I got this schematic :

Mosfet driving IR leds

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?

  • \$\begingroup\$ Why two resistors R1 and R2? They can be combined into one. Make R23 a lot bigger, perhaps 100K. Then if you want to have a gate resistor, which is optional but I recommend it to isolate the FET from the Teensy, make it 1K. Unlike BJTs, there is really nothing to calculate, as there will only be a few µA going through it. \$\endgroup\$ – tcrosley Oct 20 '14 at 22:23
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    \$\begingroup\$ "Why two resistors R1 and R2?" Board is optimized for batch production along with other 5 different designs that use a lot of 27 ohm resistors. \$\endgroup\$ – Zipporobotics Oct 27 '14 at 9:53

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:

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.

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  • \$\begingroup\$ Thank you so much for the answer. Now it is clear Regarding the leds: the have a 1.2 volt Nominal forward voltage each one. Total for the series is 2.4 volt needed and I'm driving them with 3.3. You see two resistor instead of one because the circuit is already optimized for batch production with other boards using a lot of 27 ohm resistors :) \$\endgroup\$ – Zipporobotics Oct 21 '14 at 10:31
  • \$\begingroup\$ According to datasheet; each of the pin I'm using can sink/source only 1 ma. So i need a 3.3 K resistor \$\endgroup\$ – Zipporobotics Oct 21 '14 at 10:37
  • \$\begingroup\$ This may be overly pessimistic when high gate currents are encountered; in practice only about 0.7 of that current is seen due to trance inductance etc. \$\endgroup\$ – Fizz Nov 8 '15 at 14:07

Yes you do need a gate resistor. Even if it is only 10R, one must be provided.

At the transition edges the gate-source capacitance appears as a short. At the very least you are bounding the current that will flow. This serves as a good mitigation for EMI. It is also good practice because you may not always end up in a situation where the driver of the pin has a form of "current limit"

The K20P64M72SF1 has a pin limit of +-25mA. What if you had a BJT push-pull stage that could sink/source 15A? with only trace impedance really limiting the current.

The inclusion of gate resistor also mitigates creating a pierce oscillator

150R gate resistor would provide adequate gate resistance as the initial current would be just less than the 25mA the I/O can source.

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