I am trying to use a IMS-1 H-Bridge with a 24V/20A psu and a motor 24V/20A. I already burned one of these H-bridges in one direction and I assume that was caused by a back EMF.

Specs of the H-bridge are:

Rated voltage: 5V-24V
Maximum voltage: 5V-30V
When R1 position meets with a short circuit, VCC and B + shorted. VCC output voltage is the same with input voltage of the drive.
When R1 position open, VCC and B + disconnected. VCC input voltage 3-12V..
Current Rating: 20A
Peak Current: 59A
Switching frequency: 1K to 200kHz

So this thing already got some sort of protection but probably not enough. what circuit can I add between H-bridge and motor to get extra protection? I want to make sure not more than 59A / 30V come back to it, better far below that. I read a lot about circuits for this but all describe the circuit inside the H-bridge.

I am soldering for 30 years or more but my knowledge is like limited W/V=A :-/

  • \$\begingroup\$ Have you read about flyback diodes? en.wikipedia.org/wiki/Flyback_diode \$\endgroup\$
    – Dampmaskin
    Mar 10, 2017 at 22:15
  • \$\begingroup\$ Show your current circuit please. \$\endgroup\$
    – skvery
    Mar 10, 2017 at 22:21
  • \$\begingroup\$ Yes I read about flyback diodes I am just too dumb to understand how these would work in a circuit with switching directions/poles. I cannot really show the circuit here since I don´t know the circuit of the IMS-1. So that is some sort of black box for me that outputs PWM to the motor. \$\endgroup\$ Mar 11, 2017 at 8:03
  • \$\begingroup\$ I am trying to understand the circuit with my dilettante knowledge. Once the switch is toggled, the energy flows from + to - so the diode just gets ignored at return the resistor eats up the emf and the diode blocks the back flow in its specs. The switch in the circuit is the IMS-1 of course. But thats not going to work with the other direction so I need a second diode but where and how? \$\endgroup\$ Mar 11, 2017 at 8:11

1 Answer 1


I’ve had some experience of H-bridges that were destroyed by the back-EMF when the battery was disconnected: the classical reverse diode could not help since the current could not flow back to the (disconnected) battery. I’ve started adding some TVS diodes in parallel with the motor, and since then I haven’t had an H-bridge destroyed that way. Something like a SM15T27CA should be fine.

  • \$\begingroup\$ Yup that's a classic mistake, folks forget about the power switch. \$\endgroup\$
    – Trevor_G
    Mar 10, 2017 at 22:46
  • \$\begingroup\$ Hmm that could have been my mistake since I switched off the PSU while the motor was turning. However how could a diode in parallel work when the H-bridge is switching directions? Wouldn´that just work for one direction? The SM15T27CA is a zener as I can see which clamps at 37.5V, I am not sure if the IMS-1 can stand anything above 30V so I probably need to get a zener that cuts around 24V-29.9V? \$\endgroup\$ Mar 11, 2017 at 7:51
  • \$\begingroup\$ @GemeloMolinero No, this is not a Zener diode, this is a TVS (Transient Voltage Suppressor) diode. Also it behaves somewhat like a Zener, a TVS diode is meant to support a huge (transient power) and it’s generally design to become a short in case of failure. The “CA” suffix means it is a bidirectional model, that behaves somewhat like 2 Zener diodes in series. \$\endgroup\$ Mar 11, 2017 at 11:32
  • \$\begingroup\$ @GemeloMolinero As for the voltage, you have to be sure that it won’t drain a significant current at the maximum “normal” voltage. According ST’s datasheet, the SM15T24CA may drain 1 mA at a minimum voltage of 22.8 V. This is too low. As for the SM15T27CA, its voltage won’t exceed 37.5 V at 40 A. That’s high, but your actual current won’t exceed 20 A and I’d bet the H-bridge will support the actual transient voltage. \$\endgroup\$ Mar 11, 2017 at 11:41
  • \$\begingroup\$ Okay then I found false specs for that one. I really should have been awake in physics class back then so I a try to understand the math now. The motor does 27 rpm, so 162° in one second with 20Nm (initial 90Nm) -> 2.82743 radians/s -> Eb = 2.82743/ (20Nm/20A) hmm that doesn´t really make sense to me. That would lead to lower A with higher Nm. rad/s*V=A/Nm -> 2.82743rad/1s*24V*20Nm=1.4A 2.82743rad/1s*24V*90Nm=6.1A ... hmm that´s not much there must be some mistake in my maths. \$\endgroup\$ Mar 11, 2017 at 13:47

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