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I want to make a simple fan speed regulator, but my electronic skills are failing me. I want to be able to slow down two computer fans.

I have in my hands a 12V/1A power suply, 2x 12V/0.2A fans, a B10K potentiometer and a 7805 and a 7812 ICs.

Can I make it with those? If yes, how? If not, what do I need for that circuit?

Thanks

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  • \$\begingroup\$ You want a lower dropout regulator so use a 0.1 Ohm FET and Op Amp with the pot and calc. R ratios with feedback.. \$\endgroup\$
    – Tony Stewart EE75
    Dec 22, 2018 at 19:18
  • \$\begingroup\$ Try this pcsilencioso.com/cpemma/reg.html \$\endgroup\$
    – pnatk
    Dec 22, 2018 at 19:38
  • \$\begingroup\$ Of those examples they lose 20% top speed except the MIC291xx series. from V dropout \$\endgroup\$
    – Tony Stewart EE75
    Dec 22, 2018 at 19:45
  • \$\begingroup\$ Note that some computer fans have a built-in circuit to run them. You can't 'just' lower the voltage. \$\endgroup\$
    – Oldfart
    Dec 22, 2018 at 19:50
  • \$\begingroup\$ all of them have a built-in circuit, but usually you can just lower the voltage, most will start at, or below, 5V \$\endgroup\$
    – Jasen
    Dec 22, 2018 at 20:28

3 Answers 3

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You can try this. It might not be the best but it uses the parts you have.

Range is a bit less than 6V to about 9.5V so you can't set it to full speed.

The 6V low end is more of a feature since most computer fans will stall if you drop the voltage too low.

schematic

simulate this circuit – Schematic created using CircuitLab

Pretty much any PNP transistor can be used for Q1 (well, don't use a 4GHz RF part).

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  • \$\begingroup\$ Not to ruin the fun here, but an RF transistor will work just as fine. 😂😀 \$\endgroup\$
    – paki eng
    Dec 22, 2018 at 20:30
  • \$\begingroup\$ With a nice 5mA bias it might sing a bit in an emitter follower configuration, depending on stray inductance and capacitances. \$\endgroup\$
    – Spehro Pefhany
    Dec 22, 2018 at 20:31
  • \$\begingroup\$ I didn't get your point in using a higher value resistor (15k) and limiting the voltage available to the fan to 9V. Just out of curiosity, why though?? \$\endgroup\$
    – paki eng
    Dec 22, 2018 at 20:33
  • \$\begingroup\$ @AbdullahBaig The 78xx regulator limits the voltage due to the dropout. Can't do much about that without using a different part or bypassing it with a switch. The 15K resistor just makes it so the top 60% or whatever of the pot does not go to waste. It will hit the dropout voltage of the regulator well before the pot hits the end stop. \$\endgroup\$
    – Spehro Pefhany
    Dec 22, 2018 at 20:36
  • \$\begingroup\$ Dropout is practically less than 2V, which means output a bit more than 10V. So there is still some margin that yhis circuit eats away. \$\endgroup\$
    – paki eng
    Dec 22, 2018 at 20:38
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Use this circuit. It will sure work. I have been using it. The 7805 might get a bit hot, so you would need a heat sink with it.

Use R2 as your 10k Potentiometer and R1 as any resistor from 5k to 10k. Capacitors are optional.

Any questions or comments are welcome.

schematic

(Image source)

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    \$\begingroup\$ Less than 10% of the pot electrical rotation will do anything with this circuit. 6-7mA is flowing through the rheostat and it hits full output voltage at about 4.5V, so 750 ohms to 643 ohms or about 6.5 to 7.5% of electrical rotation. \$\endgroup\$
    – Spehro Pefhany
    Dec 22, 2018 at 20:41
  • \$\begingroup\$ Use more practical measurement and less data sheet, all I would say \$\endgroup\$
    – paki eng
    Dec 22, 2018 at 20:49
  • \$\begingroup\$ Thanks. Will I get 5 to 10V range with that circuit? \$\endgroup\$
    – Dpedrinha
    Dec 22, 2018 at 21:44
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There are a few minor issues with a linear DC fan control design;

  • 1) Linear dropout of BJT type regulators is high, FET type is low
  • 2) Linear power dissipation is shared equally at half voltage, but @ 1.2W loss in the regulator is not too hard to cool yet too hot, if you make no attempt. ( hint; put heatsink near the fastest moving air )
  • 3) When the fan stalls below say 3V how will know if any active parts inside need forced air cooling at the albeit reduced current consumption
  • 4) Using a PWM approach for a Fan that has Hall sense commutation circuit may have aliasing acoustic noise in the windings with the Pulse modulated coils, can work but internal stress like cap surge current is unknown.

Nevertheless, I have used LM317 with a transistor and thermistor on a 48V supply with two 24V fans in series in production regulating at 45~55'C air temp on the power supply hotspot (XFMR).

BLDC Fans draw an average current like a resistor with voltage.

Here is my full range linear voltage regulator suitable for a 2.4W Fan.'

It works with any Op Amp as the inputs only drop down to the Vgs(th) of the NFET.

enter image description here

Adding >0.1uF decoupling cap is optional to the tap of pot. (10k~100k)

Note that since the FET inverts, negative feedback MUST use the +ve input.

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    \$\begingroup\$ Looks like an oscillator to me. ;-) \$\endgroup\$
    – Spehro Pefhany
    Dec 22, 2018 at 22:40
  • \$\begingroup\$ Not really. It is only marginally unstable with 1mF capacitance load if the Cap ESR approaches is < RdsOn of the FET. Although with positive feedback and some capacitance load on Pot, you can make a hysteretic PWM. tinyurl.com/yb3ofovl <Proof< Although one should never put a cap on the output. This is why there are lower limits to ESR on SMPS caps for single driver types,. \$\endgroup\$
    – Tony Stewart EE75
    Dec 23, 2018 at 0:30
  • \$\begingroup\$ The problem is the additional gain from the MOSFET will almost surely be more than the op-amp gain margin. That's with zero capacitive load. Will simulate it- brb. \$\endgroup\$
    – Spehro Pefhany
    Dec 23, 2018 at 0:36
  • \$\begingroup\$ I get the following: i.imgur.com/hQWYEWE.png \$\endgroup\$
    – Spehro Pefhany
    Dec 23, 2018 at 0:45
  • \$\begingroup\$ maybe I should have put a lower limit on the 0.1ΩOhm FET. Yours is RdsOn <2 mΩ and Ciss 2800pf \$\endgroup\$
    – Tony Stewart EE75
    Dec 23, 2018 at 1:09

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