Candidate motor as a generator. Some testing...

[loquin]

I've been thingking about using a permanent magnet (PM) DC motor as a generator for some time now. After a lot of poking around in the DC motors section of Burden Surplus Sales, I finally settled on this motor, and ordered one.

What was the reasoning behind choosing this motor?

1. From reviewing the specs, I estimated that the maximum available power would be in the 30-50 watt range. Contrary to some beliefs, I think that, using efficient, high output LED lamps, 30-50 watts would be more than enough power to
   • provide lighting that is more than sufficient for excellent visibility at speeds of 30-35 MPH, under almost any condition (hey - we're NOT traveling at freeway speeds - there's no need for auto intensity lighting at one-half the speed)
   • provide excellent visibility to others, with bright LED head lights, tail lights and turn signals, and
   • have some power left over for other applications.
2. This motor is reversible, so there's no worry re the motor rotation
3. The motor is rated for continuous duty - an important consideration, as it could be running for hours at a time.
4. At 4200 RPM, it's a decent match for the engines we're using.
5. While not waterproof, the motor is completely enclosed, so, I should only have to provide a shroud to keep water from directly striking it.
6. It is a face-mount design, making the generator easier to mount.
7. At 5/16th inch, the shaft is big enough to readily find bearings to support the end, and sprockets to transfer the torque from the engine. Since the motor was designed for use in a blower or pump application, it probably has bushings rather than bearings and therefore won't support much side-loading on the shaft. For this reason, you would undoubtedly want to add a bearing to the shaft before adding a sprocket or pulley, with their potentially high side thrust load.

It IS a 24 volt motor, though, so regulation will be needed. However, there's not a LOT of power to waste, and I want to keep as much efficiency as possible, therefore, a switching regulator is most likely what will be needed. More later on this.

I received the motor, and got around to testing it yesterday. For starters, I checked the unloaded voltage output from the generator; spinning it at various speeds with the drill press, ranging from 620 RPM up through 3100 RPM. Since the drill press tops out at 3100 RPM, I extrapolated the output voltage at 3600 and 4200 RPM, assuming a linear voltage to RPM output.

[FONT="Courier New"][B]RPM    Voltage[/B]
620     4.4
1100    7.6
1720   11.7
2340   16.0  
3100   20.5 
3600*  24.4
4200*  27.6
* calculated voltage at RPM[/FONT]

After this, at the highest RPM settings, I added a 10 Ohm 10 Watt power resistor as a load, then checked the voltage again. As expected, this caused the measured voltage to drop significantly, down to 16.5 volts. As a working approximation, electrically, a generator can be considered to be a perfect generator, with it's output feeding through an internal, series resistor. This internal 'resistor' is the winding resistance. By measuring the voltage drop with a known load, you can calculate the value of the internal resistance of the generator. In the case of the above generator, the internal resistance was calculated to be about 2.4 ohms.

After a minute or so of spinning the generator, the 10 watt resistor started smoking, so, I hit the power switch on the drill press. Then, after the resistor cooled down, I took a second 10 watt/10 ohm resistor, added it in parallel to the first, and powered up the drill press again. The generator purred along, the resistors got quite hot, but weren't smoking Measuring the voltage (14.2V) and recalculating the internal resistance (2.22 ohm) agreed with the first calculation within 7.5 percent. After 10-15 minutes, I turned off the drill press. The generator wasn't even warm! At 3100 RPM, it's internal resistance appears to self-limit it to a safe operating current.

At 3100 RPM, the generator was pumping out about 2.5 amps into the 5 ohm load, or about 31.25 watts. At a full 4200 rated RPM, I would estimate getting nearly 70 watts (3.7A at about 18.5V) out of this generator into 5 ohm load. Further testing, (at 4200 RPM,) will be needed to verify this, as well as to verify that the generator will not overheat at this output. In a 12 volt system, it's possible you could pull up to 6 amps at 14.4 volts (86 watts) from the generator, again, assuming no overheating. However, this also means that you would be generating over 80 watts in heat, inside the generator windings... I just can't believe that this would be good for longevity...)

Hopefully, this information will help others who are thinking about using this motor as a generator.

 

[AussieSteve]

Lou, rather than fry resistors, you can draw a load line for a given RPM by measuring:-
1. Open-circuit voltage. (I use the vertical axis)
2. Short-circuit current, (at the same RPM.). (H axis)

Effective resistance = Voc/Isc

* The hard part here is ensuring that the motor/generator is doing exactly the same speed for both of these tests - not as easy as it sounds, without a speed control on the driving motor and a tacho. Under load, especially SC, the generator will slow the driving motor heaps, as I'm sure you know.

Draw a graph and you can see the expected current or power output for a given output voltage, at those RPM. (P=ExI, of course.)

Divide both (maximums) by 2 and you will be at the most efficient output for that particular RPM.

ie If you get 28V open-circuit at a given RPM, with a SC current of 5A at the same RPM, for highest efficiency, draw 14V at 2.5A, (35W).

You could charge a 12V lead-acid most efficiently at a little over 4200RPM, when OC voltage reaches about 28VDC. 13.8V@4200 is pretty close to that. (I haven't checked your figures, but they should be similar to these.)
Once you get OC voltage and SC current for a given speed, you can extrapolate.

The OC voltage is directly proportional to speed.
The (effective) internal resistance doesn't change.
Therefore SC current is easy to calc. and the rest follows. (No fancy calc, but you'd be surprised at how close this gets you.)

My design, intended for shift-kit jackshafts only, will produce considerably more power, between 40W and 100W, or 200W peak for a short time if I run it flat out, at 14VDC. Not sure of exact RPM until my tacho arrives, and my DMM died of old age a couple of days ago, but I'm getting there.
(Using my 'scope as a DMM for now, a PITA)

How are you driving your's?

Just noticed - your extrapolations aren't quite right:
3600/3100 x 20.5V = 23.8V
4200/3100 x 20.5V = 27.8V

In a 12 volt system, it's possible you could pull up to 6 amps at 14.4 volts (86 watts) from the generator, again, assuming no overheating. However, this also means that you would be generating over 80 watts in heat, inside the generator windings... I just can't believe that this would be good for longevity...)

Generating 80W is not the same as dissipating 80W.
The 86W is only the output power.
The motor/generator's internal dissipation is far lower, effectively IR², where I = series current, R = effective series resistance. It's a little higher due to other losses, but not much.

What are your motor's specs? You can drive it to the same current as it used as a motor.
Voltage and power are secondary.
Just looked - anything over 1.8A will slowly cook it - a higher-rated motor would be needed for continuous generation of over 1.8A at 14VDC, (25W), to charge a lead-acid batt.

 

[professor]

I ordered a universal regulator for a snowmobile, thinking it would regulate power for my experimental gas/elect. bike and the thing had only 2 wires. Apparently they ground out excess current- NOT what I needed. Would this be useful to you to maintain 12 volts?

 

[AussieSteve]

That's probably a linear regulator, professor - much lower efficiency because the excess power is burned off as heat. Fine with larger charge systems, but probably too much power loss for this application.
A small switching regulator would be much more efficient and is easy to make or buy.
(I made a 0 to 15/23V, max 8A adjustable switching regulator today, to control the motor thats driving my generator during testing.)

 

[loquin]

Steve - a few points.

• I chose to add a resistance value, as this method has a much lower effect on the RPM than shorting out the generator. You're just not pulling as much power from the generator, and ultimately, the mechanical drive which spins the generator this way. When adding the 10 ohm resistor under load, I heard no change in the pitch of the generator sounds. When I added the second one, you could hear a tiny, barely noticable effect. The drill press motor is much more powerful than the generator, so you wouldn't expect much RPM change in any case.
  .
• P does not equal IR².
  
  P = IE, but E = IR, therefore, substituting the second equation into the first results in
  
  P = IIR, or P = I²R
  .
• Any non-linearity in output voltage to input RPM is likely to occur at the low end and beyond the rated RPM. So, I chose the range from 1100 RPM to 3100 RPM as probably being the most linear output response, (and probably the most respresentative of the extrapolated range I was trying to calculate) and based calculations on those speeds/voltages. This resulted in 6.45 mV per RPM Delta V constant.
  .
• Generating 80W is not the same as dissipating 80W. Absolutely. I never said it was. Calculating the potential maximum current with a load which would result in a 14.4 volt Vout and 2.5 ohm internal resistance returns 5.44 amps.
  
  P = 5.44 * 5.44 * 2.5, or 74 watts.
  
  If the internal resistance is 2.22 ohm, on the other hand, the max potential current would be 6.18 amps, and
  
  P = 6.18 * 6.18 * 2.22, or 84.8 watts.
  
  So, depending upon which value of internal resistance is correct, the potential internal max power dissipation for generator would be between 74 to 85 watts... By happenstance, a little less than the output power supplied by the generator. As I said earlier, this would NOT lead to a long life for the generator, though, even if you could potentially do so. If you hold the current out to 2 amps, for instance, the generator would be dropping approximately 5 volts internally, and at 4200 RPM, you could supply 46 watts to the regulator. If the regulator is reasonably efficient, this means 35-40 watts available at 12 volts.

 

[AussieSteve]

Sorry, Lou, I meant I²R.

How were you planning to drive it from the engine, out of interest?

Meant to add, I don't have a drill press or similar with a fixed, known speed, hence the tacho and speed control, (a 15V 8A variable switching reg). Just waiting on my tacho, an optical no-contact type.

The 74 to 85W dissipation would only occur during a full short-circuit, when the full potential is across the internal resistance. Considerably lower during normal operation, though.

 

[loquin]

Exactly - I'll be happy with 35-40 watts, so, I'll probably fuse the output with a 3A slow-blow fuse.

Under a short-circuit condition at rated speed, the generator could put out (for a very short time, I'm sure) up to about 12A. (28V/2.2 ohm)

I'm getting parts together to build a harbor freight 79cc 2.5 HP motor through a pocket bike CVT. The harbor freight engine has a fairly long shaft, so there's room for a #25 sprocket between the motor and the clutch. This sprocket will drive the generator live, before the clutch. The motor will be mounted to a piece of 2 inch by 5 inch rectangular aluminum 'box' channel, and will be mounted via apx. 3/4 inch standoffs, which will leave the end of the shaft (and clutch) slightly recessed on the opposite side of the channel, where the CVT will be mounted. The generator will be mounted on the same side of the box channel as the motor, with the generator shaft protruding inside the 'box' and a #25 chain from the motor sprocket to a somewhat smaller sprocket on the generator.

 

[AussieSteve]

I'm aiming for a similar amount of power under normal conditions, but want about double for faster charging after, for instance, running the lights without the engine running, or when stationary for a bit. (My generator will only be driven when the clutch is engaged to turn the jackshaft.)
35-40W is more than enough for lighting etc.
I'll gear appropriately. Mine's a 100-120W motor/generator.
Wish I had your long shafted engine. I'm only going to run off the jackshaft because I have nowhere else to reliably attach to, without cutting holes in or extending side-covers.
At this point, I'm planning on using a 6mm wide x 2mm pitch timing-belt drive, but the cost of pulleys is prohibitive - $18 for the generator pulley and ~$40 for the jackshaft pulley, from suppliers both in Oz and the US.
I thought about a chain drive - could almost double as an electric start, like on some ride-on mowers.

 

[loquin]

FYI. Did the 'short circuit' current test for a few seconds earlier today. Measured 9.5A at the starting RPM of 3100. I say 'starting,' because the RPM did drop noticeably when I connected the meter. The calculated internal resistance by this method was 2.15 ohms, however, since the RPM (and therefore, voltage) also dropped, without knowing the exact RPM, I couldn't calculate the actual internal resistance under this method. It would be a bit lower than the 2.15 ohms as calculated.

 

[AussieSteve]

Sounds promising. Even though the revs dropped a bit, it gives you an indication of what to expect.
I got my optical tacho a few days ago and realised that my motor/generator was rated for 18,000 rpm, (I'd spun it up to about 23,000, it seems, thinking it was doing about 10,000, before I got the tacho).
Got another motor on the way, rated for 2500rpm, 100W. Meantime, I sit on my hands again for a few days. Time to ponder the regulator. I'm leaning toward a buck-boost topology, but it might be overkill.