why control fan speed with a mosfet?

[hpibmx]

Hi,

I need short explanation please.

why controlling fan speed with mosfet in circuit? (when a pot can do the job)

Thank you.

[inse]

You can do it with a pot as well as long as its power rating matches to the fan.

[not1xor1]

The reason to use active components is to reduce power dissipation through the pot, since most common pots can withstand quite low power (100-250mW).
In any case you should look for a better circuit, the one in the picture would never allow full fan speed and in some cases would not even allow the fan to start... there are plenty of better circuits around...

[Zero999]

As mentioned above, the MOSFET the dissipate more power than a small potentiometer. The circuit will work much better with a higher power rated potentiometer, also known as a rheostat and no MOSFET.

The MOSFET circuit won't work very well, because the resistance of the MOSFET is not directly proportional to the gate voltage. This means the motor's speed will not be proportional to the position of the potentiometer's wiper.

[Circlotron]

The MOSFET circuit won't work very well, because the resistance of the MOSFET is not directly proportional to the gate voltage. This means the motor's speed will not be proportional to the position of the potentiometer's wiper.

Once the pot wiper voltage gets above about 3 volts it will work okay because the circuit is a source follower. But the voltage that gets fed to the motor will always be about 3 volts less than the pot wiper voltage.

[Zero999]

The MOSFET circuit won't work very well, because the resistance of the MOSFET is not directly proportional to the gate voltage. This means the motor's speed will not be proportional to the position of the potentiometer's wiper.

Once the pot wiper voltage gets above about 3 volts it will work okay because the circuit is a source follower. But the voltage that gets fed to the motor will always be about 3 volts less than the pot wiper voltage.

You're right. I was thinking it was common source.  The deadband can be reduced by adding another resistor.

If the OP posted a proper schematic, rather than graphics would make it easier to read.

[golden_labels]

Aside from the power rating, power losses are affecting the circuit in two more ways.

The current passed through the potentiometer must be large enough to power the motor. That means: low resistance. And since a potentiometer also passes current to the ground, you are creating low-resistance path across the power supply. A lot of energy does nothing except heting up things.

If a considerable portion of the current coming into the potentiometer is diverted through its taper, each leg sees hugely different current. At low currents that’s not an issue. But if the current is large enough to cause noticeable self-heating, each leg reaches a different temperature, slightly shifting resistances. So the output slowly drifts as the device heats up. You would need a potentiometer that can dissipate heat well for that to not happen.

The transistor may be seen as a kind of separator between the potentiometer and the load, making the output of the potentiometer independent of the motor itself and removing restrictions on potentiometer’s parameters.

If that was not a fan: a motor is not a constant resistance. Due to back-EMF the resistance seen by the source depends on the load. If that would be a simple voltage divider, any change in load would change voltage applied to the motor. The reason a fan is an exceptional case is that it is constant load after spinning up. Unless you stick your finger in it.

[Zero999]

PWM can be used to control the speed of a DC motor. Replace the incandescent lamp in this circuit, with the motor and a diode in reverse parallel: cathode to +V and anode to the transistor's drain.