Ideal way to power/wire 21 LEDs for illumination

[sjaltenb]

Hello again everyone,

On my board I will be using 21 LEDs, either high intensity blue or white (3.5-4.5 voltage drop) ~25ma for illumination.

I would like them all to be powered from a single transformer-->DC power supply. I have not selected either yet so I can use ANY voltage. However, I would like the voltage (brightness) to be variable using a Pot. I assume the best way to do this is use an adjustable power supply with like a LM317 instead of just placing a pot After a fixed supply, yes, or does it matter? (the latter seems easier)

the LEDs will be on all the time, all together, they are not indicators. What is the best way to wire these? It seems like wiring them in series is almost impossible due to the high voltage drop, but wiring in parallel will use a LOT of wire!

I planned on having the power supply run into a PCB full of the appropriate current limited resistors with the wires for each leaving from there. Just seems like a lot of wiring.

Anyway, any help or suggestions would be great! And as always, noise is an issue. Do LEDs tend to add noise? The power supply and even the trasnformer are seperate from the FX though...

thanks!

[Zben3129]

A combination of series/parallel might be good.

Say they have a voltage drop of 3v (easy math). If you wire 5 in series, you would need 15 volts, so with an 18v power supply you could power 5 in series. Now, suppose you take 4 of these "strings" and wire them in parallel. Each string needs 15v, so you can still use an 18v supply. Now you need to make sure the power supply meets the current demand of all 4 strings. 5*4 to get you 20 leds.


Could go 7 strings of 3 or 3 strings of 7 for example.


Zach

[sjaltenb]

Genious! so its lookin a little somethin like this?? (paint is great )



I still will need some type of resistor though, so would i put that on the single PSU BEFORE the pot, so the Pot goes from Max of resistor to Zero OR after pot so the pot goes from Zero to Max of resistor.

or should the PSU be variable?

[Zben3129]

Maybe you should put a resistor inline with the pot that is just enough to make it so the LED's won't burn out if the pot is at 0. From there you can set the resistance to whatever you like with the pot (wired as voltage divider). I can't imagine the pot will drop much voltage (V) but the circuit will be pulling a lot of current (A) and the wattage of the resistor and pot may come into play (W = V * A)

Zach

[jakehop]

I would use a resistor in series (remember how much is needs to dissipate!) and use a LM317 for the power-supply. I would also wire all the LED's in parallel. If you wire them in series like that, a whole chain will go if one LED burns out, and you'll then need to find out which one. No need to make it like a cheap Christmas ornament - make it great from the beginning! :-)

Kind regards, Jake

[PerroGrande]

Wire them in series -- as many as possible (based on their voltage drop) in each string.   Use one resistor per string to set the max current.

Remember -- because they are in series, the current through each series member is identical.  Thus, you'll need to supply 25 mA to each *string* to produce sufficient light.  Your power supply will have to produce n * 25mA of current, where n is the number of strings of LEDs.  If you do 4 strings, you'll need a "whopping" 100mA of current capability from the power supply. 

I do not recommend wiring them in parallel.  In parallel, you'll still have to provide each LED 25mA of current, but your supply will have to produce 21 * 25mA = 525mA of current.  You'll also need either one large wattage resistor or 21 smaller ones to set the current to the LED arrangement.  This is why LED illumination is almost always done with series connections maximized and parallel connections minimized.

So -- back to the series connection for a moment.  If you have an 18V supply, and LEDs that have a Vf of 4 volts (on average), these 4 LED's will "eat" 16 volts.  The remaining 2 volts dropped through a resistor will "program" the current. 

V = IR, so...
2v = 25mA * R
R = 2V / 25mA = 80 Ohms

Resistor wattage check: P = IV = 25mA * 2V = 0.05 Watts -- well within safety for a 1/4 watt (.250) resistor.

For 21 LED's, you could build 5 chains like this, with 1 LED done the "old fashioned" way.  Raising the voltage a bit could allow you to add an extra LED to each chain, thus reducing the number of chains.  If the LED's have a lower Vf, this too will allow you to raise the number of LEDs in each chain.

Here is a link to a nifty calculator -- although I always recommend understanding the underlying principles at work...

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

I build LED illuminators of all varieties, including infrared.  Apart from buck converters, switchers, etc, the series-parallel connection is the way to go.  I've built units with 100+ IR LEDs that will run for a LONG time on battery power via the series-parallel method. 

[mth5044]

Looks like you got plenty of good ideas here.

Out of curiosity, how are you going to mount them? Some sort of enclosure or like... a clear tube? Or something else?

[R.G.]

Building upon the suggestions:

LEDs using resistors for current limiting are inherently wasteful. For good limiting, the resistor voltage needs to be large compared to the variability of the voltage on the LED, and if you only use one LED, the power in the resistor is larger than that in the LED for power supplies of more than about 5V.

Run them in series. That way the *same* current flows in all the series ones. Find the average forward voltage of your LEDs, divide the power supply voltage by the average forward voltage, and then ensure that you have at least 1.5-2V left over; remove 1 LED from the chain to make this true.

Now, construct a constant current source from a BJT, two resistors and two diodes. For an NPN BJT, the two diodes in series go from base to ground, the base-emitter goes through a resistor to ground, and one resistor goes to a higher voltage, either higher in the LED string or all the way to the power supply. This forces the base to be two diode drops above ground, and the base-emitter junction subtracts one of these.The string of LEDs from power supply goes to the collector.  The resistor from emitter to ground has one diode drop across it, and we want it to have the LED current through it. So... yep, Ohm's law, the resistor is 0.6V/0.025A = 24 ohms. The power in the resistor is 0.6*0.25= 15mW. The voltage across the transistor is at most one LED voltage plus 1.2V, because that's how we picked it, and the power in the transistor is (4.5V+1.2V)*0.025 = 143mW, so it's just over the TO-92 threshold; a TO-220 or TO-126 device would be plenty. A gain of 100 is plenty; that would make Ib be 250uA, and a gain of 10 would only require 2.5ma from the resistor/two diodes bias string. The resistor/diodes bias string can then have a current of 4-5ma and be fine.

So the only power not spent lighting LEDs is 15mW in the emitter resistor, 143mW in the transistor, and the power supply times 4-5ma in the bias string. You should be able to keep this handily under half a watt.

More strings? More BJTs, use the same two-diode reference voltage and up the bias string current so you have enough for all of them. Efficiency goes UP as you introduce more strings.

[Sir H C]

I did something similar with current mirrors.  You can have one resistor run to the diode connected device, then run 3-4 transistors to drive the output.  Degen the transistors with about 1 volt and matching is pretty good.

[sjaltenb]

thanks for all the suggestions ya'll!

I should have known R.G. was going to make me learn something I didnt quite understand, but I will after I read it a few more times  But thats what he is here for, right?

Looks like its time to experiment. I definitely have a lot of great information here to help me out. I just need to buy the parts and experiment...The LEDs will be used to light the panels on my Cornish style board, like this:

http://www.petecornish.co.uk/gilmouralltube.html

The LEDs are fitted underneath each black panel shining up on the metal faceplate to illuminate it during shows. Do i need this? No. Do I want it? Yes! Here's my board so far:

The LEDs will come up through these holes at the bottom of each FX panel. Not sure how to mount them yet, any ideas? I'd like something where if the LED burned out I could just pop in another, like a right angle plug or something?


Also, aside from the power supply and all, will this string produce noise? It will be powered from a seperate transformer from the FX and use a good power supply. any need for shielded wire if EVERYHTING else on the board is shielded?

Thanks!

[R.G.]

I did something similar with current mirrors.  You can have one resistor run to the diode connected device, then run 3-4 transistors to drive the output.  Degen the transistors with about 1 volt and matching is pretty good.

That's another very good way!

[R.G.]

The LEDs will be used to light the panels on my Cornish style board, like this:

Neat build there!

Also, aside from the power supply and all, will this string produce noise? It will be powered from a seperate transformer from the FX and use a good power supply. any need for shielded wire if EVERYHTING else on the board is shielded?

Actually, the LEDs are running on DC, have a constant current, and should make no electrical noise whatsoever, given that the constant current thingie or the power supply don't get excited and sing - unlikely.

[sjaltenb]

Thanks- now you know what all these questions have been for  !

And the comment about you making me learn something, only kidding. I've learned so much from you and these guys, that entire board is comprised of ya'lls help. The only credit i can take for is doing the physical work, the rest is a joint effort 


Much appreciated!

[Auke Haarsma]

thank you for asking this question! It applies to my multi-fx project too.