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Question #43468

Difficulty: 8/10 | 🟢 Free

The Question

A photo-diode is modeled as an ideal AC current source iph(t)=I0sin⁡(ωt)i_{ph}(t) = I_0 \sin(\omega t) in parallel with its voltage-dependent junction capacitance CjC_j. The internal shunt resistance of the diode is assumed to be infinite. The junction capacitance varies with the applied DC reverse bias voltage VRV_R (in Volts) according to the relation Cj=C01+VR/VbiC_j = \frac{C_0}{\sqrt{1 + V_R/V_{bi}}}, where C0C_0 (in Farads) is the zero-bias capacitance and VbiV_{bi} (in Volts) is the built-in potential. The photo-diode is connected to an external series combination of an inductor LL (in Henrys) and a resistor RR (in Ohms). An ideal biasing network maintains the DC reverse bias VRV_R across the diode without affecting the AC impedance of the circuit. The intensity of the incident light is modulated at an angular frequency ω\omega (in rad/s). At a specific reverse bias voltage VR=VresV_R = V_{res}, the power factor of the equivalent AC circuit, as seen by the internal photo-current source, becomes unity. Which of the following expressions correctly gives VresV_{res}?

Options

  • A.

    Vbi[C02(R2+ω2L2)2L2−1]V_{bi} \left[ \frac{C_0^2 (R^2 + \omega^2 L^2)^2}{L^2} - 1 \right]

  • B.

    Vbi[ω4L2C02−1]V_{bi} \left[ \omega^4 L^2 C_0^2 - 1 \right]

  • C.

    Vbi[C02R4L2−1]V_{bi} \left[ \frac{C_0^2 R^4}{L^2} - 1 \right]

  • D.

    Vbi[ω2R2C02−1]V_{bi} \left[ \omega^2 R^2 C_0^2 - 1 \right]

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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

Photo-diode vs LED Bias|Driven Oscillations and Resonance|Power Factor Definition

Why This Question Is Tricky

Common Mistakes

This question tests the student's ability to correctly model an internal source in an AC circuit and apply the fundamental definition of power factor.

The primary trap lies in the visual 'series' connection of the external inductor and resistor to the diode terminals.

Students commonly misapply the series resonance condition (ωL=1/ωCj\omega L = 1/\omega C_j) without realizing that the photo-diode's internal current generation acts as a Norton equivalent source in parallel with its junction capacitance.

Consequently, the resonance (unity power factor) is a parallel tank circuit phenomenon, requiring the cancellation of the imaginary part of the total admittance, not the series impedance.

This highlights the difference between external voltage driving and internal current driving in semiconductor devices.

Quick Facts About This Question

Question Information

SubjectPhysics
ChapterSemiconductor Electronics: Materials, Devices, and Simple Circuits - Class 12
ConceptPhoto-diode vs LED Bias
Difficulty8
Cognitive Load7
Target Solve Time160 sec
AccessFree