# MOSFET push-pull low current problem

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

simulate this circuit – Schematic created using CircuitLab

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.

• 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. Oct 3, 2022 at 0:45
• 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 Oct 3, 2022 at 0:53
• 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. Oct 3, 2022 at 1:16
• ok. I ll check it out Oct 3, 2022 at 1:20

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 $$\R_{3},R_{4}\$$ may help with smooth startup from zero.
EDIT: the current draw depends entirely on the motor. The voltage applied to the motor is: $$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.