LED Power Problem

[69428scj]

Do you have a bench power supply capable 18V 1A?

Sorry but no.

 

[Helder Ferreira]

What about 2 12v batteries?

Sorry but no.

 

[69428scj]

What about 2 12v batteries?

I have batteries - please explain why?

 

[Helder Ferreira]

Just to check first if the leds light up ok connected in series/parallel like your second drawing, connect them with a 22 Ohm 5W series resistor. 20 leds @ 0,02A each equals 0,4A 2V. Eight groups in series equals 16V, the difference to 24V is 8V @ 0,4A equals 20 Ohm 3.2W, Next value 22 ohm.

This should give you an idea of the brightness of each module.

The switcher circuit I showed you, should handle 4 or 5 of these boards without a problem.

 

[69428scj]

I am sat here planning wiring for lots of different solutions - but a simple one just hit me (while looking at a purchased 24LED taillight bulb).

Surely 1 resistor can run 20 LEDs in parallel, therefore each run of LED's (48 in total) has its own resistor powering 20 LED's.

Wouldn't this simple solution work. I have just tried it and it certainly seems to be running the LED's ok.

What's your thoughts. I could always cut the rows in half and run 96 resistors running 10 LED's each!

I would need to double this up though to have the high and low output for the tail and brake light.

 

[Darth Bagel]

That's still about 4.5W dissipated assuming 20mA/LED and 11V drop across the resistor, which is a lot. I'd probably go with 10W resistors just to have the safety margin. I guess you could do it though.

This may be a silly suggestion, but you could leave some of the strips off for regular brightness and only turn them on for high brightness, eliminating the need for multiple resistors.

 

[69428scj]

This may be a silly suggestion, but you could leave some of the strips off for regular brightness and only turn them on for high brightness, eliminating the need for multiple resistors.

I thought of that too - may be an easy way of doing it.

Just been playing with putting them in serial - may be the simplest solution

 

[Helder Ferreira]

The thing about dissipation is: the less voltage you have to "absorbe" by the series resistor, tge better, so if you put, say 5 led (20led groups) in series, you only have 2 volts 0,4A equal 0,8W. If you use only one equals 10V 0,4A 4w for one group times 5 equals 20W (this is a good soldering iron )

 

[ericgibbs]

hi 69428,
If you connect LED's in parallel, without individual resistors, they will die one by one in rapid succession.

The LED with lowest ON forward drop voltage will pass ALL the current until it burns out
and the next with the lowest forward voltage drop will draw ALL the current and die................................ and so on.

As each LED fails, the current flowing in the remainder will increase and increase!!!!!!!!

 

[blueroomelectronics]

Really should have built a smaller prototype first to understand how LEDs work. I've seen the same thing with people building LED signboards.

As they say, back to the drawing board.

Here's some reading on LED driving
**broken link removed**

 

[Hero999]

If you connect the LEDs in strings of five, then it's use on fith of the power and still give the same amount of light.

 

[Oznog]

OK, I don't think you're on the right track. Pardon for yelling but you need to see this:
YOU CANNOT EVER PUT TWO LEDs IN PARALLEL, ALONE OR IN A STRING, WITHOUT A SERIES RESISTANCE.

Attempting to parallel LEDs without an appropriate series resistance for EACH LED or series string results in large current imbalances due to mfg or temp differences. Basically you put 2 or 10 in parallel one will take all the current while the others get less or fail to turn on at all. The bright one will quickly degrade and/or burn out. This is inevitable due to the sharp IV curve and the negative temp coefficient of forward voltage. Even if they appear to be of equal brightness initially, once one gets slightly warmer than another its IV curve shifts down so it draws more current for a given voltage. They all have the same voltage being in parallel so it ends up taking more current, increasing its temp and increasing the instability.

Your forward current is probably too high. A 5mm pkg can take 20mA for awhile but not all that long. Many LEDs in a small area increase the heat and the voltage source is inconsistent so let's go with 15mA (even then that's pretty high!).

With a 12V source, you can put 4x 2V LEDs in series = 8V forward voltage. Now the car's say 13.8V when running, (13.8V-8V)/0.015A = 387 ohms. 87mW per resistor. On paper it seems like you can put 5x in series, even 6x, but there's not enough ballast resistance for a poorly regulated source voltage. The current will vary widely from 12.8V with the engine off to 14.2V or more that the wires might carry under worst conditions. 14.6V is possible.

So you need to parallel 240 strings of 4 series LEDs and 1 resistor per string. That's 240 resistors. Sorry that what you have made isn't right but it's not workable.

Why 960 LEDs, anyways?? I've seen LED taillights with like 20 or 30 per side. 960 is absurd, not sure what you're going for. Doing this with 20 or 30 is nowhere near as complicated to make!

 

[audioguru]

ONLY 960 LEDs?
Have you ever been behind a huge Hummer?
It is so big that it needs thousands of LEDs.

 

[rmn_tech]

Using the calculator on the site below.

It suggests 160 rows of 6 in with a 82 ohm resistor for each group of 6.

**broken link removed**

I would also agree that if you used only half of the LED's for normal lighting and switch the other half on for braking, your whole project would be simplified.