So I've made this basic circuit to drive Vout between ~0V and ~1.65V. The default state should be ~0V, hence I use a p-type mosfet: https://docs.rs-online.com/d95a/0900766b800ae907.pdf

Now, for some reason it looks like the fet has broken down after applying a voltage to the gate the first time by closing the switch (s1). I'm having quite a hard time in fet-based circuit design so maybe somebody could help me a bit in what I'm doing wrong or how I'm mis-understanding the situation?

The source is driven by a 5V supply. R1 = 20k, R2 and R3 are 10k. Q1 is an FDV304P P-channel mosfet as given in the link above.

The circuit: the schematics of my circuit

The purpose of this circuit is to drive an IC's enable pin. By default this enable should be turned low but when the switch is closed, the enable should be high. The switch simulates a IO pin from another IC.

  • \$\begingroup\$ Please explain the purpose of your circuit. You should edit the question rather than answer in a comment so that people do not need to read the comments to find that information. \$\endgroup\$ Apr 19, 2021 at 11:53
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    \$\begingroup\$ The mosfet shunts Vout to GND as drawn \$\endgroup\$
    – tobalt
    Apr 19, 2021 at 11:54
  • \$\begingroup\$ I think you applied a positive VGS to a P MOS. In any case, to avoid mistakes put a resistance with the gate, that limits the current. E.g. 1 kohm would limit to 5 mA max. And it will not break. \$\endgroup\$
    – andrea
    Apr 19, 2021 at 11:55
  • \$\begingroup\$ @andrea thanks for the response, yes i did exactly that. the 5V source is postive 5V relative to GND. \$\endgroup\$
    – Mart
    Apr 19, 2021 at 11:57
  • \$\begingroup\$ To drive an enable pin, you do not need a mosfet with your switch. You only need your mechanical switch and a pull down resistor. If you want to limit the voltage on the enable pin to 1.65V, (with a 5V supply) you need a second resistor in series with your switch. So, is there another reason for your circuit? \$\endgroup\$ Apr 19, 2021 at 11:59

2 Answers 2


As @"Math Keeps Me Busy" says you can do this without a mosfet, but if you would like to use one you can do it like this:


simulate this circuit – Schematic created using CircuitLab

The first one uses a P-mosfet. Changes from yours that I have made:

  • turn around S/D terminals (having recently made the same mistake myself, I like showing the body diode so I can tell whether I've gotten it the right way around!)
  • Use it as a high-side switch. It passes, or not, the high rail. R3 keep Q1 turned off by pulling up the gate to the high rail
  • SW1 turns on Q1 by pulling the gate down relative to the source terminal (high rail).
  • In this schematic and yours, the gate is pulled all the way down to 0V, which means Vgs = -5V; this is within the specs of this FET but if you were using a higher voltage BAT1 you could increase R4 in order to control how low you pull the gate.

The second one uses an N-MOSFET as a low-side switch. It will invert the output (Vout* will be 1.66V when SW3 is open, and 0V when SW3 is closed)

  • \$\begingroup\$ Thanks Daniel! you say 'the same mistake'. Could you tell a bit more on what this 'mistake' then actually is / does and what causes it? that is the part I'd like to learn about a bit more. \$\endgroup\$
    – Mart
    Apr 20, 2021 at 8:21
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    \$\begingroup\$ 'the same mistake'.... just that I recently put a P-mosfet into a schematic upside-down. N-type are intuitive, P-type are tricksy/backwards/upside-down. Gotta make sure parasitic body diode in N or P type are reverse biased, unless you very deliberately intend otherwise. \$\endgroup\$ Apr 20, 2021 at 23:08

First, the use of a mosfet is unnecessary to drive an enable pin of virtually any IC. One can simply use a circuit like this:


simulate this circuit – Schematic created using CircuitLab

Assuming the enable pin does not source or sink excessive current, when the switch is open, the enable pin will be at 0V and when the switch is closed, the enable pin will be at 1.66V (close to the 1.65V you intended). If the enable pin does source or sink significant current, the voltages will be be 0 and 1.66V. R1 and R2 may be decreased to compensate, (with the side effect that more power will be used when the switch is closed).

Second, the MOSFET you have chosen (like most mosfets) has a body diode connected between the source and drain pins. As configured, the drain (and hence Vout) will never rise beyond 0.6-0.7V above the source/ground.


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