Thermal Calculations (heat rise) - Help

I'm trying to determine if a simple voltage regulator circuit will operate within its allowable temperature range.

The device package is TO-252 (0.4 deg C/W), soldered to PCB that will act as a heatsink.

Power dissipated by the device is (12-5) * 0.25 = 1.75 Watts

Using TI's thermal calculator, it appears that if I have 8"^2 of copper, the regulator will be hot but should be within it's range.



However the PCB will be mounted in a sealed (gray) metal enclosure with a surface area of 0.5ft^2 and located outdoors in full sunlight. The maximum ambient temperature in this location will be 45 deg.C

Assuming 14W/ft^2 of solar energy, the enclosure would need to dissipate a total of 1.75 + 7 = 8.75W

Based on this calculation: **broken link removed**

It appears the temperature rise in the enclosure would be about 6 deg C (seems low?)

If that's correct the maximum temperature on the board would be about 90 deg C

Does this look correct?

All thermal resistances add to Tj which is the important metric to keep below 85'C in a good design.

YOu can test it under a 300W Work Lamp using a Halogen lamp at < 0.5m distance or so to simulate solar power for 250W in a 0.25 m2 area.

A better solution would be to use a spring clip to the case inside to conduct heat directly to the case rather than indirect via the topside copper with solder mask and thermal insulating Epoxy dielectric.

The best is to eliminate the heat loss with a 3 terminal switching regulator from OKI/MURATA that avoids the VI power drop loss.

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Have you considered these? or similar? 85~90% eff at 12V at 0.25A out

http://www.digikey.com/product-detail/en/7805SR-C/811-1116-ND/1926159

I wonder if a White powder coat is better than grey?

More details on solar cooling
**broken link removed**

Hi,

Power dissipated by the device is (12-5) * 0.25 = 1.75 Watts

Click to expand...

this is the dissipation of the regulator.
and

the enclosure would need to dissipate a total of 1.75 + 7 = 8.75W

Click to expand...

But now the circuit consumes 0.25A x 5V = 1.25W, as long as the power is not pissipated outside the box, with a motor or any other outside power consuming device.

If there is no power consuming outside the box, then you can calculate with the total voltage of 12V and 0.25A = 3W.

A 85% switching regualtor will reduce this to: 5V * 0.25A / 0.85 = 1.47W

Klaus

Jester said:

"...will be mounted in a sealed (gray) metal enclosure with a surface area of 0.5ft^2 and located outdoors in full sunlight". The maximum ambient temperature in this location will be 45 deg.C

Assuming 14W/ft^2 of solar energy, the enclosure would need to dissipate a total of 1.75 + 7 = 8.75W

Click to expand...

The document above mentioned, says that:

"Over much of the United States, the approximate peak values of solar radiation striking the Earth’s surface is 97 W/ft^2"

Click to expand...

At other countries located at similar lattitudes should not vary so much.
Therefore the calculation you did seems quite optimistic, perhaps not correct.