 |
|
|
 |
New Member
|
|
Dec 18, 2009, 02:23 AM
|
|
which resistor to reduce 24v to 6v
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])
|
|
 |
Uber Member
|
|
Dec 18, 2009, 06:47 AM
|
|
Need to know LED current and Voltage, It's a basic Formula, but this is easier:
LED_calculators
|
|
 |
New Member
|
|
Dec 18, 2009, 04:55 PM
|
|
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
|
|
 |
New Member
|
|
Dec 18, 2009, 05:09 PM
|
|
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.
|
|
 |
Uber Member
|
|
Dec 18, 2009, 05:38 PM
|
|
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.
|
|
 |
New Member
|
|
Dec 18, 2009, 08:39 PM
|
|
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.
|
|
 |
New Member
|
|
Dec 18, 2009, 08:42 PM
|
|
OK I see I forgot the comma. The result would be 101.75 of course.
|
|
 |
Uber Member
|
|
Dec 18, 2009, 09:00 PM
|
|
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:
|
|
 |
Uber Member
|
|
Dec 18, 2009, 09:05 PM
|
|
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.
|
|
 |
New Member
|
|
Dec 18, 2009, 09:28 PM
|
|
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
|
|
 |
Uber Member
|
|
Dec 18, 2009, 09:59 PM
|
|
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.
|
|
 |
New Member
|
|
Dec 19, 2009, 12:48 AM
|
|
OK I will try that out and see how the heat will be. I will let you know. Thank you so far.
|
|
 |
Uber Member
|
|
Dec 19, 2009, 09:25 AM
|
|
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.
|
|
 |
New Member
|
|
Dec 19, 2009, 05:32 PM
|
|
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.
|
|
 |
New Member
|
|
Dec 19, 2009, 07:58 PM
|
|
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
|
|
 |
Uber Member
|
|
Dec 19, 2009, 08:53 PM
|
|
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.
|
|
 |
New Member
|
|
Dec 19, 2009, 11:55 PM
|
|
OK thanks, I will try this over christmas.
|
|
 |
Uber Member
|
|
Dec 20, 2009, 10:11 AM
|
|
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
|
|
 |
Uber Member
|
|
Dec 20, 2009, 10:57 AM
|
|
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.
|
|
 |
New Member
|
|
Dec 21, 2009, 02:14 AM
|
|
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
|
|
Question Tools |
Search this Question |
|
|
Add your answer here.
Check out some similar questions!
What value of resistor gives how much voltage, formula?
[ 4 Answers ]
In my home I am getting 220 AC Volts of electricity. I want to connect white color Light Emitting Diod (LED) that requires 3 volts output with the resistor which will be connected to 220 Volts power. So can any body please tell me what is the formula to find that how much voltage is output from a...
The fuse for the thermostat (24v) keeps blowing
[ 1 Answers ]
Hi, I found a blown 5amp fuse for the thermostat that was causing my heating system not to work. Put a new 5amp fuse in and 2 min. later I hear clicking noise from the thermostat and then it went out again. I may have a short somewhere. But, where do I start and what do I need to check?
Resistor switch on 98 Venture
[ 2 Answers ]
Please advise how to exchange a resistor switch on 98 Chev Venture... does not work on speed 1 & 2.. Thank you Arnold
Coil resistor wire
[ 1 Answers ]
Have problem with risistos wire on 1964 ford ware do you replace it from ?
View more questions
Search
|