• Tags: forward-bias, reverse-bias, current
• Difficulty: Moderate

Forward bias: connect positive terminal of battery to p-side, negative to n-side. The external voltage opposes the built-in potential, reducing the barrier: $V_{barrier}$ = $V_0$ - $V_{external}$. The depletion region narrows, majority carriers cross the junction, and large current flows. Current rises exponentially above the threshold voltage (~0.7 V Si, ~0.3 V Ge): I = $I_0$*(e^$\frac{eV}{kT}$ - 1). At V = 0.7 V for Si: e^$\frac{eV}{kT}$ = e^$\frac{0.7}{0.026}$ ≈ $e^{27}$ ≈ 5 x 10^11 — enormous multiplication. Reverse bias: positive terminal to n-side, negative to p-side. External voltage adds to barrier, widening depletion region. Only minority carriers (thermally generated) cross — tiny reverse saturation current $I_0$ (~nA for Si). Current is nearly independent of voltage until breakdown. The diode acts as a one-way valve: high conductance forward, negligible conductance reverse.