1
$\begingroup$

I'd noticed that for the small motors, the ones used in toy cars, there are rpm values and voltage values written on the specifications.

For example, I have a motor at a range of (12V - 24V) and rpm of (4600 - 9300) respectively.

Does this mean that I need to provide 12V to the toy car so that the wheels will rotate at 4600 rpm and 24V to rotate the wheels at 9300 rpm?

I would also like to know does the other way work? I mean if I were to use the motor as a generator.

Does this mean that if I were to provide 4600 rpm, I can generate 12V of voltage?

Any response is appreciated.

Thank you

Improve this question
$\endgroup$
1
  • $\begingroup$ I think you are confusing DC motors with AC motors... $\endgroup$
    – Jonathan
    Nov 27 '17 at 8:07
4
$\begingroup$

It means that if you were to apply 12V to the motor terminals, and the motor was not attached to the car in any way (i.e. no resistance to the shaft turning at all), then it would spin at approximately 4600 ±tolerance (and likewise ≈ 9300rpm @ 24v). As soon as you add in any resistance, then the speed will drop as the torque required to keep the motor moving increases.

Typical PMDC Motor Performance

Furthermore, it's likely that there's some sort of gear-train between the motor output shaft and the wheels, which will reduce the speed of the wheels compared to the numbers printed on the motor even further.

You won't get 12V back for 4600rpm input, due to the efficiency of the motor. Note also that peak efficiency is not measured at 'free speed'. Neglecting that for an example, however, and assuming the motor is 80% efficient in both directions, this means 4600rpm corresponds to 9.6V of ideal input, and converting back, 4600rpm could only create ~7.7V.

Improve this answer
$\endgroup$
2
  • 1
    $\begingroup$ also note just as load affects rpm in "motor mode", whatever electrical load you present will affect the voltage when running as a generator. $\endgroup$
    – agentp
    Nov 27 '17 at 20:55
  • $\begingroup$ @Jonathan R Swift Thank you for your response. $\endgroup$
    – Eric
    Apr 25 '18 at 9:00

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.