View Full Version : Power Demands of Propeller

05-24-2008, 10:53 PM
I'm working on a small hobby wind turbine using a stepper motor. I have jillions of these stepper motors and like to tinker with them in many ways. I know using one of these as a generator doesn't output much power, but I don't care. I just like to play with the proof of concept, rather than the finished device. Anyway, I've got my makeshift stepper generator up and running, but it only outputs about 200mW into a 12.6V battery. Not much power. This is the problem actually. The open circuit voltage is something like 28V after the stepper output goes through a transformer. Apparently, this isn't enough current to "charge" the battery. It's a fairly good size battery out of a gate opener - not TOO big like a car battery, 7AH.

Anyway, here's where the propeller comes in. I need to know if there is anyway to reduce the power requirements of the prop significantly. I want to be able to use it to control the charging of the battery. If I just have the rectified output straight into the battery, it doesn't charge. I think this is due to the tiny current not being enought to really CHARGE it. Instead, I want to use the propeller to control a circuit to have the generator charge up a fairly large cap bank, then discharge that bank into the battery, thereby REALLY increasing the current into the battery, but reducing duty cycle tremendously. I'm not sure if this would aid in actually charging the battery or not. In addition to this problem, if I use a simple passive circuit to feed current into the battery ALL the time, there are hardly any loses, except for the bridge rectifier. If I use the propeller in addition to that, I'm sure it will consume a good bit of my 200mW. Any ideas on how to get more bang for the buck?

Ken Peterson
05-24-2008, 11:21 PM
I think this would be a waste of a Propeller, as well as the energy to run it. You can use an op-amp to compare your voltage to a threshold, and turn on a mosfet when it reaches the set point. You can also design in hysteresis so that it shuts off at a somewhat lower voltage. I don't really have any software handy at the moment to make a schematic.


05-24-2008, 11:32 PM
That's kinda what I was thinking... using an opamp with a wide band of hysteresis. I'm not big on circuit design. Call it lazy, but I like companies like Parallax to design a complicated high level circuit(Propeller), then let me just write some code and bam - done. However, I was afraid of the power being a problem. I figured the power required wouldn't be worth it in such a tiny application.

Mike Green
05-24-2008, 11:52 PM
You need to understand the concept of charging. There are pulsed chargers for certain kinds of batteries, but these are usually AC powered and partly save money by leaving out some of the filtering for the rectified AC. The pulses are pretty high current relative to the normal steady charging current. If your charger only puts out 200mW at 12.6V, that's maybe 15mA. Usually batteries are charged at 1/10 of their capacity, so a 7AH battery needs to be charged at 700mA for at least 10 hours. Saving up some of the current and providing brief high current pulses won't help because your average current is still way too small.

05-25-2008, 12:03 AM
Doh. Again, I had a feeling that was the case, Mike. But my thinking was, lets take the charging to an extreme.... Let's get a HUGE capacitor bank, and charge that up - very efficient. It will take a LONG time to charge it up, but if it's 95% eff., ok. Now, after 100 days of charging it(extreme), lets charge the battery for 10 hours at 700mA. Sounds like a plan, right?

Now, back to reality of the current way I'm doing it.... each time the stepper rotor passes a pole inside, it pushes a TINY bit of charge across the transformer, and to the battery... SLOW charge, but constant... I just figured there was a medium somehwere in the middle...

Beau Schwabe
05-25-2008, 12:23 AM
You might be able to gain a little bit by reducing the drag on the stepper/generator by trickle charging a cap bank, and then dumping that into aˇbattery, but as Mike says... it needs to be at least 1/10 of the battery capacity in order to charge the battery adequately.

Beau Schwabe (mailto:bschwabe@parallax.com)

IC Layout Engineer
Parallax, Inc.

Ken Peterson
05-25-2008, 12:43 AM
It all comes down to the laws of conservation of energy (not in the environmental sense but in the physics sense). You are only getting so much power from your little stepper and you can't increase the power with any circuitry. The battery needs a certain minimum power to charge and that seems to be way beyond what you're generating.

Perhaps if you have a jillion of these stepper motors, there's your solution! Fill your backyard with windmills!


Beau Schwabe
05-25-2008, 03:47 AM

Absolutely right! - lol ... I was hinting that perhaps his current configuration my not beˇoptimal for getting the most out of the stepper/generator.

Beau Schwabe (mailto:bschwabe@parallax.com)

IC Layout Engineer
Parallax, Inc.

05-25-2008, 04:00 AM
Ken, I understand the LCE, but like I said in the "extremes", you build up a HUGE charge in the capacitor, assuming the charge put into the capacitor far exceeds any internal leakage. Once there is a huge charge built up, you let it flow into the battery. I'm talking about a charge large enough to allow the 700mA to charge for 10hrs. Granted, it may take a LONG time for the charge to build up in the capacitor, but assuming the charge DOES build up with minimal leakage, I would be able to charge the battery.

Now, in reality, I'm guessing there is another piece in play here - the self discharge of the battery itself. It only makes sense that the average power output from the generator would have to exceed the self discharge power dissipated by the battery. Correct?

05-25-2008, 04:49 AM
Being that this thread is leaning toward proof of concept...ie playing around with stuff...which i can resist either.ˇ even if it wont work, or practical.

Think I would take the highly variable output of the generator, run it thru a bridge rectifier and highly filter it with large capacitors allowing for long drain times during brownouts, and power a simple 555 timer chip to generate ac that I would run thru a transformer and convert to a dc level for charging.

its much easier to design and implement an over charge protection system dealing with battery voltage levels, verses trying to control your source voltages.

Also the 555 is very very cheap and operate over a wide voltage range...

nowdays the 555 in packs of 10 are dirty cheap...blow them up at will.

Post Edited (Burritoe) : 5/24/2008 10:00:34 PM GMT

Ken Peterson
05-25-2008, 09:47 AM
@Burritoe: Be careful how many components you throw at this problem. The more you use, the more inefficient it would be. Especially if you are talking about converting DC to AC and back to DC again. I think in order to charge the battery, it has an internal resistance which will dissipate a certain amount of power. If you try to charge it with an average of 200mW, the battery may be dissipating 200mW due to internal resistance, leaving no power left to actually charge the battery. 200mW is so small it would probably not even cause a perceptable increase in battery temperature.

I still think the best approach would be to get a jillion windmills going!

Theoretical time required to charge the battery with 200mW with no heat losses:

7AmpHours x 3600 seconds/hour x 12 volts = 302,400 Jules
302,400 J / 0.2W = 1,512,000 seconds = 420 Hours = 17.5 days (at 100% efficiency).

I'm guessing if you get about 30 windmills going, you should be able to charge your battery in less than a day with a good stiff breeze.


05-25-2008, 12:05 PM
Indeed, Ken. Like I said, I DO have jillions of these Nema 23 steppers. I just might do that. Maybe paint them with flamboyant colors and put them all in my front yard. Hey, maybe even get a "REGISTERED SEX OFFENDER LIVES HERE" sign and put it out there too in order to seal deal on the pervy old man look I'll have going with 100 small windmills.


Paul M
05-25-2008, 05:15 PM
To complete Ken's calculations:

Energy stored on Cap = 0.5 * C * V * V = 302400 Joules

for a 12V charge: C = 4200 Fˇ!!ˇ Quite a big capacitor!

Note also thatˇthe turbines would need to supply twice the energyˇstored on the cap, see: