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Question #43323
The Question
An engineer designs a circuit to deliver power to an ideal p-n junction diode from an AC voltage source with amplitude . Due to thermal drift, the diode stabilizes at , where the thermal voltage is . At this temperature, governed by the Law of Mass Action, the reverse saturation current is .
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 , and calculates the expected average power as .
However, the true average power is governed by the non-linear ideal diode equation .
Calculate the absolute error in the engineer's calculation.
(To find , approximate the exponential up to the cubic term: . Use standard time averages and .)
Options
- A.
- B.
- C.
- D.
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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 .
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 and are temperature-dependent, setting the operating point where this non-linearity becomes pronounced.
Quick Facts About This Question
Question Information
| Subject | Physics |
| Chapter | Alternating Current - Class 12 |
| Concept | Instantaneous vs RMS vs Peak |
| Difficulty | 8 |
| Cognitive Load | 7 |
| Target Solve Time | 160 sec |
| Access | Free |