Worked example: depletion width vs bias

Take a silicon PN junction at room temperature with built-in potential $V_0 = 0.7\,\text{V}$. Assuming the depletion-region width scales as $W \propto \sqrt{V_0 - V_{app}}$, estimate how the width changes when the applied bias goes from 0 V to +0.5 V (forward) and to −5 V (reverse), expressed as a ratio of the equilibrium width.

1. Given. Silicon PN junction, $V_0 = 0.7\,\text{V}$. Three operating points: equilibrium ($V_{app} = 0$), forward bias ($V_{app} = +0.5\,\text{V}$), and reverse bias ($V_{app} = -5\,\text{V}$).
2. Equation. Depletion width versus net junction voltage:
   $$W \propto \sqrt{V_0 - V_{app}}$$
   We'll express each result as a ratio $W / W_0$, where $W_0$ is the equilibrium width.
3. Substitution. Compute the square roots inside the proportionality:
   • Equilibrium: $\sqrt{0.7 - 0} = \sqrt{0.7} \approx 0.837$
   • Forward (+0.5 V): $\sqrt{0.7 - 0.5} = \sqrt{0.2} \approx 0.447$
   • Reverse (−5 V): $\sqrt{0.7 - (-5)} = \sqrt{5.7} \approx 2.387$
4. Result. Divide each by the equilibrium value (0.837) to get the relative width:
   • Equilibrium: $W / W_0 = 1.000$ (by definition).
   • Forward (+0.5 V): $W / W_0 \approx 0.447 / 0.837 \approx 0.534$ — the depletion region is barely half its equilibrium width.
   • Reverse (−5 V): $W / W_0 \approx 2.387 / 0.837 \approx 2.852$ — almost three times wider than at equilibrium.
5. Sanity-check. Forward bias subtracts from the built-in potential, so the barrier shrinks — and we expect the diode to start conducting once the barrier is low enough for majority carriers to climb it. Reverse bias adds to $V_0$, so the barrier grows and current is choked off. Both directions match physical intuition.