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USB generator Answered

I am building a small crank-powered generator for a college electronics class that will power a USB port. I know a USB port supplies 5V DC, and that my generator will obviously produce AC current, but I'm not sure the best way to convert AC to DC, then how to step my voltage up to 5 volts. My generator uses 6 wire coils in a circular pattern. 12 neodymium magnet rotate over the coils. This doesn't seem like the most efficient design ever, but that part is a stock university kit, the wiring is up to me. I've only been able to test each coil at a time, and each one produces .5-.7 VAC. Do any electricians out there have some tips? When all is said and done, I'll definitely be posting an instructable on this.



8 years ago

 You could can change it to DC with this AC to DC

Hey, cool project. I suggest that you take 4 shottky diodes and make a rectifier bridge. Then you could use the circuit from the MintyBoost to step up the voltage.

i think everyone forgot something here, because you guys are so used to dc. this is ac! you dont need mintyboost; you need an old fashioned CW voltage multiplier. it can not only multiply the voltage, it will rectify it and, if you use large values on the caps, filter it. install a 7805 in series with the output and voila.
patent pending

remember, cwvm parts need to be greater than 2Vin.

trickery! But it'd be less (or the same) efficiency as a minty boost

hows that? it would be more volts, less amps. also, you dont have the inefficiency of a rectifier, which then needs a cap to smoother the power (and shunt it to ground)

well, you'd still need a cap with a 7805, plus I read somewhere that 7805s are only around 50% efficient, while stuff like a boosting circuits are usually areound 95% (depends on chip) efficient. With either one you need a smoothing capacitor

no, you dont. in a cwvm, the input is ac, the output dc. if you use large value caps in the CWVM, the output has little or no ripple. and 7805s are decently efficient.

smooth enough not to cause glitches in the device you're charging? maybe use a buck circuit if avalable

if you use high value caps (not a problem, seeing as 2Vin is less than 20 volts) you can get almost no ripple.

Thanks for the help. I took another look over the later lab sections, and it looks like we will eventually be doing this exact setup. I'm too scared to try this on anything important (cell phone charger, ipod, etc), so I think I will whip some USB throwies to prove it works.

Are you using your own coils? any pictures?

They are pre-packaged coils from Tinitron. Here's a pic:


Ok, those look good. How is it powered? Wind? Hand crank? I agree with tech-king's idea

you would have to use a couple bridge rectifier (google it, the one you want is the one that consisstist of 4 diodes). Before soldering the coil together, do tests with a bread board or aligator clips. You want about 3 volts at normal speed, then you could use a minty boost circuit (google it)

Great minds think alike :P Thats almost scary...

To convert AC to DC, you'll need a rectifyer, which is essentally 4 diodes. You can make your own from 4 diodes, or buy a chip. Once you get that going, and wire up all the coils, check the voltage, you may not need to do step-up conversions. If you do, you should be in the neighborhood to use the MintyBoost's circuity.