NEET UG Physics · Chapter 24

Semiconductor Electronics

Intrinsic and extrinsic semiconductors, p–n junction, diode as rectifier, special diodes and logic gates.

NCERT Class 12NEET: 2 Qs per paper 13 min read

1. Chapter at a Glance

Semiconductor devices such as diodes, LEDs and solar cells are the building blocks of modern electronics. This chapter covers energy bands, intrinsic and extrinsic semiconductors, the p-n junction, the diode and its use as a rectifier, special diodes (Zener, LED, photodiode, solar cell), and basic logic gates. NEET usually asks two questions here, very often on logic gates, rectifiers and diode circuits.

What NEET asks most
  • Band gaps of conductors, semiconductors and insulators.
  • n-type and p-type doping; n·p = ni².
  • Forward and reverse bias; diode circuits with a 0.7 V drop.
  • Half-wave and full-wave rectifiers; output frequency.
  • Zener regulator, LED, photodiode, solar cell; truth tables of logic gates.

2. Energy Bands

  • In a solid, the energy levels of the atoms spread into bands. The valence band (VB) holds the bonding electrons; the conduction band (CB) holds free electrons. The gap between them is the band gap Eg.
TypeBand gapBehaviourNEET Note
Conductor (metal)Zero (bands overlap)Many free electronsResistance rises with temperature
Semiconductor (Si, Ge)Small (Si 1.1 eV, Ge 0.7 eV)Few carriers at room temperatureResistance falls with temperature
Insulator (diamond)Large (> 3 eV; diamond ~5.4 eV)Almost no free electrons—

3. Intrinsic and Extrinsic Semiconductors

  • Intrinsic (pure) semiconductor: electrons and holes are created in pairs, so ne = nh = ni.
  • A hole is a missing electron in a bond. It behaves like a positive charge carrier.
  • n-type: doped with a pentavalent (group 15) impurity such as P, As or Sb. These donors give extra electrons, so electrons are the majority carriers.
  • p-type: doped with a trivalent (group 13) impurity such as B, Al, Ga or In. These acceptors create holes, so holes are the majority carriers.
  • In thermal equilibrium, ne × nh = ni².
  • Both n-type and p-type materials are electrically neutral overall.

4. The p-n Junction and Diode

  • At the junction, electrons diffuse from n to p and holes from p to n. This leaves behind a thin depletion layer of fixed ions with a barrier potential (about 0.7 V for Si, 0.3 V for Ge).
  • Forward bias (p to +, n to −): the barrier and the depletion layer shrink. Above the knee voltage, current rises sharply (mA).
  • Reverse bias (p to −, n to +): the barrier and the depletion layer widen, and only a tiny reverse current (μA) flows, due to minority carriers. At the breakdown voltage, the reverse current rises suddenly.
  • A diode is therefore a one-way valve for current. Its I–V graph is non-linear, so it does not obey Ohm's law.
0.7 V forward (mA)reverse (μA) breakdown VI
Diode I–V graph: current rises sharply after the knee voltage (about 0.7 V for Si) in forward bias. In reverse bias there is only a tiny current until breakdown.

5. Diode as a Rectifier

RectifierDiodes usedOutput frequencyNEET Note
Half-wave1Same as input (f)Uses only one half-cycle
Full-wave (centre-tap)2Twice the input (2f)Both halves used; each diode conducts alternately
Bridge42fTwo diodes conduct at a time

The rectified output is pulsating DC. A capacitor filter connected across the load smooths it by charging at the peaks and discharging slowly between them.

6. Special Diodes

DeviceBiasWorkingUse
Zener diodeReverse (in breakdown)Voltage stays at VZ over a wide range of currentVoltage regulator
PhotodiodeReverseLight creates electron–hole pairs; reverse current ∝ intensityLight detection; photon energy must exceed Eg
LEDForwardRecombination of electrons and holes emits light; λ ≈ hc/EgLamps and displays; low power, long life
Solar cellNo external biasLight produces an EMF (photovoltaic effect)Power; best materials have Eg ≈ 1–1.8 eV (Si, GaAs)
Zener regulator in one line

Series resistor current IR = (Vin − VZ)/R; load current IL = VZ/RL; Zener current IZ = IR − IL. The load voltage stays at VZ as long as IZ > 0.

7. Logic Gates

ABOR (A + B)AND (A·B)NANDNOR
000011
011010
101010
111100
  • NOT gate: Y = Ā (output is the opposite of the input).
  • NAND and NOR are universal gates: any other gate can be built from them alone.
  • A NAND or NOR gate with both inputs joined acts as a NOT gate.
  • De Morgan's laws: NOT(A·B) = Ā + B̄ and NOT(A + B) = Ā·B̄.

8. Formula Master Table

QuantityFormula
Carrier lawnenh = ni²
Diode current (with knee drop)I = (V − 0.7)/R (Si)
Peak output (centre-tap)Vm = √2Vrms − 0.7
Rectifier output frequencyhalf-wave f ; full-wave 2f
Zener currentsIZ = (Vin − VZ)/R − VZ/RL
LED wavelengthλ = hc/Eg = 1240/Eg(eV) nm
Photodiode conditionhν > Eg
De MorganNOT(AB) = Ā + B̄

9. Master Summary Table

TopicKey ideaNEET Note
n-typePentavalent donorElectrons are majority carriers
p-typeTrivalent acceptorHoles are majority carriers
Depletion layerFixed ions onlyNarrows in forward bias
Reverse currentMinority carriersNearly independent of voltage
Full-wave rectifierOutput 2f100 Hz from 50 Hz mains
ZenerReverse breakdownRegulator
LED / photodiodeForward / reverseEmit / detect light
NAND, NORUniversal gatesBuild all others

10. Mind Map

Semiconductor Electronics
Materials
  • Energy bands
  • Intrinsic
  • n-type, p-type
p-n junction
  • Depletion layer
  • Forward / reverse
  • I–V graph
Applications
  • Rectifiers
  • Zener, LED
  • Photodiode, solar cell
Logic
  • OR, AND, NOT
  • NAND, NOR
  • De Morgan

11. Tips and Tricks

Shortcuts
  • In circuit questions, first decide which diodes are forward biased (p side at the higher potential); treat reverse-biased diodes as open switches.
  • OR: output 1 if any input is 1. AND: output 1 only if all inputs are 1. Put a NOT after either to get NOR or NAND.
  • NAND followed by NOT = AND; NOR followed by NOT = OR.
  • LED colour from band gap: 1240/Eg nm (1.8 eV ≈ red, 2.5 eV ≈ blue-green).
Common NEET traps
  • Saying n-type material is negatively charged (it is neutral).
  • Using a photodiode or Zener in forward bias.
  • Forgetting that a full-wave rectifier doubles the output frequency.
  • Mixing up the NAND and NOR truth tables.

12. Important One-liners

  • At absolute zero, a pure semiconductor behaves as an insulator.
  • Doping increases the conductivity of a semiconductor greatly.
  • The width of the depletion layer increases in reverse bias.
  • A Zener diode is heavily doped, so its breakdown voltage is low and sharp.
  • A photodiode is operated in reverse bias.
  • A solar cell works without any external bias.
  • NAND and NOR are called universal gates.
  • Silicon diodes have a knee voltage of about 0.7 V.

13. Solved Numericals: Average to Advanced

Five average and five advanced numericals. Take the Si knee voltage as 0.7 V and hc = 1240 eV nm. Click to reveal each answer and working.

Average level

N1Average

Pure silicon has ni = 1.5 × 10¹⁶ m⁻³. It is doped with 5 × 10²² acceptor atoms per m³. Find the hole and electron concentrations, and name the type.

Show answer & solution

Concept: nh ≈ NA; ne = ni²/nh.

  1. nh ≈ 5 × 10²² m⁻³.
  2. ne = (1.5 × 10¹⁶)²/5 × 10²² = 2.25 × 10³²/5 × 10²² = 4.5 × 10⁹ m⁻³.
  3. Holes are the majority carriers, so the material is p-type.

Answer: holes 5 × 10²² m⁻³; electrons 4.5 × 10⁹ m⁻³; p-type

N2Average

A silicon diode is forward biased with a 2 kΩ resistor across a 9 V battery. Find the current.

Show answer & solution

Concept: I = (V − 0.7)/R.

  1. Voltage across R = 9 − 0.7 = 8.3 V.
  2. I = 8.3/2000 = 4.15 × 10⁻³ A = 4.15 mA.

Answer: 4.15 mA

N3Average

A 60 Hz AC supply is rectified by (a) a half-wave rectifier, (b) a full-wave rectifier. Find the frequency of the output ripple in each case.

Show answer & solution

Concept: a full-wave rectifier uses both half-cycles.

  1. (a) Half-wave: one pulse per cycle ⇒ 60 Hz.
  2. (b) Full-wave: two pulses per cycle ⇒ 120 Hz.

Answer: (a) 60 Hz; (b) 120 Hz

N4Average

An LED is made from a semiconductor with a band gap of 2.5 eV. Estimate the wavelength of the light it emits and its colour.

Show answer & solution

Concept: λ = 1240/Eg nm.

  1. λ = 1240/2.5 = 496 nm.
  2. 496 nm is blue-green light.

Answer: about 496 nm (blue-green)

N5Average

Find the output of a NAND gate for each input pair (A, B) = (0, 0), (0, 1), (1, 0) and (1, 1). What single gate does a NAND with both inputs joined act as?

Show answer & solution

Concept: NAND = NOT(A·B).

  1. A·B = 0, 0, 0, 1 for the four pairs.
  2. NAND outputs = 1, 1, 1, 0.
  3. With A = B, Y = NOT(A·A) = Ā: it acts as a NOT gate.

Answer: 1, 1, 1, 0; a NOT gate

Advanced level

N6Advanced

A 6 V Zener diode is connected with a 200 Ω series resistor to a 10 V supply, feeding a 1 kΩ load. Find the currents through the series resistor, the load and the Zener.

Show answer & solution

Concept: the load voltage is held at VZ.

  1. IR = (10 − 6)/200 = 20 mA.
  2. IL = 6/1000 = 6 mA.
  3. IZ = 20 − 6 = 14 mA (positive, so regulation holds).

Answer: 20 mA; 6 mA; 14 mA

N7Advanced

Two identical silicon diodes are connected in series, both forward biased, with a 2.2 kΩ resistor across a 12 V battery. Find the current. What would it be if one diode were reversed?

Show answer & solution

Concept: each forward diode drops 0.7 V; a reverse diode blocks current.

  1. Voltage across R = 12 − 2 × 0.7 = 10.6 V.
  2. I = 10.6/2200 ≈ 4.82 mA.
  3. With one diode reversed, the circuit is effectively open: I ≈ 0 (only a tiny reverse current).

Answer: about 4.8 mA; practically zero

N8Advanced

A centre-tap full-wave rectifier uses silicon diodes, and each half of the transformer secondary gives 9 V rms. Find the peak output voltage.

Show answer & solution

Concept: peak = √2 × rms; one diode conducts at a time.

  1. Peak input = 9√2 ≈ 12.73 V.
  2. Peak output = 12.73 − 0.7 ≈ 12.0 V.

Answer: about 12.0 V

N9Advanced

Germanium (ni = 2.4 × 10¹⁹ m⁻³, 4.4 × 10²⁸ atoms per m³) is doped with one acceptor atom per 10⁸ germanium atoms. Find the hole and electron concentrations.

Show answer & solution

Concept: NA = (atoms per m³)/10⁸; ne = ni²/nh.

  1. NA = 4.4 × 10²⁸/10⁸ = 4.4 × 10²⁰ m⁻³ ⇒ nh ≈ 4.4 × 10²⁰ m⁻³.
  2. ne = (2.4 × 10¹⁹)²/4.4 × 10²⁰ = 5.76 × 10³⁸/4.4 × 10²⁰ ≈ 1.3 × 10¹⁸ m⁻³.
  3. Even this tiny doping makes holes outnumber electrons by about 340 times.

Answer: holes about 4.4 × 10²⁰ m⁻³; electrons about 1.3 × 10¹⁸ m⁻³

N10Advanced

Inputs A and B go into a NAND gate. Its output feeds both inputs of a second NAND gate. Write the truth table of the final output and name the equivalent gate.

Show answer & solution

Concept: a NAND with joined inputs is a NOT gate.

  1. First NAND output Y₁ = NOT(A·B): 1, 1, 1, 0 for (00, 01, 10, 11).
  2. Second NAND inverts Y₁: final Y = 0, 0, 0, 1.
  3. Y = A·B, so the circuit is an AND gate.

Answer: 0, 0, 0, 1 — an AND gate

14. NEET-pattern Questions with Detailed Answers

These questions are original and based on concepts that NEET UG asks again and again.

Q1Easy

To make an n-type semiconductor, silicon is doped with:

  1. Boron
  2. Aluminium
  3. Phosphorus
  4. Gallium
Show solution

n-type material needs a pentavalent donor. Phosphorus is in group 15, so it gives an extra electron. B, Al and Ga are trivalent and would make p-type material.

Answer: (C)

Q2Easy

Which gate gives output 1 only when all its inputs are 1?

  1. OR
  2. AND
  3. NOR
  4. NAND
Show solution

That is the definition of AND. NAND gives 0 in that case, OR gives 1 if any input is 1, and NOR gives 1 only when all inputs are 0.

Answer: (B)

Q3Medium

In forward bias, the width of the depletion layer of a p-n junction:

  1. Increases
  2. Decreases
  3. Stays the same
  4. First increases, then decreases
Show solution

The applied voltage opposes the barrier and pushes the majority carriers towards the junction, so the depletion layer becomes thinner. In reverse bias it becomes wider.

Answer: (B)

Q4Medium

The output frequency of a full-wave rectifier connected to 50 Hz mains is:

  1. 25 Hz
  2. 50 Hz
  3. 100 Hz
  4. 200 Hz
Show solution

Both half-cycles give a pulse, so the output has 2 × 50 = 100 pulses per second. 50 Hz would be the output of a half-wave rectifier.

Answer: (C)

Q5Medium

A Zener diode is used as a voltage regulator. It is connected:

  1. In forward bias, in series with the load
  2. In reverse bias, in parallel with the load
  3. In forward bias, in parallel with the load
  4. In reverse bias, in series with the load
Show solution

The Zener works in reverse breakdown, where its voltage stays constant. It is placed in parallel with the load, so the load gets that steady voltage.

Answer: (B)

Q6Medium

A photodiode made from a material of band gap 2.0 eV can detect light of wavelength:

  1. 700 nm
  2. 650 nm
  3. 800 nm
  4. 550 nm
Show solution

The photon energy must exceed Eg, so λ < 1240/2.0 = 620 nm. Only 550 nm satisfies this. The other three wavelengths are too long, so their photons carry too little energy.

Answer: (D)

Q7Hard

The output of a NOR gate is fed into a NOT gate. The combination is equivalent to:

  1. AND
  2. OR
  3. NAND
  4. XOR
Show solution

NOT(NOR) = NOT(NOT(A + B)) = A + B, which is OR. In the same way, NAND followed by NOT gives AND.

Answer: (B)

Q8Hard

As the temperature of an intrinsic semiconductor is raised:

  1. Its resistance increases
  2. Its number of charge carriers decreases
  3. Its conductivity increases
  4. Its band gap becomes zero
Show solution

Heat breaks more bonds, making more electron–hole pairs. So the conductivity rises and the resistance falls. The band gap stays about the same; it does not become zero.

Answer: (C)

15. Quick Revision Summary

Remember before the exam
  • Band gap: conductor 0, Si 1.1 eV, insulator > 3 eV.
  • n-type: group 15 donor; p-type: group 13 acceptor; nenh = ni².
  • Forward bias thins the depletion layer; Si knee 0.7 V.
  • Half-wave f, full-wave 2f; Zener regulates in reverse; LED forward; photodiode reverse.
  • NAND, NOR universal; De Morgan's laws.
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