# How do I modify a DC electric generator to get the same power at a lower speed?

This will be a theoretical question.

Suppose that I have a DC generator, and I'm getting 300VA power from it when it rotates at 2000rpm.

I want to make it run at a lower speed of about 700rpm, and I still want to get 300VA power from it.

What mechanical modifications should I do on it? I mostly prefer only modifying its rotor and doing no modifications on the stator side if possible.

Does increasing the number of turns in its rotor windings do any good? If yes, at what ratio should I increase the number of turns? Does increasing the number of turns by $\frac{2000}{700}$ do the trick?

Assumed that the load is a fixed 100$\Omega$ resistor.
Permanent magnet used as for stator.

• Wouldnt it be easier to set up a small transmission with with two cogs, or a belt? – posipiet Feb 23 '12 at 17:38
• @posipiet Yes it could be so practically, but this is only a theoretical question. – hkBattousai Feb 23 '12 at 17:51
• Please give more information on your generator, as well as the load being used (resistor? motor?). Clearly you have a wound-rotor system. Is it a permanent-magnet stator system or does the stator have windings also? Are you tapping off power from slip ring brushes connected to the rotor windings? In my opinion, the effort even to answer your question correctly is not worth it; it would be cheaper to handle the problem mechanically as @posipiet says. – Jason S Feb 23 '12 at 19:13
• @JasonS I have edited my message, details you asked are at the end of it. And yes, I will get the power from the slip ring brushes. – hkBattousai Feb 23 '12 at 19:25
• what do you mean "permanent magnet not used" ??? – Jason S Feb 23 '12 at 20:32

Does increasing the number of turns by $\frac{2000}{700}$ do the trick?

Almost. That should change the back-emf by a factor of K = 20/7, to compensate for the change in speed. The problem is that even if you manage to rewind the motor effectively, the electric machine's resistance and inductance will increase by a factor of K2 = 8.16 -- the I2R losses in the generator will increase by a factor of 8 at the same load current. And that's if you manage to rewind the motor effectively. If you can't reach a good fill factor on the rotor, the resistance will be even higher. You'll need proper equipment for this; I wouldn't try it by hand.

It's a good rule of thumb that the I2R losses in permanent magnet motors (whether synchronous or brush DC) are lowest at a given mechanical power level when the motors are running at higher speeds. Running them at lower speeds makes the torque requirements go up, and the I2R loss increases by this factor squared.

So if you can compensate for the I2R loss by making the winding ratio even higher, e.g. 21/7 or 22/7 (which causes even more I2R loss), and you don't overheat the generator, you'll meet your output power target.

This is why gears & belts are often used with electric motors at low speeds, rather than using direct-drive.

The alternative approach is to make an electric machine with more poles: higher pole count = higher electrical frequency, which brings the electric machine's operating point closer to its area of highest efficiency. But that's more involved than just rewinding the motor.

It is possible to get the same power at a lower speed without any changes to the electric generator itself.

More often than not, we connect the output wires of an electrical generator to the input of a voltage regulator, and the output of the voltage regulator to our load.

If you are lucky, a change from 2000rpm to 700rpm won't even be noticed by the load. The output voltage of the generator will, of course, drop less than half the earlier output voltage of the generator. But the voltage regulator will (if you are lucky) compensate for that and drive the same constant output voltage to the load that it always has. (Since the output voltage of the generator will be lower, an efficient voltage regulator will pull more current from the generator, in order to supply the same power to the load that it always has).

Alas, there are many ways to not be lucky:

• some poorly-designed voltage regulators won't be able to handle this lower input voltage at these power levels, and the internal power transistors will overheat and be permanently destroyed.
• some well-designed regulators are not designed for this specific voltage range, and simply shut down to prevent permanent damage.
• The internal winding resistance of some generators is so high that it's simply not possible to pull the desired amount of power out at this lower speed.
• The internal winding resistance of some generators is so high that the system may seem to work sometimes at this lower speed, but other times get stuck in an undesirable current-limited latchup at this lower speed.
• Some prime movers may not be able to rotate the generator shaft at the increased torque the generator requires at this lower speed. That would make the system spin even slower, possibly ending up with a undesirable very-low-speed torque-limited latchup very similar to the undesirable current-limited latchup mentioned before, or perhaps simply stopping.
• Since generators are less efficient at lower speeds, they will get hotter when producing the same output power at a lower speed. Some generators are not rated for these conditions -- too-high current, leading to overheating -- and will eventually be permanently damaged. (The lower self-fanning air-cooling at lower speeds makes this even worse).
• Since generators are somewhat less efficient at lower speeds, your prime mover will have to supply somewhat more power to keep the system going. Some prime movers won't be able to supply that somewhat higher amount of power.

It should be possible to find out the ratings for the prime mover, the generator, the regulator, and the load -- perhaps by reading s. Then you can calculate ahead of time whether you are lucky, and if not, exactly what thing(s) need to be replaced or modified to get the system to work.