Semiconductor electronics for NEET

Semiconductor electronics is the last unit of the physics syllabus and one of the shortest. It has very little algebra, a handful of circuits and a few truth tables. That is exactly why it deserves an unhurried read: most questions here are one-step and conceptual, and a candidate who knows the chapter well can answer them in well under a minute.

The questions tend to come in four kinds: which carrier is in the majority and why, what biasing does to the depletion layer, what a circuit of diodes or a Zener gives as output, and what a combination of logic gates does.

Energy bands: why semiconductors are different

In a solid, the energy levels of the atoms spread into bands. Two matter: the valence band (filled at low temperature) and the conduction band above it. The gap between them is the band gap, Eg.

A semiconductor's resistance falls as temperature rises, because heat lifts more electrons across the gap. A metal's resistance rises with temperature. NEET likes this contrast.

Intrinsic and extrinsic semiconductors

In a pure (intrinsic) semiconductor, every electron that jumps to the conduction band leaves a hole behind in the valence band. So the electron density equals the hole density: ne = nh = nᵢ.

Doping adds a tiny amount of impurity to change this balance.

The last row is a favourite trap. An n-type crystal has more free electrons, but each came from a neutral donor atom that is left as a fixed positive ion. The crystal as a whole stays electrically neutral.

In thermal equilibrium, the mass-action law links the two carrier densities:

ne × nh = nᵢ²

Doping raises one carrier density and lowers the other, so their product stays the same.

The p–n junction

When p and n regions are joined, electrons diffuse from n to p and holes from p to n. They recombine near the junction and leave behind fixed ions: positive on the n side, negative on the p side. This thin region with no free carriers is the depletion layer. Its fixed charges create an electric field and a barrier potential, roughly 0.7 V for silicon and 0.3 V for germanium.

Two currents flow at the junction: diffusion of majority carriers and drift of minority carriers pushed by the field. With no bias, they balance and the net current is zero.

On the I–V graph, the forward current stays small until the threshold (cut-in) voltage and then rises steeply. The dynamic resistance is r_d = ΔV/ΔI, read from the slope of the curve in that region.

Diode as a rectifier

A diode conducts in one direction only, so it can turn AC into pulsating DC.

• Half-wave rectifier: one diode. Output appears in only one half of each cycle. The output pulses repeat at the input frequency (50 Hz for a 50 Hz supply).
• Full-wave rectifier: two diodes with a centre-tapped transformer (or four diodes in a bridge). Output appears in both halves, so the pulses repeat at twice the input frequency (100 Hz).
• Filter: a capacitor across the load charges at each peak and discharges slowly between them, smoothing the output.

Special-purpose diodes

A Zener diode is heavily doped on both sides, so its depletion layer is thin and breakdown happens at a sharp, fixed voltage V_Z. In breakdown, the voltage across it stays at V_Z while the current through it changes. Placed in reverse across a load, with a series resistor to take up the extra supply voltage, it keeps the load voltage steady even when the input changes.

For an LED or a photodiode, the photon energy must match or exceed the band gap: E (eV) ≈ 1240/λ (nm).

Logic gates

A logic gate takes one or more inputs, each 0 or 1, and gives an output of 0 or 1.

NAND and NOR are universal gates: any other gate can be built from either one alone. Joining both inputs of a NAND (or a NOR) gives a NOT gate. A NAND followed by a NOT gives AND. A NOR followed by a NOT gives OR.

Worked NEET-style questions

Example 1: carrier densities after doping

A silicon crystal with nᵢ = 1.5 × 10¹⁶ m⁻³ is doped with 4.5 × 10²² m⁻³ atoms of arsenic. Find the electron and hole densities.

• Arsenic is pentavalent, so each atom donates an electron: ne ≈ 4.5 × 10²² m⁻³ (the intrinsic electrons are negligible).
• nh = nᵢ²/ne = (2.25 × 10³²)/(4.5 × 10²²) = 5 × 10⁹ m⁻³.
• Holes are now the minority carriers by a huge margin.

Example 2: a diode in a simple circuit

A silicon diode (take a 0.7 V drop when conducting) is forward biased in series with a 430 Ω resistor across a 5 V battery. Find the current.

• Voltage across the resistor = 5 − 0.7 = 4.3 V.
• I = 4.3/430 = 0.01 A = 10 mA.
• If the diode were reversed, the current would be practically zero.

Example 3: Zener regulator

A 6 V Zener diode is connected with a 200 Ω series resistor to a 10 V supply. The load across the Zener is 1 kΩ. Find the current through the Zener.

• Load voltage = V_Z = 6 V, so the load current = 6/1000 = 6 mA.
• Series resistor voltage = 10 − 6 = 4 V, so the total current = 4/200 = 20 mA.
• Zener current = 20 − 6 = 14 mA.

Example 4: which light can a photodiode detect?

A photodiode is made from a semiconductor with Eg = 2.5 eV. Can it detect light of wavelength 600 nm?

• Photon energy = 1240/600 ≈ 2.07 eV.
• This is less than 2.5 eV, so the photon cannot create an electron–hole pair. No.
• The longest wavelength it can detect is 1240/2.5 ≈ 496 nm.

Practice MCQs

1. In an n-type semiconductor: (a) ne ≫ nh and the crystal is negatively charged (b) ne ≫ nh and the crystal is neutral (c) nh ≫ ne (d) ne = nh
2. When a p–n junction is reverse biased, the depletion layer: (a) becomes narrower (b) becomes wider (c) is unchanged (d) disappears
3. A full-wave rectifier works on a 50 Hz supply. The output ripple frequency is: (a) 25 Hz (b) 50 Hz (c) 100 Hz (d) 200 Hz
4. A semiconductor with nᵢ = 10¹⁶ m⁻³ is doped so that nh = 5 × 10²² m⁻³. The electron density is: (a) 2 × 10⁹ m⁻³ (b) 5 × 10⁶ m⁻³ (c) 10¹⁶ m⁻³ (d) 2 × 10⁶ m⁻³
5. A 10 V Zener diode, a 250 Ω series resistor and a 2 kΩ load are used with a 15 V supply. The Zener current is: (a) 5 mA (b) 15 mA (c) 20 mA (d) 25 mA
6. The gate whose output is 1 only when both inputs are 0 is: (a) AND (b) NAND (c) OR (d) NOR
7. Both inputs of a NAND gate are 1. The output is: (a) 0 (b) 1 (c) undefined (d) the same as the input
8. An LED is made from a material with Eg = 2.0 eV. The wavelength it emits is about: (a) 310 nm (b) 480 nm (c) 620 nm (d) 900 nm

Answers

1. (b) Donor atoms become fixed positive ions, so the crystal stays neutral.
2. (b) Reverse bias raises the barrier and pulls carriers away from the junction.
3. (c) Both halves are used, so pulses come at twice the supply frequency.
4. (a) ne = nᵢ²/nh = 10³²/(5 × 10²²) = 2 × 10⁹ m⁻³.
5. (b) Total current = 5/250 = 20 mA; load current = 10/2000 = 5 mA; Zener current = 15 mA.
6. (d) NOR is the inverse of OR, which is 0 only for 0, 0.
7. (a) AND of 1 and 1 is 1; NAND inverts it.
8. (c) λ ≈ 1240/2.0 = 620 nm, in the red.

What to do next

• Draw the forward and reverse bias diagrams from memory, marking which way the depletion layer changes.
• Write the truth tables of all five gates on one card, and build AND, OR and NOT from NAND alone.
• Solve five Zener questions in the order: load current, total current, then Zener current.
• Revisit modern physics for NEET for the 1240/λ shortcut, then move to the NEET physics formula sheet for a full revision pass.