How do I find the voltage range for an unknown DC motor?

Question

I have a small unmarked motor with a burned out speed controller board. I'm able to figure out that the motor is a DC motor because it has two wires coming out of it and there's a rectifier on the speed controller.

I want to find out what is the maximum voltage range for the motor, but it is proving difficult. The board appears to have a triac and perhaps a diac or some diode. There's no transformer so I guess it's probably operating at full 120V. There's also a bunch of resistors and capacitors which is probably uses for PWM.

I chucked the motor on my drill press and spun the shaft at 570RPM or 59.69 radians/sec and got 16V output. The motor resistance is 39 ohms (was reading 50 before).

Is there a chart or formula for the voltage vs speed and or voltage vs torque for a DC motor?

FYI, the motor diameter is about 2 inches wide and the height of the motor is about 4 inches. The motor has a plastic worm gear attached to it and was used as a neck massager. My guess is that it needs to operate at high torque.

Here's a photo of the motor and a snap shot of the circuit:

Answer 1

It's mostly about how much power the motor can safely dissipate without getting too hot. A secondary issue is you don't want the motor to over-spin, but usually it's pretty obvious when it gets that far.

Your measurements give us some idea, but it would be additionally helpful to know the physical size of this motor. That allows for the first pass guess as to how many Watts it can dissipate.

At 570 RPM (9.5 Hz) you got 16 V out. Most motors can do at least 3600 RPM (60 Hz), so let's see how that works out. At that speed, the back EMF would be 101 V according to your measurements. If you think it might be intended to run from rectified 120 V AC, then let's see what 170 V does, since that's what you'd get if there is a capacitor after the rectifier. 170 V - 101 V = 69 V left over to drive the motor at 3600 RPM. That would deliver 95 W to the motor, which is a lot unless it's at least maybe 6 inches across.

Let's look at it another way. For the back EMF to be 170 V would require 6000 RPM (100 Hz). That would be the absolute maximum speed. Is that plausible? That's not out of line for a DC motor, not knowing anything else about it. Of course it wouldn't actually ever get that fast because there would be no EMF left to actually drive it, and no torque left to drive anything else.

At 5000 RPM, you'd have 140 V back EMF with 30 V left over to drive the motor at 170 V in, which would take 18 W. That could be quite plausible if the motor is at least fist-sized.

Answer 2

A simple model for a DC motor is \$V= R*i + e\$, where \$V\$ is the terminal voltage, \$R\$ is the motor resistance, and \$e\$ is the back-emf voltage.

R can be measured as I said above in a comment and which I'll repeat here. Resistance can vary on a DC motor due to brush contact. The best way to measure resistance is to take several measurements and average. If possible, lock the rotor and then apply a small current to the terminals. Measure voltage and current and calculate R = V/I. Typically you this test would be done at ~25% rated current. Repeat several times and average. There is a dynamic test, too, that can give even better results - do the same as I just said, but instead of locking the rotor, back drive the motor. 50 RPM would be sufficient speed at which to back drive the rotor.

\$e\$ can be determined from \$e = K_b* \omega\$, where \$K_b\$ is the back-emf constant (units of V/(rad/sec) or V/RPM) and \$\omega\$ is the speed in the same units as \$K_b\$.

You've already found \$K_b\$. It is just \$\frac{16 V}{570 RPM} = 28.07 \frac{V}{kRPM} = 0.268 \frac{V}{rad/sec}\$. As somebody else mentioned, the torque constant of a motor is equivalent to the back-emf constant, so \$K_t = 0.268 \frac{Nm}{A}\$.

At locked rotor, you know that \$V = R*i\$ because \$e=0\$. If you solve for \$i = \frac{V}{R}\$, you can find the current when the rotor is locked at different voltages. And from that current, \$i\$, you can solve for the locked rotor torque at different voltages: \$T_{lr} = K_t*i = K_t*\frac{V}{R}\$.

You can also determine the maximum speed of the motor at different voltages using \$V= R*i + e\$. If you assume \$i=0\$ when there is no load, that equation becomes \$V=e\$, which becomes \$V= K_b* \omega\$, which becomes \$\omega = \frac{V}{K_b}\$.

Once you have the stall torque at different voltages and the maximum speed at different voltages, you can plot them on a graph with speed on one axis and torque on the other axis. Connect the lines and you have various speed-torque curves at different voltages.

There are a lot of assumptions in what I wrote above. The 2 main assumptions you should be aware of are 1) that the motor stays relatively cool (so the resistance doesn't change) and 2) that the no-load current is zero (in reality it won't be).

Answer 3

Depending on how critical it is for the motor to work after testing and what equipment you have, you might slowly increase the voltage into it while keeping an eye on temperature, speed, and current. You may also include a mechanical load of some kind and measure torque, like a dynamometer. The point that you're comfortable with is up to you.

As for the formulae, it depends a lot on the geometry and how the motor is wound internally. It's basically a set of electromagnets that interact with a set of permanent magnets and are mechanically switched at the appropriate times to keep it running in the same direction. For a given strength, you can make a high-current, low-voltage electromagnet or a low-current, high-voltage electromagnet. And that's only one of many parameters. I think you're better off finding the spec sheet or doing your own experiments.

Answer 4

Assuming that it is a permanent magnet DC motor (and not a synchronous, induction, or 'universal' motor) your generator test indicates that the Kv (velocity constant) is 3.73rad/s/V or 36rpm/Volt. Therefore on 120V it should do about 4300rpm.

For a PMDC motor, Kt (torque constant) is the inverse of Kv. 1/3.73 = 0.268N-m/A or 38oz-in/A.

With a resistance of 50Ω the stall current would be 120/50 = 2.4A, so stall torque should be about 0.268*2.4 = 0.643N-m or 91oz-in. That is a lot for a 'small' motor, so I suspect that one of your measurements is out by a factor of 10. Are you sure that it was 50Ω and not 500Ω?

Here is a typical small permanent magnet motor designed for 120VDC:-

DS-5512-120-6000

Rated voltage:   120VDC
No load speed:   6000±10% rpm
No load current: ≤60mA
Rated speed:     4800±10% rpm
Rated current:   ≤100mA
Rated torque:    120g.cm
Output power:    5.9W
Stall current:   ≥260mA
Stall torque:    ≥600g.cm
Weight:          200g