Am building a simple headlight for my bike, putting a 4700 microFarads, 35 volts capacitor in parallel with supply from the rectifier, and two LEDs, on for front one for rear light. The diodes are in parallel with each other and in series with the capacitor when supply from the rectifier has been cut, so the capacitor supplies current to both LEDs. I realised that the capacitor wasn't doing much other than smoothing the current, but I'd be glad if it kept the lights kept lit for at least 5 minutes at almost full brightness after the the supply from the rectifier is cut. The solution to this would be to connect a resistor in series with the capacitor. Question is, I am not sure whether to include a voltage drop across the parallel LEDs in my calculations. I was thinking a 10k ohm resistor would do the job but after calculation, I found it would only present 5 seconds of lights minus supply. Could someone please suggest the resistannce to use and whether or not LEDs, or diodes in general drop substantial voltage. Thanks.
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1. Never put LEDs in parallel, one will hog the current over the others. LEDs are current operated devices. You always need a current limiting resistor.
2. I think you have the circuit confused, but it's difficult to tell from your description.
Are you using "diode" for both rectifier and LED? Assuming not. But further reading hints that you are.
Assuming you have an AC generator, or why would you need rectifiers. But LEDs are not rectifiers and can't be used for that purpose. Why a capacitor if you don't have rectifiers?
email me.
The rest gets too confused for me to understand. Can you post the circuit? and clear up the above points?
Yes, LEDs drop voltage. Check the Vf value in the datasheet.
The voltage of 6.37 is arrived as top much less 0.7V. The height is 5*sqrt(2) = 7.07V. The transformer offers 10V as 5+5, centre faucet. This 5V is rms. And top is 7.07V. One Vd (=zero.7) is reduced and you get6.37V. In full wave rectifier, reduce the peak ac through 0.7V. Who stated that ac rms price is equivalent to dc price?? If ac is 5V rms, dc is 6.37. As above. The peak ac is 0.7V less than sqrt(2)*rms ac. With 10V ac rms, dc will be 14.14-zero.7 = thirteen.5V approxly. All that is without heavy (approach gigantic current) masses.