Problem with LEDs and electricity generating?

Question

Here is the whole problem: I have a homemade generator that produces around 5 volts max of AC at low rpm. I want to use it to light leds so I converted AC to DC using a bridge rectifier but I sitll get some blinking so I intend to add a 1000uF capacitor to the voltage from the rectifier. Then I regulated the voltage to 2.8 volts using LM317 and two resistors. I want to make it exactly 3 volts, so I intend to use a potentiometer (which rating???) to replace one of the resistors. And I intend to connect up to 10 leds (3 volts, 20 mA). Any feedback would be appreciated. Please point out any mistakes. Thank you.

Answer 1

I like your approach. You're on the right track to maximize the light output. Problem with your design is that you start with 5VAC, drop about 0.7V to rectify, lose a bunch when you smooth the DC, then have 1.5V dropout loss through regulator. LED's flash because too little voltage left except at the waveform peaks. Assuming you are  getting 5V RMS with a 1 phase generator, looks like even with the germanium diodes and LDO below, undervoltage limits current output. If you can get about 10% or 20% higher voltage, you would get far more current out. You may not be able to get enough current for many LED's without modifying your generator.

First, a 2 or 3 phase generator will work much better than a 1phase. The reason is that smoothing the DC reduces the peaks to a lower voltage value. Of course, if you generator is already made, you don't have much choice. Then, make sure you are using a full bridge rectifier (4 leads) per phase, and that it is hooked up properly. A half bridge (generally 3 leads) makes bad ripple and loses 1/2 your power capability. Silicon in rectifier you have has OK performance, but consider using germanium diodes to get less voltage drop. You would need to get 4 with enough current capability to make full bridge. You can change this later for more current if you want to focus on getting the circuit working first.

Edit:  shottky diodes have equivalent performance, maybe better Fairchild one I just looked at was 0.4V Vfd. How to hook up diodes for rectifier: http://en.wikipedia.org/wiki/Bridge_rect...

Large electrolytic cap after rectifier (and before regulator) is good idea, will smooth out voltage a bunch. It does take some power though, so you might have to crank a little before the voltage stabilizes.

Likely big problem you are having is that the drop out voltage of your regulator is about 1.5V. This is a killer, worse than the voltage drop on the rectifier. You can see this on the Fairchild datasheet link on the bottom of the Wiki article on the LM317: http://en.wikipedia.org/wiki/LM317

You need low dropout regulator, the lower the better:

http://en.wikipedia.org/wiki/Low_dropout...

Google Low dropout regulator or ultralow to get more data.

Get a LDO regulator, set the regulator output voltage to 3V, put the LED's in parallel, and you're good to go! The reason you don't really need resistors if you are careful is that LED's have a lower effective resistance (i.e. flatter curve than the vertical slope used in the simple diode models taught to EE's) than traditional diodes when conducting in the light emitting range, so they can self-regulate within a reasonable range, see Figure 3 in the link:

http://www.ledtronics.com/ds/sml13rgb/Ds...

I would still put low value resistors in series with the LED's to check current, per my post on the other question, at least until the circuit is debugged. By the way, double check the LED datasheet (and the resistor to measure current) to make sure you are not going too high on the voltage and current. If you exceed the design max, even if you don't smoke them right away, the life will be shortened.

Bill's circuits will work too, but give much less light for amount of generator work. His design has more headroom on the voltage, so you can use the standard voltage regulator and a silicon rectifier, though you may need 1/2 bridge rectifiers for his circuits.

Can also use transformer to bump up voltage. I don't much like this option, but it will work.

By the way, you might want to add a comparitor with a small LED that lights when your voltage drops below 3V, so you know if you are spinning the generator fast enough (i.e. a low voltage indicator).

Good luck!

Answer 2

You cannot drive an  LED from 2.8 volts.  That is the operating voltage for the LED in the spec sheet at a certain temperature, but that is not the purpose of that number.  Create a higher voltage and use a dropping resistor for the simplist circuit.  The LED will automatically take what it needs and the resistor will absorb the rest.

Or better yet, use the 5v to drive a switching LED driver that can illuminate 10 LEDs in a string.  Those circuits aren't hard to find from National.com, etc.

Answer 3

I've already answered this several times.

You can't just apply a voltage across a LED without a current limiting resistor. Read my, and others, answers to your other similar questions.

I'd take the raw AC and build a voltage doubler or trippler to get a higher voltage. With 5 volts RMS AC you should get 6 VDC with a bridge, 12 volts with a doubler, and 18 volts with a trippler.

I'd go with the doubler, 2 diodes, and large caps. Then if you need to, you can add a voltage regulator set for perhaps 5 volts DC, and add a R = 2/20 = 100 ohm resistor in series with each LED.  Or set the regulator for 6 volts and use 150 ohm resistors.

I'll dig up a schematic and get back to you.