Linear voltage regulator problem lm338

[BlinkingLeds]

Yep here are the photo(s)

 

[BobK]

It looks to me like one side of the LED is connected to pin 1, and the other side is connected to the middle lug on your terminal block, which is connected to nothing. Is it actually connected to ground? Otherwise, the connections look right.

Bob

 

[BlinkingLeds]

the yellow wire of the led connects to pin 1 and to the resistor which is connected to pin 2
the white wire goes to the black negative of of the power supply.
Better photo comming

 

[BlinkingLeds]

update...

 

[BobK]

Well, that all looks correct. I would suspect the LM338 then, have you tried replacing it?

Bob

 

[BlinkingLeds]

yes i have tried replacing it with another lm338 and with an lm317. same result

 

[davenn]

I agree with Bob the layout looks OK
maybe you have killed the LED ? or ... are you sure you have it in the right way around ?
Your connections to the LED are pretty bad, you would be better off using crocodile clips on the ends of those patch leads to connect to external devices. Much tighter and secure

Dave

 

[(*steve*)]

Where did you get the LM338 and LM317 from?

It's possibly unlikely, but one of my questions was "Are you sure it's an LM338".

Did they come from the same source? Was it very cheap?

 

[Harald Kapp]

Eliminate oen unknown from the experiment. Remove the LED, insert an Ammeter instead. Set the current to e.g. 10mA. Observe the current. Is it stable?

Use a low current only (e.g. 10mA as stated) because the LM338 will now drop all the voltage (12V) from the power supply, which will result in high power loss (dissipation). At 10mA the dissipation will be 120mW which should be tolerated by the regulator without heat sink.

If the 10mA test is o.k., you may now want to increase the current up to the 800mA envisioned. To avoid high dissipation in the regulator, add a series resistor to the output. This resistor will drop some (but not all) voltage.

For example at 12V input, 800mA current, leave 2V for the LM338 and some for the Ammeter (I assume 0.5V, this value depends n your instrument). That leaves VR=12V-2V-0.5V=9.5V as the voltage dropped acrosss the resistor. At 800mA this means you need R=V/I=9.5V/0.8A=11.8 Ohm. 12 Ohm should work.


Once you have a stable circuit with a load resistor, you can go on examining the LED. DO you have a datasheet for the LED? I wonder abot the 12V LED. This is obviously a module (no single LED has 12V voltage drop). Maybe the module has an internal current limiting or regulating circuit. This would mean you could operate it on a plain 12V supply without additional circuitry. And maybe the regulator in the module and your current source interfere?
Also, with reference to the temperature dependence of an LED's forward voltage, this LED module might be happy about some cooling (heat sink).

 

[BlinkingLeds]

It's indeed a LM338 you can see it in one of the pictures where it's visible and it's from good
electronics store. The LM317 does the same.

without a led the current is stable.

I have soldered the 10w led to the wires now

You can see in the photo how the 10w led goes from 400mA to 500mA in just 1 minute

I don't think that the 10w led has any internal circuitry because if i connect it straight to 12v it draws 3 amps!!!

Ok let's forget this 10w led for a moment, why does the 1w led (see photo) behaves the same way???


Can someone pleaasee try it with an 1w led and see the results because it makes absolutely no sense

 

[Harald Kapp]

without a led the current is stable.

And is it the current you set, via the resistor or another value?

That kind of LED needs a heat sink.

 

[BlinkingLeds]

SOLVED!!!
Well not entirely ​

Because i didn't had this problem with small 20mA leds i went ahead and made extensive experiments with the 10w led to understand that strange behavior. I found that the cause is simple and almost funny .

Here is what i tried:
I used a 2.2ohm resistor on adj-out measured the amperage without any load = 0.57A this what i expected
Then i connected the 10w led, it started from 0.45A and it slowly begun to rise it also begun to heat up considerably, I decided to sacrifice a led to see what will happen. The current that it drawed continued to increase untill it reached 0.57A. I left it on for another 30sec to see if it goes any higher but it remained there. The strange thing is that when i attached a heatsink to the led, it didn't draw more than 0.47A as it remained cool all the time.

When the led drawed 0.45A it was about 25 degrees Celsius (75F).
If i use it in a flashlight what will happen in the winter when it's starting temperature will be -10C (15F) how much will it draw? 100mA? or even less? (I will put the Led in the freezer and try again)

How do i make it draw all the current that the LM338 is supplying even if the led is cold??

 

[BlinkingLeds]

well after having it in the freezer for 10 minutes i connected it in the circuit and it was drawing only 300mA it then slowly begun to rise. So in the winter + a big heatsink it will never pass 600mA even if i set it to 800mA?

 

[Harald Kapp]

Congratulations.

 

[BlinkingLeds]

So how do i solve the new problem? i don't want the led to give half of the luminosity when it's cold. I need it to draw all the current (800mA) from the beginning. Is there a way to do this?
Thanks

 

[(*steve*)]

Let me weigh in again, ask a couple of questions, and suggest a possible answer.

1) What voltage is your power supply (and what is your power supply)?
2) what is the Vf of the 10W LED?
3) what current have you designed your constant current supply for?
4) does the LED current ever exceed (3) above?

My guesses to these are that (2) is quite close to (1) -- within 3V --, and that the answer to (4) is no. (I wasn't sure of this earlier, and I'm still not, but it explains what is happening)

The possible reason is that the Vf is large compared to your input voltage and the 338 can simply not turn on enough to provide the full current for the LED. As the LED warms up and it's Vf drops, the 338 is able to supply more current, leading to more heat, lower Vf, and what is normally called thermal runaway.

In your case thermal runaway is probably limited to (3).

You say that adding a heatsink causes the eventual final current to be lower. This is what I would expect if the thermal runaway theory was true.

If this is all true, a couple of things are required:

1) A higher input voltage (you probably need only a volt or so) OR a current regulator with a lower overhead OR a LED with a lower Vf

2) A heatsink on the LED

You did have some anomalous results, but your use of a 1/4W pot set to 0.1 ohms was about three different types of mistake, so I'll ignore what happened with this.

 

[BlinkingLeds]

THANKS problem solved!
It was what you were suggesting : The supply voltage was too low (12v) and in order to draw 570mA (my current setting) the led needed 9v. I replaced the 12v power supply with an 20v power supply and it started to draw 570mA from the beginning.
Many thanks!

 

[BobK]

Hmm. I gave that answer way back, but you insisted that it happened also on an LED that required only 2.5V.

Bob

 

[BlinkingLeds]

It was happening, i don't know why, maybe because of lm317 overheating (it was veeryy hot i couldn't touch it)?