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5 Led's with LM317 - please check my design

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Firebird!

New Member
Hey guys - I need you to check my design and calculations before I put it together and burn everything up :)
I want to connect 5 1watt led's in a car. I am using LM317 in a current regulator mode.

Here is some info about components:

Led's per seller (I don't have them yet)
.Forward voltage 3.5V~3.7V
.Forward current 300 mA

R1 2 ohm
R2's 27,27,68 ohms

Total circuit current will be I = V1/R1 = 1.25/2 = 625 mA.
Divided by two strings it will make 312.5 mA per string. I know it a bit more then led's are rated for but don't think it will make a problem (or it will?)

R2 = V/I = 3.5/0.3125 = 11.2 ohm
I am thinking of connecting 3 resistors that I already have in parallel which should give me 11.264 ohm

So what do you think - should it work?

Thank You
 

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k7elp60

Active Member
Your calculations are correct. I would not run the LED's on a continous basis at or over their maximum specifications. Have you looked at the data sheet to see what the illumination in mcd is at lower current? It has been my experience that the will not notice a great change in illumination when the LED's are operated at a lower current.
 

Sceadwian

Banned
Efficiency will also be lower at higher currents. Your highest light output per ma in is likely to actually be bellow the nominal current they state. With power LED's like that the stated operating current is ONLY when it is properly heatsinked, if the LED is on a poor heat conductor you're going to have to use dramatically lower current or the LED will burn out.
 

Firebird!

New Member
Thanks for reply guys.
The Intensity Type per seller is 60~70Lm.
The Led's are premounted on a star (pic attached). That suppose to work like a heatsink too or I need to attach them to another one?
They not going to be always on - I am thinking replacing the regular bulbs with them in the middle stop signal (the bulbs are too hot for the plastics and started to burn everything inside). The led’s going to be on only when brake is pressed.
Also I used that current because I have a 2ohm resistor in my possession. Do you think I need to buy another one with the higher value?
Thank You
 

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Hero999

Banned
The voltage drop of the LM317 is also too high if you've got more than two LEDs connected in series.

Three LEDs in series gives a voltage drop of 10.5V and the regulator needs to 1.25+3 = 4.25Vto regulate properly.
 

Firebird!

New Member
The voltage drop of the LM317 is also too high if you've got more than two LEDs connected in series.

Three LEDs in series gives a voltage drop of 10.5V and the regulator needs to 1.25+3 = 4.25Vto regulate properly.
I'm sorry for my rookiness, but I'm looking at the datasheet of LM317 at the dropout graph and it shows somewhere around 1.7V-1.8V drop for 600ma load.
How do you get 4.25 out of there?
 

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Hero999

Banned
When connected as a current regulator the dropout voltage has an extra 1.25V added to it.

You're right it won't be 4.25V but it'll still be too high for what you're doing. To meet the specifications on the datasheet the voltage differential should be 3V, at lower currents it will be lower and don't expect the ripple rejection, regulation etc. to be as good as specified.

Suppose the dropout voltage is 1.8V, you need to add 1.25V across the sense resistor giving a dropout of 3.05V. Three LEDs in series have a forward voltage of 10.5V so the regulator will need 10.5 + 3.05 = 13.55V to regulate properly. Once the input voltage drops below 13.55V, the LEDs will start to dim.

I would opt for a switching regulator anyway because it's much more efficient.
 

audioguru

Well-Known Member
Most Helpful Member
The dropout voltage is when the regulator is not regulating anymore. Don't you want regulation?
The graph of the dropout voltage is for a typical LM317.
You don't know if yours is typical, its dropout voltage might be 3V.
So 3V for the dropout and 1.25V for the current regulation equals an additional 4.25V is needed.
 

Hero999

Banned
I doubt the dropout voltage will be 3V at 600mA.

If you divide the value read from the graph by 2.5 and multiply by 3 you should get a good approximation the worst case scenario which in this case is 2.16V.
 
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