Semiconductors and electronic devices for JEE Main

Semiconductors is the shortest physics chapter in the JEE Main syllabus and one of the easiest to finish completely. Questions are usually conceptual or need one line of arithmetic: which carriers dominate, which way a diode conducts, what a combination of gates does, or how much current a Zener diode carries. An evening of careful reading and two sets of practice is often enough.

The official 2026 syllabus lists semiconductors; the semiconductor diode and its I–V characteristics in forward and reverse bias; the diode as a rectifier; I–V characteristics of the LED, photodiode, solar cell and Zener diode; the Zener diode as a voltage regulator; and logic gates (OR, AND, NOT, NAND and NOR). Two practicals also belong here: the characteristic curves of a p–n junction diode and of a Zener diode.

Energy bands

In a solid, the energy levels of the atoms merge into bands. The two that matter are the valence band (filled at absolute zero) and the conduction band above it, separated by an energy gap E_g.

In a semiconductor, raising the temperature frees more electrons, so its resistance falls with temperature, unlike a metal's.

Intrinsic and extrinsic semiconductors

In pure (intrinsic) silicon, each electron that jumps to the conduction band leaves a hole behind, so n_e = n_h = n_i.

Doping adds impurity atoms:

• n-type: a pentavalent impurity (phosphorus, arsenic, antimony) donates an extra electron. Electrons are the majority carriers.
• p-type: a trivalent impurity (boron, aluminium, gallium, indium) accepts an electron, creating holes. Holes are the majority carriers.

In both, n_e × n_h = n_i² at a given temperature. Each doped crystal is still electrically neutral overall; it has more mobile carriers of one sign, balanced by fixed ions.

The p–n junction

When p and n regions meet, electrons diffuse into the p side and holes into the n side. This leaves a thin depletion layer of fixed ions near the junction and a barrier potential, about 0.7 V for silicon and 0.3 V for germanium.

• Forward bias (p to the positive terminal): the barrier is lowered. Current stays small until about the barrier voltage (the knee), then rises steeply.
• Reverse bias: the barrier and depletion layer grow. Only a tiny current of minority carriers flows, almost independent of voltage, until breakdown.

The dynamic resistance is ΔV/ΔI, read from the slope of the I–V curve at the working point.

Rectifiers

A diode conducts in one direction, so it turns AC into pulsating DC.

A capacitor across the load smooths the output by charging at the peaks and discharging slowly between them.

Special-purpose diodes

For an LED to give visible light, its band gap must be roughly 1.8–3 eV. For a solar cell, a band gap near 1.5 eV suits sunlight well.

Zener regulator. A series resistor R_s drops the extra input voltage. The Zener keeps the load voltage at V_z, and absorbs whatever current the load does not take:

• current through R_s: I_s = (V_in − V_z)/R_s
• load current: I_L = V_z/R_L
• Zener current: I_z = I_s − I_L

Logic gates

NOT simply inverts: 0 becomes 1 and 1 becomes 0.

NAND and NOR are universal gates: any other gate can be built from either one alone. Two identities do most of the work in combination questions (De Morgan's theorems):

• NOT(A·B) = (NOT A) + (NOT B)
• NOT(A + B) = (NOT A)·(NOT B)

Worked problems

Problem 1: Zener regulator. Input 15 V, Zener voltage 10 V, series resistor 250 Ω, load 1 kΩ. Find the Zener current. What happens if the input rises to 20 V?

I_s = (15 − 10)/250 = 20 mA. I_L = 10/1,000 = 10 mA. I_z = 10 mA.
At 20 V: I_s = 10/250 = 40 mA, I_L is still 10 mA, so I_z = 30 mA. The load voltage stays at 10 V; the Zener absorbs the change.

Problem 2: a diode in a circuit. A silicon diode (drop 0.7 V) is forward biased by a 5 V source through 430 Ω. Find the current. What if the source is reversed?

I = (5 − 0.7)/430 = 10 mA. Reversed, the diode blocks and the current is practically zero.

Problem 3: gate combination. A and B each pass through a NOT gate, and the two outputs feed a NOR gate. What single gate is this?

Output = NOT(NOT A + NOT B) = A·B by De Morgan's theorem, so it acts as an AND gate. Check with A = 1, B = 1: the NOR inputs are 0 and 0, so the output is 1.

Problem 4 (numerical answer): LED colour. An LED has a band gap of 2.48 eV. Find the wavelength of its light in nm.

λ = 1240/2.48 = 500 nm (green).

Practice set

1. What do two NOT gates in series do?
2. What does a NAND gate do when its two inputs are joined?
3. A full-wave rectifier is fed with 60 Hz AC. What is the output ripple frequency?
4. Intrinsic silicon has n_i = 1.5 × 10¹⁶ m⁻³. It is doped so that n_e = 5 × 10²² m⁻³. Find n_h and say which type it is.
5. Which of the special diodes are operated in reverse bias?
6. Find the output of a NOR gate for A = 0, B = 1.
7. A Zener regulator has V_in = 12 V, V_z = 6 V and R_s = 200 Ω. Find the Zener current with no load and with a 600 Ω load.
8. How can you make an OR gate from NAND gates alone?

Answers

1. They act as a buffer: output equals input.
2. It acts as a NOT gate.
3. 120 Hz.
4. n_h = (1.5 × 10¹⁶)²/(5 × 10²²) = 4.5 × 10⁹ m⁻³; n-type.
5. The Zener diode and the photodiode.
6. 0.
7. I_s = 6/200 = 30 mA. No load: I_z = 30 mA. With 600 Ω: I_L = 10 mA, so I_z = 20 mA.
8. Invert A and B with one NAND each (inputs joined), then feed both into a third NAND: NOT(NOT A · NOT B) = A + B.

What to do next

• Draw the I–V curves of an ordinary diode, a Zener, an LED, a photodiode and a solar cell from memory.
• Write the truth tables of the five listed gates and prove De Morgan's theorems with them.
• Solve 10 gate-combination questions and 10 Zener or diode-circuit questions.
• Revise modern physics, then go through common physics mistakes before your next mock.