# help for led series circuit

Discussion in 'LEDs and Optoelectronics' started by steve02, Jul 16, 2013.

1. ### steve02

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Jul 16, 2013
complete newbee I am using 3xcree xe leds in series circuit using lm317 as a regulator 12.2 v power supply I am using 2.2 1w resistor according to lm317 calculater this should run leds at 575ma when I do test I get reading of 200ma can anybody help me with this am I doing something wrong somebody told me I can run 4 cree xe in series without resistor when I do this the bulbs just glow dull

2. ### duke37

5,364
772
Jan 9, 2011
The 2.2 ohm resistor implies that you are using the LM317 as a current regulator so you are not suplying 12.2V but 570mA, or trying to.

What is the voltage required by the LEDs?
What is the input voltage when running under load?
What is the power dissipation in the LM317 and is there adequate heat sinking?

3. ### BobK

7,682
1,688
Jan 5, 2010
What is the forward voltage and current of the LEDS?

What is the input voltage to the LM317? It should be at least 14.2V. Is it capable of supplying 575mA?

You are better off to use the LM317 as a current regulator instead of as a voltage regulator because the forward voltage variables from one LED to another even of the same exact type.

Bob

4. ### steve02

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Jul 16, 2013
hi thanks for the replies f/v 2.85v @350ma [email protected] 700ma input v to lm317=12.2v power supply 12.2v 250w constant voltage led driver 20.8amps.i am trying to use 317 as a current regulator using 3 leds with 2.2 resistor on lm317 I get reading of 200ma I have tried 1ohm resistor with lm317 and I get 320ma these leds are on large heatsink with fans which is not even warm to touch.My problem is using lm317 calculator these figures do not add up voltage to 3x leds running is 2.9v hope this a bit clearer

5. ### duke37

5,364
772
Jan 9, 2011
Three leds in series will need about 8.55V , there will be 1.25V across the LM317 resistor, leaving 2.4V across the 317 from a 12.2V supply. This may not be enough, look up the details of the 317, the technical term is dropout voltage.

If you have a 12.2v stabilised supply, then a simple resistor should give sufficient stability. 12.2V - 8.55V = 3.65V.
R = V/I = 3.65 / 0.5 = 7.3 for 500mA. 10 ohm would be easy to find and a 2W resistor would be needed.

6. ### BobK

7,682
1,688
Jan 5, 2010
The dropout voltage is < 2 at 500mA, so I don't see why it is not working. Have you measured the voltage at the input, output and adjust pins when running?

Bob