4
\$\begingroup\$

Is there any way to calculate the expected voltage and current, assuming I am turning it at the specified RPM value, when using a DC motor as a generator without physically testing it?

In my case I wanted to use a DC motor with the following parameters:

DC 6V 6300 RPM Torque: 0.1Kg.cm

\$\endgroup\$
9
\$\begingroup\$

Yes, if the reverse EMF of the motor and its DC resistance are specified. Good datasheets do provide these figures.

At first approximation, a generator looks like a voltage source proportional to speed in series with the DC resistance of the windings. Put another way, you can usefully model a motor as a Thevenin source. Good datasheets tell you the voltage as a function of speed, or "reverse EMF" as a function of speed for a motor (same thing). That's the Thevenin voltage. The resistance, again from the datasheet, is the Thevenin resistance. You can then compute voltage and current as a function of electrical load from there.

Sometimes these parameters can be inferred from other specs. For example, consider a motor that puts out 100 W of mechanical power at 60 Hz rotation (3600 RPM) speed, at 24 V applied, and is 90% efficient. The electrical input power is (100 W)/90% = 111 W. That means the current is (111 W)/(24 V) = 4.6 A. The winding resistance is then (11 W)/(4.6 A)² = 518 mΩ, which is dropping (11 W)/(4.6 A) = 2.4 V. That means the back EMF at 60 Hz is 21.6 V, or 360 mV/Hz.

If you were to drive this motor mechanically at 40 Hz, it would look like a 14.4 V source with 518 mΩ in series. If you were to put a 1 Ω load on it while still keeping the rotation rate at 40 Hz, you'd get 9.5 V and 90 W.

\$\endgroup\$
  • \$\begingroup\$ Thanks for the answer. But is it possible to produce the rated voltage of the motor by turning it at specified RPM or do you need to increase the RPM? \$\endgroup\$ – 137 Feb 2 '15 at 21:55
  • 1
    \$\begingroup\$ A motor run as a generator will never put out its rated voltage at rated speed. Even with no load, you get the pure back EMF, which is always less than the applied voltage when the shaft isn't being driven externally. \$\endgroup\$ – Olin Lathrop Feb 2 '15 at 21:59

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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