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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.

Thanks

schematic

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.

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  • \$\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

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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.

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  • \$\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

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