Hi, I want to install LED lights in a boat with 24V supply. I want to use LM317T with 2 resistors, I get 3.45V which is OK but the heatsink becomes too hot. I tried the same with 6V input and the heat is OK. So I thought to reduce the 24V by a resistor to 6V as input for the LM317T. What value of resistor would I have to use? I would appreciate very much if somebody has a good suggestion. Thank you, Bernd ([email protected])

Need to know LED current and Voltage, It's a basic Formula, but this is easier:
LED_calculators

When I connect the 21 LEDs to 6 Volt it shows 1.1 A. The LED Calculator would give me a 22 Watt resistor of 18 Ohm

The planned ceiling lamps have 21 LEDs each. There are also lights below cabinets with 8 LEDs each, which in 1 room are 5 pieces on at a time (40 LEDs) . Other rooms have 1, 2, 3 or 4 pieces on at a time (8, 16, 24 or 32 LEDs). The third type of lights are "Star lights" over the bed areas with one single 8mm LED. There will be 5, 8 or 10 LEDs of this kind on at a time. The value I mentioned from the LED calculator was when I enter 6V and the total mA - may be that was wrong.

Why not just use a resistor?
Need to know LED Volts and Amps?
Say you have a 24 volt sullpy and the leds are 2 volts and 20 ma.
Subtract 2 from 24 which equals 22 then divide 22 by your led current, if it is 20 ma use .20 into 22.

I have 24V. LEDs are 3.45V. Difference is 20.35V divided by 20 mA = 1.0175. So I should use such a resistor for every LED ? Is there not a problem because the voltage of the batteries is not always constant 24V? That is why I thought a Voltage regulator like LM317T would be better. My problem is that the 24V have to be brought down to 3.45V and then the heat is too much, but if I feed the LM317T with 6V the heat would be OK.

OK I see I forgot the comma. The result would be 101.75 of course.

You guys seem to be chasing each other's tail.

#1: Are these normal LEDS or the high power things?
#2: Normal LEDS have a fwd voltage around 1.2 V and operate fro 20-50 mA
#3: What are the specs of your LED's.
#4: If they are normal LED's,20 of them in series will give you a dark light. They won't light at all. Not good.
#5: A switching regulator would be better.
#6: Best to use the LM317 in a current regulating mode, but you need to add a 3V drop for that.
#7: You could use 3 strings of 7 lights in parallel and current regulate for 150 mA. (LM317T is OK, switching regulator best, resistor poor choice.

If you took a single string of 7 at a 1.2V drop at 50 mA; R would equal ~(24-(7*1.2)/50e-3; p=(I^2)*R

You can use 3 such strings, with 3 resistors.

Now you can regulate each string with an LM317T. It would dissipate about 1 watt.

In a boat environment you would have to use the protection diodes. Transorbs would be useful too

Now I know I didn't do the design for you. I don't want too, without knowing the specs of the LEDS.

It's best to current regulate each string separately. 1 W disapation isn't too bad per string. The current regulator would maintain the brightness as battery voltage drops.
#8:

Some of the new higher current LEDS available use power supplies such as these: 3023 Wired BuckPuck - LED Supply.com

These lamps need to be heatsinked well. Using these, would probably look more professional. I think both Digikey and Mouser offer them.

The LEDs are 3.6V. I was told to make it with a LM317T 3.45V. Then it is 30mA. One ceiling lamp has 21 LEDs. If I fix them all parallel and connect it to the voltage regulator with 2 resistors 270 and 475 Ohm to get that voltage as per LM317 calculator, the light is OK. Only the heatsink has to be very big to be not too hot since they will be mounted in the ceiling. So I tried other things. When I feed the LM317T with 6 Volts the heat is acceptable and I now thought how to reduce the 24V from the batteries to 6V for the LM317T

Note on page 18, of the datasheet there is a 1A current regulator. That's what you want to use. Note that there is 1.2 V from the ADJ terminal impressed across 1.2 ohms, therfore it regulates at 1 amp. The eqn should be I= 1.2/R. Set I to 30 mA and solve for R.

OK, the LED's have a fwd drop of 3.6V and the regulator drops 3. So, to make it easy 24/3.6-1 is the max number of LEDs in a string. The max # of LED's you can have is 5.

The # of string is 21/5 = 4.2; use 4 strings of 4 and 1 string of 5. and 5 LM317T set up as current regulators.

(24-3.6*5-3) is like 21V, so the regulator is disapating about 7V.

0.21*7=.21 Watts.

In that case you can probably use the little transistor like version of the LM317T. You just need 5 of them.

OK I will try that out and see how the heat will be. I will let you know. Thank you so far.

OK, I misread you question a bit. I thought 3.45 was the voltage drop across a LED.

This is more likely 1.2 V
The voltage drop across the regulator is 3 V
The power supply is 24V
You have 21 leds

10*1.2=12
11*1.2=13.2

Add 3V

15 V drop on the 1st striing
16.2 V drop on the second

at 16.2 V or {P=VI} (0.50)*16.2 the regulator will dissipate less than 1 W.

Just so you understand, an LED is a current device and each LED uses up or drops a particular amount of voltage (typically 1.2V)

So, it looks like you can use two strings of slightly different lengths with two regulators.

The LM317L which is rated to 100 mA might make your implementation smaller: LM317L - 3-Terminal Adjustable Regulator

EDIT: It also means that it should operate down to 15-16V.

Thank you again, I am just about to fix a test configuration. I will let you know the result shortly. I have reduced the total amount of LEDs to 20 now and I have 4 strings of 5 each. I will try 2 parallel on one LM317T, and I will try all 4 on the regulator. Depending on the heat I will decide finally.

hello again, I tried it out. Even with the 4 strings connected to one LM317T (R1=420 Ohm, R2=6.2 KOhm, giving 17.4Volt) the LEDs light very good and the heatsink becomes look warm only. That is exactly what I wanted. Thank you again very much for your detailed help. Have a nice Christmas time.Bernd

Do this:

Get a single 24 ohm, 1/2 W resistor connect it in series with Vout. At the free end of the resistor (POINT A), tie it to Vadj. Connect your LEDs (+) to point A; The (-) of the LEDs to the battery (-) and the +24 V supply to Vin.

Drive a string of 10 to 11 LED's in series with that odd configuration.

You will find it on page 19, here: http://www.national.com/ds/LM/LM317.pdf

That configuration will work a lot better for you.

If you have a variable supply or even a 9V battery, connect one LED in seres with the funky regulator above. At 9V, you can connect about 1 to 3 LEDs in series.

Two such strings regulated in that fashion will be better. Trust me.

Then change to an LM317L which is about the size of a transistor to make it smaller.

The 10-11 LED's will also operate down to a much lower voltage. About 16 V.

OK thanks, I will try this over christmas.

Not sure if I missed the use for these lights. Lighting or indicators?
I believe another of your post mentions 220 volts?
http://www.paneltronics.com/PartsCom...age=3&perpage=
They have Leds That can be powered from 24 Volts DC up to 240 Volts AC.
With this arrangement you would need no regulators or resistors and would have redundancy as series meathods lack that ability.
1 out in the string will take out the others? Just a thought.
Or this:
http://www.powerstream.com/dc7.htm

My thought would have been something like this: VAOL-SW1XAX-SA VCC LED High Power (GT 0.5 W)

And one of the control modules described in one of the previous posts.

One bad LED doesn't usually cause the string to go out because the failure mode is usually a short.

Thank you again for your assistance. It helped me a lot. As I said before, I am happy with the circuits and the heat now. I got the LEDs alreay from China and the voltage regulators I will get tomorrow from Hongkong. Still I will try the last idea with the 24 oHM resistor and let you know after I tried it out.

I wisdh you a merry christmas and a happy new Year. Bernd