Electrical Engineering Stack Exchange is a question and answer site for electronics and electrical engineering professionals, students, and enthusiasts. It's 100% free.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

With DC motors, I believe the graph below depicts the typical performance:

enter image description here

Does that mean, even for a high speed motor with a gearbox for speed reduction, the same characteristics apply? That is, even if the motor can do 16,000RPM at no-load, a speed reduction of 50% is where the most power is output? Am I interpreting the graph wrong?

How is efficiency determined? The graph seems to suggest that the maximum efficiency is achieved not at where the max. RPM is produced, but at ~90% RPM.

Any help appreciated.

share|improve this question
Forget about the gearbox. The motor just sees it as part of the load. – transistor Mar 13 at 14:00
up vote 7 down vote accepted

Yes, that's right. The maximum power output is acheivable at 50% speed, for a moment. Run it there for too long and you'll melt the motor.

Motors have two principle loss mechanisms

a) windage (braking torque due to stirring the air around)
b) \$I^2R\$ losses in the conductors.

There are others, static friction, eddy currents, but windage and I2R are dominant

As the output torque is proportional to current, but the I2R to current squared, running with a high current is inefficient. This is why running at half speed with half the stall current will overheat your motor that has only been designed to dissipate the losses of running at maximum efficiency.

The simplest way to see why efficiency peaks close to the unloaded speed is to plot the losses, and total losses, along with the power output on the same graph as you have posted.

I2R losses are maximum at stall, and fall as inverse square as the current drops.

Windage losses are zero at stall, and rise as the cube of speed.

Plot the sum of those, as a ratio to the power output.

A more conceptual way is to reason thus:
a) When the motor is rotating, it is generating a reverse emf
b) the input voltage must be greater than this to make a current flow
c) only the input voltage that matches the back emf actually produces work, the excess input voltage is wasted as heat in the windings
d) this means that the maximum efficiency, from back emf consideration alone, is 0% at stall, rising linearly to 100% at synchronous speed.

So operating at 50% speed, the maximum possible efficiency will be limited to 50%, and will be less due to other losses.

Unfortunately, we can't operate a synchronous speed, no current will flow, which means zero power output. No-load speed is slightly lower than that, where all the output power goes into windage losses, and a small current flows. As we drop further below synchronous speed, the torque goes up linearly with the current, and the extra I2R losses rise as the square.

share|improve this answer
Why don't you delete the bit about synchronous speed? It makes no sense in the context of DC motors. Eddy-current losses are also absent with DC. You might mention brush losses, but they are generally lumped together with armature I squared R losses. – Charles Cowie Mar 13 at 23:28
The concept of synchronous speed is valid, the speed where back emf == input emf, so no current flows, but it's a dumb name, I'll think of a better one. If you've never noticed that the wound bit of a motor is a stack of laminations, so the armature of a PM motor, or the stator of a stepper, then I can understand your eddy current error. A motor might be DC on the battery side of the commutator, but sure as heck is AC on the winding side. The concepts are obvious to me, but didn't think long enough before dashing down a quick answer. I will be back. – Neil_UK Mar 14 at 6:42
The speed when the load torque is zero is simply the no-load speed or perhaps the idle speed. The core losses of a DC motor are difficult to determine and usually lumped together with the rotational losses or considered as part of the stray load losses. – Charles Cowie Mar 15 at 13:23
Exactly, so no load speed is not the speed at which the current drawn from the supply is zero, for when generated EMF = supplied voltage. It is the speed for which there is no output torque, for when the torque generated by the small no-load current matches the torque absorbed in (mostly) windage. So I still need to think of a better name for the synchronous speed, perhaps the 'no current' speed. Of course a motor will never run at that speed except in a powered motor test jig, or a regenerating electric vehicle. – Neil_UK Mar 15 at 13:30
I believe that when the curves are extended through that point, they are extended considerably further and that point is not labelled. – Charles Cowie Mar 15 at 13:34

Here is a representation that more accurately shows the capability of a "typical" PMDC motor. Continuous operation is limited to torque at or below the new 100% line and above 80% speed. Operation at zero speed (stall) would be limited to perhaps a second or two.

enter image description here

share|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.