I got this schematic :
In the image you see I'm driving two infrared LEDs at 40 mA each one with a FDN327N MOSFET.
Attached at D8 there is one of Teensy 3.1 digital output that has HIGH voltage of 3.3 V and a LOW of 0 V.
The R23 Resistor which you see as 0 Ohm resistor is a 10 kOhm pulldown resistor While R3 should be the gate resistor.
I included it as 0 in the case I need it and I got a doubt.
I want to drive the MOSFET as a logic gate MOSFET @ 1200 Hz with the tone library.
Do I need a resistor and if yes how do I calculate it?
Best Answer
The gate resistor on a MOSFET is really there to protect whatever is sourcing the current. Much like a discharged capacitor, the gate will initially look like a short to ground when voltage is first applied. A MOSFET with a very large gate capacitance can sink a very large amount of current for a short period of time. If you're driving the gate with, say, a MCU pin, it's usually a good idea to put a small resistor to reduce that current surge to a value the MCU can handle. If your MCU pin can handle, say, 20mA and you're driving 3.3V into the gate, then you choose a resistor that limits the current to 20mA at 3.3V:
$$R_{gate}=\frac{V}{I}=\frac{3.3V}{20mA}=165Ω$$
The digital output pins of most MCUs are current limited already, so this isn't strictly necessary. But why pound on the output drive circuitry to the point that the current limiter kicks in?
Incidentally, MOSFET drivers are ICs specifically made to drive a large amount of current into the gate of a MOSFET so as to turn on the MOSFET as quickly as possible. The MOSFET turns on quicker and the switching losses are less.
All of that said, I believe you have another issue with your circuit. I suspect 3.3V is not enough to turn on two UV LEDs in series. The voltage will need to be at least as high as 2x the foward voltage of one of those LEDs. As it is, I don't think the LEDs will turn on.