Thermal Runaway prevention

[Braxus]

Got a couple of questions for you electronic/circuit gurus.

From what I understand, resistors generally reduce the voltage from a source and the higher ohm value it is, the more voltage it shaves off.

Now with LEDs I've read that even if you have say a 3V LED and a 3V battery, it's still better to get a battery with a slightly higher voltage rating and slap on a resistor somewhere in the circuit, even if the resistance is a mere 0.10 ohm. Main reason is to prevent thermal runaway as LEDs will start to draw more amps the longer/warmer they run.

Now what I don't understand is how resistors actually prevent thermal runaway. I thought resistors didn't really do much for limiting amperage draw as whatever your powering decides on how much amps it wants to eat. I'm sure there is some easy answer for this, but I can't seem to find it (that or I'm just an idiot! lol)

Driving two 6.84v/700mA LEDs from a 7.5v 6xAA pack in parellel using two 1 ohm/5w resistors which I think is about right. Got a 2w rated resistor earlier but did the math later and looks like the 2w unit would fry and burn if I used it. Should I play it safe and get a 10w rated resistor? The circuit seems to be creating around 4.8w which is somewhat near the edge of the 5w units I got.

 

[Aluvus]

The simplest model of an LED (or any diode) is a switch with a voltage drop. If the voltage you input into it is greater than its turn-on voltage (say 3 V), then it behaves pretty close to a short circuit in terms of current draw. And it has a 3 V drop across it. Without a resistor, it will let through a tremendous amount of current and probably kill itself/the battery. This is Bad. It's worsened by an interesting property that semiconductors have: when they heat up, they become better conductors. So as they heat up, they let through more and more current.

If you have a 7.5 V source and 6.84 V LEDs, the voltage you need to drop across the resistor(s) is about 0.7 V. From Ohm's Law, that means for a current of 700 mA you would need a 1 O impedance in series (this is for each LED). You would be throwing just under 0.5 W into each resistor. And you would have some pretty beefy LEDs.

A note on resistor power ratings: those ratings are assuming you keep it at a constant temperature. Offhand, I believe the standard temperature used is 25C. If the resistor heats up (and well, it usually will if it has to dissipate any amount of heat) the rating can drop substantially.

 

[Braxus]

Ah thanks for the reply. Think the semiconductor tidbit answered my question.

You said that I'll be throwing roughly 0.5w into each resistor. Thought it would be around 4.8?

Given that the LEDs want 6.84v so...

6.84 * .7 = 4.788 [voltage x amps = watts]

Wouldn't I be throwing closer to 5w into each resistor instead of 0.5? Hence the requirement for a 5w rated resistor (10w if playing it safe).

Haven't worked with electrical circuits for ages and the class I took that did this stuff was a good 6 years ago. lol Just picking up what I read online in attempt to give my brain a refresher course.

 

[Varun]

You are calculating the power dissipated by the LEDs, not the resistor.

The resistor will only drop the remainder of the voltage past the VF of the LED.

0.7V/0.7A=1Ohm resistor needed.

0.7V*0.7A=0.49W on the resistor

 

[Varun]

I had to run before finishing my reply.

I just wanted to say that if those LED's can handle 700ma @ 7V, they are going to be bright enough to make you blind. You will likely want to use a larger resistor to keep the current down.

 

[Rubycon]

What LED runs on voltage that high? It would have to be ballasted in the first place.

 

[Braxus]

Lumileds' Luxeon V series run at 7V/700mA. They're roughly 100 times brigher than your standard 5mm LED supposedly.

They're backed my a thin aluminum heatsink but it's recommended that you epoxy another one on for extra cooling if you run them at full power.

 

[Rubycon]

A lux 5 eh? Better drive that with an upswitcher realizing regulation. Those side emitters are incredibly brilliant when mounted in a decent reflector.

 

[JimPhelpsMI]

Hi, To simplify: LEDs are designed for a certain max current. Most small LEDs are rated about 20 ma. max and will run at 2.2 volts. They make a pretty good voltage regulator if you need 2.2 V. The resistor is calculated for the proper current below or up to the Max. Max current will produce the most lite from the LED. It will still glow down to about 5 ma or less. Calculation is source voltage (5 ?) minus LED voltage (2.2) = excess voltage (2.8). Excess V (2.8) divided by desired current (10 ma ?) = 280 ohms. Wattage of the resistor is 2.8 v divided by 280 ohms = .01 watts. Use a 1/4 watt resistor nearest to 280 ohms.
Hope this helps, Jim

 

[GalvanizedYankee]

http://www.cpemma.co.uk/index.html > Eye Candy.

Hope this helps


...Galvanized

 

[Aluvus]

Originally posted by: JimPhelpsMI
Hi, To simplify: LEDs are designed for a certain max current. Most small LEDs are rated about 20 ma. max and will run at 2.2 volts. They make a pretty good voltage regulator if you need 2.2 V. The resistor is calculated for the proper current below or up to the Max. Max current will produce the most lite from the LED. It will still glow down to about 5 ma or less. Calculation is source voltage (5 ?) minus LED voltage (2.2) = excess voltage (2.8). Excess V (2.8) divided by desired current (10 ma ?) = 280 ohms. Wattage of the resistor is 2.8 v divided by 280 ohms = .01 watts. Use a 1/4 watt resistor nearest to 280 ohms.
Hope this helps, Jim

One slight correction: wattage rating of the resistor would need to be at least the voltage squared divided by the resistance. 2.8^2/280 = 0.028 W. Equally, and sometimes more simply, you can multiply the voltage and current (2.8 * 0.01) and get the same result. Again, a 1/4 W resistor would do just fine for that case.

 

[Navid]

You asked about thermal runaway.

It is explained in a very short form here or here.

You asked why a resistor can help with thermal runaway.
If you place a resistor in series with an LED (a diode) and place a constant voltage across the combination, the diode still warms up and its effective resistance still decreases due to that.
But, the resistance of the resistor does not drop with temperature (at least, not as rapidly as that of a diode). So, the increased current (due to the reduction of the diode effective resistance) will result in an increased voltage drop across the resistor since the diode current has to pass through the resistor too. The increase voltage drop across the resistor, results in a decrease in the voltage across the diode. This reduces the current of the diode.

Placing a resistor in series acts as a negative feedback and the temperature (and current) stabilizes.

 

[Aluvus]

Originally posted by: Navid
You asked about thermal runaway.

It is explained in a very short form here or here.

You asked why a resistor can help with thermal runaway.
If you place a resistor in series with an LED (a diode) and place a constant voltage across the combination, the diode still warms up and its effective resistance still decreases due to that.
But, the resistance of the resistor does not drop with temperature (at least, not as rapidly as that of a diode). So, the increased current (due to the reduction of the diode effective resistance) will result in an increased voltage drop across the resistor since the diode current has to pass through the resistor too. The increase voltage drop across the resistor, results in a decrease in the voltage across the diode. This reduces the current of the diode.

Placing a resistor in series acts as a negative feedback and the temperature (and current) stabilizes.

Resistance of a resistor will actually rise as temperature goes up, thus increasing the voltage drop and keeping down the current even more so than if it were constant. As you say, it prevents the current from going crazy.

 

[Navid]

There are resistors whose resistance increases with temperature (positive temperature coefficient). There are also resistors whose resistance decreases with temperature.


You have a constant voltage across the combination of a diode and a resistor in series.

How much of the constant voltage drops across the diode and how much of it drops across the resistor has to do with the relative resistance of the two components.

If temperature increases, the effective resistance of the diode drops.
Then, a smaller portion of the voltage will drop across the diode. A small reduction of voltage for a diode results in a significant reduction of its current (exponential IV characteristic). That is what stops thermal runaway.

So, even a resistor whose resistance decreases with temperature can still stop thermal runaway since a resistor has a linear IV characteristic.