Semiconductor Electronics: Materials, Devices and Simple Circuits

Electronics is the branch of science dealing with the flow of electrons in vacuum, gas or semiconductors and their practical applications. Modern electronics is largely based on semiconductor materials like silicon and germanium.

Energy Bands in Solids

In solids, electron energy levels form continuous energy bands:
Valence Band: Highest band that is filled with electrons at 0 K.
Conduction Band: Next higher band, normally empty at 0 K.
Forbidden Gap (E_g): Energy gap between valence and conduction band.

Classification of solids:
Conductor: Valence and conduction bands overlap; E_g = 0.
Insulator: Large E_g > 3 eV (e.g., diamond E_g = 5.5 eV).
Semiconductor: Small E_g; silicon E_g = 1.12 eV, germanium E_g = 0.72 eV.

Intrinsic and Extrinsic Semiconductors

Intrinsic semiconductor: Pure semiconductor. Electron-hole pairs generated thermally. n_e = n_h = n_i.

Extrinsic semiconductor: Doped with impurities.
n-type: Doped with pentavalent impurity (P, As, Sb). Majority carriers: electrons. Minority carriers: holes.
p-type: Doped with trivalent impurity (B, Al, In). Majority carriers: holes. Minority carriers: electrons.

In both types: n_e x n_h = n_i² (mass action law)

p-n Junction Diode

Formed when p-type and n-type semiconductors are joined. At the junction:
Diffusion of electrons from n to p and holes from p to n creates a depletion region (no free carriers).
Built-in electric field (from n to p) creates a potential barrier (V₀).

Forward Bias: p connected to +ve terminal; depletion width narrows; current flows above ~0.3 V (Ge) or ~0.7 V (Si). Current increases exponentially.

Reverse Bias: p connected to -ve terminal; depletion width increases; tiny reverse saturation current flows. At high reverse voltage, breakdown occurs.

Diode as rectifier: A rectifier converts AC to DC.
Half-wave rectifier: Uses one diode; only positive (or negative) half of AC passes.
Full-wave rectifier: Uses two diodes (centre-tap) or four diodes (bridge) — both halves of AC utilised.

Zener Diode

A specially doped diode designed to operate in the reverse breakdown region. It maintains a nearly constant voltage across it (Zener voltage V_Z) over a wide range of currents. Used as a voltage regulator.

Transistor

A transistor has three terminals: Emitter (E), Base (B) and Collector (C). Types: n-p-n and p-n-p.

Configurations: Common Base (CB), Common Emitter (CE), Common Collector (CC).

In the Common Emitter configuration:
Input: base-emitter; Output: collector-emitter.
Current gain beta = I_C / I_B (typically 20-200).
Alpha = I_C / I_E = beta/(1 + beta)
I_E = I_B + I_C

Transistor as amplifier (CE mode): Small base current change causes large collector current change. Voltage gain A_V = beta x (R_C/R_i).

Transistor as switch: In cut-off mode (no base current), transistor is OFF. In saturation mode (sufficient base current), it is ON. Used in digital logic.

Logic Gates

Basic logic gates (implemented using diodes and transistors):
NOT gate: Output = complement of input (A_out = A-bar)
OR gate: Output = 1 if any input is 1
AND gate: Output = 1 only if all inputs are 1
NAND gate: NOT of AND — universal gate
NOR gate: NOT of OR — universal gate

Boolean expressions: NOT: Y = A-bar; OR: Y = A+B; AND: Y = A.B; NAND: Y = (A.B)-bar; NOR: Y = (A+B)-bar.

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Example 1

In a semiconductor, n_e = 2 x 10¹⁶ m^-3 and n_i = 10¹⁶ m^-3. Find hole concentration.
n_e x n_h = n_i². n_h = n_i² / n_e = (10¹⁶)² / (2 x 10¹⁶) = 10³² / 2 x 10¹⁶ = 5 x 10¹⁵ m^-3.

Example 2

A Zener diode has V_Z = 6 V. It is connected in reverse bias with a series resistor R = 1 k-ohm and supply V_s = 10 V. Find current through the circuit.
Voltage across R = V_s - V_Z = 10 - 6 = 4 V. I = 4/1000 = 4 mA.

Example 3

In a transistor, I_B = 50 micro A and beta = 100. Find I_C and I_E.
I_C = beta x I_B = 100 x 50 x 10^-6 = 5 mA. I_E = I_B + I_C = 0.05 + 5 = 5.05 mA.

Example 4

Find output of AND gate with inputs A = 1, B = 0.
Y = A.B = 1.0 = 0. Output is 0.

Example 5

A half-wave rectifier uses a diode with a load resistance of 1 k-ohm. If peak input voltage is 10 V and diode forward voltage is 0.7 V, the peak output voltage is:
V_peak_out = 10 - 0.7 = 9.3 V. Average output voltage = V_peak / pi = 9.3/3.14 = 2.96 V.

Example 6

For a CE amplifier, beta = 50, R_C = 2 k-ohm, R_i = 1 k-ohm. Find voltage gain.
A_V = -beta x R_C/R_i = -50 x 2000/1000 = -100 (negative indicates phase inversion).

Example 7

Write the truth table for a NAND gate with two inputs A and B.
A=0,B=0: Y=1; A=0,B=1: Y=1; A=1,B=0: Y=1; A=1,B=1: Y=0. NAND is 0 only when both inputs are 1.

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Common mistakes

Forgetting that in n-type semiconductor, electrons are majority carriers (not in p-type).
Mixing up alpha and beta: beta = I_C/I_B (always > 1), alpha = I_C/I_E (always < 1).
Thinking the output of a NOT gate is always opposite — it is, but for NAND/NOR, remember the NOT is applied to the whole AND/OR result.

Summary

Semiconductors have a small energy gap (Si: 1.12 eV). Doping creates n-type or p-type material. The p-n junction acts as a diode (rectifier). Zener diodes regulate voltage. Transistors in CE mode give current gain beta = I_C/I_B and are used as amplifiers and switches. Logic gates form the basis of digital electronics.