In the following diagram, what value do I need for the pot (R1) so that it can run 24-7 constant duty without overheating?
Simple speed control for 12vdc 200 mA muffin fan, 0 - full RPM as listed on the motor hub:
R1 must have a wiper rated to handle 200mA, at least (or whatever the fan draws at maximum). Not all potentiometers like such a heavy wiper current. I suppose if you don't need the control to be particularly linear, a standing current around 100mA in the potentiometer would be reasonable. That would be 120 ohms. Some value close to that should be OK. At 100 ohms, you'd draw 120mA in the potentiometer plus whatever the fan load is. At 100 ohms, the potentiometer dissipates
1.44 watts with 12 volts applied, but you should use one rated for 320mA current, because that's what will be in the upper section if the fan is drawing 200mA, when the wiper is near the top. That would be a 10 watt potentiometer.(It would be a lot easier to use a simple emitter follower, base driven from a potentiometer, though you wouldn't get to quite the input voltage at the output. But it would cut down on the requirements for the potentiometer a lot.)
Cheers, Tom
I could also use an LM317 variable voltage regulator, but a single pot seemed a bit simpler.
You really don't want a pot to control the speed directly. That will overheat the pot and may cause motor stalling at very low RPM. You want a PWM (pulse width modulation) circuit.
Here's a very simple circuit to control the fan: View in Courier font
+12 ---+---+----------+--------------+-----+ | | | | | [1K] | ---------- [D2] [FAN] | | | 8 | a| | | +-----|4 | | | / | | +-----+ 25K \\
The fan load, which would have been right here if you had quoted context, is 200 mA at 12V - that's 12 * .2 watts (2.4?) and 12 / .2 ohms = 60?
At 100 ohms, the potentiometer dissipates
If all you want is a variable resistance, then you should use a rheostat. Don't wire the pot like a voltage divider - as can be seen, it will dissipate more than the fan motor. Find a 50-100 ohm rheostat that's good for 200 mA - lessee, I^2 * R = .2 * .2 * 100 = 4 watts.
Of course, this is assuming that the fan has an ordinary DC motor that's amenable to running with a resistor in series with it - if it's one of those newfangled, fancy-schmancy brushless DC things, (essentially an induction motor, or maybe stepper) I wouldn't want to try this.
Good Luck! Rich
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