I want to control the speed of a DC motor, but the chassis is connected to the negative side of the motor, making it impossible to use the MOSFET to disconnect ground.

Is there a way that I can use the MOSFET to disconnect the positive side of the motor, as in the following diagram? And does that configuration have a name, so I can read more about it?


simulate this circuit – Schematic created using CircuitLab


1 Answer 1


You need to control the 12V motor with a P channel mosfet. Using an N channel MOSFET (as per your schematic) will result in less than 12V applied to the motor due to the gate-source threshold voltage not being zero. Of course if you can make your PWM signal more like 15 Vp-p then this will probably work fine.

A P channel mosfet would work like this: -

enter image description here

The circuit above will work from logic level signals so there's an added bonus but, if you truly have a 12V PWM signal then you can drive the MOSFET gate directly but, be aware that the gate capacitance might upset your PWM voltage a bit AND the p channel MOSFET does invert the signal.

  • \$\begingroup\$ 1) Efficiency is a problem and a safety issue, in that if the PWM signal is disconnected, it would start the motor, and also keeping the motor off would cause a minor power drain. 2) If I'm willing to accept the voltage drop, could I just use an N-channel MOSFET in the same configuration? \$\endgroup\$ Commented Nov 8, 2016 at 14:40
  • \$\begingroup\$ @user280593, Andy's driver is called a "high-side switch", and is preferable to your N-channel IRF3205 circuit. While yours works, some voltage remains between drain-to-source that results in heat dissipation. Andy's dissipates far less heat. Disconnected from PWM source doesn't activate the motor. \$\endgroup\$
    – glen_geek
    Commented Nov 8, 2016 at 14:45
  • \$\begingroup\$ @user280593 Just add a 2k2 resistor from the base of Q1 to ground. If the PWM is disconnected the 2k2 will hold Q1 off (no current drain) and R2 will hold the MOSFET off (no drain). \$\endgroup\$ Commented Nov 8, 2016 at 14:45
  • 1
    \$\begingroup\$ Depending on PWM speed, increasing R2 to 10k (or larger) would help with efficiency as well. At 1k it is dissipating ~12V, meaning .144 W. Need to make sure it isn't so big that the RC constant of the Cg + R2 grows past the PWM speed. Just a thought. \$\endgroup\$
    – Jim
    Commented Nov 8, 2016 at 14:53
  • \$\begingroup\$ Now that I think about it, wouldn't the gate resistor draw zero current when the signal transistor is off? This actually seems like a very efficient design. \$\endgroup\$ Commented Nov 8, 2016 at 15:16

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.