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Question #42660

Difficulty: 8/10 | 🟢 Free

The Question

An ideal current source supplies a total current $I_{\text{total}}(t) = I_{DC} + I_m \sin(\omega t)$ to a circuit consisting of a p-n junction diode in parallel with an AC-coupled variable load resistor $R_L$ . The coupling capacitor acts as an ideal short circuit for the AC signal and an ideal open circuit for the DC signal. The diode's forward current is dominated by the diffusion of carriers over the potential barrier, following the relationship $I = I_0 \exp\left(\frac{eV}{kT}\right)$ . The DC operating point is established at a forward voltage $V_{DC} = 0.60\text{ V}$ and a DC current $I_{DC} = 2.0\text{ mA}$ . The thermal voltage is $\frac{kT}{e} = 25\text{ mV}$ , and the AC signal amplitude is $I_m = 40\ \mu\text{A}$ . To extract the maximum average AC power into $R_L$ , what must be the value of $R_L$ , and what is the corresponding maximum average AC power $P_{\text{max}}$ dissipated in $R_L$ ?

Options

A.
$R_L = 12.5\ \Omega, P_{\text{max}} = 2.5\text{ nW}$
B.
$R_L = 300\ \Omega, P_{\text{max}} = 60.0\text{ nW}$
C.
$R_L = 12.5\ \Omega, P_{\text{max}} = 5.0\text{ nW}$
D.
$R_L = 300\ \Omega, P_{\text{max}} = 120.0\text{ nW}$

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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

Reference Potential Convention|Drift vs Diffusion Current|Maximum Power Transfer|Approximation (Binomial)

Why This Question Is Tricky

Common Mistakes

This question tests the student's ability to distinguish between the static (DC) resistance and the dynamic (AC) small-signal resistance of a non-linear semiconductor device.

The primary trap lies in conflating the built-in potential barrier's macroscopic V/I ratio with the differential resistance $r_d = dV/dI$ .

Students commonly misapply Ohm's law by calculating the internal resistance as $R_{DC} = V_{DC}/I_{DC} = 300\ \Omega$ , leading to an incorrect load matching condition.

Furthermore, the problem tests the correct application of the Maximum Power Transfer Theorem in a Norton equivalent circuit and the calculation of average power for sinusoidal signals, where omitting the RMS conversion factor of $1/2$ provides a secondary trap.

Quick Facts About This Question

Question Information

Subject	Physics
Chapter	Electrostatic Potential and Capacitance - Class 12
Concept	Reference Potential Convention
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free