I need to power 108 LEDs from one 12v supply

[Brind]

string 9 leds series

I am planning to use a 12 volt regulated supply and two circuit boards, each having 54 LED's.
Each board will have 6 LED's across the width and 9 LED's in length. (simple rectangle shape)

This is where I get stuck.
I've heard LED's can vary from batch to batch so connecting too many in Parallel isn't a good idea. So I can't just wire all the LED's together and put one resistor in the circuit.
I'm hoping to use one resistor per 6 though, hopefully where my 6 LED's across (in my circuit) will finish, so each row of six will be wired in Series with a resistor on the end and then each row will be wired in Parallel until all 9 rows are powered by the 12v supply.
Am I correct in thinking this is ok and will work and will be efficient?

I don't know how many mA to supply each one either, I have the LED's and it's stated they can take 50mA, but I'm guessing that generally they are happy at running at 20mA.
1, Would there be much difference in brightness?
2, Reliability, is crucial, so is it best to stick to 20mA anyway?
3, Because I'm using 12v, if I want each LED to run at 50mA I'd end needing a transformer with big amperage output since I want to power 108 of them?
4, Am I correct in thinking that if I supply 108 LED's with 20mA each that I'd need a transformer that is capable approx 2.2 Amps?

Thanks for any help!
Mark.

 

[davidgrm]

6 led using 12v supply

The 6 LEDS in series might work depending on the votage drop across each LED. This varies with colour and type of LED. If the voltage drop amounts to > 12 volts then you might have to put fewer LEDS in series. The maximum current that the LED can handle also varies depending on type, manufacture etc. The best is to consult the manufacturers data for the particular LED that you are using.

 

[Brind]

1080ma to watts

The voltage drop is 1.85v (super bright red) so I'm guessing it should be fine as each row would equal 12v with a suitable resistor?

Would wiring so many in series and then linking each row in parallel and supplying the lot with 12v be ok?

 

[FoxyRick]

12v 108 led

Yes, that will be OK, assuming the 1.85V is the drop at 20mA. Your total voltage drop at 20mA for each string of 6 LED's is 11.1V, so you would need a 45 ohm resistor on each string to drop the remaining 0.9V.

Note that, in general, running LED's in series is to be preferred. Each LED, even from the same batch, is slightly different and will drop a slightly different voltage at a given current. By connecting them in series, they can be supplied with a regulated current and be allowed to drop their individual voltages. The differences will generaly average out in a long string to the datasheet voltage drop value.

If you were to connect them individually in parallel, then many would be slightly under driven, and many slightly over driven. Your strings of six in series will allow enough 'averaging' to occur.

So, all you need now is to supply 12 V at 180mA to the 9 paralleled strings of 6. The 180mA is because each string takes 20mA, so 9 x 20mA = 180mA.

Is the 50mA LED current given as 'maximum', or 'recommended'?`Or is the 'recommended' 20mA? If you want reliability, certainly don't exceed the recommended current.

FoxyRick.

 

[Brind]

Many thanks for your reply.

50mA is the maximum. I was aiming for the brightest, but for the sake of reliability, I'm flexible, especially if there isn't a vast difference in brightness.

I've read you need plenty of voltage overhead, is 0.9V enough for decent reliabilty?

You say that each string of 6 would pull 20mA, I thought each LED would pull its own 20mA regardless, so each string would want 120mA and thus each circuit of 54 LEDs would want 1080mA from the 12v supply?

 

[FoxyRick]

Hi,

Second Edit I've just realised the 6x9 is for one lot of two, so double the current and power requirements I give below! So, you need total 360mA current at 12V and 4.32 watts power.

First, the brighness does not go up linearly with current - so 50mA *won't* give 2.5 times the brightness of 20mA. It will be significantly brighter, but will definetely stress the components more (probably 3 times the heat).

As to the voltage overhead, yes, you are on the *only just* range with 0.9V. The problem is that if the supply is not well regulated, a small increase in voltage will cause an increase in current. With a larger overhead, and therefore a larger dropping resistor, the increase in current is smaller.

On the other hand, a small decrease in voltage will cause a large drop in current, causing the LED's to dim.

If your supply is well regulated by a proper regulator, like a 7812, or at least a power transistor and zener diode circuit, then you should be fine That means that your transformer must be at least about 15 volts, not 12, to allow for the drop on the regulator.

If your supply is just a transformer, rectifier and capacitor, you might run into problems if the mains is not well regulated, or if the circuit is run in different locations with slightly different mains voltages. For the low cost of a regulator, it makes sense to use one, then you have no problems.

As to the current needed: Current flows *through* components, for this, think of it as water. The *same* 20mA will flow through each series LED, the power the LED gets is this 20mA *combined with* the voltage drop across the LED. Just like water going through a pipe - the same amount of water flows at every point of the pipe, but its pressure (like voltage) will get less as it passes obstructions (components) and uses up some of its energy.

So, each series string of LED's only has 20mA (or whatever you decide to feed it) flowing through. But, to get the energy into them, you must have enough *volts* available for each one.

Each string (9 of them) gets 20mA and since these 9 are in parallel, they draw 9 lots of 20mA from the supply lines. So, you need 180mA total from the supply, at 11.1 volts across the strings plus the 0.9V dropping resistor.

You cannot just feed them 11.1 volts, as the slightest variation in voltage will cause large variations in current though the LED's (LED's are not linear). You *must* use a resistor, which sort of 'absorbs' a little voltage variation. The larger the resistor (and thus the 'voltage overhead' mentioned earlier) the more voltage variation can be tolerated without stressing the LED's. Even with a perfect voltage regulator, LED's are different to each other, so you still need a resistor. Higher resistors will increase the total power required and wasted though.

You could feed the LED's from a constant current, just as easy to do with a regulator, but then they really should be all in series. Otherwise, if an LED failed in one series string, the other strings would be forced to have a much higher current through them, leading to quick failure. Also, you would need to supply a high enough voltage to run them all. So this is only done with small numbers of series'ed LED's.

Finally, note that however you connect the LED's, you will need to supply the same *power* (not including any small waste in the resistors or regulator).

Assuming 12V and 180mA, you need to supply 2.16 watts of power. If you were to use 6V, therefore strings of 3 LED's and 18 strings in parallel, you would still need 2.16 watts ( 6V x 20mA x 18 ).

Cheers,
FoxyRick.

EDIT:

I just though, if you give a link to the datasheet, I'll have a quick check (probably tomorrow) and see if 20mA is OK. If you want to drive a different current, remember the voltage drop will change.

 

[Brind]

Thanks for the excellent reply!

According to the spec sheet, the typical voltage is 2.0v and maximum is 3.0v @ 20mA.
I put an LED on an LED testing multimeter, it said 1.85v but it only puts out 10mA so I misunderstood before (voltage drop changes regards amperage, now learnt)

Regards the regulator, on your advice, I will add one.
I'm happy to raise my voltage to 15v ~ 18v but I cannot change the number and placement of the LED's in any way.
Since they require a minimum of 2v @ 20mA, I'd really have to up the voltage it seems.

I do want as much brightness as possible, do you think a compromise, say, 30mA be significantly brighter than 20mA?

If I were to raise the mA and use a 7812 regulator, how much power (generally) would the regulator want?

If this was your project, what input voltage and mA supply would you choose to get the best?

Again, many thanks for your replies!

 

[FoxyRick]

Hi,

30mA will give a significant increase in brightness, and should not impact the life too much. Make sure the heat can escape somewhere though. If you put the LED boards in a small enclosure without ventilation, you'll be suprised how quickly the temperature will rise. That *will* shorten the life expectancy. The LED's alone will be dissipating at least 6.48 watts of heat - thats a lot if there's no way to remove it.

You will have to raise the voltage to run 6 LED's in series, even at 20mA, now we know the 2V to 3V at 20mA specification. I would rather rearrange the wiring and put, say, 12 LED's in series with a much higher voltage, to allow more averaging. But I assume you can't do that.

I can't really guess what the average forward voltage will be at 30mA. Any way you could test a few on a 30mA limit?

If I had to guess, I would say allow 2.5V per LED, thats 15V for 6 in series. We must allow some volts for the dropping resistor as well, especially as there could be a wide variation. So if you increase the voltage to 18V, you will probably be OK.

So, if you use an 18Vrms transformer, this will give about 23V after full-wave rectification. If you have an 18V regulator, it will thus have 5V of overhead - just enough to allow a little mains variation and not so much that you waste too much power in the regulator.

You will need to use an adjustable regulator, an LM317T in TO-220 package will be fine and they are cheap and common. It will need a small heatsink. See this site if you want a circuit for it:

**broken link removed**

In this circuit, make R1 200 ohms and R2 2700 ohms for an output of 18.125V - close enough.

You will need to drop 3V across the dropping resistors (18V-15V). That gives a 100 ohm resistor (3V / 0.03A), on each string.

Total current required is 18 x 0.03 = 0.54 amps
Total power dissipated by LED's is 108 x 0.03 x 2.5 = 8.1 Watts
Total Power dissipated by resistors is 18 x 0.03 x 3 = 1.62 Watts
Total power dissipated by regulator is 5V x 0.54A = 2.7 Watts
Total Power is 12.42 Watts

You will need an 18Vrms, 1A minimum transformer.

How does all that sound?

FoxyRick.