Please help me with my LM317-T LED driver.

A while ago I bought a 10W 12V LED on line to see if I could build a small projector. Then somewhere, I hear you need to build a driver for it, otherwise you WILL blow the thing. So I start googling and tinkering and find a few basic formulas, P=IV, ==> P/V=I ==> 10w/12V=0.833A

So I need to get a current of 833mA at 12v to run this light, next step.

According to most sites, the cheapest way of doing this is with a simple LM317-T Regulator.
The ADJ voltage on this is 1.25v, so, V=IR ==> V/I=R ==> 1.25v/0.833A= 1.5 Ohm.

So I pick up a few regulators and a fist full of resistors at my local electronics shop, dust off an old soldering iron and thow it all together.

Then I plugged it on to a 12v 1.0A Power Adapter. I'd heard the 317 had a Vdrop of about 3 volts, so I took some readings to make sure everything was ok. Got 620mA between the Power/LED, the LED/LM317, and LM317 to the Power, all is good.
Voltage was 12.13v across the Power, 9.20v across the LED, and 2.91 across the LM317v (0.95 over the resistor). All the numbers looked fine.

Here's where it got tricky. Knowing the LED was rated for 12v, and the drop over the LM317 was ~3v, I dug up a 15.2v 1.2A Power adapter, and plugged that in instead.

When I took the readings however, my current was up to 810mA, but the voltage was still down at 9.56v. The Power adapter was at 15.20v, and the 317 was dropping 5.67v.

This is the bit I don't understand.

How can I get my volts back?

I know the LM317-T regulates its current via the resistor, but I'd assumed the voltage was dependent on whatever you plugged in, minus a constant 3v drawn by the regulator. This assumption is obviously wrong.

Is there a way I can get the voltage back?


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btaskiran6 years ago
R must be 15W or more.
XCracer7 years ago
 One thing you should know is that running it at its rated max current will be as bright as it will possibly get at said current.. increasing the voltage will only mean decreased efficiency.. E.G.  if your running at 810mA @ 9.46v then that means  its burning up 7.66watts.. now lets say for instance its running 833mA and a theoretical 12v that would mean you would be getting a not even noticeable boost in light (23mA worth) but your led will be burning up 10watts (12 x 0.833) and harder to keep cool, 

so in all i think the light is doing pretty well its all about the mA (amps) with leds.. as long as you get your desired mA rating (or close to) then everything should be fine.. you could have 2 identical  leds both running at 1amp but w will be running at 3.2v and the other at 3.35 and they can be running off the same driver.. but they will be identical (or close to) light output..
hi
pls help me LM317 driver led
Fuzzy3D (author) 7 years ago
I think I've found my problem.

While the LM317-T "is" rated for up to 1.5 amps, and "is" rated up to 32 volts, it is not capable of satisfying both of those values at the same time. I hit a sweet spot at 810mA and 9.56v.

I'm gonna pick up a more powerful regulator (LM338) and see if anything changes.
lemonie7 years ago
The 9.56V isn't going to go much higher, that's what you'll get. Run it up towards the maximum design current and have it properly heat-sunk. What you are losing is the heat dissipated in your power-supply.
These things run at 9-12V, what you measure is not a problem if they they work.


L
Fuzzy3D (author)  lemonie7 years ago
So there's no way to get 12v / 850mA out of a LM317? This thing is rated for 1.2 - 25v
lemonie Fuzzy3D7 years ago
I don't see that you'll get the 12V across the LED (the regulator is different) Run the LED up to the current it's designed to take, these materials have their properties and the come as you find them to a some extent.

L
Show your circuit.
This is the correct circuit for a fixed current 317 based regulator.

Steve
LED driver.PNG
Fuzzy3D (author)  steveastrouk7 years ago
Here's a rough sketch.That blob at the bottom is the LED =)
lm317circuit.PNG
It has a voltage drop of whatever it takes to between the supply and the load to get the current demanded.

Steve