I'm trying to figure out how to switch source power from one battery to another when one is dead or disconnected. the circuit needs to be able detect the loss of power on the jettisoned battery and switch to main battery. and have a fast switching time or a capacitor for powering the device during the switch.

What its for: I want to be able to carry external batteries on my RC plane and use these batteries first. When they are discharged I want to drop them (when it is safe and yes it is legal to drop things from RC craft when proper safety is ensured. i plan on dropping them near me with small parachutes so they can be recovered and reused) the switch need to be fast so not to reset the flight controller or Electronic Speed Controller (ESC) and so not to lose GPS signal.

lowest voltage at the battery(droppable) would be 12.2V and the ESC and flight controller can opperate down to 7v The system with the motor at 0% throttle draws about 700mA(being generous OSD shows 600mA)

update: In response ton the answer by @DonJoe. Would the circuit look something like this? enter image description here

  • \$\begingroup\$ Can you arrange to throttle back and glide for a split second during the changeover? That way the changeover will occur at a modest current level instead of at 50A. This will make it much easier to hold up the voltage during the change. \$\endgroup\$
    – mkeith
    Oct 1, 2017 at 19:19
  • \$\begingroup\$ That is very do-able but I would like to have some room for error incase i needed throttle for some reason (Not WOT) I can see the current draw in my On Screen Display. \$\endgroup\$ Oct 1, 2017 at 19:47
  • \$\begingroup\$ The plane has to carry the full weight of all the batteries anyway, so what is the point of dropping them? Why not just use one larger battery? \$\endgroup\$ Oct 1, 2017 at 20:03
  • \$\begingroup\$ Are you going to use one of your receiver channels to perform the switchover? I mean, do you have the ability to generate a signal at the moment you want the switch to occur? We don't have to solve that part of it, right? \$\endgroup\$
    – mkeith
    Oct 1, 2017 at 20:25
  • 1
    \$\begingroup\$ @Bruce Abbott: yes at first the plane will have to carry the full load, but once the jettison-able batteries are dead they are just dead weight. I feel if i drop them Ill save weight and use less throttle to fly. \$\endgroup\$ Oct 1, 2017 at 22:16

2 Answers 2


You want to do some research into battery backup circuits built around an SCR (A Silicon Controlled Rectifier).

The basic principle will be something like the below diagram, but incorporating two batteries rather than a battery backing up a rectified AC power source:


simulate this circuit – Schematic created using CircuitLab

I studied this circuit in a power electronics course in college. The lamp is initially illuminated by the rectified DC from the transformer, and the battery is simultaneously charged by that same DC current. If there is an interruption in the AC signal driving the transformer, the battery will discharge through R2, driving the gate of the SCR and allowing current from the battery to flow through the SCR to the lamp.

I know this isn't exactly what you're looking for, but I think it should be a good starting place for your solution. You'll just need to do some prototyping / experimentation with SCR circuits.

  • \$\begingroup\$ What is the voltage drop in SCR1 when it is conducting 50A? With that voltage drop, how much power will it dissipate? \$\endgroup\$
    – mkeith
    Oct 1, 2017 at 22:25
  • \$\begingroup\$ Is R1 the load? the motor and system will pull about 40 amps at full throttle. \$\endgroup\$ Oct 1, 2017 at 22:36
  • \$\begingroup\$ The lamp is the load in this circuit; R1 is acting to limit charging current drawn through D3 into the battery. The forward drop on the SCR will depend on the specific SCR chosen, but somewhere near 1 V is a reasonable approximation, so 50W dissipation by the SCR should be expected. \$\endgroup\$
    – Don Joe
    Oct 1, 2017 at 22:44
  • \$\begingroup\$ can you explain how this circuit works? I know that when the AC power is lost the battery kick in but that's only because you said it was a battery backup. \$\endgroup\$ Oct 2, 2017 at 0:02
  • \$\begingroup\$ In this circuit, high voltage AC is stepped down by XFMR1 and rectified by D1/D2. In your application, this section would be replaced by your main battery. \$\endgroup\$
    – Don Joe
    Oct 2, 2017 at 0:05

If you already have some sort of programmable microcontroller on board, something like an OR-ing PMIC might be another possible solution. If you used an LM5050-1 with an external N-channel MOSFET, you could get away with much lower losses than a diode would have. Programming would be as simple as writing the shutdown pin high on your primary battery until you wanted to use it, at which point you'd write the shutdown pin low and current would start to flow out of it. The LM5050-1 on the secondary battery would protect it against reverse current until you dropped it.

You would just use your two batteries in place of the power supplies shown here


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