[Help] - Driving 1W LED

[technology]

Hi,

I'm a newbie building a simple 1w driver using http://www.instructables.com/id/Super-simple-high-power-LED-driver/.

The source is a 12v battery and LED is rated at 350mA. When I connect, the LED is not bright enough(seem like there is not enough voltage). Am I doing any wrong?

Please help!

 

[louie]

Welcome to CPF! I haven't tried that circuit, but I would ask:
-what resistor value are you using?
-what 12 volt battery are you using - are you sure it can provide enough current?
-what LED are you using? Are you sure it's not defective, or that the light output isn't normal?

 

[technology]

Thanks Louie.

I'm using 2w 3.9k resistor across vOut <---> Adj.
The source is a 12v automotive battery. Should have enough juice to power this.
This is Edison 1w(80lm) and they work perfectly(tested with 3xAA-1.2v battery).

I guess, i might be using a wrong resistor? The instructables article says 3,9 resistor. Should I use 3.9 resistor instead of 3.9k ?

 

[louie]

I guess, i might be using a wrong resistor? The instructables article says 3,9 resistor. Should I use 3.9 resistor instead of 3.9k ?

Yes! That is how I read the instructions. The writer appears to be using the Euro convention of a comma in numbers where Americans use a dot (decimal point). Ergo, you want about a four ohm resistor.

The LM317 data sheet from National Semiconductor confirms this is the appropriate range. http://www.national.com/ds.cgi/LM/LM117.pdf

 

[technology]

Thanks Louie. Yeah, changing the resistor to 3.9 ohms did the trick.:twothumbs

 

[SirJMD]

Yes! That is how I read the instructions. The writer appears to be using the Euro convention of a comma in numbers where Americans use a dot (decimal point). Ergo, you want about a four ohm resistor.

The LM317 data sheet from National Semiconductor confirms this is the appropriate range. http://www.national.com/ds.cgi/LM/LM117.pdf

Dot or not doesnt matter. The proper way to write resistor values are like this: 3900 or 3k9. If he writes 3,9 or 3.9, he means three point nine ohms.

If someone writes 3M9, its three million and nine hundred thousand ohms.


If you remember this, i can save you alot of hassle - and might even some money Just imagine if it had been the other way around.. if it should have been 3k9, and you had choosen 3,9. Bzzzzz and smoke

 

[JohnR66]

Just a tip. For using the LM317 regulator in constant current mode, you can calculate the current by dividing 1.25 by the resistance. 1.25/3.9=.321a.

Conversely, you can estimate the resistor needed by dividing 1.25 by the current: 1.25/.5=2.5 Ohms

The LM317 is really a circuit full of amplifiers and it should be decoupled from the power supply by adding a .1uf mylar capacitor from the input pin to ground.

To check if you have enough power supply voltage for your circuit, Add up the foward voltages of all your LEDs and add 3 on top of that. Ex: 3 LEDs with Vf of 3.1v is 9.3 + 3 = 12.3 volts minimum.

 

[Dave_H]

Yep, the LM317 despite being around about 30 years or so is
still a good solid and easy-to-use current regulator, up to 1A or so.
Problem can be the 2-3v overheard required. If the LED Vf is a
bit higher and/or battery voltage lower, 3 LEDs might bottom out,
and running two series LEDs would have low efficiency.

In the case of not wanting to go to switching convertors, it's
possible to use a "high-compliance" current regulator with well
under 1v overhead. I designed one some time ago using a National
LM10CN op-amp (which has a stable 200mv reference built-in)
and a power MOSFET (IRFD120 or similar). Going from memory
I recall getting ~ 0.35v drop at 100mA even using sub-optimal
components. At 350mA it should be possible to keep drop down to
0.5v , using a better MOSFET such as IRFZ30/40.

There are probably other simple designs out there, just haven't
checked for any...something with an LDO (low-dropout) linear
regulator perhaps.

 

[SirJMD]

Yep, the LM317 despite being around about 30 years or so is
still a good solid and easy-to-use current regulator, up to 1A or so.
Problem can be the 2-3v overheard required. If the LED Vf is a
bit higher and/or battery voltage lower, 3 LEDs might bottom out,
and running two series LEDs would have low efficiency.

...

I'd say, that the voltage drop across the LM317 times the current drawn, equals the power dissipated in the LM317 - aint that right?

If i remember right, the LM317 eats around 2.5-3V - with a current @ 1A, thats quite alot of power lost. Might wanna add a heat sink to it!

Efficient? Nope.
Easy to use? Indeed !

 

[Dave_H]

I'd say, that the voltage drop across the LM317 times the current drawn, equals the power dissipated in the LM317 - aint that right?

If i remember right, the LM317 eats around 2.5-3V - with a current @ 1A, thats quite alot of power lost. Might wanna add a heat sink to it!

Efficient? Nope.
Easy to use? Indeed !

That's a pretty fair summary. However at 350mA the LM317 will
probably "typically" be OK with 1.5-2v dropout. The builder is using
a heatsink, and in this case dissipation would be roughly a watt.

For three LEDs typical efficiency isn't bad (10.5v/13v = 80%), assuming
LED Vf could be up to 3.5v each.

Dave

 

[TorchBoy]

Nice thread.

 

[technology]

Thanks for you suggestions. I need further help on a problem.

Even though LM317 is with a heat sink, it heats up considerably and shuts down after 10 minutes. The current is sourced from a 12v automotive battery and the LEDs(2x350mA are connected serially) through LM315 and 3.9 ohms 2W resistor. I know LM317 can operate up to 1A quite well but It may be interesting to know why it heats fast.

 

[TorchBoy]

Even though LM317 is with a heat sink, it heats up considerably and shuts down after 10 minutes. The current is sourced from a 12v automotive battery and the LEDs(2x350mA are connected serially) through LM315 and 3.9 ohms 2W resistor. I know LM317 can operate up to 1A quite well but It may be interesting to know why it heats fast.

I'm not quite sure of your setup from that description, but you could easily be dissipating 2 W of heat with the LM317. That's enough to heat it up quite quickly. For a couple of dollars or so you could get a driver that could run them at about 95% efficiency, meaning much less waste heat.

 

[zzonbi]

"but It may be interesting to know why it heats fast."

simple, it's just an in series regulator which burns the excess energy.
for efficiency you need a converter, a switching circuit with an energy buffer(reactive elements inductors, capacitors).

 

[tonycollinet]

Here is your problem

You have a 12V supply. Assume the LEDs are running at about 6V (3V each)
There is 1.25V across the resistor.

That leaves 4.75V across the regulator. At 350mA it means you are dissipating 1.6W in the regulator.

However, the battery may be putting out as much as 14V (measure it). In which case the regulator has 6.75V - 2.36W.

Are you sure you have the leds in series - you seem to have three wires going to them - how are they wired? If you have them in parallel, there is another 3V for the regulator - taking it to as high as 3.4W

Your heatsink is quite small, probably around 30C/W - so with 2W (assuming very good thermal contact) you'll be getting a 60C temperature rise on the case of the regulator - so it will be sitting up at 80C on the case.

If you are as high as 3.4W, then your case temperature may be as high as 100C above room temperature.

Try pointing a fan at it - or get a switched mode regulator cheap from deal extreme.