I was working on the breadboard on this simple circuit to do some testing and noticed that sometimes it would not turn on as expected . Somewhat abnormal operation. I always knew to connect a resistor (say 330 ohms) to limit the current that can damage the output of a microcontroller and a 1k resistor to make sure a pull-down such that the mosfet does not turn on for some reason.

Just by simulating, however, I noticed that this resitive divider provides on the gate of the mosfet a focus no longer [0..5]V coming from an Arduino.. but [0..3.7]V This may not completely turn the mosfet on, and in any case, it is not what I want. enter image description here

To "solve" this problem I replaced 1k with 10k: enter image description here

Is this okay? Am I doing something wrong?

If I remember correctly .. the resistor toward gnd (10k) might alter something inherent in the Cgs of the mosfet, but it is not clear to me what ... if you can explain this concept a little better and the consequences I have in using 1k or 10k .. thank you very much!

  • 1
    \$\begingroup\$ I presume you meant "damage the output of a microcontroller", not input? \$\endgroup\$
    – Finbarr
    May 10, 2023 at 16:17
  • \$\begingroup\$ Yes, my fault. Just corrected and edited the thread \$\endgroup\$
    – KaleM
    May 10, 2023 at 16:19
  • 2
    \$\begingroup\$ if you put R1 on the left of R2 it will be pulled down and also go up to 5V \$\endgroup\$ May 10, 2023 at 16:30

3 Answers 3


The pull-down resistor is only to set the default state while the MCU is waking up (most go tri-state after reset). It can (indeed, should, for this reason) be placed before the series resistor, thus losing no voltage at the gate.

The series resistor serves to slow the gate risetime, against gate capacitance; this can be harmful as it increases switching losses, but also helpful as it slows switching in general, so can avoid problems like generating radio interference or needing protective circuitry to address transient currents or voltages. The value or range required, depends on the application.

Note that the MCU's pins have an approx. 20-50 ohm source resistance themselves. This isn't documented directly, but can be inferred from the VOH and VOL parameters, and what currents they are tested at. This resistance limits current some already. When driving small capacitors (other CMOS pins, trace impedance, and small MOSFETs too), this resistance alone is sufficient -- the current is not continuous so it doesn't heat up the device or cause damage. An external gate resistor is still recommended, particularly if there is a long distance between MCU and MOSFET, as the MOSFET can oscillate as it switches (another source of radio interference), and a resistor placed at the MOSFET gate prevents oscillation.

There are less common reasons you might want a gate resistor, too. For example, if you want to prevent damage to the MCU in event the MOSFET fails catastrophically. (Typically when a transistor fails, it becomes a three-way short, at least momentarily; drain supply voltage can thus flow into the gate circuit.) A series resistor, with a clamping TVS from source to gate, can be used to shunt this energy to ground, away from the MCU. (This isn't very important most times, as any electrical failure means board replacement, and survival of the MCU is irrelevant; but it can be handy while prototyping, when transistor failures are likely.)


Raising the pull-down resistor value to 10s of k works of course, as does moving it to the "input" side of the series gate resistor, which avoids the potential divider problem.


simulate this circuit – Schematic created using CircuitLab


R1 and R2 form a potential divider, such that

\$Vgate = Vin * \frac{R1}{R1+R2}\$

So if you put your values in you get

\$Vgate = Vin * \frac{1000}{1000+330} = 3.76 V\$ (approximately)

Your pullup/pulldown should be much larger than the gate resistor to avoid reducing the gate-source voltage, although with the drain-source current you're using it's unlikely to be a problem.

With the 10k resistor for R1 the equation becomes

\$Vgate = Vin * \frac{10000}{10000+330} = 4.84 V\$ (approximately)

so you can see the effect of making it much larger.

R1 has no effect on the gate capacitance of the MOSFET. R2, in this case, limits the output current of the I/O pin:

\$Iout = \frac{Vin}{330} = 15 mA\$ (approximately)

...but this will only be a brief peak as the gate will soon charge up; for this part the typical value of gate capacitance is 1350pF so the time constant will be about 446 nanoseconds.


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