1. thermal runaway?

Hi everyone,
noticed a thing here, I cant understand...

I have this quad light here at the final (mechanical) stages:

before final installation with driver and sense resistor, I have my lights run on different levels for some time, just if that might make some differences to non used emitters.

The setup is as shown (wired) + a 4-cell Li-Ion pack + a large 5W poti.
When set to any drive current, this one increases (sometines considerably) when the assembly gets warmer.

Should the emitters not get less efficient when warm?

2. Re: thermal runaway?

I assume you are talking about brightness when you say it increases...?

My guess would be that quite a small voltage is dropped by the pot; so an increase in temperature which results in lower Vf in the LEDs, will result in more current so they get brighter.

3. nono, current
say:
set to 500 mA --> just gone up to almost 520-550
its at least 20-50 mA at any current level
or, when totally cold before lighting up 500 --> 600 mA (extreme value, happend once)

PS: as to brightness: that poti has some kind of "klick" positions,
when current over the whole assembly is within 130-250 mA, I dont notice much increase between clicks, but from 300 on, each step (usually 30-40 mA more) is a noticeable increase in brightness.

4. Re: thermal runaway?

I think Jur is right: when led heats up the Vf decreases slightly. Expressing the same phenomenon in other way, the input current increas at any given input voltage level.

-N

5. Re: thermal runaway?

Yellow;

As the Cree's heat up, the voltage across them will drop.

When I built a triple Cree, the voltage dropped by 1 volt total as it heated up the 3 emitters.

You might need more heat sink, or a change to a copper model.

Fins with lots of surface area dissipate the heat.

Larry Cobb

6. Re: thermal runaway?

Yeah, from your description and voltage levels involved, there is not much voltage accross the pot; so as the LEDs warm up, their Vf drop; this puts more voltage accross the pot and consequently the current increases. Simple, really.

7. Re: thermal runaway?

Originally Posted by yellow
Should the emitters not get less efficient when warm?

The diode is a silicone semiconductor, when heated these semiconductors become more conductive, metals on the otherhand become more resistive.

Resistance is a function of voltage, so when the die warms up its Vf will drop, current will increase.

EDIT: Thats part of the reason I like to use constant current drivers

8. Re: thermal runaway?

Even a pretty small series resistor -- enough to cause a few tens of mV or more drop -- will go a long way in reducing the current rise. The larger the resistor, the more stable it will be, but you shouldn't have any trouble getting it adequately stable without sacrificing an appreciable amount of efficiency.

c_c

9. Re: thermal runaway?

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