An easy "trick" when you have a thermal "Y" structure is to start at the
"bottom" ambient and work up. There are some issues with this simplitic
approach but it will probably give you a good enough result.

NB - the following is far far farmore horrid to read and write than to
understand and to calculate.
It ends up as advanced common sense once you follow it.
E&OE - odds are I've got a subscript etc wrong below somewhere

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

Tsa = Q(x+y) . Rth_sa
so
Ts  = Ta + Tsa = Ta + Q(x+y) . Rth_sa
Now you have the common sink temperature.

Each diode or other device can now be individually calculated.

Tjx.s = Qx.Rth_jx.s

Tjx = Ts + Tjx.s = Ts + Qx.Rth_jx.s

or

Tjx = Ts + Qx.Rth_jx.s   =
Tjx = Ta + Q(x+y) . Rth_sa  + Qx.Rth_jx.s

Similarly

 Tjy = Ta + Q(x+y) . Rth_sa  + Qy.Rth_jy.s

Words - easier here

Total energy flow through sink to ambient cause sink rise based on sink to
ambient resistance.
Then
Junction to sink rise resulsts from junction energy flow through junction to
sink thermal resistance.

If Junction to sink resistances and energy flows are the same for both
devices then junction temperatures are the same.

In that case you can treat it as one devie with double the energy flow of
one device and half the thernmal junction to sink resistance.

FAR easier to understand than to read :-)


               Russell McMahon





On 24 February 2010 01:20, Moreira, Luis A <Luis.Moreira@ccfe.ac.uk> wrote:

>
>
> Hi All,
> I have an existing assembly with two power diodes in series mounted on
> an aluminium heat sink, I am increasing the power dissipated on them and
> I am trying to do the calculations, to make sure that the diodes will
> still be fine using the current setup.
> I started by modelling the assembly:
>
> I will have two branches in parallel of Rjunction-case in series with
> Rcase-heatsink and they will be in series with Rheatsink-ambient.
>
> I am just having some difficulty looking at it and need someone to check
> if I am thinking correctly. Both diodes will dissipate, in this case,
> the same power. The heat will flow from both to the heat sink and then
> to ambient.
>
> Hence if I use the formula
>
>                  Tj -(Tcase + Theatsink + Tambient) = Q*(
> Rjunction-case +                                Rcase-heatsink +
> Rheatsink-ambient)
>
> Where Q is the sum of power dissipated on both diodes
>
>
> In this case as the diodes are identical is it correct to say that the
> Tj you get out of this calculation is the junction temperature for both
> the diodes which is also the Tj each of them?
>
> How would you do it for two diodes dissipating different power? It seems
> to me that it would not matter, but at the same time I can not be sure,
> especially if this power is pulsed instead of continuous.
>
> Any thoughts welcomed.
>
>
> Best Regards
>             Luis
>
>
>
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