Concept:When two diodes with different forward knee voltages are connected in parallel, the diode with the lower threshold turns ON first and clamps the terminal voltage, keeping the second diode reverse-biased or below its turn-on threshold.
Formula:$$V_{\text{parallel}} = \min(V_{D,\text{Ge}}, V_{D,\text{Si}}) = 0.30\text{ V}$$
Solution:- As the supply voltage rises, the Germanium diode turns ON at \( 0.30\text{ V} \) and begins conducting.
- This clamps the voltage across the parallel branch to \( 0.30\text{ V} \).
- Because \( 0.30\text{ V} < 0.70\text{ V} \), the Silicon diode never reaches its turn-on threshold and remains OFF.
- Thus, the voltage across the parallel combination is \( 0.30\text{ V} \).
Why other options are incorrect:- Option A: 0.70 V would require the Silicon diode to conduct, which cannot happen while the Germanium diode clamps the voltage at 0.30 V.
- Option C: 1.00 V incorrectly adds the two diode voltages in parallel.
- Option D: 5.00 V is the open-circuit supply voltage, ignoring the forward-biased conducting diode.
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