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I’m building a device that, in part, uses two 9V batteries to power a computer fan which is controlled by a potentiometer. If the potentiometer is below 60% or so, the fan fails to move. I want to hook up an LED that is only on when the fan is spinning.

I tried using 2 AA batteries and a transistor to power and control the light, but that didn’t work. I don’t need all the calculations done for me, but I’d appreciate input on where to go from here.

The circuit is very simple. A push button controls on/off, and the potentiometer controls speed. I understand the two 9V batteries can't power the fan very long, but I will only need to run it for ~5 seconds at a time, very infrequently.

I think this diagram better represents what I'm working with. After looking into it I realized I've had the potentiometer wired as a variable resistor. enter image description here

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    \$\begingroup\$ If the fan has a "tacho" output, it is relatively easy to detect if it is spinning. In any case it is much easier to make the potentiometer part of your circuit such that the fab always has a high enough voltage such that it will spin. If you show your circuit, suggestions could be made how to do that. Also realize that using 9 V batteries for spinning a fan will mean that the batteries will be drained fairly quickly. For 9 V batteries, the current any fan needs is a bit high, 9V batteries are intended for much smaller currents. \$\endgroup\$
    – Bimpelrekkie
    Commented Feb 27, 2021 at 21:40
  • \$\begingroup\$ If you are going to show a circuit, please show a correct circuit. In the circuit you show, the pot has no effect - you have to connect the wiper to either end to make it have an effect on the motor current. Also, a normal small LED would not work as shown - polarity is wrong, and with correct polarity the LED would be destroyed by the motor current. \$\endgroup\$
    – Peter Bennett
    Commented Feb 27, 2021 at 22:31
  • \$\begingroup\$ Thanks Pete, you obviously know a lot! \$\endgroup\$
    – AbleMind
    Commented Feb 27, 2021 at 23:02
  • \$\begingroup\$ 9V batteries are horrible other than for very specific low power applications. If you want to use disposable batteries, get a 6S AA battery pack and you'll have better power density and save a bunch of money. Output voltage range is the same. \$\endgroup\$
    – K H
    Commented Feb 28, 2021 at 2:22
  • \$\begingroup\$ I've updated my answer. \$\endgroup\$
    – AnalogKid
    Commented Mar 1, 2021 at 12:59

2 Answers 2

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Does the fan have a tach output? (3-wire fan). That’s the easiest way - the tach is open-collector and can drive the LED directly.

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  • \$\begingroup\$ Yes, it is a 3-pin 12V fan. I will look into tach thank you \$\endgroup\$
    – AbleMind
    Commented Feb 27, 2021 at 22:07
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You must check your fan datasheet to see if it will work with this circuit.

Many tach fans have an open-collector tach output signal that can sink at least 10 mA. If yours is like this, then this is about as simple as it gets. A typical fan tach signal is a square wave, with two pulses per revolution, comfortably above 100 Hz to prevent perceived flicker.

Connecting the LED to the 9 V point lowers wasted energy in the current limiting resistor. Also, sometimes the tach output is rated for a max of 12 V even if the fan is running on 18 V or 24 V.

UPDATE: To eliminate the center-tap connection between the two batteries, the right side of R1 can be connected to the switch (pin 1). You will have to increase both the resistance value and the power rating. If you need more LED brightness/current than the fan can control directly, you can add one transistor and one resistor.

enter image description here

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