Class 12 Physics · NCERT notes for NEET

Semiconductor Electronics: NCERT Notes, Formulas and Tricks for NEET

Semiconductors, doping, the p-n junction diode, rectifiers, Zener and optoelectronic diodes, and logic gates. NEET usually asks a couple of quick conceptual or short numerical questions from here, mostly on biasing, rectifier output, Zener circuits and gate truth tables.

1. Intrinsic and extrinsic semiconductors
2. p-n junction and diode 🔒
3. Rectifiers and special diodes 🔒
4. Logic gates 🔒

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Open the interactive version (simulations and tap-to-answer quick checks).

Topic 1

Intrinsic and extrinsic semiconductors

Add just a few arsenic atoms per million silicon atoms and silicon conducts many times better. This deliberate 'impurity' is what makes every chip in your phone work.

By resistivity: metals $ρ \sim 1 0^{- 2}$ to $1 0^{- 8} Ω m$ ; semiconductors $1 0^{- 5}$ to $1 0^{6} Ω m$ ; insulators $1 0^{11}$ to $1 0^{19} Ω m$ .
Elemental semiconductors: Si, Ge. Compound: inorganic (CdS, GaAs, CdSe, InP) and organic or polymer (anthracene, polypyrrole, polyaniline).
Band picture: metals have overlapping or half-filled bands ( $E_{g} = 0$ ); insulators have $E_{g} > 3 eV$ ; semiconductors have $E_{g} < 3 eV$ . Si $1.1 eV$ , Ge $0.71 eV$ , diamond $5.4 eV$ .
Intrinsic (pure) semiconductor: heat breaks covalent bonds and makes electron-hole pairs, so $n_{e} = n_{h} = n_{i}$ . At $0 K$ it is an insulator.
A hole is a missing electron in a bond. It acts like a charge $+ e$ and moves opposite to electrons. Total current $I = I_{e} + I_{h}$ .
n-type: dope with pentavalent P, As or Sb (donor). The fifth electron needs only about $0.01 eV$ (Ge) or $0.05 eV$ (Si) to become free. Electrons are majority carriers.
p-type: dope with trivalent B, Al, In or Ga (acceptor). Each atom leaves a hole. Holes are majority carriers.
Donor level lies just below the conduction band; acceptor level lies just above the valence band.
Mass action law at equilibrium: $n_{e} n_{h} = n_{i}^{2}$ . Doping raises the majority carriers and lowers the minority carriers below $n_{i}$ . Both n-type and p-type crystals stay electrically neutral.
Resistance of a semiconductor falls as temperature rises (negative temperature coefficient), the opposite of a metal.

The size of the forbidden gap E_g decides the class: zero for metals, below 3\,eV for semiconductors, above 3\,eV for insulators. — The size of the forbidden gap $E_{g}$ decides the class: zero for metals, below $3 eV$ for semiconductors, above $3 eV$ for insulators.

Donor level sits just below CB, so its electrons jump up easily (n-type). Acceptor level sits just above VB, so VB electrons jump into it and leave holes (p-type).

Arsenic has 5 valence electrons: 4 bond, the 5th is nearly free. Boron has only 3, so one bond is short of an electron: that gap is a hole.

Formulas

$n_{e} = n_{h} = n_{i}$
Intrinsic semiconductor
$n_{e} n_{h} = n_{i}^{2}$
Mass action law (thermal equilibrium, also for doped crystals)
$n_{e} \approx N_{D}, n_{h} = \frac{n _{i}^{2}}{N _{D}}$
n-type with donor density $N_{D} ≫ n_{i}$
$σ = e (n_{e} μ_{e} + n_{h} μ_{h})$
Conductivity; $μ$ = mobility, $ρ = 1/ σ$
$h ν \geq E_{g}, λ_{ma x} = \frac{h c}{E _{g}}$
Photon needed to create an electron-hole pair

Tips, tricks and hacks

Penta gives NPENTAvalent gives N-type (P, As, Sb: Group 15, one extra electron). TRIvalent gives P-type (B, Al, Ga, In: Group 13, one electron short). Count: 5 - 4 = +1 electron; 3 - 4 = one hole.

Minority by divisionMinority carriers

= n_{i}^{2} / majority

. Example:

n_{i} = 1.5 \times 1 0^{16} m^{- 3}

N_{D} = 4.5 \times 1 0^{22} m^{- 3}

gives

n_{h} = 2.25 \times 1 0^{32} /4.5 \times 1 0^{22} = 5 \times 1 0^{9} m^{- 3}

1240 rule for gaps

λ (nm) \approx 1240/ E_{g} (eV)

. Si (

1.1 eV

) absorbs light up to about

1130 nm

; a

600 nm

photon has

2.07 eV

Temperature sign checkHeat up: metal resistance goes UP (more collisions), semiconductor resistance goes DOWN (more carriers). If an option says 'Si resistance increases with T', reject it.

Common mistakes

✗ n-type is negatively charged because it has extra electrons.

✓ Each extra electron came with a donor ion of charge $+ e$ , so the crystal is neutral.

✗ Doping increases both electrons and holes.

✓ $n_{e} n_{h} = n_{i}^{2}$ is fixed, so if majority carriers rise, minority carriers must fall.

✗ A hole is a positive particle like a positron.

✓ It is only a vacancy in a bond; its motion is really electrons moving the other way.

✗ At 0 K, a pure semiconductor conducts a little.

✓ At 0 K the valence band is full and the conduction band empty, so it is a perfect insulator.

NEET traps

NEET gives $n_{i}$ and $N_{D}$ and asks for holes: it is $n_{i}^{2} / N_{D}$ , much smaller than $n_{i}$ , not $n_{i}$ itself.

'Pure Si at room temperature is a good insulator' is false: it has a small but non-zero conductivity due to thermally generated pairs.

Option sets swap 'donor level below CB' and 'acceptor level above VB'; donors are near the CB, acceptors near the VB.

Quick check: Pure Si has

n_{i} = 1.5 \times 1 0^{16} m^{- 3}

. It is doped with

3 \times 1 0^{22} m^{- 3}

boron atoms. The electron density is about

$1.5 \times 1 0^{16} m^{- 3}$
$3 \times 1 0^{22} m^{- 3}$
$7.5 \times 1 0^{9} m^{- 3}$
$2 \times 1 0^{6} m^{- 3}$

Answer: C. Boron makes it p-type with $n_{h} \approx 3 \times 1 0^{22}$ , so $n_{e} = n_{i}^{2} / n_{h} = 2.25 \times 1 0^{32} /3 \times 1 0^{22} = 7.5 \times 1 0^{9} m^{- 3}$ .

Topic 2 · 🔒

p-n junction and diode

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Topic 3 · 🔒

Rectifiers and special diodes

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Topic 4 · 🔒

Logic gates

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Practise Semiconductor Electronics questions

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