Does this LED Series/Parallel array work for indicators?

[sjaltenb]

hello...

in my board each effect has 2 LED indicators... I was going to feed 18 volts in parallel to each "set" for 2 LEDs that are wired in series for each effect.

So the LED calculator shows this:



However, do these values still apply if say, only 1 effect was activated, thus only 2 of the LEDs are lit??

I know this works if they are all on at once, which is how I will do the backlighting LEDs, just wasnt sure about the indicator LEDs. Thanks!!!

[Andi]

Because they're in parallel it doesn't matter how many are on or off. But I think you'll need a slightly larger resistor value for typical LEDs.

[sjaltenb]

Yeah i thought since they were in parallel it didnt matter...but wasnt sure.

. yeah im not sure i put the correct input information as far as current/forward voltage. Just wanted to make sure before i wired the whole thing up.

[R.G.]

Special purpose calculators prevent you from learning.

If your LEDs each need a forward voltage of about 1.8V to conduct and light up, then two in series will need 3.6V. This has to come out of the available 18V, leaving 18-3.6 = 14.4V for the series resistor in series with the two LEDs. The resistor is what sets the current. If you want 10ma through the resistors, then the resistance which will drop 14.4V when it has 10ma across it is R = 14.4V/0.01A = 1440 ohms. 1.4K (1400 ohms) is a standard 5% value and would work fine. I did this little calculation in my head, no calculator involved.

Here's another - if you leave the 560R resistors in there, what current flows in each series pair of LEDs? While you're looking up the URL of the special purpose LED calculator, I'll tell you that the LEDs have the same voltage across them, leaving the same voltage across the resistor. What current flows in a 560R resistor if there is 14.4V across it? That's I = 14.4V/560R = 25.7ma. The LEDs may burn out, since 20ma is a common max current for indicator LEDs.

Hey! 560R is something close to 1/3 of 1440R. Is the current something close to three times the first calculation? Hah! It is. Maybe there is something to be said for the process of learning to understand the stuff after all.

[Cliff Schecht]

EE burn... OUCH!

[sjaltenb]

 

I always feel bad when R.G. has to explain things...

But seriously, Thank you! I have a much better understanding of what is actually going on and what the components actually do.

Much appreciated! Seriously.