Battery switching without tears

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.

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."

On Wednesday, April 15, 2020 at 6:18:08 AM UTC-7, wrote:
While I am reading these posts with interest, I confess to being an electrical illiterate. I just use two batteries, each with a fuse, and two switches. When switching, I turn on #2 before turning off #1.

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

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

A make-before-break is also called a "shorting" switch. If you use such a switch you WILL short the two batteries together, which could result in a large current flow from the battery with the higher voltage to the battery with the lower voltage. This large current could blow your protection fuse(s).. This is especially the case if you have two separate switches.

On Wednesday, April 15, 2020 at 5:57:07 AM UTC-7, wrote:

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

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

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

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

The circuit is very simply: a capacitor and a switch to a battery. In the second case, a 1.1 ohm resistor is added in series between the capacitor and the battery. You are mistaking a d.c. circuit and an a.c. circuit: when the switch closes current will flow from the battery to the capacitor. This current flow is modeled by a differential equation:

i = C dv/dt
or
dv/dt = i / C
Integrated wrt time gives you:
V = 1/C Int[i dt]

In other words, the voltage on the capacitor increases as current flows into it. If you look at it in the steady state, or d.c., you will miss this entirely.

I measured the current with a 0.0015 ohm current shunt and an oscilloscope.