I've built a H-bridge circuit to control a brushed DC motor. A schematic of the H-bridge is shown below:

enter image description here

When designing the H-bridge I didn't add any diodes in parallel to the MOSFETs as I had assumed that the intrinsic diodes of the MOSFETs would suffice to control the back EMF of the motor for the following reasons:

  1. The maximum forward current of the diode is larger than the motor current.
  2. The turn-on time of the diode is negligible (the data sheet states that the turn-on time of the diode is limited by the pin inductance, not the diode itself). This means that the diodes can immediately begin to provide a current path when the MOSFETs are off.

When testing my board, I force Q1 to be on all of the time and Q4 switched on/off at the PWM frequency. I'm assuming that when Q4 is turned off, the motor current will continue to flow through the diode of Q3 and back through the channel of Q1. I therefore expected that the back EMF of the motor to be clamped at one intrinsic diode drop and no more. However, I can see on a scope trace that the actual back EMF is more than 50 V (see below) and I'm not sure that I understand why.

enter image description here

I'm hoping that someone can explain why I'm seeing such a high back EMF and perhaps suggest a way of improving this circuit (for example, was it a mistake not to add some very fast didoes in parallel to each MOSFET?).

  • \$\begingroup\$ I think you've shown that the body diodes aren't sufficient for the currents involved. Your motor is probably too big. External diodes, with a lower forward voltage drop than the MOSFET body diode, are then needed to bypass them. Keep these diodes positioned very close to the MOSFETs. Also, there is sparking taking place when the DC motor is running. So there is often a bypass capacitor placed across the DC motor, as well. And, if you are being paranoid about it, you'd include a capacitor from each of the two terminals of the motor to its case, as well. That's all I can think of for now. \$\endgroup\$
    – jonk
    Commented Nov 6, 2022 at 22:38
  • \$\begingroup\$ What is the power source? Batteries can clamp the voltage (as long as the BMS does not trip) but power supplies usually cannot sink current, so they don't clamp the voltage. \$\endgroup\$
    – user57037
    Commented Nov 6, 2022 at 22:39
  • 1
    \$\begingroup\$ Except perhaps a capacitor to from your + rail to ground very near the circuit. (And perhaps a diode there, as well, just in case.) \$\endgroup\$
    – jonk
    Commented Nov 6, 2022 at 22:40
  • 2
    \$\begingroup\$ I have designed a few motor controllers for BLDC motors. I have never added diodes in parallel with the MOSFET intrinsic diodes. I doubt you need them. Something else is going on. No disrespect to jonk, but in this case I am dubious. \$\endgroup\$
    – user57037
    Commented Nov 6, 2022 at 22:44
  • \$\begingroup\$ Thanks all for your comments. I'm powering this from a DC power supply and not a battery. From the answer below I'm now suspecting that Q1 may actually be turning off and that could be the source of the issue. Based on that, and the fact that I'm using a bench power supply, can you think of any reason why Q1 would turn off? \$\endgroup\$ Commented Nov 7, 2022 at 10:04

1 Answer 1


In order to measure the motor voltage (forward or back EMF) a differential measurement must be made using two scope probes. When the Q4 turns off, the current is diverted through the body diode of Q3, so the voltage at the node V_bridge_B will rise to +BATT + one diode drop. This is not back-EMF. The actual back emf is behind the interval resistance of the motor and cannot be measured directly.

If you measure the voltage across the motor using the math function of the oscilloscope you will see the body diode drop + the voltage across Rdson of Q1+any voltage dropped by the wiring.

The blue arrows indicate the direction of current when Q4 is on.

The red arrows indicate the direction of current when Q4 is off.

The green arrows indicate the direction of current whether Q4 is on or off.

The voltmeter in the centre shows where the back-EMF actual exists. There is no access to the node at the junction of the motor resistance with the motor inductance so it cannot be measured directly.

The voltage that you measure is not back-emf, it is inductive "kick-back". The difference is that back emf is generated by the spinning motor, The voltage measured within the off cycle is the inductor trying to maintain the current in the motor. If the path back through Q1 is high resistance, the inductor will allow the voltage to rise as high as necessary.

That voltage should be one diode drop greater than the supply voltage. If it is greater, then something is amiss. I would suspect Q3 is damaged or inserted backwards, something like that. Or some other wiring problem. Perhaps Q1 is actually turning off as well.

I have never seen a performance improvement with external diodes.

Hope this helps.


simulate this circuit – Schematic created using CircuitLab

  • \$\begingroup\$ Thank you for your answer, very helpful. You note that Q1 may be turning off which I think could be correct. When checking the datasheet for the MOSFET, I note that VDSS has a minimum value of 55V which would appear to correlate with the magnitude of the kick back voltage. I'll check this later tonight (UK time). I'm pretty sure however that the control signal to the gate driver is held high throughout so I'm not sure why Q1 would turn off (but the values seem to suggest that it does). \$\endgroup\$ Commented Nov 7, 2022 at 10:02
  • \$\begingroup\$ @AndrewLees: The bench supply will probably not sink current which will allow the voltage to rise on the rail. If the the voltage on Q3 source is high and the supply rail is normal, then there is something wrong with Q3. If the supply rail is high also, then the problem is elsewhere. \$\endgroup\$
    – RussellH
    Commented Nov 7, 2022 at 16:18
  • \$\begingroup\$ Thanks @RussellH. Can you explain a bit more about what you mean regarding the bench supply? Are you saying that when the circuit starts sourcing current that the voltage out of the supply rises and no longer equals the preset value? Does the voltage then continue to rise until Q1 breaks down (at around 55 V) and becomes close to a short? If I understand correctly a battery would behave differently and sink the current coming from the source whilst maintaining a constant voltage at its output? Perhaps this is a basic question but it's not something I have encountered before. \$\endgroup\$ Commented Nov 7, 2022 at 19:37
  • \$\begingroup\$ And I'm also not sure why the supply needs to sink current at all? Why can it not flow through Q1 and the didoe of Q3? If Q1 was able to handle the current when Q4 is in, surely it can also handle the current when Q4 is off? \$\endgroup\$ Commented Nov 7, 2022 at 19:40
  • \$\begingroup\$ @AndrewLees: Your last comment is correct. If Q1 is on, then the supply will not need to sink current. If Q1 is off or damaged, then the motor current will pass through Q2’s diode, Q3’s diode and back through the power supply. Because most power supplies don’t sink current, decoupling capacitors are used across the bridge. Use your scope to troubleshoot. If the inductor (motor) current is blocked in any way the voltage will rise until a path is found. \$\endgroup\$
    – RussellH
    Commented Nov 7, 2022 at 19:56

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