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Question #43323

Difficulty: 8/10 | 🟢 Free

The Question

An engineer designs a circuit to deliver power to an ideal p-n junction diode from an AC voltage source V(t)=V0sin⁡(ωt)V(t) = V_0 \sin(\omega t) with amplitude V0=50 mVV_0 = 50 \text{ mV}. Due to thermal drift, the diode stabilizes at T=300 KT = 300 \text{ K}, where the thermal voltage is VT=kTq=25 mVV_T = \frac{kT}{q} = 25 \text{ mV}. At this temperature, governed by the Law of Mass Action, the reverse saturation current is Is=2.0μAI_s = 2.0 \mu\text{A}.

Applying the Maximum Power Transfer principle linearly, the engineer assumes the diode acts as an Ohmic load with a constant resistance equal to its zero-bias dynamic resistance rd=VTIsr_d = \frac{V_T}{I_s}, and calculates the expected average power as Plinear=Vrms2rdP_{linear} = \frac{V_{rms}^2}{r_d}.

However, the true average power PtrueP_{true} is governed by the non-linear ideal diode equation I(t)=Is(eV(t)VT−1)I(t) = I_s \left( e^{\frac{V(t)}{V_T}} - 1 \right).

Calculate the absolute error ΔP=∣Ptrue−Plinear∣\Delta P = |P_{true} - P_{linear}| in the engineer's calculation.

(To find PtrueP_{true}, approximate the exponential up to the cubic term: ex−1≈x+x22+x36e^x - 1 \approx x + \frac{x^2}{2} + \frac{x^3}{6}. Use standard time averages ⟨sin⁡2(ωt)⟩=12\langle \sin^2(\omega t) \rangle = \frac{1}{2} and ⟨sin⁡4(ωt)⟩=38\langle \sin^4(\omega t) \rangle = \frac{3}{8}.)

Options

  • A.

    50 nW50 \text{ nW}

  • B.

    100 nW100 \text{ nW}

  • C.

    150 nW150 \text{ nW}

  • D.

    25 nW25 \text{ nW}

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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

Instantaneous vs RMS vs Peak|Law of Mass Action|Maximum Power Transfer

Why This Question Is Tricky

Common Mistakes

This question tests the misapplication of steady-state linear AC circuit analysis (using RMS voltage and constant resistance) to a non-linear semiconductor device.

Students often assume that for small to moderate signals, the average power can be found using P=Vrms2/rdP = V_{rms}^2 / r_d.

However, the diode current is governed by the Law of Mass Action, resulting in an exponential dependence on the barrier potential.

Because the exponential function grows much faster at the peak of the sine wave than it decreases at the trough, the true time-averaged power is significantly higher than the linear RMS estimate.

The inclusion of thermal drift highlights that IsI_s and VTV_T are temperature-dependent, setting the operating point where this non-linearity becomes pronounced.

Quick Facts About This Question

Question Information

SubjectPhysics
ChapterAlternating Current - Class 12
ConceptInstantaneous vs RMS vs Peak
Difficulty8
Cognitive Load7
Target Solve Time160 sec
AccessFree