[sdiy] voltage and LEDs

[George Mattson]

The forward voltage is what the diode is going to use up across the PN
junction. It just "goes away"
Actually, that's the voltage required to get the current to cross the
barrier region. Consider it the PN road bump.

Using 5 V, subtract the 3.9V Vf from 5V to get the difference of 1.1V.
1000mA max current is 1 Amp. Using ohms law divide 1.1V by 1 Amp to get 1.1
ohms to limit the current.

That's the minimum resistance necessary to prevent burning up the LED in
regards to the specs. Anything less will just dim it down. Usually, you can
get by with half the current. Anything above a certain limit doesn't let the
LED get any brighter, just dissipates as heat at the junction and shortens
the life of the semiconductor.

But, 1 Amp is also the high limit of the 7805

You'd either need a huge heat sink and a fan to keep it cool. Or, use two in
parallel and let them split the load.

The 1.1 ohm resistor at 1 Amp is going to dissipate 1.1 Watts of power
(P=IE)

I'd suggest using a 2 to 5 W resistor and mounting them above the board to
let them get some air to cool down.


My $0.02

George Mattson


I want to drive a 3watt led
# 1000mA - Max Drive Current
# 3.90Vf - Forward Voltage

Can just use a 7805 to get 5v, and then drop it with two diode in series to 
get 3.6v?
Or do I still need resistors in series with these LEDS?


dennis.barton:skylab2000[socal]



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