In the following circuit, the LED lights up:

enter image description here

I was afraid I'd misuse the transistor symbols, so I just drew a box to make sure I communicated the right thing. This is the transistor, G/D/S means Gate/Drain/Source.

My understanding is that current does not flow if there's no circuit. From this I deduce that the line going from the positive terminal of the voltage source into the Gate pin has to at some point make its way back to ground. Since the only pin coming out of the transistor connecting to ground is the Source, I deduce that the gate is internally connected to the source pin.

But from that, I would think that if I severed the connection to Drain like this, current would still flow - but my LED switches off.

Is it that the relationship between gate and drain is sort of "symmetrical"? I was always told that current flows from D to S and the G pin just controls whether that connection is open, but is it more like an AND gate? Current flows out of S as long as it's flowing into both G and D?

  • \$\begingroup\$ You should use a ground symbol for the ground node (or a 0 volt symbol). Gate control's current in an analogue way, not digital. \$\endgroup\$
    – Andy aka
    May 14 at 11:06

2 Answers 2


Where does the current applied to the gate of a transistor "go"?

In a MOSFET transistor we don't "apply a current", we apply a voltage to the gate.


simulate this circuit – Schematic created using CircuitLab

As represented by the MOSFET symbol, the gate is insulated from the rest of the transistor by a metal-oxide layer which has very high resistance in the gigaohm region. Effectively no DC current flows through the gate but since it has a little capacitance some current flows during switching the gate high or low.


You are connecting the FET in a way which does not make a lot of sense if you want to just switch the LED on and off.

In that circuit where the FET is switching the LED positive supply, you would be better off with a P type FET.

You have an N type FET so it would make sense to have the FET switching the LED negative supply.

What happens here is that the circuit stabilizes to Vgs being equal to Vds and so the FET is somewhat halfway turned on.

So there will be current from drain to source.

If you disconnect the drain, then no current flows from drain to source and LED is off. The FET would have Vgs of approximately the supply voltage.

So no, gate is not connected to source pin. Ideally, no current flows in or out from gate, unless you are trying to change the gate charge, i.e. changing the Vgs voltage. In real world, some small leakage current may occur. If gate conducts a lot of current, it means the FET has damaged and is broken.


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