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12v fan speed control

Discussion in 'General Electronics Discussion' started by flor.27.27, Nov 15, 2012.

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  1. flor.27.27

    flor.27.27

    5
    0
    Nov 14, 2012
    Hi guys,

    I'm having some problems with making a speed controller for a 12v pc fan. I searched for it but it's not yet fully clear to me after reading the topics :(

    First some specs:

    I have a 80mm pc fan with:

    DC 12V
    0.11A
    1.32W

    And I want to connect it to an adapter which I think is not reggulated:

    output: 12V DC 500mA

    I have a pack of resistors with different values which I can use. I would like to dim the speed with a 10k ohm potentiometer I have.
    The fan will have to cool a driver so I would like to dim it between 50-100% instead of 0-100%.

    I drew a circuit which I hope will do the trick. But all my resistors are 1/4 W rated and when I calculate the power over them, the result is 7.2W :mad: is there another solution than wiring 29 resistors in parallel :confused: ?
     

    Attached Files:

  2. Harald Kapp

    Harald Kapp Moderator Moderator

    10,053
    2,144
    Nov 17, 2011
    Your fan has an equivalent resistance of ~110 Ohm (12V/0.11A). In order to set it to 50% you will need an equivalent series resistor of ~110 Ohm (assuming the speed varies linearly with the voltage, which it doesn't do exactly).
    At low speed the sum of resistances therefore is 220 Ohm. At 12 V the current then is 12V/220Ohm = 55mA. The power at the resistor then is P=I²*R=0.55²*110 W=0.333W.

    How do you arrive at 7.2W? That is more than the fan itself uses (1.32 W)!

    From the above calculation you can see that a 10k Potentiometer is way off your requirement. You will need a potentiometer of approx. 100 Ohm (e.g. 220 Ohm) and a suitable power rating.

    Harald
     
  3. flor.27.27

    flor.27.27

    5
    0
    Nov 14, 2012
    I thought I had to place a resistor in parallel with the fan, because if I only place a potentiometer in series with the fan, if the value of the potentiometer is 0 ohm, isn't the power over the fan then P=U. I= 12V. 0.5A= 6W which is more then 1.32w?

    or do I have to take the current which is in the specs? making it 12v . 0.11A =1.32W?

    Thank you very much for the fast reply! :)
     
  4. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

    25,396
    2,777
    Jan 21, 2010
    The simple solution is to use your transistors to make an astable multivibrator.

    Use your pot to change the time constant for one side.

    Tap of the output and use it to drive a transistor which switches your motor on and off.

    This will be a simple PWM controller and will work much more efficiently and reliably that a circuit using resistors.

    The resistor you have across the motor seems designed to simply waste (even more) power.
     
  5. flor.27.27

    flor.27.27

    5
    0
    Nov 14, 2012
    Thanks Steve, thanks to your info i've found some pwm circuits for 12v which I can make or at least copy :)

    Sorry for the very basic questions
     
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