I'm fairly inexperienced and while I have an understanding of classic npn and pnp transistors, I'm shaky when it comes to MOSFETs.

I'm looking for an element to act as a switch. This element would be normally ON, when no stimulus is applied. The trigger needs to be a positive voltage, and is intended to turn OFF the element (open the circuit). After a bit of searching I figured that a P-channel JFET transistor might solve my problem. There would normally be continuity between Drain and Source, unless I applied a positive voltage to the Gate.

I'm talking about a low power application:

  • Drain-source voltage: 5V or less
  • Drain-source current: a few milliamps are enough
  • Gate trigger voltage: 5V or less

Maybe hurriedly, I got a FQP27P06 MOSFET and set up a simple test circuit with a LED. I understand this is a MOSFET and not a JFET, but I thought it would work anyway.

I'm using a voltage divider to keep the gate voltage in the specified range (2 to 4V). TBH I don't understand why that's actually specified as a negative voltage: -2 to -4 V.

In my intentions, the LED should always be on unless the button is pressed.


simulate this circuit – Schematic created using CircuitLab

What I'm observing is:

  • the LED is normally ON
  • when I press the button the LED goes OFF
  • when I release the button the LED STAYS OFF. Or maybe it just glows very dimly and then turns OFF again. Sometimes it gets brighter and brighter but never reaches full brightness. This behavior seems to be randomic.
  • in order to turn the LED back ON again, I need to short the Gate to ground!

I do not understand what's going on. Can someone please help me out? Thank you


2 Answers 2


You are disconnecting the gate of a mosfet by not pressing a button - this (due to high input impedance of the mosfet) leads to picking up charges from air (it acts like capacitor - try touching gate and ground/power), and partially turning mosfet on. To overcome this issue - add a high value resistor from Gate to ground - this will keep the mosfet on all the time, and when button is pressed - it will "overwrite" the resistor and turn transistor off as before. Value of added resistor should be in such range, that when button is pressed it will not drop the gate voltage to less than mosfet threshold voltage (or any voltage that is required for exact application)


simulate this circuit – Schematic created using CircuitLab

  • \$\begingroup\$ thank you for the exhaustive explanation! it's much clearer now. And the test circuit works \$\endgroup\$
    – Valentino
    Nov 9, 2020 at 20:44
  • \$\begingroup\$ There is something else wrong with this circuit: You don;t need R3. It's just wasting energy. But you could add a 560 Ohm resistor between SWI and FPQ gate and make R2 560 ohms too. This will help debouncing the switch. \$\endgroup\$
    – Fredled
    Nov 9, 2020 at 22:30
  • \$\begingroup\$ @Fredled - what is wrong? I believe, that this works well, but I can breadboard it \$\endgroup\$
    – fifi_22
    Nov 10, 2020 at 6:30
  • 1
    \$\begingroup\$ It's wrong because it's useless, It's just a connection between + and -. Amd you waste with only 1K. Use at least 10K if you really want a connection to ground. \$\endgroup\$
    – Fredled
    Nov 10, 2020 at 18:29
  • \$\begingroup\$ I didnt touch R3, becauose IT was in orginal schematic (in question) and might be needed by other circuitry (outside schematic). I just added needed pulldown. But in this case You are right - it's useless. \$\endgroup\$
    – fifi_22
    Nov 10, 2020 at 18:32

The oxide layer in the transistor creates a capacitor with the gate as one plate and the rest of the device as the other. When you open the switch it's holding enough gate charge to deplete the channel. Try putting a pull-down resistor between the gate and ground.

  • \$\begingroup\$ amazing! a 1k resistor got it working nicely :) thank you very much! \$\endgroup\$
    – Valentino
    Nov 9, 2020 at 20:30
  • \$\begingroup\$ 1K is too little. Use at least 10K or 33K as suggested in the other answer. \$\endgroup\$
    – Fredled
    Nov 9, 2020 at 22:31
  • \$\begingroup\$ I understand that now. Initially I tried to use a 15k R and it seemed not to be working, but the reason was simply that its thin legs weren't making contact properly. N00bs' problems :) \$\endgroup\$
    – Valentino
    Nov 11, 2020 at 12:49

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