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### Topic: LM317T Driving me nuts (Read 15368 times)previous topic - next topic

#### SgtOneill

##### Jun 30, 2011, 05:45 am
LM317T is driving me nuts.

Power Supply 5v 2Amps
LED pulls 3.6v and 1Amp (works fine with 1.5amps)

I want a simple voltage drop, from Whatever voltage input (from 1.2v to 35v according ot the datasheet) to 3.5 / 3.6v output.

This doesn't work. Should be vOuting 3.5volts or so. Its fixed at 1.25v no matter what.

This works but it seems insane to leave the Adj pin disconnected from everything...

Any idea what might be wrong here?

#### dshay

#1
##### Jun 30, 2011, 05:56 amLast Edit: Jun 30, 2011, 05:58 am by dshay Reason: 1
A) Please provide values for R1 and R2

B) Please provide the datasheet's recommendation for nominal current on that LED.

Your problem lies with B. You need a current limiting resistor.

If you need help choosing one, kindly visit the following website:

http://led.linear1.org/1led.wiz

You'll need to plug in values for Voltage Source, LED Forward Voltage and LED Nominal Current, latter two you will get from the LEDs datasheet.

#### SgtOneill

#2
##### Jun 30, 2011, 06:32 amLast Edit: Jun 30, 2011, 06:50 am by SgtOneill Reason: 1
R1 = 220ohm
R2 = 470ohm

(this should make vOut 3.7v)
I used this circuit and calculator: http://www.whatcircuits.com/lm317-calculator-v2/

LED Datasheet http://www.seoulsemicon.com/en/product/prd/zpowerLEDp7.asp
3.6v 2.8amps
3.6v voltage drop

The weird thing is that, the LED lights up if i dont ground the ADJ pin, making the 470ohm useless. So the ADJ and vOUT pins are connected with teh 220 resistor, the LED is being fed by the vOut pin with 3.x volts (not sure how) and grounded.
I tried 2 LED's in series and the vOut pin cranks to 6volts wich seems correct but... should I be the one choosing the voltages?

im confused =)

#### MarkT

#3
##### Jun 30, 2011, 06:55 am
That regulator has about 2.2V drop-out at 1.5A, so the greatest output voltage you can get from 5V input is 2.8V

Also you are missing decoupling capacitors - regulators in general need them to prevent oscillation.
[ I will NOT respond to personal messages, I WILL delete them, use the forum please ]

#### dshay

#4
##### Jun 30, 2011, 06:59 amLast Edit: Jun 30, 2011, 07:02 am by dshay Reason: 1

That regulator has about 2.2V drop-out at 1.5A, so the greatest output voltage you can get from 5V input is 2.8V

Also you are missing decoupling capacitors - regulators in general need them to prevent oscillation.

He's good on drop-out, his source is 9VDC.

Decoupling caps would be highly recommended but the problem here is likely thermal shut down, no current limiting resistor is going to put this voltage regulator into shut down mode immediately as it's going to peg at 1.5 Amps.

Reconsider your power source. I recommend an LED driver made for this.

#### SgtOneill

#5
##### Jun 30, 2011, 07:03 am
my source isnt the 9volt battery xD thats just a draw i happen to make with Fritzking.

i tried with a source of 12v and a source of 5v 2amps.
the caps are optional, i dont need the filter the noise for now, ill probably add them later.

I do need to add an Heat Sink and see if its good. I'll try that later tonight see if it works.

#### dshay

#6
##### Jun 30, 2011, 07:09 amLast Edit: Jun 30, 2011, 07:28 am by dshay Reason: 1
If you want you could run this beacon at 1.4A with a big heatsink on that voltage regulator, but you're going to need to modify your circuit.

Use a 12V Source again.

Run your output on that Voltage Regulator at 5V.

Throw a 5 Watt resistor value 1.2 Ohms in Series with your LED.

http://search.digikey.com/scripts/DkSearch/dksus.dll?Detail&name=1.2W-5-ND

LEDs act as DEAD SHORTS, you have to pay attention to their specs on current and put a resistor in there after doing a little math related to LED forward voltage and Current Limiting Resistors.

Google these terms and you'll find a lot of information.

#### dshay

#7
##### Jun 30, 2011, 07:12 am

I do need to add an Heat Sink and see if its good. I'll try that later tonight see if it works.

Does not work. Maybe works for short period, then:

A) LED burns out

B) Voltage Regulator Burns out/ keeps clamping down in thermal shut down.

You're not going to get around basic solid state theory on this one.

Current limiting resistor is mandatory.

#### MarkT

#8
##### Jun 30, 2011, 07:45 am

He's good on drop-out, his source is 9VDC.

It's 5V  2A, reread the posting...
[ I will NOT respond to personal messages, I WILL delete them, use the forum please ]

#### ParallelLogic

#9
##### Jun 30, 2011, 08:30 am
Isn't the center pin of a regulator normally connected to ground?  At first glance I'd expect to see a wire going from the center of the regulator to the ground of the battery.

Also, I'm very curious, what program did you use to make these graphics?  Fritzing?

#### MarkT

#10
##### Jun 30, 2011, 11:46 am

Isn't the center pin of a regulator normally connected to ground?  At first glance I'd expect to see a wire going from the center of the regulator to the ground of the battery.

Also, I'm very curious, what program did you use to make these graphics?  Fritzing?

For a fixed voltage regulator the sense input is normally grounded, but its not a requirement even then, the regulator simply acts to keep the voltage difference between output and sense constant.
[ I will NOT respond to personal messages, I WILL delete them, use the forum please ]

#### SgtOneill

#11
##### Jun 30, 2011, 11:56 am
@ParallelLogic : yes i did use Fritzing

Ok so from what i've read in this topic, its probably Thermal shutdown and the vOut is constant 1.25v.

A heatsink wont help for long since the regulator is being hit by 2amps.
Since i want to connect 2 LED's in series, im guessing thats 2amps total but about 7 / 8volts due to voltage drop on the first LED being 3.6v.
To fix this i need a 5Watt 1.2 Ohms resistor in series with LED.

Guess i'll need to go to the electronics store before going home =)

The 5Watt 1.2ohm resistor is going to limit the current the LED is sucking so the Voltage Regulator wont shutdown due to  thermal problems.

Is there a better way to do this? Drop 12v to 3.6v to feed a 3.6v 2amp LED?
I noticed that theres alot of heat being generated, wich means the circuit is being less efficient.
LED gets hot, Voltage regulator gets Hot, and probably the 5Watt 1.2Ohm resistor will get hot too.

#### SgtOneill

#12
##### Jun 30, 2011, 03:24 pm
Something came to my mind. If its the Thermal overload protection kicking in because of the 2Amps, maybe if i use a LM318 it will work fine.
LM318 handles up to 5Amps and that should avoid the thermal overload protection =)

Damn it, cant wait to get home.

#### floresta

#13
##### Jun 30, 2011, 08:41 pm
Quote
LED pulls 3.6v and 1Amp (works fine with 1.5amps)

Your problem starts right here with a common misconception.  Unless it is a very unusual LED it does not 'pull' 3.6 v or 1Amp.  You LED has an absolute maximum current rating and a nominal current rating.  A current limiting resistor [font=Verdana]MUST[/font][/color] (I would make this flash if I knew how) be used in series with the LED.  The resulting forward voltage that appears across the LED is dependent on the current flowing through it, you can find typical values in the data sheet for the LED.   Your LED has probably been saved from destruction by the fact that the regulator has been shutting down.  If you substitute a regulator that does not shut down then you will be replacing the LED rather quickly.

Don

#### SgtOneill

#14
##### Jun 30, 2011, 08:58 pm
I see.

Well, i replaced the LM317 with the LM338 and its working now.
However, has most of you said, the fact that im not using a current limiting resistor is probably killing the LED's (i have now 2 LED's in parallel).

I need to buy that resistor, i have nothing similar to that. 5W resistor is new to me =)

Each LED Absolut Maximum Current flow is 2.8AMPS.
This ones already have a Heatsink below them wich helps with the heat.

How could i limit the current to 2amps? how many ohms on that current limiting resistor? also, since i have 2 LED in parallel, 1 resistors for both is enough? or should i have 1 resistor for each LED?

Thanks for the help everyone

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