Determine the maximum dissipation allowed for a \(100-\mathrm{W}\) silicon transistor (rated at \(25^{\circ} \mathrm{C}\) ) for a derating factor of \(0.6 \mathrm{~W} /{ }^{\circ} \mathrm{C}\) at a case temperature of \(150^{\circ} \mathrm{C}\).

Short Answer

Expert verified
The maximum dissipation allowed for the 100-W silicon transistor under the given conditions is 25W.

Step by step solution

01

Identify Given Values

We first identify and list down the given values: The rated power \(P_{rated}\) is 100 W, the derating factor is 0.6 W/°C, the case temperature \(T_{case}\) is 150°C, and the rated temperature \(T_{rated}\) is 25°C.
02

Substitute Values into Equation

We next substitute the given values into the formula for maximum power dissipation: \(P_{max} = P_{rated} - Derating Factor * (T_{case} - T_{rated}) = 100 W - 0.6 W/°C * (150°C - 25°C)\).
03

Calculate the Maximum Power Dissipation

We simplify the expression in the previous step and calculate the result: \(P_{max} = 100 W - 0.6 W/°C * 125°C = 100 W - 75W = 25W\).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Study anywhere. Anytime. Across all devices.

Sign-up for free