I have a dc motor with the following specifications:

Operating Temperature: -10°C ~ +60°C
Rated Voltage: 6.0VDC
Rated Load: 10 g*cm
No-load Current: 70 mA max
No-load Speed: 9100 ±1800 rpm
Loaded Current: 250 mA max
Loaded Speed: 4500 ±1500 rpm
Starting Torque: 20 g*cm
Starting Voltage: 2.0
Stall Current: 500mA max
Body Size: 27.5mm x 20mm x 15mm
Shaft Size: 8mm x 2mm diameter
Weight: 17.5 grams

Now, suppose I wanted to use it as a generator. How fast (in RPM's) would this motor need to rotate in order to obtain the rated voltage? What current can I expect at this speed?


A typical motor will behave as an ideal motor in series with a certain amount of inductance and resistance. At all times, the rotational speed of a motor will be a certain multiple of the applied voltage, and the torque will be a multiple of the current flowing through it. These relationships are bidirectional, so whether something acts like a motor or a generator will depend upon whether relative directions of the applied voltage and the current flow.

Based on the above description, I would estimate that your the voltage on the ideal motor will be roughly 1 volt per 1500rpm (look at no-load speed divided by rated voltage). I'm not sure whether the stall current is measured at 2 volts or 6 volts, but the short-circuit current from a motor spinning at a given speed will be roughly equal to the stall current of a motor driven at the open-circuit voltage that would correspond with that speed. To get maximum power output, you should draw less than half of the short-circuit current,.

  • \$\begingroup\$ Typo in 1st sentence dude. Motor should read generator. +1 for common sense answer. \$\endgroup\$ – Andy aka Apr 19 '14 at 22:46
  • 1
    \$\begingroup\$ @Andyaka: Not actually a typo, but I added a little more to clarify. \$\endgroup\$ – supercat Apr 19 '14 at 22:49
  • \$\begingroup\$ ok I gotcha now \$\endgroup\$ – Andy aka Apr 19 '14 at 22:54

I like how it says

No-load Speed: 9100 ±1800 rpm
Loaded Speed: 4500 ±1500 rpm

The dependencies on external factors create a 20-30% fluctuation. I would say that this isn't the best machine for use as a generator. I would guess that the excitation of this motor is a permanent magnet, so you can take it as constant.


simulate this circuit – Schematic created using CircuitLab

The equations you have to look at is $$U=R_aI_a+k\Phi n=R_aI_a+E \\E=k\Phi n \\k\Phi=const.$$ The thing is that you want to get to the second quadrant. You simply have to get to the point where the induced voltage is larger than the supply voltage so: $$E>U\\ k\Phi n>U\\ n>\frac{U}{k\Phi}$$

The problem in your case is that you don't have the parameters \$ R_a\$ and \$k\Phi\$, but they can be calculated. For $$U=RI_n+k\Phi n_n\\ U=RI_0+k\Phi n_0$$ and $$I_n=250 \mathrm {mA}, I_0=70 \mathrm {mA},n_n=4500-1000 \mathrm{rpm}, n_0=9100-1000\mathrm{rpm}$$ Well the 20-30% fluctuation is a problem, I guess it has mostly to do with temperature from the current increasing the \$R_a\$. But It can have mechanical reasons too, if you have the datasheet you can do better guesses. I reduced the speeds by 1000, but don't take my word on that.

The thing that you need most is the machine constant \$k\Phi\$, because it creates the relationship between \$E\$ and \$n\$.

I suggest you start with a lower \$U\$ of maybe 3V or something and look how it behaves. You must be aware that your motor can burn down if you have too much induced voltage or current! Start small!

  • \$\begingroup\$ What is U in this equation? \$\endgroup\$ – Paul May 4 '14 at 14:18
  • \$\begingroup\$ \$U\$ us the voltage. \$\endgroup\$ – WalyKu May 4 '14 at 14:31

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