Make:

This is a very, very bad circuit.

LED's are current devices not voltage. In this circuit, the 1N4001 will drop about 0.6 volts and the red LED about 1.7. That leaves about 0.7 volts dropped across the wire, effectively creating a short circuit limited only by the internal resistance of the batteries. You get into a weird situation that way where D batteries might fry the LED while AAA's will sort of work.

By sort of work, that means the brightness of the LED will be highly variable as the battery voltage drops. It might be a useful light on brand new batteries and uselessly dim on batteries that are only 10% used up.

This also doesn't even say what color of LED to use. It will have to be red, orange, or yellow because the shorter the wavelength the higher the voltage needed. The voltage is only high enough for longer wavelengths.

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i think it's an "ok" quickie - but i'd love to have a better one here too, send me one and i'll post it on up.

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A good "quickie" would be to replace the diode in that ciruit with a 65 ohm resistor. That would give the LED 20mA with 3 volts from the batteries.

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Hey i wana learn about making led cuircuts and everything, is there a sight or any advice for a person who wants to get into it???

Thanks heaps

Reply to this comment

Hey i wana learn about making led cuircuts and everything, is there a sight or any advice for a person who wants to get into it???

Thanks heaps

Reply to this comment