Heat Sink Thermal Capacitance

Thread Starter

qrb14143

Joined Mar 6, 2017
112
Hi everyone, I am struggling with the following question.

"A device is mounted on a heatsink. It is started from cold and run for five minutes before being switched off. The device produces 30W of losses and the thermal resistance between the device and the heatsink is 1.7C/W. The thermal capacitance of the heatsink is 100J/C. The ambient temperature is 40C. Calculate the temperature of the heatsink when the device is switched off after five minutes."

My approach so far has been to model the semiconductor device as a "current source" since thermal power is analogous to electric current. I have modelled the heatsink as a parallel RC section and I have modelled the ambient temperature as a constant voltage source, since temperature is analogous to voltage.

This yields a first order differential equation, however the differential part has a constant term on top which I am not sure how to deal with. I have even toyed with the idea of using a laplace transform to solve the equation but I don't know how to deal with the fact that the top of the differential term is d(Tj - 40) as opposed to just d(Tj)

I have attached my attempt so far.
 

Attachments

MrAl

Joined Jun 17, 2014
11,389
Hi,

Can you first calculate the final temperature of the heat sink given a long time?
In other words, the temperature as t (time) approaches infinity (not after just 5 minutes).
Does that tell you anything?
 

MrAl

Joined Jun 17, 2014
11,389
Have you been told the thermal resistance between the heatsink and ambient?
Hi

I assume that was the 1.7 deg C per watt spec because that is usually what is given on the data sheet.

So for example, with a 5 deg C per watt heatsink and 10 watts, the temperature rise is 50 deg C over long periods of time (not over a short time period though as the original question asks for).

I've given a little bit away here because i had to mention temperature "rise" and not absolute temperature, but perhaps that will function as a sort of hint for solving the original problem as i didnt see the OP come back yet.
 

crutschow

Joined Mar 14, 2008
34,285
Have you been told the thermal resistance between the heatsink and ambient?
That's a good point.
They only give the thermal resistance between the device and the heatsink, which is a red herring, as it has no effect on the answer.
So, since the heat-sink to air thermal resistance is not given, then it would appear the only factors affecting the problem are the amount of heat dissipated for the given time, and the thermal capacity of the heat sink.
In other words, the heatsink is only a sink for the heat without dissipating any of it to ambient.
So it's just a simple linear calculation to determine the heatsink temperature.

That may not have been the intention of the problem, but that's the only possible answer with the given information. ;)
 

Thread Starter

qrb14143

Joined Mar 6, 2017
112
Hi

I assume that was the 1.7 deg C per watt spec because that is usually what is given on the data sheet.

So for example, with a 5 deg C per watt heatsink and 10 watts, the temperature rise is 50 deg C over long periods of time (not over a short time period though as the original question asks for).

I've given a little bit away here because i had to mention temperature "rise" and not absolute temperature, but perhaps that will function as a sort of hint for solving the original problem as i didnt see the OP come back yet.
Over a long period of time, the thermal capacitance "charges up" so the steady state temperature is dictated by T = P * R = 30W * 1.7C/W = 91C.

Since both the semiconductor and the heatsink start off at the same temperature, ie 40C, if I think about "delta T" instead of the absolute value of T, I get this(see attachment).
 

Attachments

MrAl

Joined Jun 17, 2014
11,389
That's a good point.
They only give the thermal resistance between the device and the heatsink, which is a red herring, as it has no effect on the answer.
So, since the heat-sink to air thermal resistance is not given, then it would appear the only factors affecting the problem are the amount of heat dissipated for the given time, and the thermal capacity of the heat sink.
In other words, the heatsink is only a sink for the heat without dissipating any of it to ambient.
So it's just a simple linear calculation to determine the heatsink temperature.

That may not have been the intention of the problem, but that's the only possible answer with the given information. ;)
Hello,

I took it to mean the heatsink spec not the usual device to heatsink which is usually very very low relative to the heatsink to ambient. So it appears to be worded funny.
Without the HS to ambient we couldnt calculate anything.
 
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MrAl

Joined Jun 17, 2014
11,389
Over a long period of time, the thermal capacitance "charges up" so the steady state temperature is dictated by T = P * R = 30W * 1.7C/W = 91C.

Since both the semiconductor and the heatsink start off at the same temperature, ie 40C, if I think about "delta T" instead of the absolute value of T, I get this(see attachment).
Hi,

Yes, very good :)

I think you meant to type instead:
T = P * R +Tamb= 30W * 1.7C/W+40C= 91C.

which is in the form:
Tfinal=Trise+Tamb

but you got the right answer so congrats!
 

Thread Starter

qrb14143

Joined Mar 6, 2017
112
Hi,

Yes, very good :)

I think you meant to type instead:
T = P * R +Tamb= 30W * 1.7C/W+40C= 91C.

which is in the form:
Tfinal=Trise+Tamb
but you got the right answer so congrats!
Yes, I did intend to add the ambient temperature. Is my final solution correct? The answer sheet says it should be 48.7C but it doesn't offer any solution, just a numerical answer.
 

Alec_t

Joined Sep 17, 2013
14,280
LTspice thinks the answer is 49.1C, if the thermal resistance from device to heatsink is zero.
 
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Thread Starter

qrb14143

Joined Mar 6, 2017
112
LTspice thinks the answer is 49.1C, if the thermal resistance from device to heatsink is zero.
I now realise why the answer on the tutorial sheet is 48.7 degrees. The maximum steady state temperature of the junction is 91C as calculated in previous posts. The "delta T" term, ie the temperature drop from across the heatsink is 42.27C, take that away from the 91C gives you the answer of 48.7 degrees. BUT, since the heatsink to ambient resistance is zero, this is a contradiction since the heatsink temperature must be the ambient temperature as there can be no temperature drop across a thermal resistance of zero!

This leaves two possible explanations:
1) The thermal to electrical analogy is not perfect, since this question implies that there is a temperature drop across a zero resistance.
2) The question gives rise to a scenario which could never actually happen since there is no such thing as a resistance of zero.

@Alec_t I assume in your LT spice solution you have modelled the semiconductor device as a current source, the heatsink as a parallel RC branch and the ambient temperature as a voltage source. The junction temperature would then be the voltage at the node between the current source and the RC branch and the heatsink temperature would be the voltage at the node between the RC branch and the voltage source representing ambient. So surely the heatsink temperature must be the same as the ambient temperature if the resistance between them is zero?
 

MrAl

Joined Jun 17, 2014
11,389
I now realise why the answer on the tutorial sheet is 48.7 degrees. The maximum steady state temperature of the junction is 91C as calculated in previous posts. The "delta T" term, ie the temperature drop from across the heatsink is 42.27C, take that away from the 91C gives you the answer of 48.7 degrees. BUT, since the heatsink to ambient resistance is zero, this is a contradiction since the heatsink temperature must be the ambient temperature as there can be no temperature drop across a thermal resistance of zero!

This leaves two possible explanations:
1) The thermal to electrical analogy is not perfect, since this question implies that there is a temperature drop across a zero resistance.
2) The question gives rise to a scenario which could never actually happen since there is no such thing as a resistance of zero.

@Alec_t I assume in your LT spice solution you have modelled the semiconductor device as a current source, the heatsink as a parallel RC branch and the ambient temperature as a voltage source. The junction temperature would then be the voltage at the node between the current source and the RC branch and the heatsink temperature would be the voltage at the node between the RC branch and the voltage source representing ambient. So surely the heatsink temperature must be the same as the ambient temperature if the resistance between them is zero?

Hello,

I get 82.27 which is much different. That is after 5 minutes because the temperature RISE is 42.27 and adding 40 i get 82.27 deg C.

I basically used your circuit and calculated the equivalent voltage after 300 seconds.

Keep in mind that whatever you come up with the limiting value for t infinity has to be 51, then adding 40 we get 91.

Also, 30 watts for 300 seconds is 9000 watt seconds.
The heat sink absorbs 100 watt seconds per deg C.
That means if it was linear, the temperature rise would be 90 deg C after 300 seconds.
The "book answer" is implying that the temperature rise is only around 9 deg C.
 
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Alec_t

Joined Sep 17, 2013
14,280
@Alec_t I assume in your LT spice solution you have modelled the semiconductor device as a current source, the heatsink as a parallel RC branch and the ambient temperature as a voltage source.
Nearly. In my sim, the ground connection corresponds to 0C and the cap starts at an initial voltage of 40.
Heatsink.PNG
 
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Alec_t

Joined Sep 17, 2013
14,280
Hmm, I think that model was wrong. How about this one?
Heatsink2.PNG
Now the answer is 82.2C, in line with other answers above.
 

Thread Starter

qrb14143

Joined Mar 6, 2017
112
Hello,

I get 82.27 which is much different. That is after 5 minutes because the temperature RISE is 42.27 and adding 40 i get 82.27 deg C.

I basically used your circuit and calculated the equivalent voltage after 300 seconds.

Keep in mind that whatever you come up with the limiting value for t infinity has to be 51, then adding 40 we get 91.

Also, 30 watts for 300 seconds is 9000 watt seconds.
The heat sink absorbs 100 watt seconds per deg C.
That means if it was linear, the temperature rise would be 90 deg C after 300 seconds.
The "book answer" is implying that the temperature rise is only around 9 deg C.
I'm inclined to agree with you as your solution makes sense to me. I only posted the method in post #12 because that was the only possible way I could arrive at the same solution as the lecturer. I suspect the time has come to be bold and ask the lecturer if he's sure his answer is right ;)
 

MrAl

Joined Jun 17, 2014
11,389
Hi,

My only other guess is that the problem statement still is not quite right or something is missing.
 

Thread Starter

qrb14143

Joined Mar 6, 2017
112
Hi,

My only other guess is that the problem statement still is not quite right or something is missing.
The problem statement in post #1 is exactly as it was given to us, bar the typo I explained in post #7. I have emailed the lecturer to ask him if the final answer on the tutorial sheet is correct. If it is, I have asked him to explain how it is derived.

In the meantime, my sincere thanks to all those who have helped, @MrAl @Alec_t @crutschow :)
 

JoeJester

Joined Apr 26, 2005
4,390
"A device is mounted on a heatsink. It is started from cold and run for five minutes before being switched off. The device produces 30W of losses and the thermal resistance between the device and the heatsink is 1.7C/W. The thermal capacitance of the heatsink is 100J/C. The ambient temperature is 40C. Calculate the temperature of the heatsink when the device is switched off after five minutes."

Givens:

- Device produces 30 W of losses
- thermal resistance is 1.7 C/W
- thermal capacitance is 100 J/C
- ambient temperature is 40 C

Stem of the question (What they want to know):

- temperature of the heatsink when the device is switched off after five minutes.

How they want you to do it:

- Calculate it.

I'd go with 82.7 as the correct answer. If it asked for the rise of the temperature it would be 42.7.

 
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