ledmosfet
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:
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:
Exactly why does this happen? Would I be correct in thinking that using a P-channel mosfet would serve my purpose?
If all you need is to turn all the LEDs on/off using PWM, you can attach the N-chan MOSFET between the common cathode and ground. Here's an example circuit, ignoring LED current limiting schemes for now.
simulate this circuit – Schematic created using CircuitLab
If what you want is to control individual LED's, then you might still be able to use an N-Chan MOSFET, but your control signal will need to be at least Vthresh above the source voltage to turn the LEDs on. Alternatively, if you have use an up-stream P-Chan MOSFET, your control signal will need to be at least the level of the source voltage to turn the LEDs off (this is the MOSFET pin source voltage, not supply voltage).
If your micro is not able to achieve either of these levels, you can use a second transistor to drive the transistor in series with the LED(s).
Here's a basic example which uses a P-chan MOSFET to drive the LED's, and a N-chan MOSFET/pullup resistor to control the P-chan MOSFET: (again, ignoring any LED current limiting schemes)
Resistor values were chosen semi-arbitrarily. You can probably get away with anywhere from 1k up to 100k. Smaller values will draw more supply current when the N-chan MOSFET is on, larger values are more susceptible to noise when the N-chan MOSFET is off.
Place the load between the 7.5 Volt supply and the MOSFET drain, and you will get switch-like behavior.
simulate this circuit – Schematic created using CircuitLab
In your current layout, the source is floating depending on current through the LED and resistor. Hence, Vgs is not 0 to 5 Volts as you assumed, but much lower, depending on where Source is floated to at a point in time.
What is needed is for the "switch" to conduct when given a positive signal from the Arduino, thus pulling its Drain node to ground (or close to it), thereby expressing the desired ~7 Volts across the LED + R3.
Best Answer
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.
simulate this circuit – Schematic created using CircuitLab