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Question #43895

Difficulty: 8/10 | 🟢 Free

The Question

A Zener diode is fabricated from a semiconductor crystal rod of length LL. The rod is rigidly fixed between two immovable, perfectly insulating walls at an initial temperature T0=300 KT_0 = 300 \text{ K}. The diode is connected in series with a resistor R=100 ΩR = 100 \ \Omega and a DC voltage source VS=12.0 VV_S = 12.0 \text{ V}. At T0T_0, the diode operates in the breakdown region with a Zener voltage VZ0=6.0 VV_{Z0} = 6.0 \text{ V}.

The intrinsic temperature coefficient of the Zener voltage (at zero stress) is CT=∂VZ∂T=−4.0×10−3 V/KC_T = \frac{\partial V_Z}{\partial T} = -4.0 \times 10^{-3} \text{ V/K}. The crystal has a Young's modulus Y=1.5×1011 N/m2Y = 1.5 \times 10^{11} \text{ N/m}^2 and a coefficient of linear thermal expansion α=2.0×10−6 K−1\alpha = 2.0 \times 10^{-6} \text{ K}^{-1}. The Zener voltage also depends on the longitudinal compressive stress σc\sigma_c developed in the rod due to bandgap deformation, with a coefficient K=∂VZ∂σc=+2.0×10−8 V/PaK = \frac{\partial V_Z}{\partial \sigma_c} = +2.0 \times 10^{-8} \text{ V/Pa}.

If the temperature of the system is uniformly raised to T=350 KT = 350 \text{ K}, calculate the new steady-state Zener current IZI_Z. Assume the walls remain perfectly rigid and the diode remains in the breakdown region.

Options

  • A.

    59 mA59 \text{ mA}

  • B.

    62 mA62 \text{ mA}

  • C.

    57 mA57 \text{ mA}

  • D.

    65 mA65 \text{ mA}

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Explanation

How the Examiner Thinks

Multi-Concept_Bridge

Master problems that require linking physics and math concepts to solve complex competitive questions.

Concepts Behind This Question

Zener vs Avalanche Breakdown|Load Line Analysis|Thermal Stress in Constrained Rods

Why This Question Is Tricky

Common Mistakes

This question tests the synthesis of solid-state electronics and classical elasticity.

The primary trap lies in the student's potential failure to recognize that the mechanical boundary conditions (rigid walls) couple with the semiconductor's electrical properties via the deformation potential.

Students who only memorize that Zener breakdown has a negative temperature coefficient will calculate ΔVZ=−0.2 V\Delta V_Z = -0.2 \text{ V} and incorrectly choose 62 mA.

The problem forces the realization that the total derivative of the Zener voltage in a constrained system includes both the partial temperature derivative and the stress-induced shift: dVZdT=∂VZ∂T+∂VZ∂σcdσcdT\frac{dV_Z}{dT} = \frac{\partial V_Z}{\partial T} + \frac{\partial V_Z}{\partial \sigma_c} \frac{d\sigma_c}{dT}.

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Question Information

SubjectPhysics
ChapterSemiconductor Electronics: Materials, Devices, and Simple Circuits - Class 12
ConceptZener vs Avalanche Breakdown
Difficulty8
Cognitive Load7
Target Solve Time160 sec
AccessFree