I need to control a DC motor runs at 9v with PWM signal 3.3V enter image description here vpeak.

I found MOSFET driver noninverting MCP1401. Can this MOSFET be used directly with a Back EMF diode (diode as usual across the motor terminals)?

The motor current rating is 200mA and voltage is 9V. the MCP1401 is 500mA Is the above diagram correct?

  • \$\begingroup\$ What is the DC resistance of your motor? \$\endgroup\$ Aug 27, 2019 at 21:40
  • \$\begingroup\$ DC resistance is 17 Ohms \$\endgroup\$
    – oppo
    Aug 28, 2019 at 14:10
  • \$\begingroup\$ Stall current on 9V should be 9/17 = 0.53A, MCP1402 FET resistance would limit stall current to ~300mA, but it would still get much too hot at 200mA (and lose 2.4V). Perhaps if the actual operating current was a lot lower in your application (<100mA) you might get away with it, but I wouldn't risk it. \$\endgroup\$ Aug 28, 2019 at 21:08

2 Answers 2


There is nothing in the datasheet that indicates that the gate driver could be used to source DC, but the operation was never intended for such:

MOSFET drivers are high-speed, high-current device swhich are intended to source/sink high peak currents to charge/discharge the gate capacitance of external MOSFETs or IGBTs.

Source: http://ww1.microchip.com/downloads/en/DeviceDoc/20002052D.pdf

Mosfet gates are high impedance, and only need current while switching. Using the gate driver in a constant mode at high current could cause failure. There is nothing in the datasheet to indicate this (there are no DC specs) because the designers did not intend or test it for DC use.

I would use the MCP1402 to drive an h-bridge and use that to drive the motor.


500mA is the peak current rating. This IC is designed to drive the capacitive load of a power MOSFET, at a frequency low enough that the Gate capacitance is charged in a small fraction of the PWM cycle. To find out how much current it can deliver continuously you need to calculate the average power dissipation.

Typical FET on-resistance is 12Ω on the high side, so the power dissipation at 200mA will be 0.2*0.2*12 = 0.48W. But the absolute maximum power dissipation rating is only 0.35W, so it will burn up. Worse, at startup the motor may try to draw 5 times the normal running current or higher, so the IC could fry even before the motor gets up to speed.

The solution is simple. Use the MCP1401 to drive a power MOSFET which is rated for at least 2A (10 times the rated motor current). Alternatively you could use a MOSFET rated for 2.5V Gate drive (eg. IRLML2502), and then you won't need the driver.


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