I am trying to develop a schematic with minimal components. The board would drive a N-MOSFET from a microcontroller, and I am wondering if I can omit the gate resistors (pulldown for "default" off, series from microcontroller to gate).

The transistor is essentially charging a capacitor whose charge will be used to provide high current for a very short time (20mA at 3V for 15ms to downstream).

The rationale behind omitting the gate resistors is the following:

  • 4.7kOhm current limiting resistor (VCC from a 3V coin cell, limit high current which reduces coin cell life). This would mean low drain current.
  • Transistor very close to microcontroller, expected low parasitic noise.

My question is: would this design work?

enter image description here

  • \$\begingroup\$ Are you sure a N FET works as Q1? The more voltage there is on capacitor, the less there is Vgs to keep transistor on, right? Even an ideal P FET that turns fully on would take roughly 2.35 seconds to charge the 100uF capacitor to 99% via 4k7 resistor. Also do you think 100uF is large enough to give 20mA for 30ms? It would ideally need 6V change on capacitor. \$\endgroup\$
    – Justme
    Dec 22, 2020 at 8:34
  • \$\begingroup\$ I computed the current using I = C*dV/dt, where C=100uF, dV=3V , dt=20ms. You're right, it needs about 2s to charge after 5tau, which is fine. I'm not sure I understand the rest of the reasoning. \$\endgroup\$ Dec 22, 2020 at 8:39
  • \$\begingroup\$ EDIT: You're right, redid my calculations, it would drive 20mA for 15ms. \$\endgroup\$ Dec 22, 2020 at 8:41
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    \$\begingroup\$ @MihaiGalos When the N-FET turns on, it's source will begin rising toward VCC. With the gate voltage fixed, coming from the micro, this means the gate-source voltage will decrease, preventing the FET from fully turning on. \$\endgroup\$ Dec 22, 2020 at 8:45
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    \$\begingroup\$ you can use your MCU as a pull down, so as long as the brief power-on uncertainty period doesn't blow anything up if the FET is one state or the other, it's fine. \$\endgroup\$
    – dandavis
    Dec 23, 2020 at 0:09

1 Answer 1


No, it would not work.

And it is not about gate pull-down or series resistors, these could be left off, as nothing bad should happen even if the gate floats for a few milliseconds after connecting power and before the MCU configures the pin as output. This floating can also be pevented by if there is a large resistor, like in the order of 1 megaohm to keep the idle state.

The problem is that the FET should be of P type instead of N type to switch it on and off reliably.

  • \$\begingroup\$ Thank you for the answer. How about if the capacitor and current limiting resistor are on the high-side (connected to VCC) of the transistor, and the low-side connected to GND? - would driving via N-Channel MOSFET work then? \$\endgroup\$ Dec 22, 2020 at 9:08
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    \$\begingroup\$ Yes that would work. \$\endgroup\$
    – Justme
    Dec 22, 2020 at 9:12

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