# Thermal Resistance of Heat Sink

1. Apr 3, 2016

### StripesUK

1. The problem statement, all variables and given/known data
Calculate the thermal resistance of the heat sink required for the regulator of the question above given the information below [from the datasheet] and the data in the table.

Max Ambient Temp (Ta)= 75°C
Thermal Resistance Case to Heat (θchs)= 0.2°C/W
Control Section
Thermal Resistance Junction to Case (θjc)= 0.6°C/W
Max Junction Temp (Tjmax)= 125°C
Power Section
Thermal Resistance Junction to Case (θjc)= 1.6°C/W
Max Junction Temp (Tjmax)= 150°C

Data sheet:
http://cds.linear.com/docs/en/datasheet/108345fh.pdf

Details form previous question:
Vin=18V
Vout=1.25-15V (As calculated in the previous question. https://www.physicsforums.com/threa...tor-output-voltage-range.864877/#post-5428815)

2. Relevant equations
$ΔT=Tjmax-Ta$
$PD=Ilm(Vin-Vout)$
$θ=\frac{ΔT}{PD}$
$θhs=θ-θjc-θchs$

3. The attempt at a solution
Okay so here is what I have so far....

Control:
$ΔT=Tjmax-Ta=125-75=50°C$
$PD=Ilm(Vin-Vout)=1(18-15)=3W$
$θ=\frac{ΔT}{PD}=\frac{50}{3}=16.667°C/W$
This value is well above the value of 1.6°C/W stated on the datasheet so it will require a heat sink.
$θhs=θ-θjc-θchs=16.667-0.6-0.2=15.867°C/W$

I have done the same for the power section, but I feel I'm not on the right track due the thermal resistance values for the heat sinks being so drastically high. Especially if I'm right in thinking they then have to be added together?

Last edited: Apr 3, 2016
2. Apr 3, 2016

### StripesUK

Okay so after looking harder into this I've found this equation in the datasheet.

$Tjmax=Ta+Pd(θhs+θchs+θjc)$

Transposed

$θhs=\frac {Tj}{Ta+Pd}-θchs-θjc$

$\frac {125}{75+3}-0.2-0.6=0.8°C/W$

This seems a better answer, but I feel there's still a gap in my knowledge I'm not quite grasping somewhere...

3. Apr 3, 2016

### Staff: Mentor

The worst-case scenario should occur when there is a maximum difference between the input and output voltages. You took the best case scenario where the output is maximum at 15 V.

Check your "Transposed" equation's algebra. You're adding temperature and power units in the denominator of the first term.... that just can't be right!

4. Apr 3, 2016

### StripesUK

This is an equation supplied in the datasheets. Should I disregard it and stick with my original workings?

5. Apr 3, 2016

### Staff: Mentor

The datasheet version is fine. It's the algebra you performed on it that I have my doubts about.

6. Apr 3, 2016

### StripesUK

A FOOL I AM! the same damn equation if I'd bothered to transpose it correctly.

$Pd=1(18-1.25)=16.75W$ (Allowing for the maximum difference in voltage.)

Control
$\frac{125-75}{16.75}-0.2-0.6=2.18°C/W$

Power
$\frac{150-75}{16.75}-0.2-1.6=2.69°C/W$

Therefore the thermal resistance of the heat sink required is $2.18+2.69=4.87°C/W$

Last edited: Apr 3, 2016
7. Apr 3, 2016

### Staff: Mentor

That looks better (for the equation)

Strangely, the final paragraph of the Thermal Considerations section of the application sheet doesn't consider adding the two power contributions when they determined that their heat sink was sufficient; they just noted that it's thermal resistance was adequate for each case individually. I don't know why this is

To combine the two, if its valid to do so and their model doesn't already take it into account, I think I would combine them in parallel, not series (i.e. add thermal conductances to carry away the combined heat).

8. Apr 3, 2016

### StripesUK

I take it they should be combined in parallel as the control resistors R1 and R2 are in parallel with the LT1083?

9. Apr 3, 2016

### Staff: Mentor

The thermal "circuit" is entirely separate from the electronic one (except that the heat generated is due to the electronic side of things). It's just that you have two heat sources (control and power) dumping heat into the chip substrate which both need to be conducted away via the case-to-heatsink-to-ambient path. Think of it like adding heat currents. The total "current" is the sum of the two heat powers.

I don't understand their model if their example is correct: They seem to assume that the same total power (24 W in their case) is generated by both sections individually since they place that power across both junction thermal resistances. I'd think that the thermal circuit would look something like this:

Perhaps they're modelling some worst case scenario where it is assumed that all the heat is being generated by one section or the other. In that case you'd just pick the higher of the two heatsink resistances for your answer.