Introduction
EPIA Mini-ITX Motherboard
EPIA CIR Header
Reducing Fan Speeds
Fan Voltage Calculator
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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
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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...
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