Re: Heatsink calculation

From: John Popelish (jpopelish_at_rica.net)
Date: 07/20/04 

Date: Tue, 20 Jul 2004 18:21:29 -0400

Rok Sitar wrote:

> "John Popelish" <jpopelish@rica.net> wrote in message
> news:40FC4266.7C372B57@rica.net...
> Rok Sitar wrote:
> >
> > Hi.
> > I'm wondering if anybody can help me. I made a voltage regulator with
> LM338T
> > and everything works OK. The problem is in heatsink. In have to calc it
> and
> > I don't know how. Can somebody help me. Information from data*** of
> LM338
> > are:
> > Rjc=4°C/W
> > Rja=50°C/W
> > Rcs=0,5°C/W
> > Ta=40°C
> > Pd=25W
> > Now can somebody give me the whole calc of heatsink.
> > THX to anybody.
>
> Rja is thermal resistance junction to ambient, when the device is used
> without a heat sink, so you can forget that value.
>
> Rjc is thermal resistance junction to the mounting surface of the
> case. Rcs is thermal resistance between case and heat sink (under
> some specified mounting setup, probably just a bit of zinc oxide paste
> between case and sink and a specified clamping pressure). You still
> need Rsa, the thermal resistance of your sink to ambient to add up the
> total thermal resistance between junction and ambient. That would be
> the sum of Rjc, Rcs, and Rsa. The total thermal rise above ambient is
> the power being dissipated by the junction (difference of input to
> output voltage times load current, for a regulator) times the total
> thermal resistance.
>
> John THX.
> But I was wondering if Anybody can calculate the heatsink from these given
> information.
> Heatsink info is in format xxx °C/W.
> So The question is CAN U CALC THE HEATSINK or NOT?
>
Not without knowing the maximum allowable junction temperature. This
is often 125 to 150 C, but I usually derate to something like 100 C.

Using 100 C as the maximum allowable junction temperature, That leaves
a 60 degree rise junction to 40 degree ambient.

So, (100 - 40) degrees = 25 watts *(Rjc + Rcs + Rsa)

Rsa = 60/25 - (Rjc + Rcs) = 2.4 - 4.0 - .5 = -2.1 degrees per watt.

Obviously it is impossible to keep this device below 100 degrees C in
a 40 degree ambient while it dissipates 25 watts.

So, hang the engineering safety factor and go for the full 150 degree
junction temperature.

150-40=25*(4 + .5 + Rsa)
Rsa=-.1 degree per watt.

Nope, still not possible. You will either have to use a bigger device
with a lower junction to case resistance, lower the ambient
temperature (water cooled sink), parallel two devices, so that each
only has to get rid or 12.5 watts, or do something else. 

-- 
John Popelish
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