Battery switching without tears

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

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.

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 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.

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'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

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!

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

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.