How to Control Speed of DC Motor using Arduino, MOSFET and Potentiometer?

[@grinn585]

just curious, why do we need an arduino for this? can't you just use mosfet + pot and it will function the same?

[@Circuitdigest]

If you use the mosfet with just a pot without the PWM, you control only within the ohmic region.Even there you would be wasting energy since the mosfet will act like a resistor and will generate a lot of heat.

[@Omya_6116]

Can I use IRFZ44N Mosfet for the project? If no then can you recommend any other mosfet please.

[@Circuitdigest]

Not recommended. You must use logic level MOSFET that can fully turned on with the 5V logic level.

[@Omya_6116]

So any logic level n-channel mosfet is ok?

[@Circuitdigest]

Yes with appropriate current rating.

[@Omya_6116]

​@@Circuitdigest example STP55NF06

[@Circuitdigest]

Mosfets with VGS(th) less than 2.5V is recommended.

[@moganthkumar5831]

What are those resistors for ? What is its use ?

[@Circuitdigest]

Pull down resistor is there to make sure that the MOSFET is fully turned off and the gate resistor is used to control over-current in gate drivers and to reduce overshoot between the drain and source during switching (EMI noise-reduction). The switching time (rise and fall time) of MOSFET varies depending on the resistor of the connected gate.

[@TheHaykokalipsis]

@@Circuitdigest Why not explain it in video and make it simpler to understand?

[@drelectronics13]

Doesn't mosfet already have inbuilt diode

[@Circuitdigest]

The flyback diode is connected across the motor not across the mosfet.

[@oret2024]

10qs sir!I want to do this project how much money I have to invest?

[@Circuitdigest]

Hi, Components required are listed on the project page you may use that to check the total cost.

[@ajmal_thagzeen]

Can upload automatic motor speed decrease

[@Circuitdigest]

That what shows at the beginning of the video

[@joshuvajohnvarghese2372]

1k for gate is really high. Gate resistor should be around 20 to 50 ohms

[@Circuitdigest]

With minimum Vtgh and rise time of 10nS, Ig = (1470pF * (5V - 2V)) / (10ns) = 0.42mA. With 1K resistor the current limit will be 5mA at 5V. So the 1K resistor is more than enough and will work just fine.