How to Calculate the Derating Factor of a Semiconductor?

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SUMMARY

The discussion centers on calculating the derating factor of a semiconductor with a maximum rated dissipation of 5W at a case temperature of 50°C and a thermal resistance of θCA = 5°C/W. The user determined that at an ambient temperature of 40°C, the maximum permissible dissipation without auxiliary cooling is 2W, confirming that the device cannot operate at 5W under these conditions. The user seeks clarification on how to calculate the derating factor in watts per ambient degree, which remains unresolved in the discussion.

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  • Understanding of thermal resistance in semiconductors
  • Knowledge of power dissipation calculations
  • Familiarity with ambient and case temperature concepts
  • Basic principles of semiconductor operation and derating
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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 [tex]\theta_{CA}[/tex] = 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

[tex]P_{d} = \frac{T_{C}-T_{A}}{\theta_{CA}}[/tex]

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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