p–n Junction DiodeJEE Main

The exponential I–V characteristic — interactive Physics simulation for IIT-JEE.

Concept

Join p-type and n-type silicon and carriers diffuse across, leaving a depletion region with a built-in barrier (~0.7 V for Si). Forward bias lowers the barrier — current grows exponentially. Reverse bias raises it — only a tiny saturation current leaks. The diode is a one-way valve: the heart of every rectifier.

Key formula

I = I_s\left(e^{V/nV_T} - 1\right), \qquad V_T = \frac{k_BT}{e} \approx 26\,\text{mV at } 300\,\text{K}

Derivation

Diffusion current across the junction ∝ carrier density at the barrier edge, which Boltzmann-factors as $e^{V/V_T}$ under bias V. Drift (leakage) current is bias-independent: subtracting gives the Shockley equation.

Each +60 mV (≈ 2.3·nV_T) multiplies the forward current ×10 — that's why the 'knee' looks so sharp on a linear plot.

Scenarios to explore

p–n Junction Diode — Exponential I–V curve, depletion region & rectification.

Real-world applications

Rectifiers: half-wave (one diode), full-wave bridge (four).
LEDs (forward, radiative recombination) & photodiodes (reverse, light-generated).
Zener diodes regulate voltage in controlled reverse breakdown.

JEE exam tips

Half-wave rectifier output frequency = f; full-wave = 2f — classic MCQ.
Ideal-diode circuit analysis: assume ON (0.7 V drop) or OFF, then verify consistency.
Dynamic resistance r = nV_T/I falls as current rises.

Common mistakes

Treating 0.7 V as a hard switch — the curve is exponential, the 'knee' is a plot artifact.
Forgetting reverse saturation current doubles roughly every 10 °C.
Confusing depletion-region widening (reverse) vs narrowing (forward).

Exam traps to avoid

Diode in reverse = open circuit (except Zener breakdown).
In a bridge rectifier two diodes conduct per half-cycle.