Is this a simple Ohms law problem?

Discussion in 'General Electronics Chat' started by vonsworld, Jul 7, 2013.

  1. vonsworld

    Thread Starter Member

    Apr 27, 2011
    I think maybe this is just a simple question about ohms law, perhaps someone could help me...

    I have 12 volt, 1.8 watt cooling fan attached to a battery and therefore I understand it would draw 1.8/12 = 0.15 amps.

    Also using ohms law the resistance in this circuit would be V/I or 12/0.15 = 80 ohms.

    There is no speed control on this fan and it's noisy on full power, so I would like to attach a pot to the cable, but how do I work out from the above figures the additional resistance required to just about stop the fan? and therefore what range of resistance the pot needs to provide?

    Many thanks :)
  2. WBahn


    Mar 31, 2012
    No, it is not an Ohm's Law problem for a couple of reasons.

    First, the power specification is probably the maximum power that it is guaranteed to draw, not the amount that it actually will.

    Second, if it is a permanent magnet DC motor, then the voltage across it will be determined not by the resistance of the windings, but by the back-EMF generated by the motor, which is directly proportional to the speed it is turning at.

    Still, there is something to work with here (IF it is a simple motor and not a stepper or some other more advanced kind of arrangement). Start with your 150mA. Now figure that the current draw will drop as the fan speed slows (current is proportional to torque which will be proportional to the air load on the fan blades). Let's use a ballpark figure (and this is a real WAG to use only as a starting point) of saying that it will draw 15mA and 10% speed, which would have a back-EMF of 1.2V. So you would need to drop about 11V with about 15mA. Since this is a WAG, we can call that 10V with 10mA and call it a 1kΩ resistor.

    But a much, much better way would be to first see if the fan will run properly if powered by lower voltages. So use 9V, 6V, 3V, and 1.5V. If you have the ability, measure the current at each of these (and 12V). If not, just noting that the fan seems to run smoothly and more slowly is enough. It may stall before you get to 1.5V and, if so, that's fine and good to know.

    Next get some fixed resistors of, say, 100Ω and 330Ω and see if those produce roughly similar results. If you get a small handul of 100Ω resistors, then you can combine them in series and parallel to get a range of resistances. Try to find something that seems to slow the fan to about what it runs with 6V across it (or measure the voltage across it, if you can). Also try to find the resistance that runs it as slow as you think you would ever want to. Use that to determine what pot size you need.

    Don't forget to take into account the power dissipation in the pot.
  3. vonsworld

    Thread Starter Member

    Apr 27, 2011
    Thanks for your reply and all the good suggestions

    I've looked in my local electronics shop and all their pots are rated at 0.4 watts, and so if I used one of those to control a 1.8 watt fan I guess it would melt!?

    Do you think the best solution would be to buy a wirewound rheostat which could handle 5 watts or build a simple regulator circuit since a regulator chip can handle 1 amp?

    As you know the circuits are quite small and the low wattage pot could then simply control the regulator chip voltage output. Guess I would still need a pretty good heatsink too :)
  4. BillO

    Distinguished Member

    Nov 24, 2008
    You could use a transistor with a pot. You'd need to ensure the transistor had enough heat sinking to dissipate any heat (about .5 watts at the most).

    You could also use a small microcontroller and a little MOSFET with the pot to provide PWM speed control which would not only be more efficient (less heat to deal with) but would give you more control over the fan speed.

    Either of these solutions might be cheaper than a 5W wirewound rheostat. The 2nd one might be a lot of fun too.
  5. #12


    Nov 30, 2010
    If you use this, be sure to look up the capacitor sizes in the datasheet.
  6. WBahn


    Mar 31, 2012
    The peak power dissipated by the pot would probably not exceed about one-quarter of the rated power and usually less, so a 0.4W pot might work (though you should always oversize components when you can).

    Unless you use a nonlinear circuit, you are going to have to dissipate some power in the control circuit.

    My recommendation would be an LM317 adjustable voltage regulator, just be sure to heat sink it for at least about 1W.

    But all of this is moot if the fan won't run properly on reduced voltage, so you really need to verify that, first.