# How to Derive constant of total back/counter EMF for DC motors?

1. May 23, 2013

### Oz Alikhan

Hi,

I have been reading around about back EMF and their derivations for simple DC motors. However for some reason, the step between obtaining the total emf of the motor from summing the individual emf of the coils is not very clear. For example:

Induced emf due to single coil: $e = d∅_c/dt$

Since the flux linking a coil is: $∅_c = ∅ Sinωt$

Therefore the induced emf is: $e = ω∅ Cosωt$

Since there are several coils all around the rotor, each one has a different emf due to its position (i.e each one has a different flux change through it). Therefore, total emf is the sum of the individual emfs.

This means, $E_b = K ∅ ω_m$ <<< How?

First of all, I don't understand how the $Cosωt$ disappears to obtain the final expression. Secondly, what parameters define the constant K? In some place I read, $K = 2 N/\pi$, while in other places it stated $K = 2 N R B L$. Why is it different for each case and how is it derived in the above proof?

The above derivation is from end of page 5/start of page 6 of the following source: http://vlab.ee.nus.edu.sg/~bmchen/courses/EG1108_DCmotors.pdf.

Warm Regards,
Oz

Last edited: May 23, 2013
2. May 24, 2013

### Staff: Mentor

I guess the idea is to have sufficient windings so that no sooner are we past the peak of one sinusoid, then the commutator turns to deliver the upcoming peak of the next coil's sinusoid. So the output becomes a series of peaks, not perfectly flat.

I'll leave your more searching question for someone better placed. [Broken]

Last edited by a moderator: May 6, 2017
3. May 24, 2013

### jim hardy

Author skipped over some clever algebra.
I think you'd sum values of a series of sine functions, one for each turn in the armature winding.

Commutation makes it disappear.
The brushes pick off a segment of each winding's cycle near the sinewave peak, so voltage at the brushes is not sinusoidal but unipolar..

Nine minutes into this excellent old Army film is a graphical representation of that commutation.

Six minutes in shows from whence comes that velocity term.

I wouldn't attempt at this late hour to derive that formula for the commutated sine wave. I guess that's why your author skipped it.
I don't see how he could call equation 5 the result of integrating equation 4. Book editor shoulda called him on that shortcut, i'd say.

In my class we determined product term K$\Phi$ empirically by no-load dynamometer tests on a machine.

Last edited by a moderator: Sep 25, 2014
4. May 24, 2013

### Oz Alikhan

Hmmm, I see. Very well, that clears a lot; Thanks to both of you .

P.S. Old school videos are indeed the best