lm117 doubt

Thread Starter

imraneesa

Joined Dec 18, 2014
227
Hello Friends,

The lm117 into constant current circuit (image attached for your review).

I have set the current using 3ohm resistor. using the formula I=1.25/R.

ie. 1.25/3 = 400mA. I made the circuit and tried this.
input = 3.7 V
operating voltage of led = 2.6~3V
When I connected the circuit and checked the current with the multimeter I got 200mA which suppose to be 400mA.
When I changed my input to 7.4V I used the test load before I am afraid the led burns, I got more than 1A current. why is the current changing to the change of voltage? I am attaching my circuit made by me.
 

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MikeML

Joined Oct 2, 2009
5,444
Please post a schematic diagram.

Use the constant-current circuit shown on the LM117/317 data sheet.

The minimum input voltage needs to be Vdropout+1.25V+Vfled+ Vheadroom.

Look at the LM117/317 data sheet for Vdropout

Vheadroom should be at least 1/2V
 

bertus

Joined Apr 5, 2008
22,270
Hello,

I assume that the red thread is going to the led.
You have connected it to the wrong point on the regulator.
You have connected it to the Vout in stead of the Vadj connection.





Bertus
 

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Roderick Young

Joined Feb 22, 2015
408
I'm going to make a lot of guesses, which may not be right. I assume that the red wire at top left goes to the LED, and there is no ground on your circuit board?

I can't tell for sure, but it appears that the wire on the back is the positive supply input? If so, make sure it isn't shorting to the resistor or pin 1 of the LM117 by accident (use an ohm meter, don't trust your eyes).

Another theory is that with a 3.7 volt input, the LM117 did not have enough input voltage to satisfy its dropout voltage requirement, so was delivering less current than expected. When the input was 7.4 volts, that should have been enough. Consider that maybe you did get 400 mA out, but due to inadequate heat sinking, the LED burned out. What you could try, before burning out another expensive LED, is putting a 5 or 10-ohm resistor in place of the LED, and monitoring the actual current supplied by your circuit under various conditions. Ideally, you would choose a resistor such that at 400 mA, the voltage drop across the resistor matches what the forward voltage of the LED would have been.
 

Dodgydave

Joined Jun 22, 2012
11,284
your input voltage is too low, you need 3.6V minimum input to output,
Also i bet the red wire is to your led, if thats the case its connected to the output instead of the Adjust pin (left pin).
 
Last edited:

Thread Starter

imraneesa

Joined Dec 18, 2014
227
yes I connected the led to the output instead of the adjust. I am going to try now after correcting it. I will give you the update. thank you once again guys.
 

Thread Starter

imraneesa

Joined Dec 18, 2014
227
then what about the output capacitor. it wont be a part of circuit? isn't it? I must need a output capacitor for avoiding spikes in case.
 

Thread Starter

imraneesa

Joined Dec 18, 2014
227
Right off the data sheet where I told you to look.
View attachment 81898
so my current is 400mA. so at 25degrees and iout=500mA line crossing almost near 1.75V. but how do you arrive 1.3 from that image. lol I must be very much dummy. but until I know I have to ask instead of shying. after I messaged asking how you got 1.3V. I read all the datasheet two times. still couldn't find. I think I don't know to interpret the value of that graph.
 

Thread Starter

imraneesa

Joined Dec 18, 2014
227
PD = ((VIN − VOUT) × IL) + (VIN × IG)
in my case. vin = 7.4v vout = vf of led (3V)
IL= 400mA. what is IG?
I want to check if I need to use heatsink.
 
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