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Question #43765

Difficulty: 8/10 | 🟢 Free

The Question

A p-n junction photodiode has a built-in potential Vbi=0.8 VV_{bi} = 0.8 \text{ V}. It is illuminated by a constant photon flux, resulting in a uniform generation rate GG of electron-hole pairs per unit volume throughout the depletion region. Assume the photocurrent II is entirely due to the drift of these generated carriers across the depletion region. At a reverse bias voltage VR1=2.8 VV_{R1} = 2.8 \text{ V}, the steady-state photocurrent is I1=2.0 \muAI_1 = 2.0 \text{ \mu A} and the average drift velocity of electrons is v1=1.0×104 m/sv_1 = 1.0 \times 10^4 \text{ m/s}. If the reverse bias voltage is increased to VR2=13.6 VV_{R2} = 13.6 \text{ V} while keeping the photon flux constant, determine the new photocurrent I2I_2 and the new average drift velocity v2v_2. (Assume the electron mobility μe\mu_e is constant, and the depletion width WW follows the relation W∝Vbi+VRW \propto \sqrt{V_{bi} + V_R}).

Options

  • A.

    I2=4.0 \muA, v2=2.0×104 m/sI_2 = 4.0 \text{ \mu A}, \ v_2 = 2.0 \times 10^4 \text{ m/s}

  • B.

    I2=2.0 \muA, v2=4.0×104 m/sI_2 = 2.0 \text{ \mu A}, \ v_2 = 4.0 \times 10^4 \text{ m/s}

  • C.

    I2=2.0 \muA, v2=2.0×104 m/sI_2 = 2.0 \text{ \mu A}, \ v_2 = 2.0 \times 10^4 \text{ m/s}

  • D.

    I2=8.0 \muA, v2=4.0×104 m/sI_2 = 8.0 \text{ \mu A}, \ v_2 = 4.0 \times 10^4 \text{ m/s}

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Correct Answer

Step-by-Step Solution

Explanation

How the Examiner Thinks

Variable_Constraint

Develop pure symbolic logic skills by mastering algebraic constraints without relying on numerical data.

Concepts Behind This Question

Intensity and Photon Flux|Drift Velocity Mechanism|Depletion Region Dynamics

Why This Question Is Tricky

Common Mistakes

This question tests the student's understanding of the dynamic nature of the depletion region in a photodiode under varying reverse bias.

A common misconception is treating a photodiode strictly as a constant-current source dependent only on incident photon flux, leading students to incorrectly assume I2=I1I_2 = I_1 (Option B or C).

Furthermore, students often miscalculate the electric field by assuming the depletion width WW remains static, which would erroneously yield E∝VE \propto V and v2=4v1v_2 = 4v_1 (Option B).

By linking the generation volume and the average electric field to the voltage-dependent depletion width W∝Vbi+VRW \propto \sqrt{V_{bi} + V_R}, the problem forces a rigorous application of semiconductor physics principles over rote memorization of ideal diode behavior.

Quick Facts About This Question

Question Information

SubjectPhysics
ChapterDual Nature of Radiation and Matter - Class 12
ConceptIntensity and Photon Flux
Difficulty8
Cognitive Load7
Target Solve Time160 sec
AccessFree