Heatsink calculation

[Rok Sitar]

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 datasheet 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.

 

[John Popelish]

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 itand
I don't know how. Can somebody help me. Information from datasheet 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.

 

[Rich Grise]

probably just a bit of zinc oxide paste

Zinc oxide? Isn't that the stuff the lifeguards put on their nose?

 

[John Popelish]

Zinc oxide? Isn't that the stuff the lifeguards put on their nose?

Yes, except that the heat transfer version is silicone grease based.

 

[Rok Sitar]

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?

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 datasheet 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 Popelish]

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.