Semiconductor Electronics
Semiconductor Electronics for JEE Main & Advanced
Semiconductors and PN Junction Diode
Intrinsic and Extrinsic SemiconductorsTopic 1
Classification of Solids by Conductivity:
| Type | Conductivity (S/m) | Band gap $E_g$ | Examples |
|---|---|---|---|
| Conductor | $10^7$ | $0$ or overlap | Cu, Al, Ag |
| Insulator | $10^{-20}$ – $10^{-11}$ | $> 3$ eV | Wood, glass |
| Semiconductor | $10^{-6}$ – $10^4$ | $\approx 1$ eV | Si ($E_g = 1.1$ eV), Ge ($E_g = 0.7$ eV) |
Energy Band Theory: In solids, atomic orbitals form bands:
- Valence band: highest occupied band at $T = 0$
- Conduction band: next higher band; electrons here carry current
- Forbidden gap (band gap): energy gap between them
Intrinsic Semiconductor: Pure semiconductor (Si, Ge). At $T = 0$: behaves as insulator. At $T > 0$: thermal energy excites electrons from valence to conduction band, leaving holes behind.
In intrinsic semiconductor: $n_e = n_h = n_i$ (electron and hole densities equal).
Conductivity: $\sigma = e(n_e\mu_e + n_h\mu_h)$, where $\mu$ = mobility. For semiconductors, $\mu_e > \mu_h$.
Extrinsic (Doped) Semiconductor:
| Type | Dopant | Group | Majority carriers | Minority carriers |
|---|---|---|---|---|
| N-type | P, As, Sb | V (pentavalent) | Electrons | Holes |
| P-type | B, Al, Ga, In | III (trivalent) | Holes | Electrons |
In N-type: $n_e \gg n_h$; donor level near conduction band. In P-type: $n_h \gg n_e$; acceptor level near valence band.
Mass Action Law: $n_e n_h = n_i^2$ (always).
Both types are electrically neutral overall.
In a Si sample, $n_i = 1.5 \times 10^{16}$ /m³ at $300$ K. If doped with P giving $n_e = 5 \times 10^{22}$ /m³, find $n_h$.
Show solution
$$n_h = n_i^2/n_e = (1.5 \times 10^{16})^2/(5 \times 10^{22}) = 2.25 \times 10^{32}/(5 \times 10^{22})$$ $$= 4.5 \times 10^9 \text{ /m}^3$$
Final Answer: $n_h = 4.5 \times 10^9$ /m³.
Identify the type of semiconductor obtained when Ge is doped with: (a) Boron, (b) Arsenic.
Show solution
(a) Boron is trivalent (Group III) → creates holes → P-type. (b) Arsenic is pentavalent (Group V) → contributes electrons → N-type.
Final Answer: (a) P-type; (b) N-type.
The band gap of intrinsic Si:
In N-type semiconductor, majority carriers are:
Doping Ge with In gives:
Mass action law for semiconductors:
The conductivity of semiconductors with temperature:
PN Junction Diode, V-I Characteristics, RectifiersTopic 2
PN Junction: Formed when P-type and N-type semiconductors are joined.
Formation Mechanism:
- Diffusion of majority carriers across junction
- Recombination near junction creates depletion region (no free charges)
- Immobile ions (negative on P-side, positive on N-side) set up internal potential barrier $V_B$ that stops further diffusion
Barrier voltage: $V_B \approx 0.7$ V for Si, $0.3$ V for Ge.
Biasing:
| Bias | Connection | Depletion region | Current |
|---|---|---|---|
| Forward | P to $+$, N to $-$ | Decreases | Significant (after $V_B$ overcome) |
| Reverse | P to $-$, N to $+$ | Increases | Very small (reverse saturation, $\sim \mu$A) |
V-I Characteristics:
| Region | Behavior |
|---|---|
| Forward, $V < V_B$ | Very small current (cut-in region) |
| Forward, $V > V_B$ | Current increases sharply (exponentially: $I = I_0(e^{V/V_T} - 1)$) |
| Reverse, $V < V_{BR}$ | Small reverse saturation current |
| Reverse, $V > V_{BR}$ | Avalanche / Zener breakdown — large current |
Dynamic Resistance: $r = \Delta V/\Delta I$ at operating point.
Rectifiers: Convert AC to DC using diode's one-way conduction.
Half-Wave Rectifier (HWR): One diode; conducts only during half cycle. Output frequency = input frequency.
- DC voltage: $V_{dc} = V_m/\pi$
- RMS: $V_{rms} = V_m/2$
- Efficiency: $\eta_{\max} \approx 40.6\%$
- Ripple factor: $\gamma = 1.21$
Full-Wave Rectifier (FWR): Two diodes (center-tapped) or four (bridge). Both half cycles used. Output frequency = $2 \times$ input.
- $V_{dc} = 2V_m/\pi$
- $V_{rms} = V_m/\sqrt{2}$
- Efficiency: $\eta_{\max} \approx 81.2\%$
- Ripple factor: $\gamma = 0.48$
Filters: Capacitor or inductor smooths pulsating DC into nearly steady DC.
A Si diode has barrier voltage $0.7$ V. When forward-biased with $5$ V through $1$ kΩ, find current.
Show solution
$$I = \frac{V - V_B}{R} = \frac{5 - 0.7}{1000} = \frac{4.3}{1000} = 4.3 \text{ mA}$$
Final Answer: $I = 4.3$ mA.
Peak input voltage to half-wave rectifier is $200$ V. Find DC output voltage.
Show solution
$$V_{dc} = V_m/\pi = 200/\pi \approx 63.66 \text{ V}$$
Final Answer: $V_{dc} \approx 63.66$ V.
The depletion region in a PN junction:
Barrier voltage of Si diode:
Output frequency of half-wave rectifier with $50$ Hz input:
Full-wave rectifier efficiency (max, theoretical):
In reverse bias:
Transistors and Digital Electronics
Special Diodes and Bipolar Junction TransistorTopic 1
Special-Purpose Diodes:
| Diode | Function | Use |
|---|---|---|
| Zener diode | Operates in reverse breakdown at constant $V_Z$ | Voltage regulator |
| LED | Emits light when forward biased | Indicator, displays |
| Photodiode | Generates current when exposed to light | Light sensor |
| Solar cell | LED operation in reverse (light → EMF) | Solar power |
Zener Regulator: Used in parallel with load; maintains constant voltage across load even when input fluctuates.
LED Wavelength: Determined by band gap of semiconductor: $\lambda = hc/E_g$. Different materials → different colors.
Bipolar Junction Transistor (BJT): Three-layer device with two PN junctions. Three terminals: Emitter (E), Base (B), Collector (C).
| Type | Structure | Direction of conventional current |
|---|---|---|
| NPN | n-p-n | $I_E$ flows into emitter |
| PNP | p-n-p | $I_E$ flows out of emitter |
Three Configurations:
- Common Base (CB)
- Common Emitter (CE) ← most used
- Common Collector (CC)
Transistor Action: Emitter is heavily doped, base is thin and lightly doped, collector is moderately doped. Junctions: emitter-base forward biased, collector-base reverse biased.
Current Relations: $I_E = I_B + I_C$.
Current Gain Parameters:
| Configuration | Definition |
|---|---|
| Common Base $\alpha$ | $\alpha = I_C/I_E$ ($\sim 0.95 - 0.99$) |
| Common Emitter $\beta$ | $\beta = I_C/I_B$ ($\sim 20 - 500$) |
Relations: $\beta = \alpha/(1-\alpha)$, $\alpha = \beta/(1+\beta)$.
Voltage Gain in CE: $A_V = \beta \cdot R_L/R_i$.
Transistor as Amplifier: Small input AC at base → large output AC at collector (CE). Phase reversal of $180°$ between input and output.
Transistor as Switch: Operates between cut-off (both junctions reverse-biased; $I_C = 0$) and saturation (both forward-biased; $V_{CE}$ minimum). Basis for digital logic.
A transistor in CE has $\beta = 50$. Find $\alpha$ and $I_C$ when $I_B = 100\,\mu$A.
Show solution
$$\alpha = \beta/(1+\beta) = 50/51 \approx 0.98$$ $$I_C = \beta I_B = 50 \times 100\,\mu\text{A} = 5 \text{ mA}$$
Final Answer: $\alpha \approx 0.98$; $I_C = 5$ mA.
An LED has band gap $E_g = 1.9$ eV. Find emitted wavelength.
Show solution
$$\lambda = hc/E_g = 1240/1.9 \approx 653 \text{ nm}$$
This is red light.
Final Answer: $\lambda \approx 653$ nm (red).
Zener diode operates in:
The emitter of a transistor is:
In an NPN transistor in CE:
Relation between $\alpha$ and $\beta$:
In common emitter amplifier, output is:
Logic Gates and Boolean AlgebraTopic 2
Logic Gates are digital circuits performing Boolean operations. Inputs/outputs are binary: $0$ (low/false) or $1$ (high/true).
Basic Gates:
| Gate | Symbol | Operation | Truth Table |
|---|---|---|---|
| NOT | $\bar A$ | Inverter | $A: 0,1 \to Y: 1,0$ |
| AND | $A\cdot B$ | $Y = 1$ only if both inputs $= 1$ | $00\to 0$; $01\to 0$; $10\to 0$; $11\to 1$ |
| OR | $A+B$ | $Y = 1$ if any input $= 1$ | $00\to 0$; $01\to 1$; $10\to 1$; $11\to 1$ |
Combination Gates:
| Gate | Operation | Truth Table | |
|---|---|---|---|
| NAND | $\overline{A\cdot B}$ | $00\to 1$; $01\to 1$; $10\to 1$; $11\to 0$ | |
| NOR | $\overline{A+B}$ | $00\to 1$; $01\to 0$; $10\to 0$; $11\to 0$ | |
| XOR | $A\oplus B$ | $1$ if odd number of $1$s | $00\to 0$; $01\to 1$; $10\to 1$; $11\to 0$ |
| XNOR | $\overline{A\oplus B}$ | $1$ if even number of $1$s | $00\to 1$; $01\to 0$; $10\to 0$; $11\to 1$ |
Universal Gates: NAND and NOR can each produce any other gate combination.
NOT from NAND: $\bar A = \overline{A\cdot A}$ AND from NAND: $A\cdot B = \overline{\overline{A\cdot B}}$ (NAND followed by NOT) OR from NAND: $A + B = \overline{\bar A \cdot \bar B}$ (De Morgan)
Boolean Laws:
| Law | Statement |
|---|---|
| Commutative | $A + B = B + A$; $AB = BA$ |
| Associative | $(A+B)+C = A+(B+C)$ |
| Distributive | $A(B+C) = AB + AC$ |
| Identity | $A + 0 = A$; $A \cdot 1 = A$ |
| Null | $A + 1 = 1$; $A \cdot 0 = 0$ |
| Idempotent | $A + A = A$; $A\cdot A = A$ |
| Complement | $A + \bar A = 1$; $A\bar A = 0$ |
| Double Negation | $\bar{\bar A} = A$ |
| De Morgan's | $\overline{A+B} = \bar A\bar B$; $\overline{A\cdot B} = \bar A + \bar B$ |
Find output $Y$ of a circuit with two AND gates and inputs $A, B, C$ such that $Y = (A\cdot B) + C$, with $A = 1, B = 0, C = 1$.
Show solution
$A\cdot B = 1\cdot 0 = 0$. $Y = 0 + 1 = 1$.
Final Answer: $Y = 1$.
Use De Morgan's law: $\overline{A + B\bar C}$.
Show solution
$$\overline{A + B\bar C} = \bar A \cdot \overline{B\bar C} = \bar A \cdot (\bar B + C)$$
Final Answer: $\bar A(\bar B + C)$.
NAND gate output is $1$ when:
Universal gates:
Output of XOR gate for $A = 1, B = 1$:
De Morgan's law states:
NOR gate is logically equivalent to:
Ready to test yourself?
Attempt the full timed mock tests — Main & Advanced level.
Start Mock Test 1 →