I found this circuit in an application note on P-Channel Power MOSFETS.


The note explains that modern technology has made it possible for P-Channel MOSFETS to approach N-Channel performance in power applications. I would like to make a DC motor driver that uses high-side P-Channels, but there is very little info out there on driving them at high speeds. This circuit looks promising, but I do not understand all of it.

Here is what I DO understand:

  • The two resistor pairs (Rh1/Rh2 and RL1/RL2) are use to set the amount of time each respective FET spends in the linear region (the time to charge up the gate capacitance). The diodes Dh/DL make it possible for the Rx2 resistors to set the rise time and Rx1 + Rx2 to set the fall time. The diode is orientated this way because it is almost always desirable to have a longer fall time than rise time.

  • The zener diode Dz somehow clamps the gate-to-source voltage within an acceptable range, regardless of the supply voltage.

Questions I have:

  • How does the capacitor Ch and the resistor Rz switch the FET on and off, and will this work for a range of supply voltages?

  • How does the zener Dz clamp the voltage in the acceptable range for gate drive?

  • Can this circuit keep the MOSFET enhanced or depleted indefinitely, or does it require a pulse every-so-often to deal with Ch?

  • What would the gate drive IC look like? Is this application referring to a micro-controller or something more current-capable? If a specialized driver IC is required, what are some example parts that could work, and what voltage range must they provide to Ch on the high side?

  • 1
    \$\begingroup\$ The App Note explains perfectly well. Yes U need a continuous pulse where Rz Ch=T is shown by Fig 4. negative slope while +slope Zener is open so it is fast. Thus both turn off faster than ON. So there is dead time. Read the App note more carefully \$\endgroup\$ Commented Dec 12, 2019 at 7:24
  • \$\begingroup\$ To be frank, such simplistic passive drive circuits are always an approximation. If you want tight control over switching losses, you'll want a closed-loop gate controller that slews the channel current in a controlled fashion. \$\endgroup\$ Commented Apr 2, 2022 at 4:22

1 Answer 1


The zener is there for to stop the negative pulse through Ch from exceeding the maximum gate to source voltage, on positive pulses it is simply a diode. The circuit load is capacitive coupled therefore it will only energise the load while there are pulses. Both Inh and Inl require pulses not a constant DC voltage, also Inl must be low if sending a pulse to Inh, hence if pulsing both pins they should be 180 degrees out of phase with each other, otherwise you will destroy the ouput stages. Check the device datasheet to confirm it's operation as I am asuming that the gate drive IC is to enable pulses from a microcontroller. If this is a single device then you should have included the part number in your description.


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