I am trying to drive a motor accelerator. For that, I am using a push-pull amplifier. Sketch below.

The left part of the sketch is driven by a micro controller. It generates +/- ~19V for the MOSFETs to turn on. The right part should drive the accelerator by giving negative and positive voltages respectively.


When I had it on the breadboard there was no problem, now I transferred it to PCB. Now when I try to accelerate the motor it works as expected on the first try, but later on it does not work.

I measured the Rload signal after it broke down and it just gives positive pulses and no negative, it used to give both as expected. I replaced the IRF530 a few times and it worked as expected on the first try but same thing happens for the other runs and it doesn't work. What might be the problem here?

Edit: I changed the schematic since there was a sketch mistake. + Sketch fix.

  • \$\begingroup\$ PCB or solder issue? Post photos. \$\endgroup\$
    – winny
    Aug 10, 2021 at 10:24
  • \$\begingroup\$ Download LTSpice and simulate it. Your schematics is weird to me, I guess a shoot through. \$\endgroup\$ Aug 10, 2021 at 11:05
  • \$\begingroup\$ I already made the simulations even before testing it on breadboard! \$\endgroup\$
    – Zed K.
    Aug 10, 2021 at 11:42
  • \$\begingroup\$ You have an NFET on the high side and a PFET on the low side? Is that right? \$\endgroup\$
    – Aaron
    Aug 10, 2021 at 13:42
  • \$\begingroup\$ Put two channels of a scope on the two gates and see how they overlap (or don't overlap). I think you'll find your answer there. \$\endgroup\$
    – Aaron
    Aug 10, 2021 at 13:45

2 Answers 2


Read (1) then read (2) then read (3): -

enter image description here

I expect that your original build had source and drain reversed hence it worked. Do you see why having 1N4148 diodes is probably pointless?

EDIT following schematic update by OP: -

Those diodes (the 1N4007s) in series with each MOSFET are going to cause problems with back-emfs from your inductive load and damage your MOSFETs - you want the internal MOSFET bulk (parasitic) diodes to act as inductive flyback clamps to the power rails. You don't want the 1N4007 diodes to try and block these back-emf currents. To do so is asking for trouble because where does the flyback energy go? Think about it and short them out.

  • \$\begingroup\$ Thank you for your response. I know there exists parasitic diode. As you pointed, there was a sketch mistake in the schematic, I updated it. Drawing on a website is hard (:D). Normally sources are connected together. The circuit was not working even before I added 1N4148 diodes. Normally there is no freewheeling diodes. \$\endgroup\$
    – Zed K.
    Aug 10, 2021 at 11:46
  • 1
    \$\begingroup\$ @ZedK. I've amended my answer to indicate what I now believe to be a significant problem with the 1N4007 diodes in series with each MOSFET. \$\endgroup\$
    – Andy aka
    Aug 10, 2021 at 11:49
  • 1
    \$\begingroup\$ You have 2 x 19.5V voltage sources in series powering that LM358! The opamp is rated for 32V max, and you are providing 39V minus the two diode drops. What are those diodes for, BTW? \$\endgroup\$ Aug 10, 2021 at 13:11
  • 1
    \$\begingroup\$ Opamp is LM358B and it is rated 36V, and absolute maximum rating is ±20 or 40V. Voltage is now 19.5 but I will drop it approximately 0.5 to 1V with SMPS's pot, with the diodes there will be drops as well. Thanks for your concern. \$\endgroup\$
    – Zed K.
    Aug 10, 2021 at 14:38
  • 1
    \$\begingroup\$ I guess the input signal is digital PWM pulses, not a linear voltage. The 47k resistor R3 turning on the IRF530 Mosfet will turn it on very slowly causing the Mosfet to get hot. The old LM358 also switches slowly. \$\endgroup\$
    – Audioguru
    Aug 10, 2021 at 15:34

Here are some suggestions:

  1. You might consider a design with the P-channel MOSFET on the high side, and the N-channel on the low side. As drawn, the MOSFETs both have a voltage drop equal to the gate threshold voltage when "on".
  2. The low-side MOSFET is driven directly by the LM358 output. It will turn "on" and "off" relatively quickly, while the high side MOSFET gate is pulled down by the LM358 but pulled back up by a 47K resistor. As drawn, the high side MOSFET will turn on more slowly.
  3. As @Andy aka says, when the power supply is turned off, the inductive load will mean that you have voltage on the MOSFET outputs higher than the rails. Adding a current path back to the supply by removing the series diodes is a good idea. As drawn, the Vgs(max) may be exceeded when shut off if the inductive spike exceeds 20 volts, a sure way to damage a MOSFET. This is likely what is causing your "one time use" issue.
  4. There is no way to turn this circuit off. No matter what your microcontroller does, either the high side or the low side is energized whenever power is present.

Good Luck!

  • \$\begingroup\$ Thanks for your valuable comments. I use a relay for both power supply to the circuit and for feeding to accelerator so as to turn the circuit off. I removed the series diodes and for now it pulses are normal but there are some huge spikes I will need to deal with.You are right about 47k, I am considering to lower it. Hopefully, will test it thoroughly in a little while. Why should I consider moving P-channel MOSFET to the high side? \$\endgroup\$
    – Zed K.
    Aug 11, 2021 at 5:26
  • 1
    \$\begingroup\$ In a design with the P-channel on the high side, when you pull the gate down the MOSFET will turn fully on and the voltage drop from drain to source will be only caused by the 'on' resistance of 0.3 ohms. In your design, the N channel MOSFET on the high side must be about 4 volts higher than the source. So, when you let the gate drive go up the the rail, the source voltage will have to be about 4 volts below the rail, meaning that you are losing 4 volts across the Drain-to-Source. Any current will heat up the part in addition to the reduced voltage making your motor run slower. \$\endgroup\$ Aug 11, 2021 at 18:23
  • \$\begingroup\$ Thank you for the explanation. I will check it out. \$\endgroup\$
    – Zed K.
    Aug 12, 2021 at 5:05

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