# How can I limit the RPM of an electric motor by using resistors or trimpots? (cheap/simple way)? Answered

I've done it with a small trimpot that almost ignited by heating. What the problem? The current is to high? To lower the RPM must i lower the current or the voltage?

And what exactly do the resistance lower? voltage or current?

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## Comments

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Using a series resistor is a bad idea, because the motor not only runs slower, but the ability to stay running under load is screwed up. Much better to put a very  simple variable power supply together, using an LM317 for current < 1A or an LM150 for upto 3A.

Whatever, you'll still need to dissipate a lot of power in the regulator, whether its a fixed resistor or a semiconductor.

Mathew's idea is conceptually sound, but he has omitted any power handling circuits.

A REALLY easy way to build a PWM motor supply is with the Micrel Mic502

Steve

The current is much too high for what you used. Drop the voltage supplied, or add in something like a filament light-bulb for resistance.

Increasing the resistance of the circuit will reduce the current flow, but in adding a resistor the voltage across the motor will also be reduced.

L

You can use a relaxation oscillator to PWM the motor. Instead of a single feedback resistor, you can use a potentiometer and two diodes.

Attach the wiper pin of the potentiometer to the output of the Op-Amp, with a diode pointing in each direction, (as shown, hopefully).

This will give the full range of mark/space ratios from 1:0 to 0:1 (ish).

the pot is probably trying to pass to much current for its power rating, which is why it's burning up.

The potentiometer inline with a load is likely part of a voltage divider.

Assuming  a DC source for simplicity

let R1= instantaneous resistance of the pot
let R2=load resistance presented by the motor (maybe not exacty correct,but for illustration purposes)

The voltage seen by the load is

VR2 = Vin x (R2/(R1+R2))

IFF your intention is to simply use Resistance as a control, then you need to invest in power resistors or potentiometers that have high power ratings.

I forgot to include it in my last post, but the power can now be calculated (from the above) as

P = V x I, but  I = V/R, so

PR2 = (VR2)2 / R2