Battery switching without tears

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

Observation of real life events makes this seem to be as frightening as
a comet hitting the earth.Â* I have a degree in electrical engineering
and I don't see the problem.Â* What I do see is people quoting theory.Â*
They are, of course, absolutely correct and their elegant and sometimes
complicated solutions will also work.Â* 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.Â* Nothing bad is going to happen with your setup.Â*
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).

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

Well, Tom,Â* I don't have a scope and you do, so I ask again:Â* Why don't
you measure it?

But I will say that it's absolutely impossible to have a 5 volt
difference between the batteries.Â* I've never seen a 12 volt SLA higher
than 13.6 volts and that's fresh off the charger.Â* By the time you plug
it into your system, it's closer to 12.2 or 12.4 volts.Â* An 11.4 volt
battery will run a variometer, but likely won't transmit over your
radio.Â* 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.Â* 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.Â* Come toÂ* 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?Â* What's the voltage
difference between the two batteries?Â* What's the total circuit
resistance, including the internal resistance of the batteries?
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

--
Dan, 5J

I'm not disagreeing with you, Tom.Â* You are absolutely correct about the
resistor.Â* 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

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

On Thursday, April 16, 2020 at 12:09:32 PM UTC-7, Dan Marotta wrote:
Well, Tom,Â* I don't have a scope and you do, so I ask again:Â* Why don't
you measure it?

But I will say that it's absolutely impossible to have a 5 volt
difference between the batteries.Â* I've never seen a 12 volt SLA higher
than 13.6 volts and that's fresh off the charger.Â* By the time you plug
it into your system, it's closer to 12.2 or 12.4 volts.Â* An 11.4 volt
battery will run a variometer, but likely won't transmit over your
radio.Â* 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.Â* 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.Â* Come toÂ* 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?Â* What's the voltage
difference between the two batteries?Â* What's the total circuit
resistance, including the internal resistance of the batteries?
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

--
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