I am trying to make a push-pull amp to drive a dc motor with a dual power supply. I have this circuit and it works but I can only get about 1A of current for each rail. Is it possible to raise this current? I have tried several configurations in falstad simulator but I haven't found a way to do it, except when swapping sources to drains but then I think the circuit is a short circuit between the rails.



simulate this circuit – Schematic created using CircuitLab

enter image description here enter image description here

EDIT: I have swapped the diodes with resistors and tried several values but I don't see any difference. The current climbs to +- 0.85A.

  • \$\begingroup\$ Check the way your supply voltages are drawn. The first schematic either has V2 the wrong way around or it should be 12V not -12V, in the others the bottom battery symbol should be the other way around and should be defined as a positive voltage, otherwise it's confusing since the side with the longer line is usually positive and there are no labels on them to indicate otherwise. \$\endgroup\$
    – GodJihyo
    Oct 3, 2022 at 0:45
  • \$\begingroup\$ Thanks. I thought to flip the negative PS vertically but I think this is the correct way to depict it. Ground is + in the negative rail. Please correct me if I am wrong \$\endgroup\$
    – John Am
    Oct 3, 2022 at 0:53
  • \$\begingroup\$ If you make a voltage source negative it reverses the marked polarity, the - will be positive and the + will be negative. If you need a negative supply you either make it a positive voltage and make the + common, or make it a negative voltage and make the - common. You've made it a negative voltage with the + common so if this was in a simulator it wouldn't work correctly. \$\endgroup\$
    – GodJihyo
    Oct 3, 2022 at 1:16
  • \$\begingroup\$ ok. I ll check it out \$\endgroup\$
    – John Am
    Oct 3, 2022 at 1:20

1 Answer 1


The diodes are preventing proper gate bias. Replace them with resistors (\$R_{3},R_{4}\$) so that the voltage drop across each resistor just under \$V_{T}\$. Also a resistor from the junction of the FET source to the junction of \$R_{3},R_{4}\$ may help with smooth startup from zero.

The circuit will dissipate a lot of power. You will need big heatsinks on the transistors. A pulse width modulation (pwm) driver is better.

EDIT: the current draw depends entirely on the motor. The voltage applied to the motor is: $$V_{motor}=12V-V_{GS}~=8V$$ If the motor resistance is about \$10\Omega \$, then the current will be \$0.8 A\$. The circuit is operating properly. The reason for replacing the diodes with resistors is to remove the dead band near zero.

  • \$\begingroup\$ Thanks. Please see my edit. \$\endgroup\$
    – John Am
    Oct 2, 2022 at 23:41
  • \$\begingroup\$ Looks like the circuit is behaving correctly in both cases. The most voltage that you can apply to the motor is about 8 volts. \$\endgroup\$
    – RussellH
    Oct 2, 2022 at 23:52
  • \$\begingroup\$ See my edit. You can't just get more current. The current is determined by the motor voltage and the motor resistance. \$\endgroup\$
    – RussellH
    Oct 3, 2022 at 0:00
  • \$\begingroup\$ If I use the mosfets seperately I can make them provide 2-3 amps. I think it is the push pull configuration that limits the current. \$\endgroup\$
    – John Am
    Oct 3, 2022 at 0:56
  • \$\begingroup\$ MOSFETS don't provide current. The voltage to the motor must be increased. The motor draws current based on its resistance. \$\endgroup\$
    – RussellH
    Oct 3, 2022 at 0:59

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.