I have a doubt about an electronic circuit. Below is the circuit diagram of the fan controller in my 3d printer controller ( Printrboard Rev D ). enter image description here enter image description here This circuit can control the speed of a fan.

I know that MOSFET can be used as a switch. When sufficient Vgs is applied, MOSFET will turn on and so the load (fan).

This is my understanding of this circuit, please correct me if I’m wrong or missed anything?

5V PWM is a digital signal and by adjusting the duty cycle, I can control the average value.

For 100% duty cycle Average voltage = 5V For 50% duty cycle Average voltage = 2.5V For 0% duty cycle Average voltage = 0V

So if I apply a 5V PWM with 50% duty cycle to the gate of MOSFET it will turn the MOSET On and OFF depending upon the Ton and Toff value(Ton= Toff since duty cycle is 50%). Since this ON and OFF happen so fast, if I check the output voltage using a muti-meter it will read 6V (50% of 12V), and in a CRO or DSO I can see the 12 V PWM. Therefore, the fan will operate at 6V and will run at half the speed (if the rated speed is 500RPM, it’ll run at 250RPM).

I tried to explain with a picture also

enter image description here Is this how this circuit changes the speed of fan?

I'm not an electronics guy so I'm expecting simple answers


  • 1
    \$\begingroup\$ Yep, sounds like you understand it. You can use the same technique to dim an LED and on many other electromechanical circuits. \$\endgroup\$
    – Dan Laks
    Aug 1, 2016 at 7:19
  • \$\begingroup\$ You stated: " Below is the circuit diagram of the fan controller ....". I don't see it in your post. Also, what type of motor is used for the fan? Is it a DC permanent magnet type or a stepper motor? Most fans these days are actually stepper motors with built in drivers. Each type of motor will react differently to a PWMed supply voltage. \$\endgroup\$
    – FiddyOhm
    Aug 1, 2016 at 8:52
  • \$\begingroup\$ @FiddyOhm Oh sorry I thought I uploaded it . Fan motor is a brushless DC motor 12V \$\endgroup\$
    – Athul
    Aug 3, 2016 at 7:14
  • \$\begingroup\$ Finally someone who draws nicely \$\endgroup\$
    – user76844
    Aug 5, 2016 at 20:24

2 Answers 2


The analysis is correct. The main period of the duty cycle is higher than the reaction time of the motor mechanic, so the motor reacts only on the average voltage applied by the PWM output signal.

Vo(average)= Vsupply * DutyCycle

Basically it works like a low pass filter.


What you have to remember, is that you need certain frequency of PWM. Normally something like 5kHz to 20kHz. And that means some serious gate current, so you can't do it with a microcontroller output. Use some gate driver.

Also be sure to use a fan without built in controller.

  • \$\begingroup\$ How to calculate current based on PWM frequency? \$\endgroup\$
    – Athul
    Aug 8, 2016 at 8:49
  • \$\begingroup\$ You can't. I = (Vin - Vbemf) / R \$\endgroup\$
    – user76844
    Aug 8, 2016 at 8:51
  • \$\begingroup\$ But in your comment you said What you have to remember, is that you need certain frequency of PWM. Normally something like 5kHz to 20kHz. And that means some serious gate current What does that mean? anad how gate current is affected by PWM frequency? \$\endgroup\$
    – Athul
    Aug 8, 2016 at 8:58
  • \$\begingroup\$ Gate curren! So roughly I = 2P/V=2*(fCV*V/2)/V=fCV, where f is frequency, C is gate capacitance and V is gate driver voltage. \$\endgroup\$
    – user76844
    Aug 8, 2016 at 13:13
  • 1
    \$\begingroup\$ @Athul The gate of a MOSFET has a capacitance that causes it to charge up like a capacitor. Until the gate is "fully charged" the MOSFET is in a partially on state, and its internal resistance is much higher than it's ideal RDSon. The higher the current you use to switch the MOSFET's gate, the quicker the gate "charges up" and the quicker the MOSFET turns fully on. If you do not provide enough current to charge the gate quick enough at a given frequency, the MOSFET will stay in a partially on or off state all the time, essentially doing nothing but expending energy as a heat. \$\endgroup\$ May 8, 2017 at 4:59

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