Hi Wouter/Russell,

I am not getting the half junction to sink resistance... I can see it
from the electrical model analogy, but when I think about it, it does
not fell right. If I have one diode on the heatsink, then it only needs
to dissipate the heat generated by that diode, but when I have two
diodes on the same heatsink generating double the heat then it will be
harder for the heatsink to dissipate that heat hence Tj will be higher.
Hence in real terms you would say that the junction-to-sink resistance
is higher.
Sorry if I am missing something basic, but I am having one of those
cases of using these calculations for years for a single component and
now faced with more than one, things started not making sense.

Best Regards
            Luis
 

-----Original Message-----
From: piclist-bounces@mit.edu [mailto:piclist-bounces@mit.edu] On Behalf
Of Wouter van Ooijen
Sent: 23 February 2010 12:41
To: Microcontroller discussion list - Public.
Subject: Re: [EE] Heat Dissipation calculations

> I am just having some difficulty looking at it

without giving a specific answer: translate the heat-stuff to 
electronics. The thermal resistance is a resistor, a heat sources is a 
current source, and a temperature is a voltage.

It will surely matter when the diodes dissipate different amounts. Take 
the extreme cases: when all heat is dissipated by one diode, the 
junction-to-sink resistance of the other one is out of the equation. 
When they both dissipate the same (taking your word for it) you can 
replace them with one diode which dissipates the sum, and has half the 
junction-to-sink resistance.

-- 

Wouter van Ooijen

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