so i have a power window motor with spec the voltage 12V, continues torque 3 N.m, max load current 12A, and 90 RPM. the resistance is 2.5 ohm i understand that torque is correlated with current and the speed with voltage. since the ideal equation for motor is


and if have load so the torque is 1 N.m and i apply 12V, the motor will run with constant 90 RPM. from the equation(suppose Vemf is 2) i get the current


but if i add the load so the torque is 3 N.m and my voltage still 12V, my motor still rotate at 90 RPM but with higher current. i'm lost in this point, i assume my current will be 9A according to the spec, but from the equation it doen't make sense

9 x 2.5+2=24.5V ??

the voltage is changed to 24.5V, but my voltage supply still 12 V

how does this happen?

thank you

  • \$\begingroup\$ At higher load, either the motor will slow down, or you will increase the driving voltage to maintain the same RPM. You have solved the equations involving Kv and Kt for the new voltage at constant speed. If voltage is constant, solve for the new speed instead. \$\endgroup\$
    – user16324
    Commented Aug 17, 2017 at 11:49
  • \$\begingroup\$ A motor running at constant speed and with zero friction would draw zero current. Motor developed torque is proportional to current, and if the motor is not accelerating the only torque (hence current) needed is to overcome friction. \$\endgroup\$
    – Chu
    Commented Aug 17, 2017 at 11:56
  • \$\begingroup\$ Vemf is proportional to RPM, thus available torque reduces with RPM since available I is reduced . Torque load is never constant with RPM unless ideal friction load. Normally loaded motor may produce max power at ~ 75~80% no load RPM like MPT. \$\endgroup\$ Commented Aug 17, 2017 at 12:00

2 Answers 2


DC motor specifications are actually a mixed bag of numbers which do not correlate to each other directly quite the way you think and the boilerplate numbers can be very confusing.

In your case it states continues torque 3 N.m. and a speed of 90RPM. Whether that implies it will run at 90RPM with a 3Nm load is open to interpretation. 90RPM could equally well be a no shaft load speed at 12V applied voltage. Torque and current at that speed would be negligible.

12A is the max current you are allowed to drive it with without burning out the coil wire.

Continues torque 3 N.m is really more of an indication of the power rating of the motor. That is it should not overheat when run continuously at that current.

When designing with motors it is prudent to ignore most of those marketing numbers and look at the published rating curves instead. Those will, once you understand how to read them, tell you far more about how the motor will perform and how to drive it.


I assume you think a DC motor that has

  • ideal non-saturating (=linear) magnetic path
  • perfect commutation
  • permanent magnets
  • some resistance R

Let's rewrite your idealized equation without divisions which are not handy without math typesetting:

U = I * R + A * W

where U=the input voltage, I =the current drawn by the motor, R = total circuit resistance, A = the structure and material constant of the motor, W = the rotation speed as radians/second.

Note: A * W is the induced voltage.

Elementary electrodynamics give to us another equation between the torque T and the current:

T = A * I

This torque is needed for 2 purposes:

  • to accelerate the rotation speed of the motor and possible external mass joined to the axis
  • to win the frictional losses inside the motor and in the load

If there's no frictional losses, the rotation speed increases until the induced voltage is as high as the input voltage. The motor draws no current and the rotation speed is constant (= U/A). R means now nothing because I = 0.

If you have some friction, the behaviour of the system depends greatly on how the friction is related with the rotation speed W.

If the friction is a constant torque Tf that does not depend on the speed, the motor takes a current I = Tf/A. The achieved speed is less than U/A, it's (U - I * R)/A

In practice the friction torque Tf is at first highest in the startup, then drops when the rotation has started and finally increases when the rotation speed increases. This makes it hard to calculate exactly the achieved speed and the current. But it's surely possible if the relations are known.

Practical motors have some internal tricks that try to reduce the speed loss caused by the friction and resistance. One elementary trick is to add an extra magnetization coil and put the motor current go through it. This makes A non-constant but designing it carefully the drop of the rotation speed can be kept substantially smaller at moderate torques.

It's well possible that the speed loss caused by external 3 Nm torque is so small that you have not noticed it. This is possible even if there's no tricks applied to reduce the loss of the speed. Unfortunately I have no proper motor specs to check it.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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