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The thermal resistance for the thyristor – sin k combination are QJC = 0.18 and ACS = 0.09 °C/W for a heat sin k temperature of 75°C and maximum junction temperature is 130°C . In case the heat sin k temperature is brought down 70°C by forced coaling find the percentage increase in the device rating.
  • a)
    3.48%
  • b)
    4.45%
  • c)
    4.40%
  • d)
    4.9%
Correct answer is option 'B'. Can you explain this answer?
Most Upvoted Answer
The thermal resistance for the thyristor – sin k combination are...
The thermal resistance for a thyristor refers to the ability of the device to dissipate heat generated during operation. It is typically measured in degrees Celsius per watt (°C/W) and indicates how much the device's temperature will rise for every watt of power it dissipates.

Thermal resistance is an important parameter for thyristors, as excessive heat can lead to device failure or reduced performance. It is influenced by various factors, including the device's package type, material properties, and the effectiveness of its heat sink or cooling system.

Thermal resistance can be calculated using the formula:

Thermal Resistance (°C/W) = (Tj - Ta) / P

Where:
- Tj is the junction temperature of the thyristor
- Ta is the ambient temperature
- P is the power dissipated by the thyristor

By knowing the thermal resistance, designers can estimate the maximum power dissipation that the thyristor can handle without exceeding a specified temperature rise. This information is important for selecting appropriate heat sinks, cooling systems, and operating conditions to ensure reliable and efficient operation of the thyristor.
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Community Answer
The thermal resistance for the thyristor – sin k combination are...

Thus total average power loss in the thyristor – sink combination is 203.70 W. with improved coaling

Thyristor rating is proportional to the square root of average power loss
∴ Percentage Increase in thyristor rating
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The thermal resistance for the thyristor – sin k combination are QJC = 0.18 and ACS = 0.09 °C/W for a heat sin k temperature of 75°C and maximum junction temperature is 130°C . In case the heat sin k temperature is brought down 70°C by forced coaling find the percentage increase in the device rating.a)3.48%b)4.45%c)4.40%d)4.9%Correct answer is option 'B'. Can you explain this answer?
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The thermal resistance for the thyristor – sin k combination are QJC = 0.18 and ACS = 0.09 °C/W for a heat sin k temperature of 75°C and maximum junction temperature is 130°C . In case the heat sin k temperature is brought down 70°C by forced coaling find the percentage increase in the device rating.a)3.48%b)4.45%c)4.40%d)4.9%Correct answer is option 'B'. Can you explain this answer? for Electrical Engineering (EE) 2024 is part of Electrical Engineering (EE) preparation. The Question and answers have been prepared according to the Electrical Engineering (EE) exam syllabus. Information about The thermal resistance for the thyristor – sin k combination are QJC = 0.18 and ACS = 0.09 °C/W for a heat sin k temperature of 75°C and maximum junction temperature is 130°C . In case the heat sin k temperature is brought down 70°C by forced coaling find the percentage increase in the device rating.a)3.48%b)4.45%c)4.40%d)4.9%Correct answer is option 'B'. Can you explain this answer? covers all topics & solutions for Electrical Engineering (EE) 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The thermal resistance for the thyristor – sin k combination are QJC = 0.18 and ACS = 0.09 °C/W for a heat sin k temperature of 75°C and maximum junction temperature is 130°C . In case the heat sin k temperature is brought down 70°C by forced coaling find the percentage increase in the device rating.a)3.48%b)4.45%c)4.40%d)4.9%Correct answer is option 'B'. Can you explain this answer?.
Solutions for The thermal resistance for the thyristor – sin k combination are QJC = 0.18 and ACS = 0.09 °C/W for a heat sin k temperature of 75°C and maximum junction temperature is 130°C . In case the heat sin k temperature is brought down 70°C by forced coaling find the percentage increase in the device rating.a)3.48%b)4.45%c)4.40%d)4.9%Correct answer is option 'B'. Can you explain this answer? in English & in Hindi are available as part of our courses for Electrical Engineering (EE). Download more important topics, notes, lectures and mock test series for Electrical Engineering (EE) Exam by signing up for free.
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