Calculate LED voltage drop

Hello all,

I have a very simple question but studying by myself I count not find a way to calculate it. To use a LED with a 9Volt battery I calculated it is needed a resistor of 350 omhs, I just have a 1k omhs resistor and when i put them on circuit the LEDs light was poor (as expected).

Simulating with EveryCircuit showed the power dropped to 1.89V after the 1k resistor.

How to find this 1.89V ?

Thanks ! :grin:

That informations is on the datasheet of your LED. You must find a graph or a table that must say that for a given current the voltage drop of the LED is x.

It makes sense ! I used all formulas unsuccessful !

For that's reason the Multimeter values were different... Thank you !

If you used two of those 1k resistors in parallel, you'd get a lot closer.

Hmm, maybe you have a dead-ish battery? (9V - 2V)/1000 ohm = 7mA That should be fairly bright. I use 1K with 5mm LEDs, and they are pretty bright. Or maybe you have or a low output LED.

The forward voltage for an LED may be given in the datasheet, but its trivial to measure - just test with a reasonable value of resistor such as 1k.

The forward voltage won't change much for a wide range of current, since its a diode, and you shouldn't be depending on its exact value anyway as it changes with temperature and batch / make of LED.

The biggest factor in forward voltage is the colour, since this determines the bandgap of the semiconductor used. white LEDs count as blue as they are blue + flourescent dye.