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Question #43152

Difficulty: 8/10 | 🟢 Free

The Question

An intrinsic semiconductor component has a temperature-dependent resistance given by R(T)=R0exp⁡(Eg2kBT)R(T) = R_0 \exp\left(\frac{E_g}{2k_B T}\right), where EgE_g is the energy band gap, kBk_B is the Boltzmann constant, and TT is the absolute temperature. The component is connected to an AC voltage source V(t)=V0cos⁡(ωt)V(t) = V_0 \cos(\omega t). The frequency ω\omega is sufficiently high such that the temperature TT of the component remains effectively constant at a steady-state value. The component dissipates heat to the surroundings at a constant temperature T0T_0 according to Newton's law of cooling, with a heat transfer coefficient KK (in W/K\text{W/K}). As the peak voltage V0V_0 is gradually increased, the steady-state temperature TT increases. However, beyond a critical voltage V0,cV_{0,c}, thermal runaway occurs and no steady state is possible. Assuming Eg≫kBT0E_g \gg k_B T_0, which of the following expressions best approximates the critical peak voltage V0,cV_{0,c}? (Here ee is the base of the natural logarithm).

Options

  • A.

    4KkBT02R(T0)eEg\sqrt{\frac{4 K k_B T_0^2 R(T_0)}{e E_g}}

  • B.

    2KkBT02R(T0)eEg\sqrt{\frac{2 K k_B T_0^2 R(T_0)}{e E_g}}

  • C.

    4KkBT02R(T0)Eg\sqrt{\frac{4 K k_B T_0^2 R(T_0)}{E_g}}

  • D.

    KkBT02R(T0)eEg\sqrt{\frac{K k_B T_0^2 R(T_0)}{e E_g}}

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Correct Answer

Step-by-Step Solution

Explanation

How the Examiner Thinks

Intuition_Breaker

Train your brain to resist "obvious" answers that are designed to deceive your intuition.

Concepts Behind This Question

Average Power over Cycle|Energy Band Gap Temperature Dependency|Approximation: Taylor Series

Why This Question Is Tricky

Common Mistakes

This question tests the misconception that a system with a negative temperature coefficient of resistance will always reach a stable thermal equilibrium.

Students often incorrectly assume linear resistance or fail to recognize that thermal runaway is a bifurcation point where the derivative of heat generation with respect to temperature equals the heat dissipation coefficient (dPgen/dT=dPdiss/dTdP_{gen}/dT = dP_{diss}/dT).

Furthermore, a common trap is forgetting to use the RMS voltage for average power in an AC circuit, leading to a missing factor of 2, or failing to linearize the exponential function correctly around the ambient temperature T0T_0.

Quick Facts About This Question

Question Information

SubjectPhysics
ChapterAlternating Current - Class 12
ConceptAverage Power over Cycle
Difficulty8
Cognitive Load7
Target Solve Time160 sec
AccessFree