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

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?

sort by: active | newest | oldest
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

lemonie6 years ago
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.

mathews6 years ago
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).
Sandisk1duo6 years ago
build a variable power supply, then you can limit the voltage, which would reduce the RPMs
seandogue6 years ago
 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