I'm trying to determine at what speed my brushed dc motor will run at, given a certain power dissipation. The background is I've built a lathe, using this motor, and for a given depth of cut and feed rate (which is dependent on motor speed and how fast I turn the z axis handle), there is a certain power requirement to remove that much material. Calculating the power is easy, if I make an assumption about the motor rpm, but if I'm running the motor at 12v, I don't have direct control over the speed, just the range of speeds, correct?

So, if I set the voltage, then attempt to stall the motor (huge depth of cut or very high feed rate), I will draw max current, and have zero speed, or I can take no cut, and have essentially the no load speed. For a given cut though, I can find the power, but there are two points along the torque speed curve (other than max power) that give that power, so how do I know which point I'm at? The high torque low speed point, or the high speed low torque point?


  • \$\begingroup\$ Please elaborate on this statement: "...if I'm running the motor at 12v, I don't have direct control over the speed, just the range of speeds, correct?" Specifically, what you mean by "range of speeds". \$\endgroup\$
    – FiddyOhm
    Aug 20, 2016 at 19:57
  • \$\begingroup\$ At 12v, there is a specific torque-speed curve, right? or 24v, or 9v, or any voltage. So I could run 12v at w = no load speed, or 12v at w = o at stall torque, or anywhere in between, but it's dependent on the torque from my load. \$\endgroup\$ Aug 20, 2016 at 20:32
  • \$\begingroup\$ Do you need an encoder to tell you what your shaft speed is? \$\endgroup\$ Aug 20, 2016 at 21:59
  • \$\begingroup\$ The encoder would make all of this very easy, but the intent is to determine the speed and torque from analysis. \$\endgroup\$ Aug 20, 2016 at 22:00
  • \$\begingroup\$ This is an interesting question. I assume you are talking about a brushed motor. If you know the instantaneous power as a function of time, you should be able to get the frequency from a Fourier transform of the power. I expect however that you are talking about the average power. If that is the case, then no, you can't get the torque and the speed, you need more information. You might be able to get it from the power factor: essentially the phase between the current and the voltage. \$\endgroup\$ Aug 21, 2016 at 2:43

4 Answers 4


Looking at the datasheet, while it doesn't lie, it is definitely on the edge of misleading.

Note that the "max power" stated occurs at almost exactly half the unloaded RPM and half the stall current. This is indeed the "max power" point for such a motor, but the datasheet fails to mention that it is also the nominal 50% efficiency point, thus dissipating 12V*68A-337W = 479W in that tiny motor - destroying it, probably in minutes.

(Ideally, exactly half the power would be delivered, about 400W shaft and 400W heat, but the motor isn't ideal).

The motor is probably suitable for 100-150W continuous output and 200-250W short term.

So practically you must operate the motor at the upper end of the speed range, and if the speed falls below (say) 70% of the unloaded speed (or the current rises to 30% of the stall current) then - unless this is strictly temporary, like starting a heavy load or hitting a chilled spot while machining a cast iron surface, you need to cut the current and protect the motor.

Then the question of which side of the torque speed curve doesn't apply - unless the protection has tripped, you should be on the high speed side.

You can get circuit breakers that will allow short-term overcurrent. These are "motor rated" or Class C breakers for the AC motors used in most machine tools. I don't know of anything suitable for 12V DC though. I'd be looking for a 12V DC supply that can be set to trip if its output exceeds 40A for more than a couple of seconds. And as Olin says, if you want to monitor it yourself, measuring the current is definitely the way to go.


Can you sense and filter the voltage spikes or current switching on the motor to trigger a One shot and then make a tach speed speed output for servo control.

Or limit the current pulses , LPF and input to a typeII PLL and monitor the VCO control,voltage for tach speed?

1st option is better if you can detect back EMF with a clean pulse conditioner.

  • \$\begingroup\$ I can't perform any measurements on the lathe anymore because I had to give it to my professor. The general question is, from a given power, and a voltage, what will the torque and speed of a motor be? Will it be the high torque, low speed solution, or the high speed high torque solution? \$\endgroup\$ Aug 21, 2016 at 1:18
  • 1
    \$\begingroup\$ It will be the speed closest to the initial condition when the load is applied. Either zero speed or load applied at full speed. \$\endgroup\$ Aug 21, 2016 at 2:21
  • \$\begingroup\$ This comment seems to be the answer that the OP was looking for. All of the other answers and comments missed the mark because of difficulty understanding what was really the question. \$\endgroup\$
    – user80875
    Aug 21, 2016 at 3:00

Probably the easiest way to tell the motor operating point is to put a ammeter in series with it. You can measure the stall current and no-load currents up front. Any actual operating current should fall within that range.

Since you apparently have a fixed voltage source (12 V in your case), the power is directly proportional to the current. In fact, the power in Watts is the current in Amps times the voltage in Volts. Note, however, that this is the power going into the motor. The mechanical power coming out will be different. One obvious way to see this is to consider the input and output powers when the motor is stalled. The electrical input power will be maximum, but the mechanical output power will be zero. In that case, all the power goes into heating the motor, which the motor may not be rated to take indefinitely.

  • \$\begingroup\$ Ah, I should have clarified, I am measuring the mechanical power output, but I have no access to the device anymore, so I can't physically measure anything at the wall. From the mechanical output, defined as T*w, can I determine, with no other measurements except my data sheet and the knowledge it's running at 12V, what those two values (T,w) will be? Thanks for the help. \$\endgroup\$ Aug 21, 2016 at 1:23

For a given cut though, I can find the power, but there are two points along the torque speed curve (other than max power) that give that power, so how do I know which point I'm at? The high torque low speed point, or the high speed low torque point?

You can tell by the smoke :)

At the high rpm low torque point the motor should be working relatively efficiently (delivering more power than it dissipates) but at the low rpm high torque point the opposite occurs. Most motors are not designed to dissipate a lot more power than they output, so running them in this region will cause them to overheat.

A permanent magnet or shunt wound DC motor has a linear torque vs current and inverse rpm vs torque response to varying load, delivering maximum power output at around 50% of no-load rpm.

enter image description here

Your motor has the following specs:-

Voltage: 12 volt DC
No load RPM: 5,310 (+/- 10%)
Free Current: 2.7 amps
Maximum Power: 337 Watts (at 2655 rpm, 172 oz-in, and 68 amps)
Stall Torque: 2.42 N-m, or 343.4 oz-in
Stall Current: 133 amps

At maximum power output it is doing exactly 50% rpm, producing 337W mechanical output from 816W electrical input for an efficiency of 41%. At this operating point it has to dissipate 479W, which is probably already above its continuous capacity. Any higher load will rapidly increase power dissipation, up to 1596W at stall! You do not want to operate the motor in this region.


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