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Reducing Fan Speeds

Fan Voltage Calculator


Fan speed adjustments...
...or how to prevent your computer sounding like a jet engine!

DISCLAIMER:
The fans in your computer are there for a reason - I will not be held responsible for any damage to you or your system caused by correctly or incorrectly following the advice contained in this article - you carry out these modifications at your own risk...

Fans are usually the noisiest items found in a computer, and often run unnecessarily fast - a 1.5GHz Athlon may well need the fan to be turning at 4,500 rpm, but a lesser machine with a good quality heatsink probably doesn't.

One situation where a slower fan is definitely beneficial is when a CPU fan in an older machine needs replacing.  Socket A CPU fans are often quite reasonably priced, and can often be jerry-rigged to work on older machines even if they're too big - but they also tend to be noisy.

So what are the options for slowing down a fan?

One option that's been quite popular recently is to run a 12 volt fan on 7 volts, by wiring it between the 12V and 5V lines on a disk-drive power connector, instead of between 12V and ground.  It has also been suggested, however, that this is hazardous to the health of your power supply - though the reasoning has been somewhat spurious...

Whatever the truth of this situation, a safer option for slowing down a fan is to place a resistor in series with the fan.
In this way, you can achieve any voltage drop you like, provided you don't mind doing some heavy math to get there...

Here's how we would calculate the resistor value required to run a fan at 7V from a 12V power source, dropping the voltage by 5V.

Original fan voltage: 12V

Original fan current: 0.2A (Read from sticker on fan)

    V=IR (Ohm's Law), so 12 = 0.2R   =>   R = 12 / 0.2   =   60

Fan resistance: 60 Ohms

Target voltage: 7V

    V=IR (Ohm's Law), so 7 = 60I   =>   I = 7 / 60  =  0.12 (2 d.p.)

Target current: 0.12A

Voltage drop across resistor: 12V - 7V  =  5V

    V=IR (Ohm's Law), so 5 = 0.12R    =>    R = 5 / 0.12   =  42

Target resistance:  42 Ohms

Power dissipated by resistor:  5V * 0.12A  =  0.6W

So to achieve a voltage drop of 5V, we would need a resistor with value of 42 Ohms, and capable of dissipating 0.6W of power.  This highlights one potential pitfall of this method - you can't safely use 1/4W resistors for this, unless you make up the target resistance using several smaller resistors (of equal value) in series...

Have a play with this little calculator, and try out some values of your own...
Reduce  volts,  amps, to  volts.