I'm designing a system that has to sit out in a field for days with no access to power mains. It contains a computer, low speed motors and a few sensors. Power info is 29V nominal input (8S Lipo), 15A continuous draw. It takes me quite a while to set up all of the software every time it's booted (power on and off) and I don't want to do that in the field. It's also cumbersome to transport back and forth between the office and the operation location.

Ideally, I want to set it up once, deploy it and leave it there and just visit every few hours to swap the battery.

I've been looking into "hot swap" circuits using diodes so that I can plug a second, fully-charged battery in when the first battery is depleted without losing power to the computers.

A lot of resources mention Schottky diodes but at my required voltage and current, I'm not sure how to approach this circuit.

There are some controllers like the LTC4225 from Linear Technology but they don't meet my voltage requirements. If it's not feasible then I can look into boosting the voltage down the line from this controller. There's also some literature from TI here but it's a bit overwhelming.

The second power source will either be another 8S or an equivalent power benchtop power supply.

Any advice on where to start would be greatly welcomed. Thanks!


1 Answer 1


A lot of resources mention Schottky diodes but at my required voltage and current, I'm not sure how to approach this circuit.

The Schottkys won't be reverse polarized with high voltage, so any reasonable voltage rating should be fine. You can always add say 1k resistor across each branch's diode, to ensure the stray charges won't zap them.

Current-rating-wise, you can parallel as many of those as needed. Say, if you got 5A diodes, put 6 of them in parallel - that'd give you 50% derating.

So, the circuit with 5A Schottkys would look as follows:


simulate this circuit – Schematic created using CircuitLab

The only reason to use diodes at all is to avoid cross-charging currents flowing between two batteries.

With just a bench power supply limited to the operating current of the load + charging current, the diodes wouldn't be necessary at all. But they are necessary when two battery packs are in use.

You may want to try and measure the charging current between a fully charged and a fully discharged battery pack. Then, instead of diodes, you could use a PMOS-based current limiter, whose only role would be to limit reverse-charging current.

So, something as follows:


simulate this circuit

With suitable PMOS, this will keep the reverse (charging) current limited to 5A, while the forward current is not limited.

If you don’t want the PMOS to be on a large heatsink, then limit the reverse current to a small one, like 250mA or less. Or, since this is a one-off circuit where the op-amp offset can be trimmed out, you can make it into an "ideal" diode and set max reverse current to say 10mA.

  • \$\begingroup\$ Thanks for all of the detailed info! I have a question on the first circuit. How are you calculating the Schottky values and resistors based on the 29V/15A requirement? (What’s the math behind this) Can the circuit handle such a high continuous current flow? And second, what is the overt benefit of the PMOS circuit over the other? Thanks! \$\endgroup\$
    – ardoknow
    Aug 19, 2022 at 6:08
  • \$\begingroup\$ @ardoknow In the first circuit: if each diode can handle 5A, then you only need three to "handle" 15A. But paralleled diodes have to be derated since they don't share currents identically. So take 2x as many as "needed", i.e. 2*15/5=6. As for the voltage rating: the most reverse voltage the diodes will see is the difference in battery voltages, so no problem there. \$\endgroup\$ Aug 19, 2022 at 14:34
  • \$\begingroup\$ The benefit of the PMOS-based circuit is that it is an ideal diode. And it's dirt cheap: the only real cost is the PMOS, and that will be about the same as the cost of Schottky diodea. \$\endgroup\$ Aug 19, 2022 at 14:34
  • \$\begingroup\$ This is super helpful. I’m going to make this and test it out. So based on this logic, I can hot swap batteries and not loser power. But, based on what you’ve described, does this circuit also provide reverse polarity protection? \$\endgroup\$
    – ardoknow
    Aug 30, 2022 at 4:10

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