CIM generator

Technical Motors
#1

I built a gas generator using 2 cim motors driven off the shaft of a Small 29cc 4 stroke gas trimmer engine. I am able to get a fairly decent, stable 18v out of it running in parrallel. Or at a lower rpm i can easily get 18v in series.
Just wondering. I want to run 2 dewalt 18v DC925 motor/transmission combos off of this. Just wondering if the cims will produce enough amps to power both.

I have a separate regulator to make the 12vs for the rest of the robots electircal system.

Does anyone see any problems with this? Will the cim bearings/coils stand up to this?

And does anyone know what kind of amperage i can expect out of this setup.

Thanks in advance.

#2

the only problem I see is hooking generators in parallel. They will not be perfectly balanced, so one of them will motor the other to some extend.

Hooking them in series is not ideal either, because what you get from a generator is not really pure DC. Its more like a rectified AC waveform. This is due to the commutators/brushes, and the many windings. It would be very difficult to attach two motors to one drive shaft such that the windings and commutators are perfectly synchronized. If you have a two channel scope, and you hook each generator to separate channels you will be able to see the two rectified AC waveforms, and most likely you will see they are out of phase (out of synch).

But that can be fix to some extend by putting caps across each generator - even better if you can put power diodes infront of the caps so they dont back feed the windings.

by now you are saying YIKES! this is more complicated than I thought. Best answer is to use one generator.

The other part of your question: how much power can they put out? Quick answer is to try it and see. Voltage is only half the power equation. If you can load the generator up with power resistors or a bunch of 12 light bulbs (like headlights) you will get a quick indication of when the generator starts to droop its voltage, and how hot its going to get after pumping out real power for a while.

DC motors tend to be in the 90% effeciency range, so a motor that is rated at 1/4 HP (180W) will require 200W to run, and you might get 162W out when you use it as a generator.

#3

I am getting about 90% efficiency out of these motors when hooking one to another. I would assume that means 95% efficiency per motor.
when you say put caps across each generator are you talking about a capacitor so that it equalizes the voltage? If i use one generator i dont think i will get enough amperage.
Does anyone know the comparison of amperage use between the 12v dewalt and the cims? (200w version) from 2005 i believe.

And the diode thing i dont quite understand. I know exactly how a diode works. i understand what you mean by ac waveform, But is the diode just so the capacitor doesnt try to run the motor?

and where would i find a diode of that size?

Thanks

edit: im going to try to get up a picture of my current setup. As of right now its setup with only one motor to the engine, its direct driven off the shaft. Right now its all mounted to a board of wood for testing before i manufacture aluminum mounts(altho its all mounted using custom aluminum mounts.)

#4

yes, since two generators would be putting out rectified AC waveforms ( like upside down U’s UUUUUUUU ), the charged cap will try to backfeed the generator when it is at the low point of its AC waveform

the diodes would let the generator pump the cap up to 12 or 18 volts, but the cap would not be able to send current back into the generator as the brushes are switching from one set of windings to the next on the commutator.

to find diodes, try the search function on www.digikey.com

like this 30A 50V rated power diode for example: http://www.irf.com/product-info/datasheets/data/30ctq050pbf.pdf

#5

Dream on, Ken.

Larger machines, rated for industrial duty at, say, 20 HP and higher typically achieve 90% or better efficiency.

Fractional horsepower, intermittent-duty purpose built motors like the CIM (link is to the datasheet provided by the manufacturer) typically run at efficiencies around 65%. For example, operating as a motor, the CIM’s rated output at maximum efficiency is 154 Watts at 4614 RPM, and at this load it draws 19.8 Amperes from a 12V supply. So the electrical power input is 12 x 19.8 = 238 Watts and the efficiency is Pmech/Pelec = 154/238 = 65%.

The CIM could be operated as a generator at the same speed, current and torque (with the current and torque in the opposite polarity), and under those conditions its power losses would be similar; i.e., about 238 - 154 = 84 Watts. This operating point would require 154 Watts input (mechanical) power and the electrical output would be 154 - 84 = 70 Watts, so the efficiency would be Pelec/Pmech = 70/154 = 45%. Efficiency could be improved by running at higher RPM (i.e., at higher output voltage), and might get back up to about 65% at speeds above ~5700 RPM.

Of course the bearings won’t last long running like that. But remember, the CIM was designed for intermittent duty as a winch motor, and its endurance is rated (again, refer to the manufacturer’s data) for 6000 seconds total operating life at nominal rated power; i.e., at 4 lbf-in torque, 4320 RPM. That’s roughly equivalent to 100 minutes (50 FRC matches) of relatively hard (27Amp) driving. We all know that the CIM typically lasts longer, but those are the ratings.

#6

Except that Cirrus said that he has them running at 90+% efficiency, unless that is inaccurate, which it very well could be.

How exactly did you test efficiency, Cirrus? If you had them hooked up by connecting the leads on one motor to the leads on the other, spinning one, then counting revolutions on both motors, and comparing them; then I believe that that is an accurate representation of efficiency… But I would like to learn, so if anyone has an answer, I’d be more than happy to hear it :slight_smile:

#7

to measure efficiency the motor part of the system would need to have a significant load.

Letting the motor spin with no load doesnt tell you much.

#8

I didn’t think so… that is why I’m wondering how he measured efficiency.

#9

you would also need a way to measure the torque and rpm’s on the generator shaft, to know how much power you are putting into the system.

#10

I suppose I wonder… why? Why not just use an alternator from a car?

-Mike

#11

I would like to add to Richard’s excellent discussion that the brushes in the CIM motors are not positioned or rated for generator use. The diminishing returns on this project are weighed against the Chalupa as a generator. Remember also that there is heat generated within the motor during this process. The efficiency losses are turned to heat and this motor is designed for intermittant duty (Read: No provisions for cooling).
Mike,
An alternator is an often used device in this type of application. To use an alternator, you need a variable current supply to power the field coil. The output voltage/current is directly affected by input speed and field current. Many people just remove the alternator and voltage regulator from a junk car. They use a battery to both supply current to the field through the regulator and to smooth the pulsating DC voltage from the alternator.

#12

Actually the other day i started concidering the alternator as a more plausible solution. The only problem is then my motors will be under driven at 14v instead of 18v. Is there any way to use an alternator but jump up the voltage to 18v rather than the 13.5-15 that usually come from them?

Also i dont quite understand the field current stuff. Like i have to supply field current to it, then the current/ voltage output depend on my supply?

Thanks so much everyone for your help.
Its really appreciated :slight_smile:

Oh and about my checking efficiency. Im almost certain my method is flawed, It doesnt seem right at all. All ive done is run one motor into the other and check the input voltage against the output voltage with the motor acting as a generator running a halogen lamp.
I got output voltages from the generator motor that were 90% of the feed voltage running into the first motor.
I know i know. Its not that easy.
But i was just mostly trying to get an idea of rpm needed to produce certain voltages. And this was just making sure i could produce a good voltage before i ordered the engine. (which i now have in my possession)

Thanks again

Craig

#13

The field is easy to understand. Instead of using permanant magnets like the motors, the alternator from a car uses an electromagnet. This allows the voltage output to be adjustable: strong field= higher voltage

with an alternator, when you first start it up, if there is no field at all then it cannot produce any output (its like the magnets are missing), so you power the field from a battery. Once the alternator has an output voltage you can tap off part of its output to drive the field winding and you dont need a battery anymore.

Im not sure about the voltage output levels from an alternator. Can you get 18V from it? usually the voltage regulator is built into the alternator itself, and its designed to charge a car battery, and to power your car

also they are usually potted in epoxy, so I dont think you could open one up and change a resistor to increase the output.

#14

Nice explanation of the field excitation, Ken.

As for the automotive alternator’s output voltage – yes, it is regulated (usually by means of built-in power semiconductor circuits) to 14V.

And the power tool is designed for use with an (unregulated) 18V battery. So under heavy loading the alternator will actually deliver more voltage to the power tool motor terminals that the battery would, and therefore enable the tool motor to deliver more mechanical power to the application. The motor will not get up to the same free speed, but that may not be important.

Cirrusport, can you tell us more about your application?

#15

its for a medium sized off road robot with 3 wheels on either side. The wheels are all completely independant suspension. Each wheel is connected to eachother by a chain and the drill motors are direct driven to the drive axles at one wheel each. It is almost all custom aluminum construction. And fairly light in weight(considering size/engine motor combination)
It is differential driven by 2 victor 883s (may upgrade if needed)

Thanks about it

ask any further questions

craig

#16

looking away from the generator setup, could other methods be used to reduce energy wastage and complexity due to conversion?
How about 2 clutches (to control power to left and right sides of wheels) and auto-tensioned chains to link the drives to the wheels. Even better if you can find CVT to adjust rpm/torque going to each side.
Then your alternator could be smaller to provide power (and charge a backup 12v batt) for the electronics. The ‘alternator’ could also be used to re-start the engine if you require it.

#17

well the problem with the clutch idea is then i need to create a reversing gear box for each side. It would probably give me the absolute best torque and speed. but with the added gearing it will severely increase the weight.