Concept:Forward bias is established whenever the anode potential is algebraically more positive (higher) than the cathode potential: \( V_A > V_K \).
Formula:$$V_D = V_A - V_K = (-1.0\text{ V}) - (-4.0\text{ V}) = -1.0 + 4.0 = +3.0\text{ V} > 0 \implies \text{Forward Bias}$$
Solution:- Although both potentials are negative with respect to ground, \( -1.0\text{ V} \) is higher (less negative) than \( -4.0\text{ V} \).
- The net voltage across the diode is \( V_D = +3.0\text{ V} \), which exceeds the \( 0.7\text{ V} \) barrier potential of silicon.
- Therefore, the diode is forward-biased and conducts forward current.
Why other options are incorrect:- Option A: Negative absolute potentials do not imply reverse bias; the relative difference determines the bias state.
- Option B: The potentials differ by \( 3.0\text{ V} \), so the diode is not zero-biased.
- Option D: Avalanche breakdown occurs only under large negative reverse voltages (\( V_D \ll 0 \)).
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