Luxeon current regulator on 6V and LM317

[jrn]

Hi to everybody,

I new in this forum

I want to power a 1.2 Watt Luxeon diode (3.6 Volt @ 350 mAmp) from a constant 6V DC battery. I found a website suggesting using an LM317 with only R1 and no capacitors. Since the formula is:
Vout = Vref (1 + R2/R1)
By setting R2 =0, Vout = Vref = 1.25 Volt.
In my design, R1 = 3.6 Ohm, (1/2 watt) drives the Luxeon diode @ 0.34A very close to the 350 mAmp suggested by diode manufacturer.

The design works fine in 12 Volt the only problem is that the LM317 is heating up. I’m loosing too much energy. Wasted power on the LM317 is 8.4 Volt x0.35 Amp = 2.94 Watt!!!!!!

To reduce the power loss and the heating problem, I decide to use a 6 Volt battery. This option works only if the battery is fully charged (6.5 Volt), because of the 3V dropout for the LM317. As soon as the battery get under 6.3 Volt, the current starting to decrease so does the light intensity.

My questions are:

1- Can I find a regulator who can do this EXACT SAME JOB with a lower dropout (around 1.5-2.5 Volt or less) to be able to use a 6V battery?

2 - Since R2=0 and I don’t use any capacitors, can that be a problem?

3 – I’m not very familiar with voltage regulator, it seems like some of them have “automatic shutdown” protection device. Is low input voltage can activate this shutdown? This would be great to protect the battery from deep discharge.

I need something very very very very simple, like the regulator and a resistor.

Thanks

[evan9162]

jrn,

First, the Vout formula doesn't apply to using the '317 as a constant current source. For the one-resistor, constant current source, the output current will equal 1.25/R.

Second, to solve your current dropping problem, get an LMS1585-ADJ, and replace your LM317 with it. It's a drop-in replacement for an LM317 in the constant current circuit. At the current level you're aiming for, the regulator will have about 0.8V of dropout. It'll stay in regulation until your battery drops to 5.6V.

[rdshores]

I would say that if the dropout voltage of the LMS1585-ADJ is .8 volts, then it would stay in regulation until the battery dropped to 4.4 volts(3.6 volt Luxeon + .8 volt regulator dropout).

[evan9162]

rdshores,

You still need to take into account the 1.25V drop across the current sense resistor (R). So 1.25+0.8 = 2V of dropout.

[rwolff]

Haven't tried this myself, but how about using the LM317 in current-source mode to provide the base current for a power transistor wired in common-emitter configuration, with your Luxeon in the collector circuit? You'd have to play with the resistor value due to beta on the transistor varying from one unit to the next (I'd suggest getting a couple of 10 ohm 10 watt resistors - try the circuit with one in place of the Luxeon, then add a second one in parallel). Once you've got the resistor "tweaked", you can install the Luxeon (after all, you want to run it in LED mode, not SED mode [img]/ubbthreads/images/graemlins/ooo.gif[/img]).

Another possibility (one that I'm considering for a project) is to use the regulator circuit as above and a 12 volt SLA battery to drive 3 series-connected Luxeons. After all, the power for the extra Luxeons is "free", since you don't draw any additional current, and the voltage drop would otherwise be wasted on your regulator. Why 3 and not 2? Because there's a triple optic available.

[evan9162]

rwolff,

Even with an external pass transistor, you're still dissipating the same amount of power. May as well keep things simple and just use the regulator and single resistor.

A '317 is right on the edge for being able to drive 3 Luxeons in series from 12V. The LMS1585-ADJ has about 2V less headroom, and works exactly the same way the '317 does, but will stay in regulation down to about 11V with 3 Luxeons in series.

[lux_manish]

how can a 6volt lead acid battery be regulated to work for a 3.6volt 350ma 1watt LED.Typically lead acid battery provides 6.8volt while fully charged and around 5.4volt before set in for deep discharge.If LM 317 drops 3volts ,then it won't be usefull.Some says it drops2volts.