How to Calculate the Derating Factor of a Semiconductor?

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The discussion focuses on calculating the derating factor of a semiconductor with a maximum rated dissipation of 5W at 50°C. The user confirms that the device cannot operate at 5W without auxiliary cooling when the ambient temperature is 40°C, calculating the maximum permissible dissipation to be 2W. There is uncertainty regarding the correct approach to determine the derating factor in watts per ambient degree. Clarification is sought on the temperature scale for the derating factor calculation. The conversation emphasizes the importance of precise wording in technical questions.
shaiqbashir
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Okz my question is like this:

A certain semiconductor has a max rated dissipation of 5W at 50 degree Celsius case Temperature and must be derated above 50 degree celisus case temperature \theta_{CA} = 5 degree C/W

1) Can the device be operated at 5W of dissipation without auxiliary cooling (without heat sink or fan) when ambient temperature is 40 degree celsisus?

2) if not, then what is the max possible dissipation, with no auxiliary cooling at 40 degree celsius?

3) What derating factor should be applied to the device in watts per ambient degree?


okz so now this is what i have solved myself:

1) we can find

P_{d} = \frac{T_{C}-T_{A}}{\theta_{CA}}

THis give me

Pd = 2 W

which clearly is less then the required 5W. So it it means that the device cannot be operated without auxiliary cooling.

2) The maximum permissible dissipation at 40 degree celsius will be 2W as calculated above.

3) no idea

Please tell me have i solved the first two parts correctly or not. If not then what is the correct way to solve them. Secondly, please tell me how to solve this third part.

I shall be thankful to u for ur precious and quick replies.

Take carez!

Good Bye!

SB--
 
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You are fine on 1 & 2. I think 3 is poorly worded. What temperature scale are they asking for? Celsius? Kelvin? Could you post the exact wording of the question please?
 
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