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Question #43898

Difficulty: 8/10 | 🟢 Free

The Question

A biasing circuit for a transistor uses a voltage divider consisting of a fixed ideal resistor R1=2.0 kΩR_1 = 2.0 \text{ k}\Omega and a temperature-sensitive resistor R2R_2 connected in series across a constant DC supply VCC=10.0 VV_{CC} = 10.0 \text{ V}. The output voltage VoutV_{out} is taken across R2R_2. To provide thermal compensation, R2R_2 (initial resistance R0=2.0 kΩR_0 = 2.0 \text{ k}\Omega at T0T_0) is a thin-film resistor perfectly bonded to the top surface of a bimetallic strip. The strip consists of two metals of equal thickness with linear expansion coefficients α1=3.0×10−5 K−1\alpha_1 = 3.0 \times 10^{-5} \text{ K}^{-1} (top metal) and α2=1.0×10−5 K−1\alpha_2 = 1.0 \times 10^{-5} \text{ K}^{-1} (bottom metal). When the temperature increases by ΔT\Delta T, the strip bends, inducing a longitudinal tensile strain ϵ=34(α1−α2)ΔT\epsilon = \frac{3}{4}(\alpha_1 - \alpha_2)\Delta T on its top surface. The resistor R2R_2 has an intrinsic temperature coefficient of resistance αρ=1.0×10−4 K−1\alpha_\rho = 1.0 \times 10^{-4} \text{ K}^{-1} and a strain gauge factor G=ΔR2/R0ϵ=4.0G = \frac{\Delta R_2 / R_0}{\epsilon} = 4.0. Assuming the total fractional change in resistance is the sum of the intrinsic thermal and mechanical strain contributions, what is the rate of change of the output voltage with respect to temperature, dVoutdT\frac{dV_{out}}{dT}, at T=T0T = T_0?

Options

  • A.

    0.40 mV/K0.40 \text{ mV/K}

  • B.

    0.25 mV/K0.25 \text{ mV/K}

  • C.

    0.10 mV/K0.10 \text{ mV/K}

  • D.

    0.45 mV/K0.45 \text{ mV/K}

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Correct Answer

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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

Transistor Biasing Stability|Thermal Expansion of Area and Volume|Bimetallic Strip Curvature|Cell EMF vs Terminal Voltage

Why This Question Is Tricky

Common Mistakes

This question tests the student's ability to cross-couple thermodynamic deformation with electronic circuit analysis, specifically focusing on non-linear feedback in a biasing circuit.

The primary trap lies in ignoring the mechanical strain induced by the bimetallic strip's bending and only accounting for the intrinsic thermal coefficient of the resistor (αρ\alpha_\rho), which leads to Option B.

Students often misapply the resistance-temperature formula by failing to recognize that physical mounting geometry (stress-strain) acts as a secondary, coupled mechanism that shifts the Q-point.

Additionally, sign errors can occur if the student incorrectly assumes the top layer undergoes compression rather than tension when α1>α2\alpha_1 > \alpha_2, leading to Option C.

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Question Information

SubjectPhysics
ChapterSemiconductor Electronics: Materials, Devices, and Simple Circuits - Class 12
ConceptTransistor Biasing Stability
Difficulty8
Cognitive Load7
Target Solve Time160 sec
AccessFree