing to worry about.



As I wrote, I tested it a while ago and do not remember all the

details. Should someone express interest, I can redo the experiment

and post the waveforms.



I do not understand where would the "voltage spike," mentioned earlier

in this thread, come from. Maybe if one had really long wires with a

bunch of distributed inductance...



Bart

It's perhaps worth mentioning that the LFP batteries becoming popular now have much lower internal resistance than SLA batteries. They also have a flatter voltage versus state of charge characteristic *but* if you connect in parallel an LFP battery than is fully discharged ( 10 V) with one at full charge ( 13.9 V) I wouldn't be a bit surprised to see blown fuses, and boy, that would suck on a long ridge/wave mission.

Diodes would be cheap insurance here. Agreed that they are overkill for SLA batteries.

T8