2
\$\begingroup\$

I am using a stepper motor as a generator in a proof of concept experiment for something. The motor came from an old printer so do not have specs for it, but I believe it's unipolar. It has 5 wires, one of which appears to be common to the other 4. That is, there is 80 ohms from this common to the other 4, and 160 between any other 2 leads. I drove it with a DC motor to view the outouts of all channels, shown below. Waveform is a bit jagged I think due to a slight vibration in the coupling

4 channel output

I have previously used a DC brushed motor and want to compare outputs, DC vs AC. My first question is, are the following calculations/conclusions correct?

So, here is the outout of the actual experiment:

output into 80 ohm load

I have each of the 4 coils driving an 80 ohm load (a resistor) and get this outout (just showing 2 channels) , giving a 10v peak to peak sinewave. To convert this to the equivalent DC I calculate RMS, which for a sinewave is Vpeak * 0.707 (or 1 over squareroot 2) which works out at 3.5V and power = (3.5V*3.5V)/80R = 0.15W

So, is my 10V peak to peak sinewave generating the same power as a 3.5V DC supply would, for example will it heat up the resistor the same amount?

Second question is, as I have 4 coils, can I say the motor is generating 0.6 W? Without going into details, I assume it is possible to combine the 4 out of phase outputs into one?

\$\endgroup\$
1

1 Answer 1

1
\$\begingroup\$

Answering your first question: Yes, it will. Your calculation is correct, but you always have to make measurements under some load (which you did). Your generator will always have some internal resistance/impedance and you have to take it into account.

Answering your second question: Yes. Easiest way to combine power would be to create four phase diode rectifier. If you work with AC you always have to remember about phase alignment (treat every winding as AC source with some phase shift).

\$\endgroup\$
2
  • \$\begingroup\$ Thank you! Yes, I used an 80R load on each winding, I recall matching internal and load gives max power transfer? As an aside, my 0.15W, is that technically W RMS? I have never seen that written but since the voltage is RMS would the power not be too? \$\endgroup\$
    – KevInSol
    Commented Feb 24, 2022 at 7:40
  • \$\begingroup\$ @KevInSol Yes, you are 100% correct about load-source resistance matching. Regarding power, it is basically always treated as RMS unless stated otherwise. You can have instantaneous power as a product of: W(t) = U(t)*I(t), but I don't see it used very often in calculations. \$\endgroup\$
    – Felix
    Commented Feb 25, 2022 at 8:21

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.