heatsink calculations

[metebalci]

Hi,

I want to use LM1084 as a voltage regulator. To use its full 30W output, a heatsink is required. Thermal resistance junction-to-case of LM1084 is specified as 0.65/2.7 C/W in datasheet. Does that mean its thermal resistance is 0.65/2.7 = 0.24 C/W ?

Tmax = 100 (maximum junction temperature is 125)
Tambient = 50 (it will be used in a closed space)

Thermal compound will be used between IC and heatsink. (its thermal resistance is assumed to be 0.2 C/W)

Then, thermal resistance of heatsink is to be (100-50) / 30 - 0.24 (junction-to-case) - 0.2 (case-to-heatsink) = 1.22 C/W

Isnt 1.22 C/W a very low value for heatsinks ? Heatsinks in Mouser catalog (they are aavid thermalloy) have minimum 3.4 C/W.

PS: power dissipation * thermal resistance = the temperature we will measure on device, is it right ?

Thanks in advance.

MB

 

[jneutron]

Does that mean its thermal resistance is 0.65/2.7 = 0.24 C/W ?

No.

They spec the output pass transistor and the control circuitry separately. the control chip is .65 C/W, while the output is 2.7.

It appears that this is a two chip IC, as the control is spec'd to 125 and the output at 150...but it could be just two chip processes have been used on the same wafer.

Low drop regulators biggest problem is the gain of the output pass element with low C-E voltages, meaning the base drive goes up a lot, stressing the drive circuits more. (typical sat voltage tests assume hfe of 10, and drive that ratio to view Vcesat)

The national app note does not clearly spell out how one determines the control dissipation vs the output. The schem shows the control drive current to substrate connection, so you can't derive control dissipation by external measurement.

Also, the numbers look backward...why would a control chip have smaller thetajc than the pass element??

National has some work to do to bring that spec page up to their normal excellent standards..meanwhile, we all have to grope.

Cheers, John.

 

[cpemma]

I want to use LM1084 as a voltage regulator. To use its full 30W output, a heatsink is required...

It's a 5A regulator, so will waste 30W @ 5A when the voltage drop input to output is 6V. If you can reduce the drop, heat produced will be reduced proportionally. It is a low drop-out regulator after all.

Isnt 1.22 C/W a very low value for heatsinks ? Heatsinks in Mouser catalog (they are aavid thermalloy) have minimum 3.4 C/W.

A low value, but not a very low one.

 

[metebalci]

Hi cpemma,

I will try to reduce the drop and test it again.

Could you give me an example of an heatsink having a low thermal resistance (like 1 C/W) ?

Thanks in advance.

MB

 

[cpemma]

Could you give me an example of an heatsink having a low thermal resistance (like 1 C/W) ?

In the UK I'd be looking at Farnell or RS Components, at the RS site navigate to All Products | Electronic | Semiconductors - Sockets/Heatsinks/Accs. | Heat Sinks | High Power Heatsinks for a wide range.

Farnell's grouping is more haphazard, but http://uk.farnell.com/jsp/endecaSearch/partDetail.jsp?SKU=150016&N=401 is a nice price 1C/W.