Why is the MOSFET in this fan control circuit overheating?

I’ve designed a fan control circuit using an N-channel MOSFET (STN3NF06L.) The fan operates at 12V, and I’ve isolated the 12V side from the 3.3V microcontroller side using an optocoupler (ACPL-W314-000E.)

I’m encountering an overheating problem with the MOSFET. After turning it on, it reaches a maximum temperature of 120°C.

Can anyone explain why the MOSFET might be overheating? My hypothesis is that the gate driver may not be able to provide enough charge to overcome the Miller plateau, so the drain-source channel might not be fully open. I'm giving well above the gate threshold voltage. Will the gate driver not be able to drive enough charge into the gate capacitor?

The current from the key switch to the MOSFET is around 1A. The load initially pulls 1.7 A which then settles to 0.9 A.

Furthermore I've seen that the fan only turns on when the gate pull down resistor R95 is removed.

The voltage across the gate and source when operational is ~3V.

• MOSFET: STN3NF06L
• Optocoupler: ACPL-W314-000E
• Keyswitch= 12V from battery

• If you don't have a diode or something to deal with the current pulse from the fan when you turn off the pass element, you will burn it out quickly.
– vir
Commented Jul 18 at 19:37
• In the future, please use standard symbols instead of blocks. Makes schematics much easier to read. Commented Jul 18 at 20:47
• @fgrieu: Looks like it is two different grounds though. Commented Jul 20 at 8:52

It looks like you have an N-channel MOSFET operating as a "source-follower", otherwise known as "common-drain". That means the source is always a few volts lower than the gate (by more than $$\V_{GS(TH)}\$$, which you can find in the datasheet), and is never fully switched on:

simulate this circuit – Schematic created using CircuitLab

Even though the MOSFET's gate is at +12V, its source (in this example) is at +9.4V, leaving 2.6V across the MOSFET's channel, between drain and source. That channel is passing nearly 1A here, and the resulting power being dissipated by the MOSFET is:

$$P = I \times V = 1A \times 2.6V = 2.6W$$

That will overheat the MOSFET after a few seconds.

You must connect the MOSFET with common-source (source grounded) to ensure that it's fully switched on, when the gate is at +12V:

simulate this circuit

With the load (R1 here, which would be your fan), and the source (node S) held at 0V, now the voltage across the MOSFET is only 0.1V or so, and with current at 1.2A, the power being dissipated by the transistor is:

$$P = I \times V = 1.2A \times 0.1V = 0.12W$$

That's much more reasonable. Apart from the transistor's different behavior in this configuration, the only other difference is that the load (fan) is at the high-side of the transistor.

The opto-coupler you are using has a push-pull output, quite capable of driving the gate directly, but R93 in your design is far too large to take advantage of that. If you are controlling fan speed with PWM, then you'll need a much smaller gate resistance. If you're simply switching the fan on and off for longer periods, like a second or more, than I suppose 1.5kΩ would be fine, but I would still recommend a smaller resistance.

Without knowing more about the fan, I would advise a diode across it, to protect the MOSFET from voltage spikes caused by the inductive load.

R95 only serves to reduce gate potential, by forming a voltage divider with R93, which in your original circuit would further decrease source potential, the very reason for your MOSFET heating in the first place. Removing R95 would certainly improve performance in your original circuit. It isn't necessary at all (for either common-drain or common-source) since this particular transistor model will tolerate $$\V_{GS}\$$ up to 16V.

R94 would be useful if the opto-coupler's output were ever able to float, and would prevent the MOSFET from accidentally switching on in such a condition. The push-pull output of the opto-coupler cannot float, and is always unambiguously high or low, rendering R94 pointless, especially considering that R95 would have the same effect.

Here's what I would recommend:

simulate this circuit

Lastly, don't forget supply decoupling capacitors, near the motor positive and MOSFET source, and also close to the opto-coupler's $$\V_{CC}\$$ and $$\V_{EE}\$$.

The problem is you have the fan in the wrong spot - your MOSFET is operating in its ohmic region.

To correct your circuit, tie the source of the MOSFET to ground. And put the fan between the Key switch supply and the MOSFET drain.

• I’m a bit confused with the wording here. The MOSFET is operating in the ohmic region before or after the proposed circuit correction? Commented Jul 19 at 6:30
• The current circuit has the MOSFET operating in the ohmic region (linear operation). if the MOSFET were tied to ground, it would be operating in the saturation region @SteKulov Commented Jul 19 at 8:46
• @RaphaelTreccani-Chinelli I believe your terminology is wrong. You're using BJT terminology with FETs. The saturation regions in BJTs vs FETs are opposite. See here: electronics.stackexchange.com/a/410957/254890 Commented Jul 19 at 16:47
• @SteKulov OK. Triode Region, according to your reference. There is more than one way to describe the operating regime and the experienced engineers will immediate recognize that. "Ohmic region", "Linear Region", and "Triode region" all mean the same thing. The salient point wasn't the terminology, rather, it was the fact that the original circuit didn't configure the transistor as a switch. And no, "The saturation regions in BJTs vs FETs are opposite" is just not true. Commented Jul 19 at 21:09
• The terminology was exactly what stood in the way of me properly understanding your answer. So no, not a salient point at all. I suggest reading the link I posted one more time (maybe two or three times). Anyway, I’ll leave it at that. Commented Jul 20 at 13:53