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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).

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Schilive
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    In both instances measure the voltage between the source and gate pins on the mosfet. The answer should be self evident. – Kartman Nov 25 '22 at 04:56
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    Please take a good look at [these guidelines for drawing good schematics](https://electronics.stackexchange.com/questions/28251/rules-and-guidelines-for-drawing-good-schematics). – brhans Nov 25 '22 at 16:28

2 Answers2

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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.

schematic

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.

schematic

simulate this circuit

Spehro Pefhany
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    Thank you very much for answering! I'll try to use a 12 V source and see what happens. By the way, I thought the other pins did not matter for the MOSFET to turn on mostly because the animations showed electrons "popping in" with the other pins disconnected. Again, thanks. – Schilive Nov 25 '22 at 05:24
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    Have fun. Remember that voltage is always measured between two points, and the MOSFET does not know anything beyond what is applied to its pins. – Spehro Pefhany Nov 25 '22 at 05:25
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    @Schilive Something that's useful to know is that the voltage at just one pin, all by itself, **never** matters. The only thing that ever matters is the difference in potential between two pins. (The difference in potential is sometimes referred to as the "voltage difference," but the correct term for it is just "voltage." Similarly, people often say "voltage" when the correct word is "potential.") – Cassie Swett Nov 25 '22 at 15:45
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    Yes, the source could be at +1000V with respect to ground, but if the gate is at +1000 as well the MOSFET will be off, and if it is at +1010 it will be on. And if the gate is at +1030V or +970V it could be destroyed (because +/-20V is the maximum Vgs). Similarly you'd want the drain to be more positive than the source for normal operation, but not by more than the Vds(max) rating. – Spehro Pefhany Nov 25 '22 at 16:01
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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.

DKNguyen
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