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Hi,

under "absolute maximum ratings"
there is
* 2W @ TA = 25°C and
* 30W @ TC = 25°C

--> If you keep on this, then you are on the safe side ... What "rule of thumb" do you expect?

I agree: the usual way is to calculate with junction temperature. But this is only "calculation", because usually you can´t measure junction temperature.

***********
The datasheet also gives the "Power derating chart" in Figure 4.

--> here you simply can see how much power you may dissipate at the given case temperature and vice versa.

It tells you the max. case temperaure of 150°C with "no power dissipation" --> thus (independent of any thermal resistance) the junction temperature is 150°C, too.
--> so you may expect TJ(max) to be 150°C.

The chart also tells you that you may dissipate 30W at 25°C case temperature.
Now assume the TJ(max) is 150°C... then you have a delta t of (150°C -25°C) at 30W.... which gives a R_th_JC of 4.2 K/W.

I think this is all you need to know.

Klaus
 
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KlausST said:
Hi,

under "absolute maximum ratings"
there is
* 2W @ TA = 25°C and
* 30W @ TC = 25°C

--> If you keep on this, then you are on the safe side ... What "rule of thumb" do you expect?

I agree: the usual way is to calculate with junction temperature. But this is only "calculation", because usually you can´t measure junction temperature.

***********
The datasheet also gives the "Power derating chart" in Figure 4.

--> here you simply can see how much power you may dissipate at the given case temperature and vice versa.

It tells you the max. case temperaure of 150°C with "no power dissipation" --> thus (independent of any thermal resistance) the junction temperature is 150°C, too.
--> so you may expect TJ(max) to be 150°C.

The chart also tells you that you may dissipate 30W at 25°C case temperature.
Now assume the TJ(max) is 150°C... then you have a delta t of (150°C -25°C) at 30W.... which gives a R_th_JC of 4.2 K/W.

I think this is all you need to know.

Klaus
Click to expand...


Why 4.16 kelvins not centigrade?
 

Hi,

Delta_K is the same as delta_°C.

But Kelvin is referenced to the absolute zero temperature
And °C is referenced to the temperature when water freezes.

Klaus
 
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KlausST said:
Hi,

Delta_K is the same as delta_°C.

But Kelvin is referenced to the absolute zero temperature
And °C is referenced to the temperature when water freezes.

Klaus
Click to expand...

Just seems a bit more useful to go along with centigrade since every thermally related spec in the datasheet is in depicted in centigrade. 4.16 C/W seems almost useless without gang arrangements for power applications - package size for BJTs seems to inhibit optimal thermal performance far more than for MOSFETS.
 

If you look thoroughly at semiconductor or heat sink specifications, you'll notice that the commonly used unit of thermal resistance is K/W rather than "C/W". Generally, it's common practice in science and engineering to express temperature intervals in K.
 
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FvM said:
If you look thoroughly at semiconductor or heat sink specifications, you'll notice that the commonly used unit of thermal resistance is K/W rather than "C/W". Generally, it's common practice in science and engineering to express temperature intervals in K.
Click to expand...

The thermal charts are typically in centigrade.
 

It looks like European manufacturers are preferring K/W while US vendors stick to °C/W. It's an arbitrary choice in any case.
 
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Hi,

Sadly the datasheet gives no additional informations.
Thus we need to calculate with r_th_jc of 1.38 °C/W (or K/W).

Klaus
 
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How about verifying junction to case thermal impedance for diodes?
**broken link removed**
The 0.25c/w appears unlikely because

Vf = 0.96v
Imax = 90A per devices @ 25c
Total Power = 0.96v * 90A = 86.4W
Max Operating Temp = 175c

Thermal resistance would be (175c - 25c) / 86.4w = 1.74C/W

Im not sure if this is the proper method to obtain it for diodes.
 
Last edited:

Hi,

You misread the datasheet.
The given 0.25°C /W is "case to heatsink", not junction to case.

Thus the value is only a small part of the total thermal resistance.... and btw not included in r_th_jc.

Additionally you forgot to include the "50% duty cycle" information in your calculation.

Thus the dissipated power is only half of your calculated value:

Klaus
 

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