Diodes & the PN Junction

The diode is the simplest semiconductor device — a one-way valve for current. Learn doping and the PN junction, forward and reverse bias, the Shockley equation and 0.6–0.7 V drop, I–V characteristics, diode types (signal, Schottky, LED, photodiode, varactor), and practical uses such as LED current limiting and reverse-polarity protection.

Joining p-type silicon (doped with acceptors, holes as carriers) and n-type silicon (donors, electrons as carriers) creates a depletion region with a built-in potential. Forward bias shrinks it and current flows; reverse bias widens it and blocks current.

Every rectifier, LED, solar cell, transistor and IC builds on this junction.

An analogy

A diode is a one-way check valve in a water pipe: water flows easily one way (after a small push to open the valve) and is blocked the other way.

Forward and reverse bias

Silicon diodes conduct noticeably above ≈ 0.6–0.7 V (germanium 0.3 V, Schottky 0.2–0.4 V, LEDs 1.8–3.3 V depending on colour). In reverse, only a tiny leakage flows until breakdown.

Diode models

Ideal switch; constant-drop model (0.7 V); piecewise-linear (0.7 V + series resistance); exponential Shockley model for accurate analysis. Temperature: V_F falls ≈ 2 mV/°C — used in silicon temperature sensors.

Diode types

Signal (1N4148, fast), rectifier (1N400x), Schottky (low drop, fast, higher leakage), Zener (controlled breakdown), LED (light emission), photodiode (light detection), varactor (voltage-variable capacitance for tuning), TVS (surge protection).

Dynamic behaviour

Small-signal resistance r_d = nV_T/I_D (26 Ω at 1 mA). Junction capacitance and reverse-recovery time limit switching speed; Schottky diodes have almost no recovery charge — preferred in switching supplies.

Formulas

• Shockley equation: I_D = I_S (e^(V_D / nV_T) − 1)
• Voltage for a given current: V_D = n V_T ln(I_D / I_S)
• Per decade of current: ΔV ≈ 2.3 n V_T ≈ 60 mV
• LED series resistor: R = (V_S − V_F) / I_F

Worked example: Diode voltage at 1 mA

Given: I_S = 10⁻¹⁴ A, n = 1, V_T = 25.85 mV

Find: V_D at 1 mA and at 10 mA.

• V_D = 0.02585 × ln(10⁻³/10⁻¹⁴) = 0.655 V
• At 10 mA: + 0.0595 V = 0.714 V

Answer: 0.655 V and 0.714 V

Worked example: LED resistor

Given: 5 V supply, red LED V_F = 2.0 V; I_F = 20 mA

Find: R and its power.

• R = (5 − 2)/0.02 = 150 Ω
• P = 0.02² × 150 = 60 mW

Answer: 150 Ω, 1/8 W part is fine

Common mistakes

• Wrong: Connecting an LED directly to a supply. — Right: Always limit current with a resistor or driver.
• Wrong: Forgetting the diode drop in supply budgets. — Right: Subtract 0.3–0.7 V from the available voltage.

Applications

• Reverse-polarity protection: Series diode or P-MOSFET on power inputs.
• Flyback diodes: Clamp relay and motor coil spikes.
• Signal clamping: Protect ADC inputs from over-voltage.

Key takeaways

• PN junction conducts one way.
• I = I_S(e^(V/nV_T) − 1).
• ≈ 0.7 V Si, 0.3 V Schottky, 2–3 V LEDs.
• Limit LED current with R = (V_S − V_F)/I.

Going further

• Reverse recovery: t_rr and charge in fast rectifiers.
• Diode temperature sensors: −2 mV/°C and PTAT circuits.
• Ideal diode controllers: MOSFET-based low-loss OR-ing.