I'm driving a 12V, 1.5A (no load) motor from a 12V lead acid battery. From this same 12V source, I connected a 5V boost/buck power regulator behind a few diodes to drop the nominal 12.9V from the battery to ~11.5V to keep the voltage on the regulator in spec. Off the 5 volts, I power a Raspberry Pi 2 (drawing about 200mA,) which then closes a relay to power the motor.

Schematic of motor, relay, microcontroller and power regulator:

Schematic of motor, relay, microcontroller and power regulator

The RPi2 starts great off the regulated voltage, and independently the motor works great off the battery. When I hook it all up, and the relay is closed, the RPi2 shuts off. The 2nd time this happened during a test, it appears the power regulator got fried, as it no longer supplies 5V.

I'm guessing the motor draws too much current to start, dropping the voltage too much on the regulator, but why would this burn out the regulator? How can I prevent this from happening? Would a large capacitor across the power regulator input save it?

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    \$\begingroup\$ Because your battery couldn't absorb all the back EMF. \$\endgroup\$ Jul 4, 2016 at 23:35
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    \$\begingroup\$ It would help a lot if you took voltage readings at the motor and voltage regulator and battery, before and after switching on the motor. We need some numbers to work with. The relay should have a clamping diode across the coil, or back EMF could destroy the voltage regulator. Another diode across the motor would help as well. \$\endgroup\$
    – user105652
    Jul 4, 2016 at 23:37
  • \$\begingroup\$ The relay does have the diode - it's part of the relay module. The motor is connected so that it can switch voltage and reverse (there are 4 relays total, the diagram is simplified), so the clamping diode won't work. The static voltage readings are all as expected: 12.9v, 11.5v and 5.15v. \$\endgroup\$
    – codekaizen
    Jul 5, 2016 at 0:14
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    \$\begingroup\$ Are you asking us to determine why your circuit failed by giving us a schematic of another circuit? \$\endgroup\$
    – Transistor
    Jul 5, 2016 at 6:38
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    \$\begingroup\$ Can you afford a small resistor in series with and before your LDO? In that case you might pull this off with an zener across the input to clamp the voltage. Anti-paralell zener-diodes and snubber across the motor comes to mind too. \$\endgroup\$
    – winny
    Jul 5, 2016 at 7:55

2 Answers 2


If you reverse the motor before it stops spinning, you may be placing more than 12 volts at the input of the power regulator. Try adding diodes from the battery toward the motor (before the polarity switching relays). You want to stop any voltage larger than you are expecting from reaching your regulator and especially your processor.

  • \$\begingroup\$ Guess I should ask: Did the regulator fail while switching the motor directly from forward to reverse? \$\endgroup\$
    – st2000
    Jul 5, 2016 at 0:44
  • \$\begingroup\$ I'll do this, thanks. The motor failed before it reversed, in fact it didn't even spin up, just barely started. This shut down the processor and opened the relays, stopping the motor. \$\endgroup\$
    – codekaizen
    Jul 5, 2016 at 1:20
  • \$\begingroup\$ Ok, thanks for the insight, and thanks for understanding the principles involved rather than needing an exact schematic. A snubber diode did the trick. \$\endgroup\$
    – codekaizen
    Jul 8, 2016 at 22:09

I know it's been 5 years since the question, but I would have suggested a large capacitor across the battery, such as 100-1000uF, 16V, unless you already have it. You should also add one at the input and one at the output of the voltage regulator, with values from 1uF to 100uF (even 0.1uF could do the trick, see what is the lowest that solves the problem, and then use 2 to 5 times larger capacitance). They would greatly help by absorbing most of the voltage spikes.


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