powering a large number of LEDs

[Gouf]

Looking to make a bank of 78 3mm 3volt LEDs.

Any idea what may be the best way to do/power this?

[USS Atlantis]

First off - combo series/parallel

Second - use a wall-wart

4 LED's in series is 12v
78/4 is 9.5

So you'll have 10 runs; 9 with 4 LED's each and one with 2 LEDs

10 runs at 25ma per run is 250 ma

So your wall-wart needs to supply at LEAST 250 ma - I'd go with a 500ma or even a 1A unit, less heat generation and longer life from the wall-wart - also make sure it's a Regulated unit

Note that even though you are using all the voltage in the 9 runs of 4 - I'd still put a minimal resistor on it - say a 10-20 ohm - to prevent any minor fluctuations - there will be near zero effect on the light output

Then just use one of the calculators that's linked and compute the needed resistor for the 2-LED run

[WarpeD]

This was helpful when I did calculations for my cockpit LED's (didn't follow through....didn't have right kind of backlit panels made. )....

http://led.linear1.org/led.wiz

[tetsujin]

Note that even though you are using all the voltage in the 9 runs of 4 - I'd still put a minimal resistor on it - say a 10-20 ohm - to prevent any minor fluctuations - there will be near zero effect on the light output

A minimal resistor is also going to have minimal ability to linearize the voltage/current curve... The effectiveness of a resistor in evening out the performance of a LED circuit over voltage fluctuations depends entirely upon the fact that you're wasting a bunch of power to do it.

Let's say you went with a 10 Ohm resistor. At 25mA it would drop a quarter volt. A "minor voltage fluctuation" of, say, one tenth of a volt (quite a big fluctuation if you're expecting a regulated voltage, but nothing if your source is unregulated) - that 0.1 volt is actually a pretty large value compared to the amount your resistor is already dropping (0.25V) - so you can expect to see something like a 40% increase in the current through the circuit (given that LEDs exhibit huge increases in current for small changes in voltage, the resistor soaks up almost all of the additional voltage...) - so from 25mA to about 35mA. You'd have to start a lot lower than 25mA if you want to safely handle voltage surges with a 10 Ohm resistor.

If you were using chains of 3 LEDs with one resistor each (12V source, 9V dropped on LEDs, 3V dropped on a 120 Ohm resistor), a 0.1V surge from the source would result in about 0.8mA increase in current through the chain - by wasting a quarter of your power on the resistor you gain a circuit that can reasonably handle a fair amount of fluctuation in voltage.

I had my doubts initially but these days I'd say a current regulator (such as the Madman Lighting board) is the way to go, unless you're prepared to waste a big chunk of your power on a resistor and possibly a voltage regulator as well...

[Madman Lighting]

For that many LEDs you'll need current regulation of some kind, esp if you're wiring them in long serial strings.

Why? Tolerances. There's tolerances in the voltage drop of the LED, and it gets worse if you mix and match colors. Different colors have different voltage drops.

Why is this so bad? Because. Lets say you use a regulated wall wort and one resistor to power your long LED string. Now add up 10% tolerance on 6 LED voltages. This could be off as much as 1.2 volts assuming a 2V drop per LED. Now the current set by the resistor will be off by that much and that does not include tolerance from the wall wort, often 5 to 10% for good ones.

In short, you can easily blow your LED string by providing too much current, even if your math is perfect.

OK, so what do you do instead? You need a regulated constant current source to drive long LED strings. You can buy these as parts from guys like Mouser and digi-key if you have an electronics background or you can buy my card and be done now.

New Delux-24 card can drive up to 4 strings of 6 LEDs (any size or color) from two 9V batteries. (18VDC). All Madman Lighting kits now come with new Delux-24 cards.