Concept:The standard I-V curve of a PN junction exhibits an exponential increase in forward current above the knee voltage \( V_k \) (\( \approx 0.7\text{ V} \) for Si) and a small reverse saturation leakage current that breaks down at \( V_Z \).
Formula:$$I = I_0 \left( e^{\frac{q V}{\eta k T}} - 1 \right)$$
Solution:- In the forward quadrant, current remains negligible until \( V \ge V_k \), after which it rises exponentially.
- In the reverse quadrant, only a tiny saturation leakage current flows until reverse breakdown is reached.
- This non-linear asymmetric profile is the characteristic signature of a semiconductor diode.
Why other options are incorrect:- Option A: An ohmic resistor produces a linear line through the origin (\( I = V/R \)).
- Option C: A constant current source produces a horizontal line independent of voltage.
- Option D: An inductor exhibits frequency-dependent AC reactance and a linear DC response, not an exponential I-V curve.
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