Practice question
Question
A silicon diode has a threshold voltage of approximately:
Explanation
**p-n junction formation** diffusion of holes from p to n and electrons from n to p leaves ionized donors positive on n-side and acceptors negative on p-side, forming space-charge depletion region with electric field directed from n to p (positive to negative), barrier potential V_b ≈0.3 V Ge, 0.7 V Si, opposes further diffusion, drift current due to minority carriers swept by field balances diffusion at equilibrium net current zero. The threshold or cut-in voltage for a silicon diode is about 0.7 V, beyond which the forward current increases significantly. Substituting values gives 0.7 V, which matches expected behaviour for this semiconductor device configuration, confirming doping, depletion and rectifier principles.
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