Battery switching without tears

On 4/13/20 11:25 PM, Andy Blackburn wrote:
I used Shottky diodes plus power resistors plus capacitors. I'm no EE but I took enough circuits courses to handle this problem. The Shottky diodes keep the batteries from cross-discharging each other, the capacitors keep the instruments powered when the switch is disconnected from battery 1 and before it is connected to battery 2 and the resistors keep the capacitors from drawing too much current when you power them up since they make the circuit (even with the diodes) look like a direct short initially.

Andy

On Monday, April 13, 2020 at 1:08:16 PM UTC-7, kinsell wrote:
That could be a significant voltage drop across the resistors. I'd use
a couple power Schottky diodes instead.

You might want to take a serious look at the IR drop across the
resistors. You have a complex solution to a simple problem that could
actually be causing more problems than it solves.

On Tuesday, April 14, 2020 at 1:25:55 AM UTC-4, Andy Blackburn wrote:
I used Shottky diodes plus power resistors plus capacitors. I'm no EE but I took enough circuits courses to handle this problem. The Shottky diodes keep the batteries from cross-discharging each other, the capacitors keep the instruments powered when the switch is disconnected from battery 1 and before it is connected to battery 2 and the resistors keep the capacitors from drawing too much current when you power them up since they make the circuit (even with the diodes) look like a direct short initially.

Andy

KISS. Just the two diodes (and no switch) should be enough. Whichever battery is stronger (higher voltage) would take the load. Automatically. No switching needed. With the higher voltage of LiFePO4 batteries (relative to lead-acid) the voltage drop in the diode is acceptable, especially if it's the Schottky type.

Or, if you really want to remove the voltage drop in the diodes, add an SPDT switch (perhaps one with also a center-off position) IN PARALLEL to the diodes. No matter which position that switch is in, both batteries will still be connected. But the battery the switch leads to will feed the avionics with no voltage drop since the switch bypasses the diode on that side. The other diode will meanwhile prevent current from going INTO the other battery. The middle-off position (or no switch at all) is the safest though, since if either battery develops a shorted cell (or shorted or loose wiring, blown battery fuse, etc) without your knowledge, it won't affect the other battery, thanks to the diodes.

On Tuesday, April 14, 2020 at 9:31:13 AM UTC-4, wrote:
On Tuesday, April 14, 2020 at 1:25:55 AM UTC-4, Andy Blackburn wrote:
I used Shottky diodes plus power resistors plus capacitors. I'm no EE but I took enough circuits courses to handle this problem. The Shottky diodes keep the batteries from cross-discharging each other, the capacitors keep the instruments powered when the switch is disconnected from battery 1 and before it is connected to battery 2 and the resistors keep the capacitors from drawing too much current when you power them up since they make the circuit (even with the diodes) look like a direct short initially.

Andy

KISS. Just the two diodes (and no switch) should be enough. Whichever battery is stronger (higher voltage) would take the load. Automatically. No switching needed. With the higher voltage of LiFePO4 batteries (relative to lead-acid) the voltage drop in the diode is acceptable, especially if it's the Schottky type.

Or, if you really want to remove the voltage drop in the diodes, add an SPDT switch (perhaps one with also a center-off position) IN PARALLEL to the diodes. No matter which position that switch is in, both batteries will still be connected. But the battery the switch leads to will feed the avionics with no voltage drop since the switch bypasses the diode on that side. The other diode will meanwhile prevent current from going INTO the other battery. The middle-off position (or no switch at all) is the safest though, since if either battery develops a shorted cell (or shorted or loose wiring, blown battery fuse, etc) without your knowledge, it won't affect the other battery and the avionics, thanks to the two diodes.

- Clarification: I meant a diode between each battery and the avionics bus as a whole. Not separately for a specific instrument.

On Tuesday, April 14, 2020 at 6:38:00 AM UTC-7, wrote:
On Tuesday, April 14, 2020 at 9:31:13 AM UTC-4, wrote:
On Tuesday, April 14, 2020 at 1:25:55 AM UTC-4, Andy Blackburn wrote:
I used Shottky diodes plus power resistors plus capacitors. I'm no EE but I took enough circuits courses to handle this problem. The Shottky diodes keep the batteries from cross-discharging each other, the capacitors keep the instruments powered when the switch is disconnected from battery 1 and before it is connected to battery 2 and the resistors keep the capacitors from drawing too much current when you power them up since they make the circuit (even with the diodes) look like a direct short initially.

Andy

KISS. Just the two diodes (and no switch) should be enough. Whichever battery is stronger (higher voltage) would take the load. Automatically. No switching needed. With the higher voltage of LiFePO4 batteries (relative to lead-acid) the voltage drop in the diode is acceptable, especially if it's the Schottky type.

Or, if you really want to remove the voltage drop in the diodes, add an SPDT switch (perhaps one with also a center-off position) IN PARALLEL to the diodes. No matter which position that switch is in, both batteries will still be connected. But the battery the switch leads to will feed the avionics with no voltage drop since the switch bypasses the diode on that side.. The other diode will meanwhile prevent current from going INTO the other battery. The middle-off position (or no switch at all) is the safest though, since if either battery develops a shorted cell (or shorted or loose wiring, blown battery fuse, etc) without your knowledge, it won't affect the other battery and the avionics, thanks to the two diodes.

- Clarification: I meant a diode between each battery and the avionics bus as a whole. Not separately for a specific instrument.

I measured the inrush current once again and found that the vertical of the scope was set for a 1X probe instead of the 10X actually being used. This meant that the peak current was 90A instead of 9A, which is a bit high. I added a 1.1 ohm resistor and the peak current dropped to 6A. A simulation shows that a 2 ohm resistor drops it to 3A. This is a good value to use if you have a 1A current drain as the voltage drop will be 2V. The wattage of resistor is unimportant because so little energy is being dissipated by the resistor. The energy transferred remains constant regardless of the resistor value as it is the energy required to charge the capacitor (the current pulse lengthens for larger resistor values).

Tom

I measured the inrush current once again and found that the vertical of the scope was set for a 1X probe instead of the 10X actually being used. This meant that the peak current was 90A instead of 9A, which is a bit high. I added a 1.1 ohm resistor and the peak current dropped to 6A.

My understanding of that test and the numbers don't add up. I'm thinking you are charging a cap with a 12 V battery and measuring the current with and without and added 1.1 ohm series resistor?

If I combine the two measurements to find the open circuit voltage of the battery and overall wiring resistance I get 7.07 volts and 0.078 ohms.

Probably I don't understand what you are testing. We seem to have lots of time to kill. Could you send a picture of the circuits with the probe attached?

While I am reading these posts with interest, I confess to being an electrical illiterate. I just use two batteries, each with a fuse, and two switches. When switching, I turn on #2 before turning off #1.

If these circuits with diodes, resistors, make-before-break switches and so on are superior, please explain why, and if the case is compelling, a circuit diagram would be appreciated so that I might take advantage of the information.

After all, in aviation "R & D" actually stands for "Ripoff and Duplicate."

Your procedure generally works, but while both batteries are connected, battery #2 will be recharging battery #1 and powering your panel, so there's a chance you could blow the fuse on battery #2. That leaves you to finish your flight with only one nearly depleted battery.

Cheers,
...david


On Wednesday, April 15, 2020 at 9:18:08 AM UTC-4, wrote:
While I am reading these posts with interest, I confess to being an electrical illiterate. I just use two batteries, each with a fuse, and two switches. When switching, I turn on #2 before turning off #1.

If these circuits with diodes, resistors, make-before-break switches and so on are superior, please explain why, and if the case is compelling, a circuit diagram would be appreciated so that I might take advantage of the information.

After all, in aviation "R & D" actually stands for "Ripoff and Duplicate."

David- Thanks for the advice. However, I rarely have both battery switches ON for more than a second. Is this still a potential problem? I can add diodes if necessary.

No, two similar batteries have virtually no ability to cross charge like
that. People take introductory EE classes, learn about ideal voltage
sources, and assume batteries are like that. They're not.

Years ago, in a previous incantation of this same discussion, I
suggested someone take a charged and discharged battery, connect them
through an ammeter, and report the results. They did, and couldn't even
see a flicker of the needle. They concluded that ampere hours of charge
had instantly transferred from one battery to the other, before the
needle had a chance to twitch. I pointed out the ammeter would be
nothing more than a smoking hole in the table if that were true.

Mark can use his procedure if he wants, or better yet just keep both
switches on an forget about flipping switches. Unless he's a former 747
captain, who likes to fiddle with lots of switches.

Dave

P.S. Assumes batteries of the same chemistry



On 4/15/20 8:38 AM, David S wrote:
Your procedure generally works, but while both batteries are connected, battery #2 will be recharging battery #1 and powering your panel, so there's a chance you could blow the fuse on battery #2. That leaves you to finish your flight with only one nearly depleted battery.

Cheers,
...david


On Wednesday, April 15, 2020 at 9:18:08 AM UTC-4, wrote:
While I am reading these posts with interest, I confess to being an electrical illiterate. I just use two batteries, each with a fuse, and two switches. When switching, I turn on #2 before turning off #1.

If these circuits with diodes, resistors, make-before-break switches and so on are superior, please explain why, and if the case is compelling, a circuit diagram would be appreciated so that I might take advantage of the information.

After all, in aviation "R & D" actually stands for "Ripoff and Duplicate."

Unless he's a former 747 captain, who likes to fiddle with lots of switches.

Thanks for that advice, also. It's pretty much what I suspected. But, while I don't mind flipping switches, some people tell me I am better at pushing their buttons.