LEDs, LED Drivers, Thermal Management

[EM_Guy]

I'm trying to wrap my head around thermal management of LEDs and LED driver circuits.

http://blogs.indium.com/blog/amanda...ure-of-leds-with-thermal-management-materials

A concern has been expressed that a lighting supplier needs to demonstrate (not sure how) that the LEDs of the lighting fixtures being supplied are properly thermally managed. I'm not sure that this is necessary, and if it is, I'm not sure what exactly I would require. The required performance of the lights (lumen output, CCT, etc.) has been specified. If the light fixture fails to perform as required, that would constitute a technical failure.

So, I guess I'm looking for some direction here. Does anyone know the basics of proper thermal management of LEDs and/or LED driver circuits?

The data sheet of the LED says that its absolute maximum junction temperature rating is 120 degrees Celsius, while the absolute maximum operating temperature rating is 100 degrees Celsius. It can also handle up to 180 mA of current and can dissipate up toe 594 mW of power and has a typical forward voltage rating of 2.88 V. It's thermal resistance can be as high as 19 degrees Celsius per watt. And it appears that its typical solder temperature is around 25 degrees Celsius.

$Tj = Ts + R \times P$

But something doesn't add up on the data sheet. If Tj = 120 degrees Celsius when Ts = 25 degrees celsius, then either the power dissipated needs to be more than 0.594 W or the thermal resistance needs to be higher than 19 degrees Celsius / watt, or both.

 

[Jeff Rosenbury]

The junction temperature is a maximum, making your equation an inequality (<=). Is the light supposed to operate at 100º C.? That seems high.

• Q Power to be dissipated
• T_j Junction temperature
• T_aAmbient temperature
• R Thermal resistance

R = ∆T/Q; You claim R = 19º/W. Is that given by the data sheet? If it is, it's low because there are several steps to get to ambient. They might include the heat sink compound and the case/heat sink. But I'll assume those are already figured in. If not, add the appropriate figures to the total resistance. (I'll cover that later just in case.)

So ∆T = RQ = 11.3º C.

For 100º ambient your Tj is 111.3º C. which is less than 120º so you are good.

Now, let's suppose you didn't include all the sources of resistance:

• ∆T_jc is the between the junction and the case (presumably 19º/W).
• ∆T_ch case to heat sink.
  
• ∆T_ha heat sink to ambient.

Your heat sink compound will have a thermal resistance value that is area dependent. 1º/in²W is generous. (You can get 1/10^th that if you pay lots for it.) So say your diode is 0.05 in², that adds another 11.9º (0.594/0.05). (A bigger diode or better heat sink could help here.):

∆T_ch = 11.9º C.

Next select a heat sink. To do that we first need to know what sort of air flow we have. Basically there's high (forced), medium (convection), or low (in a closed area). In addition you need to derate for altitude. I'll assume medium (natural convection) and a 0.9 derate for up to 1500 meters. I'll assume a stamped metal heat sink (likely the case to which the diode is mounted).

For a flat plate of area A in cm², R ≈ 1/0.0025A. So for a 5 cm², it is perhaps 80º/W. Derated call it 90º/W. Or ∆T_ha=47.5º

So over all we have T_j ≤ 120º; ∆T_jc = 11.3º; ∆T_ch=11.9º; ∆T_ha=47.5º.

∆T_ja=70.7º C.

120º is the absolute maximum. So the circuit should avoid temperatures over about 50º C. or 122º F. This doesn't seem acceptable to me. But I guessed at a lot of values. You might end up needing a real heat sink and more expensive compound. Stamped heat sinks run down to about 30º/W. Extruded ones go lower than 10.

 

[EM_Guy]

R = ∆T/Q; You claim R = 19º/W. Is that given by the data sheet? If it is, it's low because there are several steps to get to ambient. They might include the heat sink compound and the case/heat sink. But I'll assume those are already figured in. If not, add the appropriate figures to the total resistance.

R = 19º/W. That is given in the data sheet as a maximum value for thermal resistance (between the junction and the Ts measuring point). The typical value is R = 13º/W. The data sheet also says that this characteristic is for Ts = 25ºC (which I have to think is an error).

Now, let's suppose you didn't include all the sources of resistance:

• ∆Tjc is the between the junction and the case (presumably 19º/W).
• ∆Tch case to heat sink.
  
• ∆Tha heat sink to ambient.

Your heat sink compound will have a thermal resistance value that is area dependent. 1º/in2W is generous. (You can get 1/10th that if you pay lots for it.) So say your diode is 0.05 in2, that adds another 11.9º (0.594/0.05). (A bigger diode or better heat sink could help here.):

∆Tch = 11.9º C.

I'm trying to make sense of your units here, and I don't think it works. The thermal resistance is inversely proportional to the cross-sectional area of the thermal conductor.

$R = \frac{x}{A \times \kappa}$

So, it doesn't work to discuss thermal resistance area density.

 

[Jeff Rosenbury]

T_s might refer to the surface temperature of the diode. The plastic's thermal resistance may be a function of temperature. Of course 25º may not be the most useful choice of measurement temperature test values for your application, but it's what they gave. I would likely just use the 19º/W lacking other data. (But I'm lazy. Perhaps run some tests of your own? You can measure junction temperature by corralating it with charge mobility.)

Sorry I messed up that calculation. You are right. I've been playing fast and loose with my terms. I've used "thermal resistance" both as a per area and a per device/situation. I led myself astray. Then I got really confused.

∆T_ch = 1 º/in²W ÷ (0.05 in² x 0.594W) = 33.67º

I hope I misjudged some of your needs. You might get a better thermal connection with a polished and press fit hole or something. Thermal paste is designed to be both an electrical insulator and a thermal conductor. If you don't need the insulation (due to plastic parts perhaps) you might find a better thermal bonding agent. But it also fills micro gaps, so I don't think it can be left out without some other solution. (Metal paste?)

 

[EM_Guy]

$1 º/in^2W$

I still don't think you can do this.

$∆Tch = 1 º/in^2W ÷ (0.05 in^2 \times 0.594W) = 33.67º$

Dimensional analysis here shows that this doesn't work.

$R = \frac{x}{A \times \kappa}$

Thermal resistivity is inversely proportional to the cross-sectional area of the thermal resistor. So, if you divide both sides by A, you do indeed get the units you are using, but look what happens on the right side of the thermal resistivity equation!

 

[EM_Guy]

The bottom line here is that our lighting supplier just needs to demonstrate that the junction temperature of the LED never rises more than 20 degrees Celsius above the ambient temperature. I've been given the thermal resistivity between the junction and the case, and I have been given a maximum rating for the power dissipation in the LED. So, I just need the thermal resistivity between the case and ambient.

However, would other electronics in the lighting fixture (driver electronics) also cause the junction temperature to increase?

 

[Jeff Rosenbury]

Clearly I have no idea what I'm doing. I can only blame incipient dementia. You probably want to use your own figures.

I was finding the temperature, not the resistance. (The resistance was given, but in odd units by the website I cribbed it from.) But I should have multiplied, not divided. So that's ~ 0º and possibly incorrect as well.

Perhaps you should start over with an actual heat sink compound using its datasheet value. I'm feeling particularly stupid today, so I'll leave you to it.