Thermostatic applied on electronic chip

AI Thread Summary
The discussion revolves around the unexpected temperature behavior of an ethernet PHY component on a PCB when subjected to a higher ambient temperature in a climatic chamber. The engineer initially expected the component's case temperature to rise linearly with the ambient temperature but observed a lower than anticipated temperature of 90°C instead of the calculated 100°C. This discrepancy raises questions about the thermal dynamics at play, suggesting that factors such as heat dissipation, thermal resistance, and the component's power consumption may be influencing the results. The engineer seeks clarification on these thermal behaviors and references a calculator for PCB trace temperature calculations, indicating a need for deeper understanding of thermal management in electronic components. The conversation highlights the complexities of thermal behavior in electronics and the importance of accurate modeling.
schnuber
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Hi everybody
I am an electronic engineer and I have a question concerning thermostatic behaviour of an electronic component.

I have a printed circuit board (PCB) with electronic chips on it. At room temperature 25°C I measure with the infrared themometer the case temperature of one component which is 40°C (It is an ethernet PHY).
Now I made the following assumption: If I put the PCB in a climatic chamber and adjust the chamber temperature to 85°C degree then the case temperature of the component should be 100°C = (40+(85-25)). But that is not the case, it is lower, only 90°C.
So what is going on there I don't understand this.
Please help
 
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I don't believe its a linera rise to ambient temperature as I would have to study all the calculations involved in PCB I'll instead direct you to a claculator that I've used in the past.

http://home.comcast.net/~pcb.george/trace.html
 
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This is a calculator that computes the temperature and resistance of a copper trace on the PCB in relation to its mechanical dimensions and electrical current.
I don't see how that gives me an answer to my question.
 

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