Semiconductor Electronics - Materials, Devices and Simple Circuits

The chapter moves from the crystal to the circuit in five stages. First, energy bands explain why some solids conduct freely, some never do, and semiconductors sit in a controllable middle ground. Second, intrinsic and extrinsic semiconductors show how thermal excitation and, far more powerfully, deliberate doping create and control mobile electrons and holes. Third, the p-n junction — its depletion region, barrier potential, diffusion and drift — is built up as the physical engine behind every device that follows. Fourth, the junction diode's forward and reverse V-I characteristics are used to build rectifier circuits, and the same junction physics is specialised into the Zener diode, the LED, the photodiode and the solar cell. Fifth, a second junction turns the structure into a bipolar transistor, whose current relations and characteristic curves make it a current-controlled amplifier and switch, and cascades of such switches implement the logic gates of digital electronics.

Energy bands: why metals, insulators and semiconductors differ

In an isolated atom an electron's energy is fixed by its orbit, but when $N$ atoms pack into a crystal their outer orbitals overlap and each atomic level splits into a band of $N$ closely spaced levels. The highest band normally filled with valence electrons is the valence band; the band above it, where electrons move freely once they get there, is the conduction band. Whether a solid conducts, insulates, or sits in between is decided by the size of the gap between these bands and how completely the conduction band is occupied.

where: $\rho$ is electrical resistivity. Metals have overlapping or partially filled conduction bands, so electrons are available at no energy cost — hence very low resistivity. Insulators have a large gap $E_g > 3\ \text{eV}$, so room-temperature thermal energy (order $0.025\ \text{eV}$) cannot lift electrons across it. Semiconductors have a small gap, roughly $0.2$ to $3\ \text{eV}$ ($1.1\ \text{eV}$ for silicon, $0.7\ \text{eV}$ for germanium), small enough that a useful number of electrons are thermally promoted into the conduction band, leaving an equal number of vacancies in the valence band.

Intrinsic semiconductors: electrons and holes born together

Pure silicon or germanium crystallises in a diamond-like lattice where each atom shares its four valence electrons with four neighbours, forming covalent bonds. At absolute zero every bond is intact and the crystal is an insulator. As temperature rises, thermal vibrations occasionally break a bond, freeing an electron into the conduction band and leaving a vacancy — a hole — in the valence band, behaving as an apparent particle of charge $+q$. Because every excited electron leaves exactly one hole behind, an intrinsic semiconductor always has equal numbers of both:

where: $n_e$ is the free electron concentration, $n_h$ is the hole concentration, and $n_i$ is the intrinsic carrier concentration (about $1.5\times 10^{16}\ \text{m}^{-3}$ for silicon at room temperature). A hole does not physically travel; it is a bookkeeping device for a chain of bound electrons hopping into successive vacancies, indistinguishable in effect from a positive charge drifting through the lattice. Both carrier types contribute current in the same direction under an applied field, so total current is their sum:

where: $I_e$ is the conventional current due to free electrons and $I_h$ is that due to holes migrating toward the negative terminal. Generation and recombination proceed simultaneously; at thermal equilibrium the two rates balance, keeping $n_e$ and $n_h$ steady at a given temperature.

Extrinsic semiconductors: doping tips the balance

An intrinsic semiconductor's conductivity is too low and too temperature-sensitive for reliable devices. The fix is doping: introducing a controlled trace, typically a few parts per million, of an impurity whose atomic size closely matches silicon or germanium. A pentavalent dopant (P, As, Sb) contributes four electrons to bonding and leaves its fifth only weakly bound (ionisation energy $0.01$-$0.05\ \text{eV}$, far below the host gap), free at room temperature — this donor impurity gives an n-type semiconductor with electrons as the majority carrier. A trivalent dopant (B, Al, In) completes only three of the four bonds, leaving a vacancy a neighbouring electron readily fills, creating a mobile hole — this acceptor impurity gives a p-type semiconductor with holes as the majority carrier. In each case,

where: the product law holds regardless of doping level, since heavier doping on one side speeds recombination and suppresses the other carrier proportionally. Crucially, neither n-type nor p-type material carries any net charge: each dopant's mobile carrier is exactly balanced by its fixed, oppositely charged ionised core, so the crystal stays electrically neutral even though it now conducts far better than the pure material.

The p-n junction: diffusion, drift and the depletion region

A p-n junction is formed when a p-type region and an n-type region meet within a single continuous crystal, never by physically pressing two separate slabs together (no real surface is flat at the atomic scale). Two processes compete once the junction exists. Because electrons are far more concentrated in the n-region and holes far more concentrated in the p-region, both carriers diffuse down their concentration gradients — electrons from n to p, holes from p to n — a diffusion current. Every electron that leaves the n-side behind an ionised donor core, and every hole that leaves the p-side behind an ionised acceptor core, so a thin layer of exposed positive charge builds on the n-side and exposed negative charge builds on the p-side. Together these layers, stripped of mobile carriers, form the depletion region, only about a tenth of a micrometre thick. The exposed charges set up an electric field from the positive n-side layer to the negative p-side layer, pushing any stray electron on the p-side back to the n-side and any stray hole on the n-side back to the p-side — a drift current opposing diffusion. As diffusion proceeds the depletion region widens, the field strengthens, and drift current grows until it exactly balances diffusion current: net current is zero and the junction reaches equilibrium. The accumulated charge separation appears as the barrier potential $V_0$ (about $0.3\ \text{V}$ for germanium, $0.7\ \text{V}$ for silicon), n-side positive relative to p-side, opposing any further net flow of majority carriers.

The junction diode: forward and reverse bias

A semiconductor diode is a p-n junction with metal contacts for an external voltage $V$. Connecting the positive terminal to the p-side is forward bias: the applied voltage opposes the built-in field, so the effective barrier drops to $(V_0 - V)$, the depletion layer narrows, and once $V$ approaches $V_0$ a large number of majority carriers cross (minority-carrier injection), giving a current in the milliampere range that rises steeply, essentially exponentially, with voltage. Reversing the battery, n-side positive, is reverse bias: the barrier rises to $(V_0 + V)$, the depletion layer widens, and only the small population of minority carriers near the junction is swept across, giving a tiny, almost voltage-independent reverse saturation current (microampere range). Push the reverse voltage far enough and it reaches the breakdown voltage $V_{br}$, beyond which current rises sharply and, unless externally limited, destroys the junction by overheating. Forward current does not begin appreciably until $V$ crosses a threshold or cut-in voltage — about $0.2\ \text{V}$ for germanium, $0.7\ \text{V}$ for silicon — after which it rises very rapidly. Because the V-I curve is not a straight line, a diode's resistance is best described locally, as the dynamic (or ac) resistance:

where: $\Delta V$ and $\Delta I$ are small changes about an operating point. Forward dynamic resistance is small, tens of ohms; reverse dynamic resistance is enormous, order $10^{6}$-$10^{7}\ \Omega$, reflecting how little the tiny reverse current changes even for large voltage swings.

The diode as a rectifier

Because a diode conducts easily in one direction and blocks the other, it converts alternating current into unidirectional current, a process called rectification. In a half-wave rectifier, a single diode conducts only during the half-cycle that forward-biases it, so the output frequency equals the input frequency, $f_{out} = f_{in}$. A full-wave rectifier uses two diodes with a centre-tapped transformer (or four diodes in a bridge): one diode conducts on each half-cycle, so the output frequency doubles, $f_{out} = 2f_{in}$. The raw rectified output is a series of one-signed pulses, not steady dc; a capacitor across the load charges toward each pulse's peak and discharges slowly between pulses (time constant $R_L C$), smoothing the output toward a nearly steady voltage close to the peak — hence a larger filter capacitor gives a flatter, more usable dc output.

Special-purpose diodes: Zener, LED, photodiode and solar cell

The same junction physics, tuned by doping level and packaging, yields a family of specialised devices. A Zener diode is a heavily doped junction engineered for a sharp, non-destructive reverse breakdown at a chosen voltage $V_Z$; held deliberately in that breakdown region, its terminal voltage stays almost constant over a wide current range, exactly the behaviour needed for voltage regulation (derived below). A light-emitting diode (LED) is a forward-biased junction in a direct-band-gap compound semiconductor (GaAs, GaP, not silicon) where recombination releases energy as a photon rather than heat:

where: $h$ is Planck's constant, $c$ the speed of light, $\lambda$ the emitted wavelength — a wider gap gives a shorter, bluer wavelength, which is why red, green and blue LEDs use different compounds. A photodiode, reverse biased in the dark, sees its reverse current rise almost linearly with light intensity, since absorbed photons generate extra pairs the field sweeps across — a fast, linear sensor. A solar cell uses the same photogeneration with no bias at all: light generates pairs, the built-in field separates them, and the charge drives current through an external load, converting light directly into power.

The junction transistor: structure and transistor action

A bipolar junction transistor sandwiches a thin, lightly doped base between two more heavily doped regions of the opposite type, the emitter and the collector, giving an n-p-n or p-n-p device with two back-to-back junctions. Normally the emitter-base junction is forward biased, injecting carriers from emitter into base, while the base-collector junction is reverse biased. Because the base is thin and lightly doped, almost all injected carriers diffuse straight across before recombining and are swept into the collector; only a small fraction recombines, forming the small base current. Charge conservation at the three terminals then requires

where: $I_E$, $I_B$, $I_C$ are the emitter, base and collector currents, with $I_C$ close to but a little less than $I_E$, and $I_B$ the small remainder, typically a few per cent of $I_E$. This near-total pass-through of emitter current into collector current, controlled by a much smaller base current, is what makes the transistor a current amplifier (quantified in the derivation below).

Transistor characteristics, the amplifier and the switch

Plotting a transistor's terminal currents against its terminal voltages gives its characteristic curves: the input characteristic (base current against base-emitter voltage, resembling a forward diode curve) and the output characteristic (collector current against collector-emitter voltage for a family of fixed base currents — a set of almost-flat curves stacked by base current, since collector current is nearly independent of collector voltage in the active region). In common-emitter mode, a small variation in base current produces a much larger variation in collector current — exactly the leverage an amplifier needs: a weak signal at the base controls a strong replica of itself at the collector, drawing extra power from the supply rather than the input. Biased instead to its extremes — saturation (fully "on") or cut-off (fully "off") — the same transistor acts as an electronic switch, the building block of every logic gate described next.

Digital electronics: logic gates

A logic gate is a switching circuit, typically built from one or more transistors, that produces a well-defined output (logic $1$, high voltage, or logic $0$, low voltage) for every combination of binary inputs, summarised in a truth table. The simplest is the NOT gate (an inverter), with one input $A$ and output $Y = \overline{A}$: it outputs $1$ when $A$ is $0$ and vice versa. The OR gate outputs $Y = A + B$, which is $1$ if either input (or both) is $1$; the AND gate outputs $Y = A \cdot B$, which is $1$ only if both inputs are $1$. Combining a NOT with an OR or an AND gives the NOR gate ($Y = \overline{A+B}$) and the NAND gate ($Y = \overline{A \cdot B}$) respectively.

NAND and NOR each have a special status: either one, used repeatedly and wired appropriately, can reproduce the behaviour of a NOT, an AND and an OR gate, which is why they are called universal gates and are the standard building blocks of integrated logic circuits.

Board level — carrier concentration in a doped silicon crystal

A pure silicon crystal has $5 \times 10^{28}$ atoms per cubic metre. It is doped with a $1$ ppm concentration of pentavalent arsenic. Given the intrinsic carrier concentration $n_i = 1.5 \times 10^{16}\ \text{m}^{-3}$, find the number of electrons and holes per cubic metre after doping.

Step 1. Find the donor concentration. A doping level of $1$ ppm means one arsenic atom per million silicon atoms: $N_D = 5 \times 10^{28} \times 10^{-6} = 5 \times 10^{22}\ \text{m}^{-3}$.

Step 2. Compare with the intrinsic concentration. Since $N_D = 5\times 10^{22}\ \text{m}^{-3} \gg n_i = 1.5\times 10^{16}\ \text{m}^{-3}$, the thermally generated carriers are utterly negligible next to the donated ones, so essentially every donor is ionised and contributes one free electron: $n_e \approx N_D = 5\times 10^{22}\ \text{m}^{-3}$.

Step 3. Use the equilibrium product law to find the (very small) hole population: $n_e n_h = n_i^2$, so $n_h = \dfrac{n_i^2}{n_e} = \dfrac{(1.5\times 10^{16})^2}{5\times 10^{22}} = \dfrac{2.25\times 10^{32}}{5\times 10^{22}} = 4.5\times 10^{9}\ \text{m}^{-3}$. Answer: $n_e \approx 5\times 10^{22}\ \text{m}^{-3}$ and $n_h \approx 4.5\times 10^{9}\ \text{m}^{-3}$ — doping has raised the electron population by many orders of magnitude while crushing the hole population, exactly the signature of an n-type semiconductor.

Board level — dynamic resistance of a diode from its V-I graph

The V-I characteristic of a silicon diode, taken as a straight line between $I = 10\ \text{mA}$ ($V = 0.7\ \text{V}$) and $I = 20\ \text{mA}$ ($V = 0.8\ \text{V}$) in forward bias, and giving $I = -1\ \mu\text{A}$ at $V = -10\ \text{V}$ in reverse bias, is used to find the diode's resistance in each region.

Step 1. Forward dynamic resistance uses the slope of the forward segment: $r_{fb} = \dfrac{\Delta V}{\Delta I} = \dfrac{0.8 - 0.7\ \text{V}}{20 - 10\ \text{mA}} = \dfrac{0.1\ \text{V}}{10\ \text{mA}} = 10\ \Omega$.

Step 2. Reverse resistance is read directly from the single reverse point, treating it as an ordinary Ohm's-law ratio since the reverse current is essentially constant (a straight line through the origin is a reasonable approximation over this small a current range): $r_{rb} = \dfrac{10\ \text{V}}{1\ \mu\text{A}} = 1.0\times 10^{7}\ \Omega$. Answer: forward dynamic resistance $\approx 10\ \Omega$; reverse resistance $\approx 1.0\times 10^{7}\ \Omega$, a ratio of about a million, which is precisely the property — very low resistance one way, enormous resistance the other — that makes the diode useful as a rectifying element.

Competitive level (JEE / NEET) — transistor current gains and amplifier voltage gain

A transistor operating in common-emitter mode carries an emitter current $I_E = 10\ \text{mA}$ and a base current $I_B = 0.2\ \text{mA}$. Find (a) the common-base gain $\alpha$ and common-emitter gain $\beta$, and (b) the voltage gain of a CE amplifier built with this transistor if the collector load is $R_C = 4\ \text{k}\Omega$ and the small-signal input resistance is $r_i = 800\ \Omega$.

Step 1. Find the collector current from charge conservation: $I_C = I_E - I_B = 10 - 0.2 = 9.8\ \text{mA}$.

Step 2. Compute the two gains from their definitions: $\alpha = I_C/I_E = 9.8/10 = 0.98$, and $\beta = I_C/I_B = 9.8/0.2 = 49$ — consistent with $\beta = \alpha/(1-\alpha) = 0.98/0.02 = 49$, exactly as derived above.

Step 3. Apply the voltage-gain formula from the derivation: $A_v = -\beta \dfrac{R_C}{r_i} = -49 \times \dfrac{4000}{800} = -49 \times 5 = -245$.

Step 4. Interpret the sign and magnitude, and note the power gain a full analysis would add: the magnitude $245$ means the ac output voltage swing is $245$ times the input swing, and the negative sign means the output is inverted (a rising input gives a falling output). The approximate power gain, current gain times voltage gain, is $A_p \approx \beta \times |A_v| = 49 \times 245 \approx 1.2\times 10^{4}$. Answer: $\alpha = 0.98$, $\beta = 49$, and $A_v = -245$ (magnitude $245$, phase-inverted). Timing note: about two to three minutes. The fast path is $I_C = I_E - I_B \to \alpha = I_C/I_E \to \beta = I_C/I_B \to A_v = -\beta R_C/r_i$; do not waste time deriving $\beta = \alpha/(1-\alpha)$ from scratch in the exam — quote it directly and use it only as a cross-check on your arithmetic. A common slip is plugging $\alpha$ instead of $\beta$ into the voltage-gain formula, or forgetting the minus sign when a question asks for the phase relationship rather than just the magnitude.

Treating n-type or p-type material as electrically charged. The names refer only to the majority carrier type; every doped crystal stays overall neutral because each mobile carrier is balanced by an immobile, oppositely charged ionised dopant core.

Swapping which junction is forward biased in a transistor. Normal (active-region) operation always has the emitter-base junction forward biased and the base-collector junction reverse biased — reversing either kills the amplifying action.

Confusing $\alpha$ (common base, $I_C/I_E$, close to $1$) with $\beta$ (common emitter, $I_C/I_B$, tens to hundreds). Using the wrong one in a gain formula gives answers wrong by an order of magnitude; check the configuration before picking the symbol.

Getting forward and reverse bias backwards for the LED and photodiode. An LED must be forward biased to emit light; a photodiode is reverse biased to detect light. Mixing these up is a very common recall error.

Forgetting the output frequency doubles in a full-wave rectifier. Half-wave: $f_{out} = f_{in}$. Full-wave: $f_{out} = 2 f_{in}$, since each of the two conduction paths contributes one pulse per input cycle.

Assuming any reverse bias destroys a diode. A diode is safe under reverse bias up to its rated breakdown voltage; damage occurs only if reverse current is not externally limited once breakdown is reached, or forward current exceeds its rated value.

Reading the diode V-I curve as linear. Both branches are markedly non-linear, which is exactly why local dynamic resistance $r_d = \Delta V/\Delta I$ is used instead of a single fixed value.

• Classify a solid as a conductor, insulator or semiconductor from its resistivity range or its energy band gap $E_g$.
• Explain how thermal excitation creates electron-hole pairs in an intrinsic semiconductor and why $n_e = n_h = n_i$.
• Distinguish n-type from p-type doping, identify the dopant valency in each case, and name majority and minority carriers.
• State and apply $n_e n_h = n_i^2$, and explain why doped semiconductors remain electrically neutral overall.
• Describe how diffusion and drift build the depletion region and the barrier potential $V_0$ of a p-n junction.
• Sketch and interpret the forward and reverse V-I characteristic of a junction diode, locating threshold voltage, reverse saturation current and breakdown.
• Analyse half-wave and full-wave rectifier circuits, including the role of a capacitor filter, and compute output frequency from input frequency.
• Explain the working principle of a Zener diode, LED, photodiode and solar cell, and match each device to its correct bias condition and application.
• Derive and apply $I_E = I_B + I_C$ and the relation $\beta = \alpha/(1-\alpha)$ for a bipolar transistor.
• Compute the voltage gain of a common-emitter amplifier from $\beta$, $R_C$ and $r_i$, and construct truth tables for NOT, AND, OR, NAND and NOR gates.

Ask the tutor: if you want to work through why a specific transistor biasing circuit sits in saturation, cut-off, or the active amplifying region, ask the tutor to walk that load-line analysis with you.