I have a Honda GX25 mini 4-stroke engine that I want to built a battery charging generator for 12 V lead acid batteries.

My first idea was to use an old car alternator as the generator part, but now I am not so sure. How would I go about selecting the most efficient (and hopefully compact) generator for this motor?

Also as a side note, I will need to make an automatic starter for this motor. Would it be possible to combine the starter motor with the generator somehow?

  • With batteries at about 10% efficient, generator efficiency will probably not be an issue. – Optionparty May 4 '15 at 1:51
  • 10% sounds low. Can you back that up? My thinking is that the best thing would be to buy a 3-stage charger which runs from main power, and run it with a generator whose output is a good match for the required input power. – mkeith May 4 '15 at 2:42
  • 10% is unreasonably low for batteries, that doesn't sound right. – pjc50 May 4 '15 at 8:29
  • I can't run of mains, my application does not allow it :-) – Lennart Rolland May 4 '15 at 9:07
  • Expect battery efficiency more like 70% over a charge/discharge cycle. – Brian Drummond May 4 '15 at 10:58
up vote 1 down vote accepted

As an alternative, in keeping with the light weight, look at model aircraft BLDC motors - specifically outrunners, with a relatively low Kv (RPM/volt).

For example if you turn it at 6000rpm and want 12V, you would be looking for 500rpm/volt. This refers to the open-circuit voltage, and you will lose some voltage across the motor's resistance. So if you want a slightly higher open-circuit voltage, pick a motor with slightly less than 500 RPM/volt.

There's a huge variety available and it shouldn't be difficult to find a few that match your required power and voltage levels. For reasonable efficiency at the expense of slightly more weight, pick one that's rated for higher current than you expect - it will have a lower winding resistance, and lose less voltage internally. For example, at 1hp (0.746 kw) you might expect 746/12 = 62A, so a motor rated for 70A or more would be worthwhile.

Note however that these motors may not be rated for continuous operation, or may have much lower ratings for continuous operation, or the ratings may assume air cooling from a propellor! Again, picking a motor with higher ratings will only help, though you may need to arrange air cooling via a fan fitted to the coupling shaft.

Output from such a motor will be 3-phase AC, which you must rectify to DC with a 3-phase rectifier, made from suitable diodes.

Power levels and Kv I'm finding for such motors so far don't quite meet your requirements (Kv around 1000, currents up to 40A), but they are fairly close. I'll update with links if I find a suitable motor.

EDIT : this motor is the closest so far, with Kv=380 (so producing 12V at 4600rpm) and with a (presumably short term) rating of 90Amps and power output of 2600W. Continuous power rating is not stated, but given the margin, likely to meet your requirements given adequate cooling.

Or this motor with specs: RPM: 290kv Max current: 78A Internal resistance: 0.022 ohm comes even closer to the ideal specs.

And finally - though it's more expensive - this motor claims continuous power ratings higher than your requirements. It's also interesting that the continuous current is 75% of the short term peak rating, so these motors may need less de-rating than I thought.

  • Awesome answer thanks! My best jab at calculating this: 14.4 is optimal charging voltage for 12v lead-acid batteries. Optimal rpm for gx35 motor is 5 - 7 krpm. Thus I am looking for motor with 6000/14.4~=416kv (rpmvolt). Is that correct thinking? Then the rest is just amp. Engine is 1.1 HP so i should make sure generator handles 1kw + margins. – Lennart Rolland May 18 '15 at 4:01
  • You're thinking along the right lines, but (a) you need to allow for some voltage loss inside the generator. That'll be IxR = (say) 60A x winding resistance or 1.32V for the second (0.022 ohm) motor. (b) Plus some loss in your wiring. (c) Plus some (couple of diode drops) in the rectifier. (d) Minus something for the fact that rectifying AC gives you the peak voltage - theoretically 1.4* the RMS voltage, assuming sinewaves (which may not be accurate). (d) probably just about cancels out a,b,c but some experiment may be required, and you can tune the motor speed. – Brian Drummond May 18 '15 at 9:24

Your Answer

 
discard

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Not the answer you're looking for? Browse other questions tagged or ask your own question.