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Resistor to reduce voltage for a mini fan

Discussion in 'General Electronics Discussion' started by Steveggz, Mar 19, 2018.

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  1. Steveggz

    Steveggz

    15
    0
    Oct 7, 2017
    I need to power a small 20mm mini 5v fan. I have a 5v mean well power supply but I want to use a resistor to lower the voltage to 2.80v. I tested the fan at 2.80 volts (with a vaiable power supply) and it was the quietest while giving enough air. Is it possible to just put a resistor before the fan to lower the voltage to 2.80v? Everything online talks about using a resistor for LED's. I don't get it, can you only use a resistor with constant current?

    My power source is 5v 6A
    My mini fan is rated at 5v .150 but I want it to run at 2.80v (Its draws around 50mA at that voltage according to my variable ps)
     
  2. Bluejets

    Bluejets

    4,151
    879
    Oct 5, 2014
    You have the current draw, and the voltage required on the motor and any resistor connected in series.

    With ohms law you can calculate the resistor size and just as important, the wattage of the resistor.
     
  3. hevans1944

    hevans1944 Hop - AC8NS

    4,521
    2,104
    Jun 21, 2012
    There are many reasons to use a resistor to impede the flow of electricity. Dropping excessive voltage to a load that is essentially constant, like your fan, is just one reason. It is not very efficient to lower voltage in this manner, but for small loads (your fan qualifies as a "small" load) that remain constant, a series resistor is an effective solution.

    You want to drop (5 V - 2.8 V) = 2.2 V across a resistor in series with a fan drawing around 50 mA at 2.8 V. The value of that resistor is R = V / I = 2.2 V / 0.05 A = 44 Ω. The closest standard value 5% tolerance resistor is 43 Ω. It will dissipate P = V² / R = (2.2)² / 43 = 0.11 W. A 1/4 watt resistor should be adequate.
     
    BobK and Steveggz like this.
  4. Steveggz

    Steveggz

    15
    0
    Oct 7, 2017
    Thank you so much hevans1944! Thanks for the formula and breaking it down. I've seen the formula before but you'd be surprised how many questions and concerns arise for someone that is a complete novice to electronics. Having it broken down specifically using my values makes me finally understand it. THANK YOU!
     
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