# LED forward voltage for 700mA LED at 350mA

1. ## LED forward voltage for 700mA LED at 350mA

Hi
For a 700mA LED run at 350mA, would I be right in saying the nominal Vf of the LED would be generally as follows…(at 85degc)
Red= 2.3v
White = 3.3V
Blue = 3.8V
Green = 3.6V

Ive looked at many datasheet and this seems to be the general situation, would you agree?

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2. ## Re: LED forward voltage for 700mA LED at 350mA

if your LED is looking for 100 watts, you want to pick a 100 watt driver.
you will find them in 20V, 32v, 38v, 42v, 48v... etc.

the voltage range is often in the 50% to maximum, meaning you cannot use them to power less than the voltage that is 50% of the maximum. some are more like 10volts minimum, some a little less.
the idea is that there has to be come juice left over to use to measure !

as was stated your LED is probably using 3 to 6 volts.
let's use 6 volts.. if you have 6 LED, at 6 volts, the total voltage drop would be 36 volts.
a 38v driver would probably have a lower voltage level of 18 volts, meaning you would have to use 3, but no more than 6.

also, you do not want to run near or at maximum. typically a safe area is up to, but not more than 80%

the more you spend on the driver, the closer to the maximum you can push your circuit. I have had some of the cheap chinese versions fail at a 70% load. the Fulham drivers, the ThoroLED can be pushed a lot closer to maximum, but cost 3 to 4 times more

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3. ## Re: LED forward voltage for 700mA LED at 350mA

The figures look reasonable per junction but many higher power LEDs have several junctions in series. Also consider that the dynamic resistace of junctions add together too so you should expect more Vf variation with current for multi-junction packages.

Brian.

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4. ## Re: LED forward voltage for 700mA LED at 350mA

The I-V graph for a semiconductor junction can be modelled reasonably well by a simple exponential graph.

This approximation is often used in many simulation packages (SPICE models). So we have an equation something like I (current)=(exp(k*V(forward voltage)) -1)

the constant k will depend on temp and the material (nature of the junction).

So I think it is better to forget about the voltage and control the current. YMMV.

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