Why connecting the LED to the source instead of the drain does not work?

Question

I am trying to understand why I cannot put the LED afterwards. As I believe the drain is positive and the source is negative and that both connect when the gate has enough voltage, I thought there would be the same result, but the LED did not shine. There actually was reverse voltage between the LED, and there was voltage between the LED and the 10 kΩ resistor, both of around - 2 V (from the positive leg of the LED to the “ground” leg of the resistor).

The first circuit works, and the second one does not. And, by the way, the transistor is a MOSFET TF12N60, which I took from a UPS (Uninterruptible Power Supply).

Answer 1

The MOSFET is controlled by the voltage between the MOSFET gate and the MOSFET source. Further, you have a high voltage MOSFET that takes about +4V to start to turn on, and +10V (from gate to source) to fully turn on.

With only 5V you get a bit of current through the MOSFET (typically) but in the second arrangement, just a few uA flowing through the LED will raise the source voltage high enough that the MOSFET barely, barely conducts (uA).

In the third case, below, I've increased the supply voltage to 12V. The MOSFET is still dropping more than 3V, but now you have substantial LED current.

If you want this MOSFET to act as an efficient switch you need to have about 10V from gate to source.

^{simulate this circuit – Schematic created using CircuitLab}

Now consider this final example, where I've moved the resistor to the drain side- the MOSFET is fully turned on and the LED is conducting quite a bit of current (almost 20mA). That's because the voltage drop across the LED is a nonlinear function of current and is only about 2V at 20mA. It won't work on a 5V supply because 5V - 2V = 3V is not enough to turn the MOSFET on, but for >6V it sort-of works, and is fully functional with a 12V supply.

^{simulate this circuit}

Answer 2

An NMOS is controlled by the voltage DIFFERENCE between the gate and source terminal. It is not controlled by the gate voltage relative to ground because there is no way it can know what ground is. It only knows the voltage difference between its pins.

When you put the LED at the source pin you are allowing the LED to affect the source voltage relative to ground. This in turn affects the gate-source voltage difference of the MOSFET because you are still applying a voltage to the gate relative to ground (not relative to the source pin).

The more current flows through the LED and resistor, the higher the voltage drop across them are, which means the higher the voltage of the source pin relative to ground. Since you are still applying a gate voltage referenced to ground, you are reducing the MOSFET's Vgs. The more the LED turns on, the more it opposes the MOSFET from conducting more. Negative feedback.