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I recently bought a potentiometer to regulate the speed of two 12V/1.1A DC fans in parallel, powered by a 12V source. Its specifications are the following:

Resistance= 10k Ohm, MaxPower = 0.2W.

Can I connect the fans, the potentiometer and the power source in series, or do I need additional circuitry to avoid circuit damage? I feel like the potentiometer's maximum power is too low, but I may be wrong.

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  • \$\begingroup\$ You want a smaller pot to start with. If you try to push 2A through 10K, you'll need 20,000V. \$\endgroup\$
    – Cristobol Polychronopolis
    Commented Jun 6, 2018 at 16:10
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    \$\begingroup\$ This is what you want to control your fans. A suitable potentiometer for the job is crazy big. \$\endgroup\$
    – lakeweb
    Commented Jun 6, 2018 at 16:17
  • \$\begingroup\$ Thanks @CristobolPolychronopolis, that was one of my main concerns. Unfortunately that leads to another problem, which is: The cheapest potentiometers had a lot of resistance to begin with. I imagined that for two fans in parallel, which is a pretty simple circuit, potentiometer prices would correspond to the cheapest and not the way around, so I'm kind of confused. \$\endgroup\$
    – madsantos
    Commented Jun 6, 2018 at 16:20
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    \$\begingroup\$ No, no, no this won't work - smoke will erupt. Pots are used to control milli amps at most and not amps. \$\endgroup\$
    – Andy aka
    Commented Jun 6, 2018 at 16:20
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    \$\begingroup\$ I guess I will try the PWM controller that @lakeweb suggested, instead. Thanks for saving my life :) \$\endgroup\$
    – madsantos
    Commented Jun 6, 2018 at 16:23

2 Answers 2

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If you have a 10 kohm pot rated at 0.2 watts then, end-to-end you could apply 44.7 volts across it - that would take a current of 4.47 mA. Reason: -

Power = Voltage\$^2\$/Resistance hence voltage = \$\sqrt{P\cdot R}\$ = 44.7 volts.

That's the maximum current the pot can be used for.

Even if you only used one-tenth the range of the pot (1 kohm) i.e. you used it like a rheostat, the power is also reduced by one-tenth to 20 mW and the maximum voltage across one tenth is 4.47 volts. Guess what... the current is still 4.47 mA.

4.47 mA is the maximum current you can pass through a 10 kohm 200 mW pot.
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  • \$\begingroup\$ Thanks Andy, but isn't the pot supposed to have variable resistance? I thought the 10kohm were the maximum resistance, when there's almost no current and the fans are turned off. \$\endgroup\$
    – madsantos
    Commented Jun 6, 2018 at 16:40
  • \$\begingroup\$ My answer has shown that no matter what position the wiper is, the maximum current that can be passed through a small section is 4.47 mA. If you are only using one-tenth the range of the pot, 90% is unused and therefore the power rating that applies to the section that is used MUST BE one-tenth of 200 mW i.e. 20 mW. \$\endgroup\$
    – Andy aka
    Commented Jun 6, 2018 at 16:43
  • \$\begingroup\$ Apart from anything else a 10 k pot (even if it had a decent power rating) would be wholly unsuitable to regulate the fans. You would be looking for something like a 10 ohm pot to regulate to a bit below half speed and this is wire wound and is therefore costly and if you do the math you'll find that when you are pushing 2 amps through a 10% section (fans at high speed) that power dissipation is 4 watts. Thus the fan would need an overall rating of 40 watts! \$\endgroup\$
    – Andy aka
    Commented Jun 6, 2018 at 16:46
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In today's world electronic controls for this kind of work are the only way to go. And thanks to cheap products on Ebay, there is a solution. I'm sure this is cheaper than a pot that would work. And, it is way more efficient.

Ebay Product to do the job

enter image description here

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