Chapter 14: Semiconductor Electronics — Materials, Devices and Simple Circuits

In the second half of the twentieth century, electronics moved from vacuum tubes — bulky, hot, fragile — to silicon, the second most abundant element in Earth's crust. A solid that conducts neither well nor poorly, but whose conductivity can be tuned by impurity doping over fifteen orders of magnitude, turned out to be the perfect material for switches, amplifiers, and ultimately every computer. This chapter develops the band picture, distinguishes intrinsic from extrinsic semiconductors, builds the p-n junction (the keystone of all semiconductor devices), and then assembles diodes, rectifiers, Zeners, LEDs, photodiodes, solar cells, transistors, and logic gates from this single building block.

Concept Map

Energy bands
  Conductors (overlap) / Insulators (Eg≫kT) / Semiconductors (Eg~1 eV)
        │
        ├── Intrinsic (pure Si/Ge)  n = p = n_i, thermally generated
        │
        └── Extrinsic
              n-type (donor P, As, Sb) — electrons majority
              p-type (acceptor B, In, Al) — holes majority
                        │
                        └── p-n junction (depletion region, V_b ~ 0.7 V)
                                   │
                                   ├── Diode (forward / reverse / breakdown)
                                   │       ├── Half-wave rectifier
                                   │       ├── Full-wave (centre-tap, bridge)
                                   │       └── Zener regulator
                                   │       └── LED, Photodiode, Solar cell
                                   │
                                   └── Bipolar Junction Transistor (npn / pnp)
                                              CE amplifier, switch
                                                       │
                                                       └── Logic gates
                                                              (NOT, AND, OR, NAND, NOR, XOR)

14.1 Energy Bands and Classification

Definition

When isolated atoms come together to form a solid, the discrete atomic energy levels of each atom split and overlap, forming continuous energy bands. Two bands matter for electrical behaviour:

• Valence band (VB): highest band that is completely (or partially) full at $T = 0$.
• Conduction band (CB): next higher band, normally (mostly) empty at $T = 0$.

They are separated by a forbidden energy gap $E_g$ in which no electronic states exist.

At room temperature ($kT\approx 0.026\ \text{eV}$), thermal excitation across $E_g\sim 1\ \text{eV}$ in Si is rare but non-zero, giving semiconductors their intermediate (and temperature-sensitive) conductivity.

Derivation — carrier concentration vs temperature

The probability of an electron being in a CB state of energy $E$ is given (in the non-degenerate Maxwell–Boltzmann limit) by $f(E)\approx e^{-(E - E_F)/kT}$. The number of electrons in CB is approximately

with $N_C$ the effective density of states. For intrinsic material, $E_F$ sits roughly midway in the gap and

The exponential factor explains the dramatic conductivity rise with $T$ in semiconductors — opposite to metals, where resistivity rises with $T$ (lattice scattering).

Worked Example

At $T = 300\ \text{K}$, intrinsic Si has $n_i = 1.5\times 10^{10}\ \text{cm}^{-3}$. At $T = 600\ \text{K}$, estimate $n_i$.

Assume $n_i\propto T^{3/2}\,e^{-E_g/2kT}$. The exponential dominates:

The exponential argument: $-\dfrac{1.12}{2\cdot 8.617\times 10^{-5}}\cdot \dfrac{-1}{600} = +10.83$. So $n_i$ grows by factor $\sim 2.83\cdot e^{10.83}\approx 1.5\times 10^5$, giving $n_i(600\ \text{K})\approx 2\times 10^{15}\ \text{cm}^{-3}$.

Pitfalls

• A conductor has $E_g = 0$ (or partially filled CB), not negative.
• For insulators vs. semiconductors, the only sharp difference is $E_g$; there is no fundamental physical line, just the operating temperature range.
• Semiconductor resistivity decreases with $T$, opposite to metals.

14.2 Intrinsic Semiconductors

Definition

An intrinsic (pure) semiconductor contains only the host atoms (Si or Ge) — no foreign impurities. At any $T > 0$, thermal energy excites some electrons from VB to CB, leaving behind holes (vacant states in VB that behave as positive carriers).

Charge-neutrality: $n = p = n_i(T)$, the intrinsic carrier concentration.

For silicon at $300\ \text{K}$: $n_i = 1.5\times 10^{10}\ \text{cm}^{-3}$ (compared with $\sim 5\times 10^{22}\ \text{cm}^{-3}$ Si atoms — only $1$ in $10^{13}$ Si atoms is ionized).

Conductivity

Conductivity has contributions from both electrons and holes:

where $\mu_n$, $\mu_p$ are the electron and hole mobilities (m$^2$/V·s). For Si at 300 K: $\mu_n \approx 1350\ \text{cm}^2$/V·s, $\mu_p \approx 480\ \text{cm}^2$/V·s.

Worked Example

Compute the intrinsic conductivity of Si at 300 K.

Wait — convert $n_i$ to m$^{-3}$: $1.5\times 10^{10}\ \text{cm}^{-3} = 1.5\times 10^{16}\ \text{m}^{-3}$, and mobilities to m$^2$/V·s: $\mu_n = 0.135$, $\mu_p = 0.048$.

Resistivity $\rho = 1/\sigma \approx 2.3\times 10^3\ \Omega\cdot\text{m}$. (Intrinsic Si is a poor conductor — but very different from an insulator.)

Pitfalls

• The intrinsic carrier density depends exponentially on $T$ — even a 10 K change matters.
• Holes are not anti-particles of electrons; they are missing electrons in the VB, but conveniently treated as positive mobile carriers.
• For Ge ($E_g = 0.72\ \text{eV}$), $n_i$ at 300 K is $\sim 2.4\times 10^{13}\ \text{cm}^{-3}$ — 1500$\times$ larger than Si, because of the smaller gap.

14.3 Extrinsic Semiconductors — n-type and p-type

Definition

By deliberately adding tiny amounts ($1$ ppm or less) of pentavalent or trivalent impurities to tetravalent Si/Ge, the carrier concentrations can be controlled by factors of $10^5$–$10^{10}$. This is doping.

n-type (donor doping)

Add a pentavalent impurity (group V): phosphorus, arsenic, antimony. Four of its valence electrons make covalent bonds with neighbouring Si; the fifth is loosely bound — at room temperature this fifth electron is freed into the CB with only $\sim 0.045\ \text{eV}$ activation (Phosphorus in Si). The donor atom becomes a fixed positive ion.

• Majority carriers: electrons (concentration $n\approx N_D$, the donor density).
• Minority carriers: holes (very few).
• Mass-action law: $np = n_i^2$.

Energy diagram: a donor level $E_D$ appears just $\sim 0.045\ \text{eV}$ below the conduction band edge $E_C$.

p-type (acceptor doping)

Add a trivalent impurity (group III): boron, aluminium, indium, gallium. It has only three valence electrons, so when bonded with four Si neighbours one bond is incomplete — a hole sits at the impurity. At room temperature this hole easily detaches into the VB; the acceptor atom becomes a fixed negative ion.

• Majority carriers: holes ($p\approx N_A$, the acceptor density).
• Minority carriers: electrons.
• Mass-action law: $np = n_i^2$.

Energy diagram: an acceptor level $E_A$ appears just above the valence band edge $E_V$.

Mass-action Law

Even after doping, the product $np$ remains constant at thermal equilibrium:

So in heavily-doped n-type with $n = N_D = 10^{16}\ \text{cm}^{-3}$ and $n_i = 1.5\times 10^{10}\ \text{cm}^{-3}$, the minority hole concentration is $p = n_i^2/n = 2.25\times 10^{4}\ \text{cm}^{-3}$ — twelve orders of magnitude less than electrons.

Worked Example

A Si sample is doped with $5\times 10^{16}\ \text{cm}^{-3}$ of P. Find majority and minority carrier densities at 300 K. ($n_i = 1.5\times 10^{10}\ \text{cm}^{-3}$.)

Majority (electrons): $n \approx N_D = 5\times 10^{16}\ \text{cm}^{-3}$.

Minority (holes): $p = n_i^2/n = (1.5\times 10^{10})^2/(5\times 10^{16}) = 4.5\times 10^{3}\ \text{cm}^{-3}$.

Pitfalls

• The doped semiconductor remains electrically neutral — the donor (or acceptor) ions cancel the mobile carriers' charge.
• "n-type" does not mean negatively charged — only that electrons are the majority carriers.
• "Doping" with $\sim 1\ \text{ppm}$ is enough to change conductivity by $10^5$–$10^{10}$; precise control is what makes semiconductor industry possible.

14.4 p-n Junction — Formation, Depletion Region, Barrier

Definition

When a single crystal of Si is doped p-type on one side and n-type on the other (by diffusion or implantation), a p-n junction forms at the metallurgical interface.

Three things happen immediately upon junction formation:

1. Diffusion: electrons from n-side (high concentration) diffuse into p-side; holes from p-side diffuse into n-side. They recombine near the junction.
2. Depletion region: a thin layer (width $W\sim 0.1$–$1\ \mu\text{m}$) on either side of the junction becomes depleted of mobile carriers, leaving behind the fixed donor (+) and acceptor ($-$) ions.
3. Built-in potential / barrier: the fixed ions set up an electric field $\vec E$ pointing from n to p, opposing further diffusion. At equilibrium, drift (driven by $\vec E$) balances diffusion, and a built-in potential $V_b$ (also called barrier potential) is established:

Typical: $V_b \approx 0.3\ \text{V}$ (Ge), $\approx 0.7\ \text{V}$ (Si). Note no current flows in the open-circuit equilibrium.

Drift vs. Diffusion Currents at Equilibrium

In the depletion region, the field $\vec E$ drives a drift current $J_{\text{drift}}$ of minority carriers (electrons in p, holes in n, both pushed across the junction). The diffusion current $J_{\text{diff}}$ of majority carriers (electrons from n to p, holes from p to n) flows in opposite direction. At equilibrium they cancel exactly:

Worked Example

For a Si p-n junction with $N_A = N_D = 10^{16}\ \text{cm}^{-3}$ at 300 K, compute $V_b$.

(Right at the "0.7 V" rule of thumb.)

Pitfalls

• The depletion region has no mobile carriers but does have fixed ionized donors/acceptors — it is not "empty."
• $V_b$ exists at zero external bias — you cannot measure it directly with a voltmeter (the metal-semiconductor contacts have their own built-in potentials, which cancel out).
• The electric field in the depletion region can be enormous ($\sim 10^5$ V/cm) despite tiny voltages — the region is just very thin.

14.5 Semiconductor Diode — I-V Characteristic

Definition

A p-n junction with external leads is a diode. Its terminals are conventionally the anode (p-side) and cathode (n-side).

Forward Bias ($V > 0$, p positive)

External voltage reduces the barrier from $V_b$ to $V_b - V$. The depletion width shrinks; majority carriers diffuse across. Current grows exponentially (Shockley equation):

with $I_0$ the reverse saturation current and $\eta$ (1–2) the ideality factor. For $V\gg kT/e\approx 26\ \text{mV}$, current rises quasi-vertically once $V$ exceeds the cut-in (knee) voltage $V_\gamma\approx 0.3\ \text{V}$ (Ge), $\approx 0.7\ \text{V}$ (Si).

Reverse Bias ($V < 0$, p negative)

External voltage adds to the barrier. Depletion region widens; majority diffusion is suppressed. Only minority drift remains, giving a tiny reverse saturation current $I_0\sim\ \text{nA}$ (Si) to $\mu\text{A}$ (Ge) — almost independent of $V$.

Breakdown

At a high reverse voltage $V_Z$ (the Zener voltage or breakdown voltage), the junction breaks down by one of two mechanisms:

• Zener breakdown (heavily doped, narrow depletion, $V_Z\lesssim 5\ \text{V}$): the field tears electrons directly from covalent bonds (tunneling).
• Avalanche breakdown (lightly doped, $V_Z\gtrsim 6\ \text{V}$): a minority carrier gains enough KE between collisions to ionize the lattice; carriers multiply geometrically.

The current rises steeply at $V = -V_Z$ but the voltage clamps to $-V_Z$ — exploited in voltage regulators.

I-V Characteristic (sketch)

        I
        ▲
        │              ⎢ steep
        │              ⎢
        │            ╱
        │          ╱
   ─────┼──────────┼────────────► V
        │ 0.7 V    ↑
   Vz ◄─┘          cut-in
        │
        ▼
     reverse breakdown

Worked Example — Dynamic Resistance

A Si diode has $I = 10\ \text{mA}$ at $V = 0.70\ \text{V}$ and $I = 20\ \text{mA}$ at $V = 0.72\ \text{V}$. Compute the AC (dynamic) resistance.

Pitfalls

• The Shockley equation gives a non-linear I-V — most JEE problems use the simplified "ideal diode" approximation: $V = V_\gamma$ in forward, $I = 0$ in reverse.
• Dynamic resistance $r_d = dV/dI$ is not the same as static $V/I$.
• Reverse saturation current doubles roughly every $10\ \text{K}$ rise — important for thermal stability.

14.6 Diode as a Rectifier

Definition

A rectifier converts AC to DC by allowing current in only one direction. The diode is the simplest rectifier.

Half-Wave Rectifier (HWR)

Circuit: a single diode in series with the load $R_L$, fed from the secondary of a step-down transformer.

• During the positive half-cycle, the diode is forward-biased, conducts, and the output across $R_L$ follows the input (minus $V_\gamma\sim 0.7\ \text{V}$).
• During the negative half-cycle, the diode is reverse-biased, blocks current; output is zero.

Output waveform: a series of positive half-sine "humps," DC + ripple.

Full-Wave Rectifier (FWR) — Centre-Tap

Uses two diodes and a centre-tap transformer. The centre-tap is the common ground; the two ends of the secondary are 180° out of phase.

• Positive half: $D_1$ conducts, $D_2$ blocks.
• Negative half: $D_2$ conducts, $D_1$ blocks.

Both halves produce positive output across $R_L$ — every half-cycle is used.

Full-Wave Rectifier — Bridge

Uses four diodes in a bridge arrangement; no centre-tap required.

• Positive half: $D_1, D_3$ conduct; $D_2, D_4$ blocked.
• Negative half: $D_2, D_4$ conduct; $D_1, D_3$ blocked.

Output is again a stream of positive half-sines but every half-cycle is captured. The bridge rectifier is dominant in modern power supplies (no expensive centre-tap transformer; only one diode drop per half-cycle drawback is compensated).

Filter Capacitor

A capacitor $C$ in parallel with $R_L$ smooths the pulsating DC. During the conducting interval $C$ charges to nearly $V_m$; when the diodes block, $C$ discharges into $R_L$ with time constant $R_L C$. Choosing $R_L C \gg T$ (period of the input) gives nearly pure DC with small ripple voltage

Worked Example

A FWR feeds $V_m = 12\ \text{V}$ peak into $R_L = 1\ \text{k}\Omega$ with a $470\ \mu\text{F}$ filter capacitor at $50\ \text{Hz}$ mains. Find the ripple voltage.

Ripple factor $\approx \Delta V/(V_m\sqrt 3)\approx 6\times 10^{-3}$ — quite smooth.

Pitfalls

• HWR's ripple frequency is $f$; FWR's is $2f$.
• Centre-tap FWR uses 2 diodes; bridge uses 4. Bridge has 2 diode drops per half-cycle, centre-tap has 1.
• "Efficiency" here means ratio of DC power to AC input power; the maximum theoretical is 100% (not achieved).

14.7 Special Diodes

Zener Diode

A heavily-doped p-n junction designed to operate in reverse breakdown at a sharp, well-defined voltage $V_Z$. In the breakdown region, current changes by orders of magnitude with negligible voltage change — so the Zener clamps voltage to $V_Z$.

Voltage Regulator Circuit:

Vin ─[R_s]─┬─── V_out (= V_Z)
           │
        [Zener]
           │
          GND

The series resistor $R_s$ drops the excess voltage. Provided $V_{in}>V_Z$ and the load current does not exceed $I_{Z,\max}$, the output stays at $V_Z$ even as $V_{in}$ varies. Common values: $V_Z = 3.3, 5.1, 6.2, 9.1, 12, 15, 24$ V.

Photodiode

A p-n junction in reverse bias, exposed through a transparent window. Incident photons with $h\nu > E_g$ create electron-hole pairs in the depletion region, which are swept apart by the built-in field — adding to the reverse current. Reverse current rises linearly with light intensity (over wide range).

Used in optical sensors, light meters, fibre-optic receivers, fast-camera image sensors.

LED (Light-Emitting Diode)

A forward-biased p-n junction made from a direct-bandgap semiconductor (GaAs, GaP, GaN, …). Recombination of injected electrons and holes releases the energy $E_g$ as a photon of frequency $\nu = E_g/h$:

White LEDs combine a blue GaN chip with a yellow phosphor.

Solar Cell

An unbiased (open-circuit), large-area p-n junction. Sunlight photons with $h\nu > E_g$ create electron-hole pairs throughout the depletion region (and within a diffusion length on either side). The junction's built-in field separates them — electrons drift to n-side, holes to p-side — generating a photovoltage. Connected to a load, it drives current.

Operating quadrant: fourth quadrant of I-V (positive $V$, negative $I$ in diode convention) — the device delivers power.

Open-circuit voltage $V_{OC}\sim 0.5$–$0.7\ \text{V}$ (Si); short-circuit current density $J_{SC}\sim 30\ \text{mA/cm}^2$ (under AM1.5 sunlight); efficiency $\sim 20\%$ (commercial Si cells, ~26% lab record).

Pitfalls

• Zener: operates in reverse breakdown. LED: forward bias. Photodiode/solar cell: reverse / open.
• LED needs direct-gap material — Si (indirect gap) makes a poor LED.
• The photon energy in solar cells must exceed $E_g$ — energy above $E_g$ is wasted as heat (Shockley–Queisser limit $\sim 33\%$ for single-junction).

14.8 Bipolar Junction Transistor (BJT) — Brief

Definition

A BJT is a three-terminal sandwich of doped layers — npn or pnp — with regions called Emitter (E), Base (B), Collector (C). The base is thin and lightly doped; emitter is heavily doped; collector is moderately doped, larger area.

   npn:  n+ │ p │ n
          E   B   C

Operation (npn, common-emitter)

• E-B junction is forward-biased ($V_{BE}\approx 0.7\ \text{V}$). Electrons flood from emitter into base.
• B-C junction is reverse-biased ($V_{CB}\gg 0$). The (thin) base allows most electrons to diffuse through and be swept into the collector by the strong field. Only a small fraction recombine in the base, giving the base current.

Currents satisfy: $I_E = I_B + I_C$, with $I_C \gg I_B$. Define:

• $\alpha = I_C/I_E < 1$ (common-base current gain, typically 0.98–0.998).
• $\beta = I_C/I_B = \alpha/(1-\alpha) \gg 1$ (common-emitter current gain, typically 50–500).

A small base-current change produces a large collector-current change — this is current amplification.

Transistor as an Amplifier (CE configuration)

In a common-emitter amplifier: input AC voltage $v_{in}$ is applied between base and emitter (causing small $i_B$); output is taken across a collector resistor $R_C$.

Voltage gain:

where $r_{in}\approx h_{ie}$ is the input resistance ($\sim 1$ k$\Omega$ typical), and the minus sign indicates a $180°$ phase inversion. Typical $A_V \sim 100$.

Transistor as a Switch

In a CE circuit:

• Cut-off region: $V_{BE} < V_\gamma$, $I_B = 0$, $I_C\approx 0$, $V_{CE}\approx V_{CC}$ → output HIGH, transistor OFF.
• Saturation region: $I_B$ large enough that $I_C = \beta I_B$ would exceed $V_{CC}/R_C$; the transistor saturates, $V_{CE}\approx 0.2\ \text{V}$ → output LOW, transistor ON.

Cut-off and saturation correspond to logic 1 and 0 respectively — the basis of all digital electronics.

Worked Example

For a CE amplifier with $\beta = 100$, $I_B = 20\ \mu\text{A}$. Find $I_C$, $I_E$.

$I_C = \beta I_B = 100\cdot 20 = 2000\ \mu\text{A} = 2\ \text{mA}$. $I_E = I_B + I_C = 2.02\ \text{mA}$.

Pitfalls

• $\beta\ne\alpha$, but $\beta = \alpha/(1-\alpha)$.
• $I_E = I_B + I_C$ — the base current is small, not zero.
• pnp transistor: all polarities and current directions reverse, but $\beta$ formula is identical.

14.9 Logic Gates — Symbols and Truth Tables

Definition

A logic gate implements a Boolean function of its binary inputs ($0$ = low voltage, $1$ = high voltage).

NOT Gate (Inverter)

Inverts the input. Symbol: triangle with circle.

Boolean: $Y = \bar A$.

AND Gate

Output 1 only when all inputs are 1.

Boolean: $Y = A\cdot B$.

OR Gate

Output 1 when at least one input is 1.

Boolean: $Y = A + B$.

NAND Gate

NOT-AND: complement of AND. Universal gate (any other gate can be built from NANDs).

Boolean: $Y = \overline{A\cdot B}$.

NOR Gate

NOT-OR. Also universal.

Boolean: $Y = \overline{A + B}$.

XOR Gate

Exclusive-OR: output 1 when inputs differ.

Boolean: $Y = A\bar B + \bar A B$.

Universal Gates

Both NAND and NOR are universal: any logic function can be built using just NAND or just NOR.

Examples (NAND-only):

• NOT: connect both inputs together: $\overline{A\cdot A} = \bar A$.
• AND: NAND followed by NAND: $\overline{\overline{A\cdot B}\cdot \overline{A\cdot B}} = A\cdot B$.
• OR: $\overline{\bar A \cdot \bar B} = A + B$ (De Morgan).

Pitfalls

• NAND $\ne$ AND with negative supply. Read symbols carefully — the bubble denotes inversion.
• Truth tables for 2-input gates have $2^2 = 4$ rows; for 3-input, $2^3 = 8$ rows.
• XOR is not the same as OR — XOR(1,1) = 0, OR(1,1) = 1.

Solved Problems

Problem 1 — Wavelength emitted by GaAs LED

Find $\lambda$ for GaAs ($E_g = 1.43\ \text{eV}$).

$\lambda = \dfrac{hc}{E_g} = \dfrac{1240\ \text{eV·nm}}{1.43\ \text{eV}} = 867\ \text{nm}$ (IR).

Problem 2 — Minority carrier density in n-type Si

Si at 300 K, $N_D = 10^{17}\ \text{cm}^{-3}$. Find $p$.

$n_i = 1.5\times 10^{10}$. $p = n_i^2/N_D = 2.25\times 10^{20}/10^{17} = 2.25\times 10^{3}\ \text{cm}^{-3}$.

Problem 3 — Diode in series with a resistor

A Si diode ($V_\gamma = 0.7\ \text{V}$) is in series with a 1 k$\Omega$ resistor and a 5 V battery. Find the current.

Forward bias: $V_R = V - V_\gamma = 5 - 0.7 = 4.3\ \text{V}$, so $I = 4.3\ \text{mA}$.

Problem 4 — Zener regulator

A 9.1 V Zener is connected through a $200\ \Omega$ series resistor to a $12\ \text{V}$ input, with a $1\ \text{k}\Omega$ load. Find the load current and Zener current.

Output across load = $9.1\ \text{V}$. Load current: $I_L = 9.1/1000 = 9.1\ \text{mA}$.

Input current: $I_{in} = (12 - 9.1)/200 = 14.5\ \text{mA}$.

Zener current: $I_Z = 14.5 - 9.1 = 5.4\ \text{mA}$.

Problem 5 — Half-wave vs full-wave ripple

If $V_m = 10\ \text{V}$, find $V_{DC}$ for HWR and FWR.

HWR: $V_{DC} = V_m/\pi = 10/3.14 = 3.18\ \text{V}$.

FWR: $V_{DC} = 2V_m/\pi = 6.37\ \text{V}$.

Problem 6 — Transistor in CE configuration

A npn transistor has $\beta = 200$, $V_{CC} = 10\ \text{V}$, $R_C = 2\ \text{k}\Omega$. For $I_B = 25\ \mu\text{A}$, find $I_C$, $V_{CE}$, and check if transistor is in active region.

$I_C = 200\cdot 25\ \mu\text{A} = 5\ \text{mA}$. $V_{R_C} = 5\cdot 2 = 10\ \text{V}$. $V_{CE} = V_{CC} - V_{R_C} = 10 - 10 = 0\ \text{V}$.

Transistor is saturated: it cannot deliver $5\ \text{mA}$ with $V_{CE}\approx 0$. Actual $I_C\approx V_{CC}/R_C = 5\ \text{mA}$ at $V_{CE,\text{sat}}\approx 0.2\ \text{V}$.

Problem 7 — Build XOR from NAND

Show XOR can be built using four NAND gates.

Let $A, B$ be inputs. Define:

• $C = \overline{A\cdot B}$ (NAND 1)
• $D = \overline{A\cdot C} = A + \bar C\cdot \text{(...)}$ — easier: let $P = \overline{A\cdot \overline{A\cdot B}}$, $Q = \overline{B\cdot \overline{A\cdot B}}$, $Y = \overline{P\cdot Q}$.

Then $Y = A\oplus B$. Four NANDs total. ✓

JEE/NEET Edge Cases

1. Mass-action law $np = n_i^2$ holds in thermal equilibrium — not under bias or illumination.
2. Hall effect distinguishes n- vs p-type: same direction of current $I$ in opposite-sign material gives opposite Hall voltage.
3. Avalanche vs. Zener mechanism: dopant density and breakdown voltage indicate which. $V_Z< 5$ V = Zener; $V_Z>6$ V = avalanche; $5<V_Z< 6$ is mixed.
4. Direct vs. indirect gap: Si and Ge are indirect (no LED emission); GaAs, GaN are direct (LED, laser).
5. Diode ideality: assume "ideal" unless told otherwise — drop $V_\gamma$ in forward, $I=0$ in reverse.
6. Peak inverse voltage (PIV):
   • HWR: PIV $= V_m$.
   • FWR centre-tap: PIV $= 2 V_m$.
   • FWR bridge: PIV $= V_m$.
7. Ripple factor:
   • HWR: $r = 1.21$.
   • FWR: $r = 0.482$.
8. Transistor $\alpha$ vs $\beta$: $\alpha = \beta/(1+\beta)$. If $\beta = 99$, $\alpha = 0.99$.
9. CE output is 180° phase-shifted from input; CB and CC are in phase.
10. NAND and NOR are universal — both can implement NOT, AND, OR independently.
11. Photon energy for LED/solar: $\lambda(\text{nm}) = 1240/E_g(\text{eV})$.
12. Static vs dynamic resistance:
    • Static $R = V/I$.
    • Dynamic $r = dV/dI$.

Quick Recap

• Bands: conductor ($E_g = 0$), insulator ($E_g > 3$ eV), semiconductor ($E_g \sim 1$ eV).
• Intrinsic: $n = p = n_i$. Extrinsic: n-type (donor, group V), p-type (acceptor, group III).
• Mass action: $np = n_i^2$.
• p-n junction: depletion region, $V_b\approx 0.7\ \text{V}$ (Si), $0.3\ \text{V}$ (Ge).
• Diode: Shockley $I = I_0(e^{eV/kT}-1)$; forward $V_\gamma$ knee; reverse small $I_0$; breakdown $V_Z$.
• Rectifiers: HWR ($V_{DC} = V_m/\pi$, ripple $f$), FWR ($V_{DC} = 2V_m/\pi$, ripple $2f$).
• Zener: reverse-breakdown regulator.
• LED: forward, direct-gap, $\lambda = hc/E_g$. Photodiode: reverse. Solar cell: open.
• Transistor: $I_E = I_B + I_C$, $\beta = I_C/I_B$. CE amplifier, CE switch.
• Gates: NOT, AND, OR, NAND (universal), NOR (universal), XOR.

Formula Sheet