How to work out the Resistors needed to get 25% & 50% speed on a 6v motor using a rotary switch?

Hi there, I have a small motor rated from 1.5 to 9 V DC that operates off a 6 volt lead acid battery.
I need to connect a rotary switch (1 pole 3 way) so that in position 1=off  position 2= 25% speed and position 3 = 50% speed.

How can I work out what Resistor to use to get the motor to operate at roughly 25% & 50% speed?
No Load @ 6V   =   Current = 0.021 A    Speed = 2790 rpm
At Max. efficiency @ 6V = Current = 0.084 A    Speed = 2234 rpm  Torque  = 12.9g.cm
Maximum Load @ 6V = Current = 0.180 A Speed = 1395 rpm Torque = 32.5g.cm

Any help will be greatly appreciated.

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Agree with Rick, you can't realistically control motor speed by adding resistors - it destroys the "regulation" of the motor - small changes in torque make variable changes to the speed of the motor depending on the resistor.

You should ALWAYS use a voltage regulator (or a variable DC power supply)
MrGreenFingers (author)  steveastrouk4 years ago
Thanks for the replies guys.
I have put together a DC Motor Speed Controller before which changes speed using a Potentiometer.
Unfortunately im pressed for space for this project (as small and simple as possible)
and as you'll see in this pic the motor controller is too big for what i need.

I might even get away with just running it at say 50% speed so i was hoping I could possibly just use a simple ON/OFF switch and a Resistor to lower the speed of the motor. which would work right? just not efficiently.

But after hearing your comments, the wasted power & Heat buildup worries me.

What would the simplest smallest thing be to work the motor @ roughly half speed off a 6 volt battery?

I will look into the LM317 - Just don't have much space at all.

Thanks again guys.
I usually get my hands dirty in the Garden and workshop, Electronics is not my strongest point unfortunately. But learning more everyday. #Trial&error

A 317 and two resistors, is all you need.
MrGreenFingers (author)  steveastrouk4 years ago
Thanks Steve,
So If I understand the LM317 correctly,

Using a 6V battery, to drop the voltage to 4.3V with the LM317 ~ R1=240ohms and R2 = 586ohms.

Yes, very nicely done. Simple isn't it ?
rickharris4 years ago
Not the best or most economical way to regulate speed - The resistors use power and will get warm

You will be better off controlling the voltage LM317 would be better.

However what you need information wise is ohms law.

V=I x R So if you know V and I then you can calculate R - Put similar resistor in series to halve the speed.
Use voltage dividers to give you 1.5V (25%) and 3V (50%). Do an internet search for a voltage divider calc and start playing around with values. But any 2 matching resistors in a voltage divider configuration will half the voltage. Just make sure you use resistors that can handle the amount of power you'll be passing through them. The lower the resistance value the better.

You could also go a little more complex and have the selector switch rout the power through an LM317 with the right resistors values so the LM317 is regulating the voltage to where you need it.