Semi-conductor junctions always drop a small voltage, but unlike a resistor it's fairly constant over a range of currents. A silicon diode such as the 1N4001 drops about 0.75 volts with a typical fan drawing, say, 150-200mA, and only about 1V at 1amp.
Using a chain of diodes mounted on a rotary switch, a series of voltage steps can be produced giving adequate control for fans.
On the first switch position, the 12-volt supply goes directly to the fan. Next notch, a diode in series will reduce this by 0.75V giving 11.25 volts, then a further diode produces 10.5 volts on position #3 and so on. Outputs possible are 12V, 11.25V, 10.5V, 9.75V, 9V, 8.25V, 7.5V, 6.75V, 6V, etc.
This control method has several advantages;
I used a 6-way, two-pole switch (Maplin code FF74R, £1.29). Contacts are rated at 5A continuous, switching 350mA at 110V AC/DC so it will stand a fair bit more down at 12V. The six diodes on Pole 'A' are 1N4001 (Maplin's QL73Q, 6p each), rated at 50V, 1A. The two on pole "C" are 1N5817 Schottky diodes (Maplin JA46A, 9p each).
The diodes must all be arranged with the banded-end away from the 12-volt supply connection (Band = cathode = pointed end of diode symbol).
I used 2 diodes on the first step, giving me outputs of 12V, 10.5V, 9.75V, 9V, 8.25V and 7.5V.
If you don't want lights, just use the left-hand section of the schematic, for pole 'A'. Using a 12-way 1-pole switch (FF73Q) would allow the range to be extended, using single diodes between each position. You could then use 1A Schottky diodes, they drop under 0.5V per diode so would give smaller steps, and/or have an unconnected "off" position at one end.
First step is to check your rotary switch contact system. My Lorin switch is marked 1-12 round the outside, with "A" and "C" contacts in the middle. "A" makes with 1-6 in turn, making with 6 when turned fully clockwise, "C" makes with 7-12, C-12 being connected at full clockwise travel, so the following instructions apply to similar switches.
Then prepare 3 lengths of wire to go from control to fan. You need to cut the fan lead, inserting a 3-way connector to give a 12V supply (I used a yellow wire), the controlled supply back to the fan (red) and a ground (black).
Connect the red wire to A. Bend the first diode's lead so the lead goes through contacts 6 & C, taking care the band is near the un-bent lead. Also connect the yellow wire to C. Bend a second diode lead at the banded end and pass through contact 5, twist the free ends together and solder.
Now make a "U" bend on each of the other diodes' anode lead (away from the band) close to the body (bending round a small drill bit gives a neat result), cut the ends level, make a sharp bend near each end, and connect between 5 & 4, 4 & 3, etc, soldering as you go.
The right-hand half of the schematic is the lighting switching. Various options are possible.
(b) Single leds, all same type
(c) Single leds, different types
Single leds could be used, arranged in a part-circle round the knob to show the switch position. If they're all the same Vf, a single current-limit resistor can be used, as in the left-hand diagram (b). Value depends on the leds used; knowing the forward voltage (Vf) and required forward current (If),
eg, for red leds with Vf=2.2V, If=10mA, on a 12V supply, resistor = (12 - 2.2) / .01 = 980R. The nearest standard resistor value is 1k.
If different colours are wanted (with different Vf and/or desired If) calculate each required resistor value and use one in series with each led as in (c).
Remember the led cathode (pointed end of the symbol) is usually the shorter lead and near a flat on the case.
The layout shown in the main schematic lights more leds as the switch is turned, so at 12V all 6 are lit. This could be used to show a bargraph display or again arranged round the control knob.
With only 12V available, running 6 leds in series is impractical so after a leg of four, a second leg is started. (With leds of Vf higher than 3V (eg blue) limit the leg to 3 leds, or the total Vf per leg to under 10V.)
|R2||4.2V||780||750R or 820R|
|R4||8.4V||360||330R or 390R|
|R6||4.2V||780||750R or 820R|
|R7, R8||8.9V||310||300R or 330R|
A separate connection from the switch keeps the first leg lit, with a rectifier diode to stop premature lighting of leg #2. These diodes should be low Vf Schottky types such as the 1N5817 or 1N5819 (assume Vf = 0.5V) to leave maximum headroom for the current-limit resistor.
The current-limit resistors are calculated by adding the forward voltages of all the lit leds (and diode for R7 & R8) then using the above formula.
With 2.1V leds, aiming for 10mA, and assuming the Schottky diodes drop 0.5V, this gives values as shown in the table (right).
I've been asked about making a bargraph-style display using blue leds, which have a forward voltage of around 3.7V, though some are over 5V. You can do it using strings of leds as described above, with two strings of three or three strings of two for the highest voltage types, but I think this is an easier way to assemble the circuit if you don't have three hands and good burn tolerance. The system also works for 2V/2.1V leds.
The connections 7-12 go to the switch SW1b 7-12 as in the first diagram above.
The diodes D7-D12 ensure that only the leds below and including the switch position indicator are lit. (Schottky types are used for minimum voltage loss, look for any with a Vf of 0.5V or less that can carry 100mA or more.)
Using this system, the leds don't all get the same current when several are lit.
In the worst case, with all six lit, LED6 and its resistor see the full 12V, so current is (12 - 3.7)/820 = 10mA.
About 0.5V is then lost through each schottky diode, so LED5 gets 9.5mA, LED4 8.9mA, LED3 8.3mA, LED2 7.7mA, down to LED1 with 7.1mA. (That's the worst case, the top led in the bargraph always gets 10mA and the ones below 9.5mA, 8.9mA, etc.)
In practice, this makes very little difference to the apparent brightness of each led, particularly when they're in a row, as any minor change is gradual along the line. I've shown 820R resistors (its an indicator light, not a searchlight ;), but if you want a brighter average any value down to 430R is safe for most blues, or use the led resistor formula for those you have to be sure. (If you really want to dim leds, look in the Eye Candy section.)