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I need to control an LED with a N-channel mosfet, but need the cathode to be connected to ground. I learned to do it this way:

Textbook

but when I tried to connect it differently to keep the cathode of the LED grounded like this, there was a big voltage drop across the mosfet for some reason:

Imgur

Exactly why does this happen? Would I be correct in thinking that using a P-channel mosfet would serve my purpose?

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2 Answers 2

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Would I be correct in thinking that using a P-channel mosfet would serve my purpose?

Yes, you would be absolutely right.

Since MOSFET's are turned on by applying a voltage between Source and Gate, the diode in your second image acts as that very voltage drop that is stolen from your Source-Gate potential. That pretty much means, that the transistor is no longer turned on by the full battery voltage, but rather \$V_{bat}-V_{diode}\$.

What you are looking for is what's called a high-side switch. This usually requires a P-channel MOSFET.

schematic

simulate this circuit – Schematic created using CircuitLab

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    \$\begingroup\$ You have omitted the base resistor and I'm sure you know the consequences of that... \$\endgroup\$
    – alexan_e
    Jan 24, 2014 at 10:24
  • \$\begingroup\$ Frick, I knew I forgot something :D ... And CircuitLab has never worked for me. \$\endgroup\$
    – Dzarda
    Jan 24, 2014 at 10:55
  • \$\begingroup\$ @alexan_e As an EE noob, I don't :D why are all the extra components necessary? What would be the consequences of me hooking gate and drain to v+ and source to load then ground? \$\endgroup\$ Jan 24, 2014 at 21:37
  • \$\begingroup\$ @user3052786 The base resistor I mentioned is needed to limit the current of the base-emitter diode, if it is not used then the transistor will be damaged from a high current flowing from base to emitter. Regarding your question about the extra components, you can use just a switch from ground (supply -) to the mosfet gate and ommit the transistor, that will work fine. \$\endgroup\$
    – alexan_e
    Jan 24, 2014 at 21:46
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    \$\begingroup\$ Kind of a pushbutton-suicide circuit if V1 was high enough for it to do anything at all. \$\endgroup\$ Jan 25, 2014 at 3:03
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The led introduces a voltage drop equal to the forward voltage drop. That means that the source of the mosfet is not at 0v but to 2v or 3v (depends on the led).

What controls the mosfet conductivity is Vgs (the voltage difference between source and gate) so by raising the mosfet source voltage you effective decrease the Vgs voltage difference which reduces the conductivity of the mosfer (it obviously depends on the Vgs threshold of the particular mosfer and supply levels used).
A lower conductivity means a higher ON resistance between the source and drain which leads to a higher voltage drop.


I'm adding the following schematic because the OP asked in a comment why there is a need for the additional components and why not drive the gate directly.
The circuit can work fine with just a pull-up resistor that keeps the gate positive (mosfet off) and a switch that pulls the gate down when pressed and turns the mosfet on.
Note that the Vgs voltage should be high enough to turn the mosfet on and within specs for the max Vgs allowed for the mosfet used (usually 15-20v).

schematic

simulate this circuit – Schematic created using CircuitLab

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