Hi guys,

This is a real simple question, can someone just double check my workings for a LED project I'm working on;

Basically I have a panel that consits of 35 LEDs - all the same type.
I have wired them in parallel running off a single resistor (I know not ideal but I'm really tight on space)
So the Source Voltage is 12v
The LEDs have a votage range of 2~2.2v (I'm using 2v for this project to give me some headroom) & a Max Forward Current of 20ma so my math is:

R = (12 - 2) / 0.7
R = 14.285

So this would mean I'd need a 10w 15ohm resistor?

Thanks guys!

 

Your math is correct but you're about to find out why 35 LEDs in parallel won't work.

 

Your math is correct, but your setup is not.

I think you need:

2v x 35 LED = 70 vdc for this to work.

 

#12 - please explain - I dont want to trash 35 good leds.

As I say I know I should have used a resistor per led or use a array but I'm far too tight on space - so I'm remote mounting the resistor

 

The LED with the lowest breakover voltage will hog the current. Several LEDs won't light at all. After the brightest LED pops, the next brightest LED wil pop, and all down the line.

If you have the time to mess with this, you can do 5 LEDs in series and add a resistor to limit them to .02 amps like, [12-(5 x 2.1V)] /.02 = 75 ohms but you'll have to double check each string of 5 to make sure you don't go overcurrent.

The other way is to make strings of 4 LEDs and expect your "resistance needed" to be more predictable.

There is also the option of not driving the LEDs to the brink of the absolute maximum current they can survive. Half current works for me. Pick something between 10 and 20 ma and have something reliable and long lived.

 

LEDs should NEVER be connected in parallel unless they are all tested and sorted so that all have EXACTLY THE SAME forward voltage.

A few of your LEDs might be 2.0V and many will have a higher forward voltage. Then the LEDs with the 2.0V forward voltage will hog most of the current (maybe 6 LEDs at 2.0V draw 100mA each). They will quickly burn out then the remaining 29 LEDs will also have some burn out and the remaining LEDs will try to share the current which will be too high for them so they will also burn out.

Try it. All LEDs might last a few seconds or an hour. They all might burn out in an hour or in a week.

 

But keeping within their designed parameters (which I am?) they shouldn't fail, I mean yes your right they will cascade but they're designed to last 25+ years so unless theres a underlining manufactoring fault within the led itself it 'should' be fine?

Maybe I should have made clear that the 35 LEDs are all identical - colour, manufacturer, specifications etc. So as long as each led itself have been made withing the manufacturer's own tolorances....?

 

Wrong. We are talking about manufacturing variations that are within specs.

 

No two devices are "identical" and, in this case, the minor differences are critically important. As your first post indicates, the manufacturing tolerances allow for the actual voltage to range between 2V and 2.2V at the same current. I'm not sure about LEDs, but in silicon diodes you get a factor of 10 change in current for about 60mV change in forward bias voltage near room temperature. If that holds for LEDs, then with a 200mV range in forward voltage you might have some LEDs that are operating at well over 1000x the current of others.

Now, let's assume you put all 35 in parallel with a 15Ω resistor and 12V across the whole thing. Imagine that, at 20mA, five of have a forward votlage drop of 2V and the rest have voltage drops of 2.18V. You will still have a total current of about 700mA, but almost all of it will be shared by those five LEDs, so they would each have something like 140mA while the others might average a few hundred microamps.

In reality, it is almost guaranteed that one of the LEDs will have a slightly lower voltage than the others so it will initially take slightly more current than the others. This means that it will be dissipating slightly more power than the others and will therefore heat up slightly more than the others. But as the temperature of the LED rises, the forward voltage drops slightly and, so, it now takes a bit more of the share of the current and the process repeats itself. This is known as "thermal runaway" and is what is meant by saying that one LED will "hog" the current. As each LED fails, the average current in all the remaining LEDs goes up, so that, even if they shared the current equally, they are being stressed.

 

Ok all sounds reasonable.
Everythings sealed in now so I'll just have to hook up the resistor & see. I think in my favour is the leds are not going to be used in a constant 'on' status so that may well extend lifetime. If they suffer from cascading or termal runaway I'll have to write the cost off to experiance & remake them using remoted mounted resistors for each series of leds.

My original question was to confirm my maths that I'd have the right resistor to run the leds safely in the configuration that they are in.

So thanks for the replys its all useful stuff that I'll keep in mind for my next - bigger - led project. (That will be for sure wired in series in a number of arrays as I have much more room to play with)

 

Ok all sounds reasonable.
Everythings sealed in now so I'll just have to hook up the resistor & see.

Please reply back with how well things go. It will be a useful data point for people to store away in the back of their mind.

 

I'd like to see what happens too

 

I also request information about the results.
as somebody has for their signature line, one real data point is worth a hundred theoretical somethings.

 

LEDs, 555s, Flashers, and Light Chasers