> 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.

You have it correct.
What you are saying is what I tried to show
ie note here fo the sink calculations I have Q(x+y) - the total heat
flux from all sources 9two in this case) flowing through the common
Rth_Sa

I said:
>>
Here the total power flows from sink to ambient so
Call diodes x & y.

Tsa =3D Q(x+y) . Rth_sa
so
Ts  =3D Ta + Tsa =3D Ta + Q(x+y) . Rth_sa

Now you have the common sink temperature.
>>

Q(x+y) is the combined heat flux.
Rth_sa is the sink to air resistance shared by both of them.

So, you have it right, "go round power please" . :-)
Just keep going from there


    R



> 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
> =A0 =A0 =A0 =A0 =A0 =A0Luis
>
>
> -----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
>
> -- -------------------------------------------
> Van Ooijen Technische Informatica: www.voti.nl
> consultancy, development, PICmicro products
> docent Hogeschool van Utrecht: www.voti.nl/hvu
>
> --
> http://www.piclist.com PIC/SX FAQ & list archive
> View/change your membership options at
> http://mailman.mit.edu/mailman/listinfo/piclist
>
> --
> http://www.piclist.com PIC/SX FAQ & list archive
> View/change your membership options at
> http://mailman.mit.edu/mailman/listinfo/piclist
>

-- =

http://www.piclist.com PIC/SX FAQ & list archive
View/change your membership options at
http://mailman.mit.edu/mailman/listinfo/piclist