I'm using my custom circuit in the figure for driving dc motors. I'm providing a PWM (25khz). While trying the activation threshold (minimum duty cycle that activates my motor) experimentally I have noticed (with an oscilloscope) that my circuit I have a problem: if the duty cycle it is 0, the voltage on the motor is 0, but just after I increase my duty cycle, even 1%, I have the VCC voltage on the motor, without any pulse width modulation.enter image description here I don't really know what can be the problem, anyone have any idea of the problem? thanks

p.s. The PWMU header (ULN_OUT wires) are connected to the single PWM1, PWM2, PMW3 and PWM4 headers on the mosfets'gates


1 Answer 1


It's pretty straightforward. Your 10k gate pullup resistor is too large. If the gate is pulled to ground, the MOSFET is turned on. When the gate drive is released, the gate voltage is pulled towards +12 by the 10k resistor. With a nominal Ciss of 1400 pF, the time constant is about 14 usec. To turn off the MOSFET gate will need to rise to ~2-3 volts below +12, which will take ~3 time constants, or roughly 40 usec. Since your PWM frequency is 25 kHz, the period is - you guessed it - about 40 usec. Just about the time the MOSFET is getting ready to turn off, it gets turned on again.

You can do one of 3 things. First, you can reduce your pullup to 100 ohms or so. This will work pretty well, but the resistor will get hot (worst-case power will be about 1.5 watts).

Second, you can reduce your PWM frequency a lot, while reducing the pullup to about 1k or so. You will have more MOSFET power dissipation than you expect, but it may be within limits.

Third, (recommended) get a proper gate driver. I personally like the MAX4427/4428 series, but there are lots of others which will work just fine. You can even roll your own and get adequate results.

  • \$\begingroup\$ Thank you very much! I understand the problem, I will use one of the option that you presented me. In option one, which current value you have used to estimate the worst-case power to 1.5 watts? Thanks @WhatRoughBeast \$\endgroup\$
    – Daniel
    May 3, 2015 at 7:39
  • \$\begingroup\$ At 100% duty cycle the gate is held low. Voltage across resistor is 12 volts. Power dissipated by resistor is (12 x 12) / 100, or 1.44 watts. \$\endgroup\$ May 3, 2015 at 13:22

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