Another sparkies question about generators

Yes, about the same probably ... but .... Be aware using a motor as a generator is only really practical when considering DC machines. The common induction motor is not easily adapted to generation and is almost never done in a home-shop environment.

Bob Swinney

Not true! Look here:

-Carl

Well, no. The power it consumes is the mechanical power it puts out, plus electrical and mechanical losses. If you put the same mechanical power in, using it as a generator, you have losses going the *other* way, too. So the difference in power in versus power out, for a fixed value of mechanical power in and out, is the sum of those two sets of losses -- the losses it has as a motor, plus the losses it has as a generator.

The efficiency of the motor depends on its design and size. A small DC motor used as a generator starts off in the hole, because its efficiency as a

*motor* isn't all that great (maybe 70%). And, at the very best, you have equal losses working the other way. Big, quality motors are much more efficient -- into the 90%+ range.

To make an efficient generator out of a brush-type permanent-magnet DC motor, you may have to shift the phase angle of the brushes slightly. I have here in the mess I call my shop an old DC motor that was modified by someone to allow fine adjustments of the brushes for that express purpose. Unfortunately, it's a series/parallel-wound electromagnet motor, which requires a lot of knowledge to get it properly excited.

My understanding is that small, modern, permanent magnet DC motors run at something close to their motor efficiency when run as generators, but not quite.

Ed Huntress

I knew a woman like that once, only it was cash, not knowledge.

-Carl

It will be good for the same amount of current and therefore the same amount of wattage if the voltage is the same. What are you trying to do and what kind of motor are you thinking of using? Induction motors will work well especially if you use the grid to excite them.

Dan

If you can just figure out where to hook the wires, and how much voltage to use...

Ed Huntress

Well. I was thinking about using a brushless DC motor to act as a generator for boat lights. The ones that have magnets in the center and the rotating field is produced by the drive electronics. If I spin the magnets then I'm hoping to get usable power out of the thing. 100 watts should be plenty. Eric

Those old DC servo motors (like old DASDI drive motors) make good generators.

I thought about using a big stepper I have but it saturates at the rpm that it would be running at. At least I think that's what happens. If all the wires are twisted together and the shaft spun the resistance increases and then at a certain rpm drops dramatically. I suppose if there were less poles in the rotor this wouldn't happen. ERS

Why do you think you can count the same resistance twice? The current only goes through it once, regardless of whether it is being used as a motor or a generator.

Gary

Only if it's 100% efficient. The simplest case to look at is a low speed permanent magnet DC motor where the armature resistance is the major loss component. With an 80% efficient motor, 20% of the input power is lost in armature resistance. Operated at the same speed as a generator, 80% of the original motor input power is available as generated output. However it now has to travel through the armature resistance before it reaches the output terminals and this loses a further 20% of the available power. For the same internal heating, an 80% efficient machine used at the same speed as a generator will deliver about 0.8 x 0.8 = 64% of its rated motor input power.

Jim

Gary sez:

"Why do you think you can count the same resistance twice? The current only goes through it once, regardless of whether it is being used as a motor or a generator."

He is not counting anything twice. Go back and read the post carefully. You can rely on anything that Jim says as gospel.

Bob Swinney

Ah, I read it carefully, and I read it the same way that Gary does. It isn't entirely clear why "Operated at the same speed as a generator, 80% of the original motor input power is available as generated output," unless he's talking about armature resistance losses. There are other losses but that seems to be the subject. Then he counts armature resistance losses again, no?

Ed Huntress

Greetings Bob, Maybe you can tell me what is going on with a stepper I just measured. It is rated at 5 volts, 5 ohms, and 7.5 degrees. When spun by hand the resistance to turning is quite high. Spun faster by hand it of course gets way easier to turn. Spinning the shaft in the lathe, with no load, gets voltage readings from 3 volts @ 330 rpm to 14 volts @ 1800 rpm. Those voltage readings are no load. Using a lamp as the load the current measured @ 1800 rpm is .5 amps. At 550 rpm it is about.3 amps. The lamp is a twelve volt lamp. It doesn't glow at all. When the voltage is measured with the lamp connected it drops to .2 volts. Id s it because there are so many poles on the rotor that it behaves this way? If all the coils were wired in parallel and the rotor replaced with one with just 4 poles would this improve the situation? Thanks, Eric

You are getting AC out of the servo - and poly phase at that. You need some diodes to make them work properly. Tying wires together makes the phases work against each other.

Eric,

Sorry, but any thing I could offer would only be a guess. Steppers aren't my cup of tea but I do know they are classed as synchronous motors. Here's a guess though - your stepper rated 5 volts, 5 ohms; this implies the motor is maybe rated at only 5 watts, if I'm "guessing" correctly.

In this case I'm not ready to grant Jim deity status. Perhaps you'd better go back and read it twice, Bob. Here it is.

Now this is true.

This is at best misleading. There is no "original motor input power" here. The motor isn't being used as a motor. It is being used as a generator. There's *external* mechanical power driving the motor shaft. This could be equal to 50% of the "original motor input power", or it could be 80% of the "original motor input power", or it could be

100% of the "original motor input power", or it could be 200% of the "original motor input power". In other words, it is *totally independent* of whatever power the motor may have drawn when being used as a motor. *Now* he can account for armature losses. If they are 20%, then the generated output will be 20% less than the mechanical input power applied to the generator.

Gary

Check out this site.

Richard W.

I respect both contributors, agree with Gary here. Loss is I^2 R regardless of which way the current is going. Power out is power in minus I^2 R, whether power "in" is electrical or mechanical.

I think the reason the stepper motor wimps out at higher speeds has to do with coil inductance. Reactance grows with frequency which is proportional to speed. The same issue limits how fast they can be driven with given available voltage.

I wonder if a brushless DC motor would commutate without external excitation. Something has to supply the electronics that makes it brushless.

What are you going to spin it with, Eric? Most electric-start engines (including outboards) already have alternators capable of a good deal more than 100 watts.

My local surplus store sometimes has DCPM brush-type 12-volt motors. I tested a couple in my collection today. One has resistance of 0.121 ohms, so at 8.33 amps (100 watts at 12 volts out) the loss would be

8.4 watts and the efficiency would be 100/108.4 or 92%. The price sticker on it says $5.95 but I've had it a while. Ya never know what the Ax-Man will have on any given day, but if you'd like I'd be glad to make the selfless sacrifice (snort!) of visiting the Ax-Man to see what he might have for you. This particular motor weighs 3 lb. I'm near Minneapolis.