Semiconductor Electronics — p-n Junction, Diode, Transistor, Logic Gates

From Sand to Smartphones

Semiconductors are materials with conductivity between metals and insulators. Silicon and germanium — with 4 valence electrons — are the workhorses. By adding tiny amounts of impurities (doping), we create p-type and n-type semiconductors. Put them together and we get the p-n junction — the building block of diodes, transistors, and all modern electronics.

CBSE Class 12 boards give this chapter 5-7 marks. JEE Main occasionally tests logic gates and transistor configurations.

graph TD
    A[Semiconductor] --> B{Doping type?}
    B -->|Group 15 element| C[n-type: electrons majority]
    B -->|Group 13 element| D[p-type: holes majority]
    C --> E[p-n Junction]
    D --> E
    E --> F[Diode]
    F --> G{Application?}
    G -->|Rectification| H[Half/Full wave rectifier]
    G -->|Voltage regulation| I[Zener diode]
    G -->|Light emission| J[LED]
    G -->|Amplification| K[Transistor]
    K --> L[Logic Gates: AND, OR, NOT, NAND, NOR]

Key Concepts

Forward bias: p-side to positive terminal. Barrier potential decreases. Current flows.

Reverse bias: p-side to negative terminal. Barrier increases. Negligible current (only leakage).

Knee voltage: ~0.7 V for Si, ~0.3 V for Ge (threshold for conduction).

Configuration	Input	Output	Current gain	Use
Common Emitter (CE)	Base	Collector	$\beta = I_C/I_B$ (20-200)	Amplifier
Common Base (CB)	Emitter	Collector	$\alpha = I_C/I_E$ (0.95-0.99)	High-frequency
Common Collector (CC)	Base	Emitter	High	Impedance matching

$I_E = I_B + I_C$ and $\beta = \frac{\alpha}{1-\alpha}$

Gate	Symbol	Output
AND	A.B	1 only if both inputs are 1
OR	A+B	1 if at least one input is 1
NOT	A’	Inverts input
NAND	(A.B)‘	0 only if both are 1
NOR	(A+B)‘	1 only if both are 0

NAND and NOR are universal gates — any logic function can be built from either alone.

Solved Examples

Example 1 (Easy — CBSE)

In a CE transistor, $I_B = 20\ \mu$ A and $\beta = 100$ . Find $I_C$ and $I_E$ .

$I_C = \beta I_B = 100 \times 20 = \mathbf{2000\ \mu\text{A} = 2 \text{ mA}}$

$I_E = I_B + I_C = 20 + 2000 = \mathbf{2020\ \mu\text{A} \approx 2.02 \text{ mA}}$

Example 2 (Medium — CBSE)

Construct an AND gate using NAND gates.

Use three NAND gates. Feed both inputs $A$ and $B$ into the first NAND gate. The output is $(AB)'$ . Feed this output into a second NAND gate with both inputs tied together (acts as NOT). Output: $((AB)')' = AB$ . Done — that’s an AND gate from two NAND gates.

Example 3 (Medium — JEE Main)

A Zener diode with breakdown voltage 6.8 V is used with a 1.2 k $\Omega$ series resistor. Input voltage is 16 V. Find the current through the Zener and a 4 k $\Omega$ load.

Voltage across load $= 6.8$ V. $I_L = 6.8/4000 = 1.7$ mA.

Voltage across series resistor $= 16 - 6.8 = 9.2$ V. $I_S = 9.2/1200 = 7.67$ mA.

$I_Z = I_S - I_L = 7.67 - 1.7 = \mathbf{5.97 \text{ mA}}$

Common Mistakes to Avoid

Mistake 1 — Confusing forward and reverse bias. Forward: p to positive, n to negative. Reverse: opposite. Students mix these up frequently.

Mistake 2 — Using $\alpha$ instead of $\beta$ in CE configuration. CE uses $\beta = I_C/I_B$ . CB uses $\alpha = I_C/I_E$ . Mixing them gives wrong current values.

Mistake 3 — Wrong truth table for NAND. NAND output is 0 only when BOTH inputs are 1. For all other combinations, output is 1.

Practice Questions

Q1. If $\alpha = 0.98$ , find $\beta$ .

$\beta = \alpha/(1-\alpha) = 0.98/0.02 = 49$ .

Q2. Draw the truth table for NOR gate.

A	B	A+B	(A+B)‘
0	0	0	1
0	1	1	0
1	0	1	0
1	1	1	0

Q3. Why is NAND called a universal gate?

Any Boolean function (AND, OR, NOT, XOR, etc.) can be constructed using only NAND gates. NOT: connect both inputs together. AND: NAND followed by NOT. OR: NOT each input, then NAND.

Q4. A silicon diode is forward biased with 3 V through a 1 k $\Omega$ resistor. Find current.

$V_{diode} = 0.7$ V. $I = (3 - 0.7)/1000 = 2.3$ mA.

FAQs

What is the depletion region?

At the p-n junction, electrons from n-side and holes from p-side recombine near the junction, creating a charge-free zone called the depletion region. It acts as a barrier to further charge flow.

Why is silicon preferred over germanium?

Silicon has a wider band gap (1.1 eV vs 0.67 eV), making it more stable at higher temperatures. Silicon oxide (SiO $_2$ ) is a natural insulator, useful for fabrication.

What is the difference between a diode and a transistor?

A diode has one p-n junction and acts as a one-way valve for current. A transistor has two junctions (either npn or pnp) and can amplify signals or act as a switch.

Advanced Concepts

Energy bands — why semiconductors are special

In a solid, atomic energy levels broaden into bands. The valence band is the highest filled band; the conduction band is the next higher (empty or partially filled).

Material	Band gap	Conductivity
Conductor (Cu, Ag)	0 eV (bands overlap)	Very high
Semiconductor (Si, Ge)	1–3 eV (Si: 1.1 eV, Ge: 0.67 eV)	Moderate, increases with temperature
Insulator (diamond)	> 3 eV (diamond: 5.5 eV)	Negligible

At absolute zero, a semiconductor’s valence band is full and conduction band is empty — it behaves like an insulator. At room temperature, thermal energy excites some electrons across the gap into the conduction band, creating conductivity.

“Why does the resistance of a semiconductor decrease with temperature?” is a 2-mark CBSE board question. Answer: higher temperature gives more electrons enough energy to jump the band gap, increasing charge carriers. More carriers mean lower resistance.

Rectification — from AC to DC

Half-wave rectifier: Uses one diode. Only the positive half-cycle passes through. Output frequency = input frequency. Ripple is high.

Full-wave rectifier: Uses two diodes with a centre-tapped transformer (or four diodes in a bridge). Both half-cycles are utilised. Output frequency = 2 $\times$ input frequency. Smoother output.

For a full-wave rectifier with peak voltage $V_0$ :

V_{DC} = \frac{2V_0}{\pi} \approx 0.636 V_0

For a half-wave rectifier:

V_{DC} = \frac{V_0}{\pi} \approx 0.318 V_0

Zener diode as voltage regulator

A Zener diode operates in reverse bias at its breakdown voltage $V_Z$ . The voltage across it remains constant ( $V_Z$ ) even if input voltage or load changes.

Given: $V_{in} = 15$ V, $V_Z = 9$ V, series resistance $R_s = 600$ $\Omega$ , load $R_L = 1.8$ k $\Omega$ .

Voltage across load = $V_Z = 9$ V.

$I_L = V_Z/R_L = 9/1800 = 5$ mA.

$I_s = (V_{in} - V_Z)/R_s = (15 - 9)/600 = 10$ mA.

$I_Z = I_s - I_L = 10 - 5 = 5$ mA.

The Zener absorbs the excess current to maintain constant voltage across the load.

Transistor as a switch

In the cutoff region ( $I_B = 0$ ), no current flows through collector — the transistor is OFF (switch open). In the saturation region ( $I_B$ is large enough), $V_{CE} \approx 0$ and maximum current flows — the transistor is ON (switch closed).

This ON/OFF behaviour is the basis of digital electronics. Every logic gate, processor, and memory chip uses transistors as switches — billions of them.

Boolean algebra basics for logic gates

De Morgan’s theorems are essential for simplifying logic circuits:

\overline{A \cdot B} = \bar{A} + \bar{B} \quad \text{(NAND = OR of NOTs)}

\overline{A + B} = \bar{A} \cdot \bar{B} \quad \text{(NOR = AND of NOTs)}

To verify De Morgan’s theorems, draw truth tables for both sides — they will be identical. This verification is a 3-mark CBSE question.

Additional Solved Examples

Example 4 (CBSE Board): Construct an OR gate using NAND gates only.

First, NOT each input using NAND gates (tie both inputs together): $\bar{A}$ from one, $\bar{B}$ from another. Then NAND the two NOT outputs: $\overline{\bar{A} \cdot \bar{B}} = A + B$ (by De Morgan’s theorem). Three NAND gates total.

Example 5 (JEE Main): A photodiode is operated in which bias? Why?

Reverse bias. The reverse saturation current increases proportionally with light intensity. In forward bias, the photocurrent would be swamped by the large forward current, making light detection impossible.

Additional Practice Questions

Q5. What is the band gap of silicon in eV?

1.1 eV. Germanium has a band gap of 0.67 eV, which is why Ge is more sensitive to temperature changes than Si.

Q6. In a bridge rectifier, how many diodes are used and what is the output frequency for 50 Hz input?

Four diodes. Output frequency = 2 $\times$ 50 = 100 Hz, because both halves of the AC cycle are rectified.

Q7. Verify De Morgan’s first theorem using a truth table.

A	B	A.B	$\overline{A.B}$	$\bar{A}$	$\bar{B}$	$\bar{A}+\bar{B}$
0	0	0	1	1	1	1
0	1	0	1	1	0	1
1	0	0	1	0	1	1
1	1	1	0	0	0	0

Columns 4 and 7 are identical. Theorem verified.

Q8. An LED emits light of wavelength 600 nm. What is the band gap energy?

$E = hc/\lambda = \frac{6.63 \times 10^{-34} \times 3 \times 10^8}{600 \times 10^{-9}} = \frac{1.989 \times 10^{-25}}{6 \times 10^{-7}} = 3.315 \times 10^{-19}$ J $= 3.315 \times 10^{-19} / 1.6 \times 10^{-19} \approx 2.07$ eV.