Fan power can be reduced by dropping some unwanted volts from the 12V line across a resistor, but the method has two disadvantages.
With a rheostat you can turn up the juice to get the fan going, but you can't do that with a fixed resistor, so semi-conductor solutions are recommended.
To calculate the required resistor, you need to know Ohms Law (Voltage/Current = Resistance) and the fan resistance, which can be calculated from the fan wattage or current rating, usually marked on the fan label.
Let's say it's rated 2W at 12V. Resistance is 72R (12*12/2). To run the fan at 9V from the original 12V supply, the resistor will need to drop the spare 3V. The same current flows in both fan and resistor, so by Ohms Law, Fan voltage/Fan resistance = Resistor voltage/Resistor value.
9/72 = (12–9)/R, or R = 3/9 of 72 = 24 ohms. The nearest common value is 22R.
To run at 7V, the required value is 5/7 of 72, 51.4 ohms, 51R being the nearest common value.
Power rating is V*V/R = 3*3/22 in the first example, 0.41W. A 0.5W would be running near its limit and get very hot, so a 1W resistor is better.
One problem – many fans have a diode in series with the +12V input to guard the electronics against an accidental reverse polarity connection. That will lose about 0.75V. Then the electronic switching system will have an inevitable voltage drop, perhaps 0.25V. So a fan connected to a 12V supply may only see 11V at the motor winding, and at 7V get only 6V.
So, in such cases, Ohms Law is not obeyed. Check the actual voltage dropped across your chosen resistor and adjust to suit.
Better than just a resistor, but still limited in the range of fan powers a single setup can handle.
In the standard circuit, the zener diode is connected across the fan with its banded cathode end pointing to the positive supply. Below the zener voltage chosen, it behaves like an ordinary diode, blocking current flow, and does nothing, but above the zener voltage it starts to conduct in its reverse direction, keeping the voltage across the fan constant.
The unwanted volts are dropped across the series resistor, and this is usually calculated for the worst case, when the load is open-circuit and the zener is carrying its maximum current.
Let's say we've chosen to run the fan at 9V. The nearest zener diode is 9.1V, and we'll use a commonly-available 1.3W type.
By Ohms Law, 9.1V produces 1.3W at 0.143A (1.3/9.1). With a 12V supply, to remove the unwanted 2.9V at 0.143A we need a 20.3R resistor (2.9/0.143), rated at 0.4W (2.9*0.143). A 20R is a close standard value, and back-tracking, 2.9V over 20R will give 0.145A.
So the largest fan we can use will draw 0.145A at 9.1V, equivalent to 0.19A at 12V or a 2.3W model. With this, the zener is doing nothing, all the current is going through the resistor, then the fan.
If we use a less powerful fan, things change. Taking a 1.2W model, the fan draws 0.1A at 12V, reducing to 0.076A at 9.1V. The resistor is still carrying 0.145A, so the other 0.069A is passing through the zener diode , keeping the fan voltage steady at 9.1V and generating 0.63W of heat (9.1*0.069), well within the 1.3W rating. Should the fan fail open-circuit, the zener will have to carry the full 0.145A from the resistor, heat produced then is 1.32W, its top limit.
However, if we start with the regulated 12V supply from a computer PSU, we can use the simpler circuit shown left.
Now the Zener is in series with the fan, again with the banded end pointing towards positive, but now the fan acts as the resistor to use the spare volts (Supply Voltage - Zener Voltage). On the 12V supply, a 4.7V zener diode will have that voltage dropped across it, leaving 7.3V for the fan.
The 1.3W BZX85 Series zener diodes come with voltages of 2.7V, 3.0V, 3.3V, 3.6V, 3.9V, 4.4V, 4.7V, 5.1V, 5.6V, 6.2V, and so on, so there's a good selection of under-voltages possible.
To work out the zener power rating, check the fan rating and convert to current. A 2W fan will pass 2/12 = 0.167A at 12V, and around 7.3/12 of that at 7.3V, say 0.102A. The 4.7V zener will then produce 4.7*0.102 = 0.48W of heat, which the 1.3W model will easily handle. In fact, this zener should be able to regulate up to 5W in fan power.
Semiconductor junctions always drop some voltage, and diodes are no exception. A 1N4001 silicon diode drops about 0.75V at typical fan currents (100-350mA), so one or more diodes in series with the load can be used to take the edge off a noisy fan. Two diodes will reduce the fan voltage to about 10.5V.
(The voltage drop does rise with current, but not by much – the specification's maximum drop for a 1N400x (1A) or 1N540x (3A) diode is below 1.1V at the full rated current. And the minimum drop is over 0.6V at very low currents.)
One advantage is that no circuit board is needed – a few diodes can be soldered into the fan leads, or to a 2-way switch to give a high/low option.
The diode band (the cathode end) should point to 0V (black) on the supply (or to the red lead's terminal on the fan). Cover the joints with heat-shrink tubing to prevent shorting.
A few major plusses for diodes are,
If the 7-volt Trick is too drastic a power reduction, a semiconductor regulator can be used for a higher voltage. Fixed output 8V and 9V designs are made, but may be hard to find and/or expensive. However, the cheap and common 7805 5V 1A regulator can be wired to give other voltages.
In the standard circuit, the "common" terminal goes straight to ground and the output is 5V.
Note the diode across the regulator is optional – it protects the regulator against reverse voltage from C2 or the load if the supply side is shorted to ground.
The regulator should also be fitted with a small heat-sink, such as Maplin's RN84F (20degC/W).
If the voltage at the "common" pin is raised above ground, this is added to the 5V regulator output, up to a limit about 2V below the input voltage – so a 12V supply can give 5-10V output to fans.
Text-book method shown is with a two-resistor potential divider on the output. Voltage across R1 is 5V, which produces a current of 5/R1 amps. This, plus the regulator's current from the "common" terminal (quiescent current Iq), flows through R2 producing a voltage across R2 of R2(5/R1 + Iq) volts. This voltage across R2 is added to the 5V across R1 to give the raised output voltage.
Formula for the output voltage Vo is
which rearranges to
Iq is typically about 4mA (.004A), so if R1 is 240 ohms and R2 is 150 ohms,
To minimise the effect of Iq variations between regulators, the divider current needs to be significantly higher than Iq, so R1 should preferably be under 330 ohms.
Check the output with a multimeter, as small changes in Iq from one regulator to another have a significant effect. Or use a 220-ohm preset potentiometer for R2 and adjust as required.
Another simpler method is to use a zener diode to raise the ground terminal voltage, then the output is
The BZY88/BZX55 series 0.5W zeners come in a range of values between 2.7V and 5.1V giving 7.7-10.1V from the regulator.
The resistor R1 is present to produce enough current to make sure the zener is well on, and many regulators will work fine without it (Iq is enough).