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Question #43466

Difficulty: 8/10 | 🟢 Free

The Question

A p-n junction diode is forward-biased at a steady DC operating point with a current $I_{DC} = 50 \text{ mA}$ and a voltage $V_{DC} = 0.7 \text{ V}$ . The diode's current-voltage relationship is governed by the Shockley equation $I = I_0 (e^{qV/nkT} - 1)$ , where the ideality factor is $n = 2$ and the thermal voltage is $\frac{kT}{q} = 25 \text{ mV}$ . A small-signal AC photocurrent $i(t) = I_m \sin(\omega t)$ with peak amplitude $I_m = 2 \text{ mA}$ is superimposed on the DC current due to a modulated light source. Assuming $I_{DC} \gg I_0$ , what is the peak amplitude of the AC voltage signal across the diode?

Options

A.
$1 \text{ mV}$
B.
$2 \text{ mV}$
C.
$14 \text{ mV}$
D.
$28 \text{ mV}$

Check Your Answer

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

Photo-diode vs LED Bias|Dynamic Resistance (AC Resistance)|Average vs. Instantaneous Calculus

Why This Question Is Tricky

Common Mistakes

This question tests the student's ability to distinguish between static (DC) resistance and dynamic (AC) small-signal resistance in non-linear components like diodes.

A common trap is applying Ohm's Law macroscopically ( $R = V_{DC}/I_{DC} = 14 \ \Omega$ ) to find the AC voltage, which incorrectly yields $28 \text{ mV}$ .

Students must recognize that for small AC variations around a DC bias point, the response is governed by the local derivative of the V-I curve ( $r_d = dV/dI$ ), requiring calculus applied to the Shockley diode equation.

Quick Facts About This Question

Question Information

Subject	Physics
Chapter	Semiconductor Electronics: Materials, Devices, and Simple Circuits - Class 12
Concept	Photo-diode vs LED Bias
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free