Concept:A varactor diode operates in reverse bias as a voltage-controlled capacitor: \( C_T \propto (V_0 + V_R)^{-1/2} \). Varying \( V_R \) changes the tank capacitance and tunes the resonant frequency \( f_0 = 1 / (2\pi\sqrt{L C}) \).
Formula:$$f_0 = \frac{1}{2\pi \sqrt{L \cdot C_T(V_R)}} \implies \text{Increasing } V_R \implies C_T \downarrow \implies f_0 \uparrow$$
Solution:- Increasing the reverse voltage \( V_R \) widens the depletion layer, decreasing the transition capacitance \( C_T \).
- Because resonant frequency is inversely proportional to \( \sqrt{C_T} \), adjusting the DC control voltage provides electronic frequency tuning.
Why other options are incorrect:- Option B: Varactor diodes do not emit light and cannot alter coil inductance.
- Option C: The varactor is operated in reverse bias; its forward resistance is not part of normal operation.
- Option D: Varactors operate purely via solid-state electrostatics.
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