# How to reduce the RPM of a DC motor

I need a small 3-pole commutator motor with 4000 RPM @ 12 V and the best I can buy is 8000 RPM @ 12 V. I am considering buying an 8000 RPM motor and replacing the existing windings with thinner wire.

I have calculated that I can double the number of turns if I cut the cross-sectional area of the wire in half. At the same time, the resistance of the winding will quadruple. Consequently, the magnetic moment of a winding (at the same voltage) will halve.

What are the effects of such a replacement? Will the RPM be halved? How will this affect the torque?

• Are you talking about a BLDC motor? Commented Feb 1, 2022 at 12:18
• @JonathanS. No, motor a commutator with three poles. Commented Feb 1, 2022 at 12:23
• what frequency do you mean? 12V is not a common AC source and DC has no frequency that you can depend on. If you are talking about essentially building your own motor, there are a lot of factors at play, possibly more so on a brushed motor. 8k gears to 4k much easier than building your own.
– Abel
Commented Feb 1, 2022 at 12:50
• @Abel Rotational frequency, 8000RPM = 133Hz. Commented Feb 1, 2022 at 12:52
• I think there is some confusion here. A DC motor does not use any frequency (Hz.) Just a DC voltage. Regardless, speed control of any motor is best suited to some sort of speed feedback, whether that be a generator, resolver, encoder, etc. You can get close with a BLDC or AC servo motor, but even those have speed feedback in demanding applications. Commented Feb 1, 2022 at 13:06

Look carefully at the motor specifications to determine if 8000 RPM is the rated speed when delivering rated torque or the no-load speed. Most motors of this type are intended for toy and hobby use and sales information is poor. If 8000 RPM is the rated speed, consider operating at half of the rated voltage. That should provide something close to half the rated speed and 100% of rated torque.

What are the effects of such a replacement? Will the RPM be halved? How will this affect the torque?

Doubling the number of turns doubles the back EMF for a given RPM. Cutting the wire cross section in half along with doubling the turns will quadruple the resistance. Half the current with twice the turns will result in about unchanged torque capability. The increased back EMF will result in about half the speed at the same torque and supply voltage. Half current and quadruple resistance will result in about double the previous voltage drop in the winding but the same power loss. The increased voltage drop will have some effect on the speed, but the back EMF is the predominant factor.

As pointed out by @Spehro Pefhany, doubling the full-torque voltage drop in the resistance with half the full-speed voltage will result in 4X change in percent speed for a given change in torque. Also the stall current is completely determined by the winding resistance since the back EMF is zero. That means stall torque of 1/4 the original. That will likely still be more than rated torque. Also in toy and hobby use, stall torque is often limited by power supply capability.

• This is what I was going to say, if its just a standard DC motor, then decreasing the voltage will decrease the speed. Commented Feb 1, 2022 at 16:02
• I use these motors for model railroading, which has a fixed top voltage of 12V. Usually the locomotives work at almost no-load speed, so this information is most important. As for the rest, I assume that you confirm my assumption that the speed is halved while claiming that the torque remains the same, right? Commented Feb 1, 2022 at 20:48
• Yes, that is what I am saying..
– user80875
Commented Feb 1, 2022 at 23:54

I have calculated that I can double the number of turns if I cut the cross-sectional area of the wire in half. At the same time, the resistance of the winding will quadruple. Consequently, the magnetic moment of a winding (at the same voltage) will halve.

What are the effects of such a replacement? Will the RPM be halved?

Yes.

How will this affect the torque?

Torque constant is the inverse of velocity constant, so it will be doubled. However as the resistance is quadrupled the stall current will be quartered, so the stall torque will be halved.

To get the original stall torque you have to double the voltage, and then the speed will be doubled too. So you end up with the same motor specs (power, torque, rpm) at 24 V as you had at 12 V before rewinding.

• Thank you. That is exactly my thinking. Knowing that power is torque times angular frequency, P = M * omega (omega = RPM /60*2*pi) and assuming that the power of the motor does not change when rewinding, the working torque should be doubled. On the other hand, the stall torque is definitely halved. I was not able to reconcile these two facts which seem contradictory at first sight. Commented Feb 1, 2022 at 20:34

If you were successful in rewinding the motor, I think you'd basically end up with a higher voltage motor given the other fixed parameters.

At 4x the resistance the stall torque will be 1/4 and the open-loop change in speed with torque change will be 4x worse.

So you might as well just run the motor as-is at 1/2 voltage.

You could consider a closed-loop or compensated control. The former requires an RPM sensor, but you could also do compensation by sensing motor current and jacking the voltage up proportionally to compensate for the winding resistance. This might work well for a modest decrease in RPM like 2:1. The constant of proportionality should be chosen to almost null out the winding resistance (effectively inserting a negative resistance in series with the positive winding resistance).

Given the added information in comments, if voltage is unidirectional only you could try a ~6V Zener diode in series with the motor and a 1N5819 diode directly across the motor (both reverse biased in normal operation)

simulate this circuit – Schematic created using CircuitLab

The required current will be doubled, the power dumped in D1 mostly, which should be appropriately rated.

• Well, I use these motors for model railroading, which has a fixed top voltage of 12V. I just want all locomotives to run at about the same speed at maximum voltage. Running at half voltage with a special controller is not an option. Commented Feb 1, 2022 at 20:38

Get a pulse-width modulated (PWM) speed controlLer. This techniques changes the duty cycle of the input voltage to give an "average" voltage that is lower. There is a huge advantage to PWM vs simply decreasing the voltage in that you get better Low-speed control because the motor is getting the higher voltage for short bursts. Typically. PWM frequency is 8 to 20k Hz and on vs off duty cycle is controlled. For very low speed control, you can bring the PWM frequency down to 60Hz or less in larger motors that have lots of inertia that may not start well at low duty cycles. You can find low priced pwm controllers on Amazon or eBay.

Finally, just because someone calls a 12v motor a 12v motors, doesn't mean it can't be run at 8 or 9v at a lower speed just fine. Some motors are rated in rpm per volt.

• I use these motors for model railroading, which has a fixed top voltage of 12V. PWM is already used to control speed via DCC control. I just want all locomotives to run at about the same speed at maximum voltage (PWM = 100% duty cycle). Commented Feb 1, 2022 at 20:39
• @Pygmalion Better DCC controllers have throttle curves to match locos. I have bought many small motors from eBay to re-power my N scale locos. At this smaller scale 9000rpm at 6V is perfect, so I would think 6000-9000rpm at 12V would be fine for HO/OO. However the motor should have carbon brushes for long lifespan. Try searching eBay for eg. "DC 6V-24V 12000RPM Micro FK-130PH" or "FK-130PH-11405 DC6V 3200rpm" or "FK-130SH-09450 DC24V 15000RPM". Commented Feb 1, 2022 at 22:33
• Motors are rated a given RPM under no load. Adding the weight of the loco with its gears will definitely slow the motor. The internal design of the motor (length of armature, number of windings, wire resistance, strength of magnet and resistance of brushes and more all play a role in how fast the motor will spin under load. My recommendation is to install the motor, then measure speed and current draw at the power supply. Then add a low-value power resistor to slow the loco to an acceptable speed. I can help with resistor selection once you measure the current and speed (and target speed). Commented Feb 2, 2022 at 4:40
• @BruceAbbott The problem is that my antique locomotives have very low gear ratios, 1:13 or 1:8.5. So I really need low RPM in order to get good performace of the motor. Running a motor at less than 6V all the time is not very efficient. Commented Feb 2, 2022 at 11:40
• @GTElectronics This is a very inefficient solution as half of the energy will be lost on the resistor as a heat and performance of the motor will be much worse. Plus there is a problem of heat dissipance. Commented Feb 2, 2022 at 11:42