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I've been working on this for a while now and I'm sick of it so I figured its time to ask real people...
I'm building a diode bridge to convert ac current to dc current for a rechargeable flashlight. I am connecting a magnet tube wound with magnet wire to the ac + and - and I have a capacitor connected to the + and - side of the dc

I am using this image here
http://upload.wikimedia.org/wikipedia/commons/2/29/Diodebridge2.png
to figure out how to connect the bridge. I have hooked this up to a Voltimeter and when I hook the dc to the - and + and shake it to generate current I still get some - current when I shake it and I also get some ac where the dc is. what is wrong?
please help me
and if it helps you I am using an 1n4007 diode and a Aerogel Super Capacitor

2007-02-13 11:53:04 · 3 answers · asked by ? 2 in Science & Mathematics Engineering

3 answers

Your generator will give pulses of current each time the magnet enters, and leaves the coil. The rectified pulses will be of short duration, and may not read properly on a meter(especially a digital one). As long as you wire the polarity of the bridge as you have described, the capacitor should integrate the pulses and will charge up. However, to have a constant light output, you will need to regulate the voltage to the light source(bulb or led). Use a low drop out 3 terminal regulator selected for either the bulb voltage, or as a current sink to regulate the led.
Hope this was helpful. Any questions, just email me. Good luck.

2007-02-13 17:29:56 · answer #1 · answered by charley128 5 · 0 0

Bridge rectifiers do create DC voltage from an AC signal, but this is not all they generate. The diode bridge will basically flip the negative portion of the waveform so that everything is a positive voltage (neglecting diode drops in the rectifier). If your applied power is a sine-wave, then the rectified sine-wave at the output of the bridge will contain a DC component, and a strong 2nd harmonic. This is because the rectified sinewave looks like a signal that has twice the frequency of the original sinewave.

To get a close to Pure DC Voltage, you'll need some kind of filter or regulator to block the non-DC components from getting to your load.

DC power supplies have a ripple spec that tells you how much non-DC signal will feed-through to the output.


Hope this helps.

2007-02-13 12:08:10 · answer #2 · answered by Jess 2 · 0 0

It sounds like everything is working, but you are not making enough power to light your bulb from the capacitor.
In place of the bulb try an LED (be sure the + goes to + on the bridge output).
Maybe the "super cap" is too big and it is taking too long to charge. Try the circuit without the capacitor at all, and the LED should blink on every time you shake the coil. Try a capacitor with smaller capacity (fewer microfarads, fewer uF)

2007-02-14 03:07:11 · answer #3 · answered by Roy C 3 · 0 0

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