Electronic Devices | ISC Class 12 Physics Notes

This note covers energy bands, conductors and insulators, intrinsic and extrinsic semiconductors, electrons and holes, p-n junction formation, diode biasing and characteristics, half-wave and full-wave rectification, light-emitting diodes, photodiodes, solar cells and Zener voltage regulation.

How do energy bands explain electrical conduction?

What changes when isolated atoms form a solid?

An energy level is an allowed energy of an electron. In an isolated atom, electrons occupy distinct levels. When many atoms come together to form a solid, interactions produce closely spaced levels grouped into energy bands.

The valence band contains the energy levels associated with valence electrons, the electrons involved in bonding. The conduction band lies above it; electrons occupying available states in this band can contribute to electrical conduction through the solid.

A forbidden energy gap separates allowed bands where no electron energy states are available. Let E₁ denote the energy at the bottom of the conduction band and E₂ the energy at the top of the valence band. The gap is E.

E = E₁ − E₂. This is an energy difference, not a physical distance between layers of material. Band diagrams arrange electron energies vertically; their horizontal bands do not represent separate slabs inside a semiconductor.

How do the three classes differ?

Electric current is the rate of flow of electric charge. Conductivity describes how readily a material carries current. Resistivity describes its opposition to conduction. Semiconductor conductivity lies between that of metals and insulators, but energy-band structure explains this distinction more fully than a resistivity value alone.

The conductor panels show partially occupied or overlapping bands. The insulator panel shows a large gap and an empty conduction band. The semiconductor panel shows a smaller gap with some conduction-band occupation. The vertical arrows indicate increasing electron energy.

See Fig. 14.2 in your NCERT textbook

Why are silicon and germanium semiconductors while diamond is an insulator?

How large are their energy gaps?

Silicon, represented by Si, and germanium, represented by Ge, are elemental semiconductors. Carbon, represented by C, in its diamond form has a much larger forbidden gap. All three have four valence electrons, so valence-electron count alone does not determine their electrical behaviour.

Potential difference means energy transferred per unit charge; its unit is the volt. The electronvolt, abbreviated eV, is the energy acquired by an electron when it moves through a potential difference of one volt. It is an energy unit, whereas the volt is a unit of potential difference.

The S.I. unit of energy is the joule, symbol J.

These values give the order diamond, silicon, germanium from largest to smallest gap. The smaller gaps of silicon and germanium allow a significant population of conduction electrons through thermal excitation; the corresponding population in diamond is negligibly small under ordinary conditions.

What does temperature change?

At absolute zero, the lowest thermodynamic temperature, a pure semiconductor has a filled valence band and an empty conduction band. Write temperature as T and measure it in kelvin, symbol K. At T = 0 K, it behaves as an insulator.

At higher temperatures, some electrons gain energy and enter the conduction band. Each leaves an electron vacancy in the valence band. Increasing the available thermal energy generates more mobile charge carriers and increases the conductivity of an intrinsic semiconductor.

The useful distinction is therefore between occupied and available energy states, the gap separating them, and the energy available to excite electrons. Simply describing a substance as solid, or counting its valence electrons, does not establish whether it conducts well.

How do electrons and holes carry current in an intrinsic semiconductor?

What is a hole?

An intrinsic semiconductor is a pure semiconductor. A covalent bond is a bond formed by sharing electrons. Thermal energy can release an electron from a bond, leaving a vacancy called a hole, which behaves as a carrier with effective positive charge.

A neighbouring bound electron can fill the vacancy, leaving a new vacancy at its former position. This successive movement of bound electrons appears as hole motion. The conduction electron originally released moves independently; it need not be the electron that fills the hole.

Carrier concentration means the number of carriers per unit volume. Let n be the conduction-electron concentration, p the hole concentration and nᵢ the intrinsic carrier concentration. Their unit is m⁻³, meaning per cubic metre. The symbols n and p here denote concentrations, while n-type and p-type name the two doped material types.

n = p = nᵢ for an intrinsic semiconductor. Each thermal generation event produces an electron and a hole together. Equality of their concentrations does not mean that their physical motions are identical.

How do their currents combine?

The S.I. unit of electric current is the ampere, symbol A. Let I denote total conventional current, Iₑ its electron contribution and Iₕ its hole contribution. Conventional current follows the direction assigned to positive-charge flow.

I = Iₑ + Iₕ. An electric field describes electric force per unit positive test charge. In an applied field, electrons and holes move in opposite directions because their effective charges have opposite signs. Their contributions to conventional current nevertheless add.

Recombination occurs when a conduction electron fills a hole, removing the pair as mobile carriers. Generation and recombination occur together. At thermal equilibrium, with no continuing change in temperature or carrier populations, their rates are equal, so the carrier populations remain steady even though individual carriers are continually generated and recombine.

Note: A hole is an electron vacancy with effective positive charge. It is not a proton moving through the semiconductor, and hole motion does not require atoms to travel through the material.

How does doping produce n-type and p-type semiconductors?

What do donor and acceptor impurities do?

Doping is the deliberate addition of a small amount of suitable impurity to a pure semiconductor. The impurity is a dopant; the resulting material is an extrinsic semiconductor. Suitable dopants increase the number of mobile carriers without substantially disrupting the host structure.

A pentavalent atom has five valence electrons. Phosphorus, arsenic and antimony are examples of donor impurities in silicon or germanium. Four electrons take part in bonding and the fifth is weakly bound, so it can become a conduction electron with a small energy supply.

A trivalent atom has three valence electrons. Boron, aluminium and indium are acceptor impurities. They leave a bonding vacancy that can accept an electron from a neighbouring bond. This produces a mobile hole and an effectively negative, immobile acceptor ion.

Majority carriers are the more numerous mobile carriers; minority carriers are the less numerous kind. Donor-doped material is called n-type, with electrons as majority carriers. Acceptor-doped material is p-type, with holes as majority carriers. Both kinds remain present.

Where are the impurity energy levels?

The donor energy level lies slightly below the conduction-band bottom. The acceptor energy level lies slightly above the valence-band top. Small energy supplies can therefore free donor electrons or allow valence electrons to occupy acceptor levels, leaving holes behind.

At room temperature, most donor or acceptor atoms become ionised. The doped material still maintains overall charge neutrality because the charge of the additional mobile carriers is balanced by the opposite charge of the immobile ionised impurities.

The n-type panel places the donor level just below the conduction band and shows many conduction electrons. The p-type panel places the acceptor level just above the valence band and shows holes in the valence band.

See Fig. 14.9 in your NCERT textbook

How are electron and hole concentrations related?

What changes after doping?

Doping increases the concentration of one carrier type, but it does not generate equally large populations of both types. Additional majority carriers make recombination with minority carriers more likely. Consequently, the minority-carrier concentration falls below its intrinsic value at the same temperature.

For a semiconductor in thermal equilibrium, the carrier concentrations obey np = nᵢ². The intrinsic concentration nᵢ belongs to the same material at the same temperature. This condition matters because nᵢ changes with temperature.

Let N denote the concentration of donor atoms. When donors are effectively ionised and their contribution greatly exceeds intrinsic generation, n ≈ N. The symbol ≈ means approximately equal to. The minority concentration then follows from p = nᵢ²/n.

Derivation: How does donor concentration determine the hole concentration?

Assume thermal equilibrium at a fixed temperature, with ionised donors supplying far more electrons than thermal generation.

1. The equilibrium carrier relation is $np = n_i^2$, where $n$ and $p$ are electron and hole concentrations and $n_i$ is the intrinsic concentration at that temperature.
2. Divide both sides by the electron concentration: $p = \frac{n_i^2}{n}$.
3. Each ionised pentavalent donor supplies one conduction electron. When donor electrons dominate, the electron concentration is approximately the donor concentration $N$, so $n \approx N$.
4. Substitute this approximation into the equilibrium relation to obtain $p \approx \frac{n_i^2}{N}$.

Result: $p \approx \frac{n_i^2}{N}$. At fixed temperature, increasing the donor concentration lowers the minority hole concentration within this approximation.

Worked example 1. A pure silicon crystal contains 5 × 10²⁸ atoms m⁻³ and is doped with one part per million of pentavalent arsenic. Its intrinsic carrier concentration is 1.5 × 10¹⁶ m⁻³. Assume the donors supply their extra electrons and dominate intrinsic generation. Find both carrier concentrations.

Answer: One part per million means a fraction of 10⁻⁶. The donor concentration is (5 × 10²⁸) × 10⁻⁶ = 5 × 10²² m⁻³. Hence n ≈ 5 × 10²² m⁻³.

Using the equilibrium relation, p = (1.5 × 10¹⁶)²/(5 × 10²²) = 4.5 × 10⁹ m⁻³. Electrons are the majority carriers and holes remain present as minority carriers.

How should the result be checked?

The electron concentration is much greater than the given intrinsic concentration, supporting the approximation used. Multiplying the calculated electron and hole concentrations returns nᵢ². The units are consistent: squaring a concentration and dividing by another concentration leaves m⁻³.

Do not replace p by nᵢ after donor doping. Equal electron and hole concentrations describe the intrinsic material, whereas the doped material must satisfy the equilibrium product and its appropriate majority-carrier concentration.

How does a p-n junction develop a depletion region and potential barrier?

What happens when the two regions meet?

A p-n junction is the boundary between p-type and n-type regions formed within a semiconductor. It is not obtained simply by pressing separate pieces together: surface roughness prevents the continuous atomic contact needed for an ordinary junction.

Diffusion is carrier movement caused by a concentration difference. Electrons diffuse from the electron-rich n-side to the p-side, while holes diffuse from the hole-rich p-side to the n-side. Recombination near the boundary removes mobile electrons and holes.

1. Electrons leaving the n-side expose positive donor ions near the junction.
2. Holes leaving the p-side expose negative acceptor ions near the junction.
3. These immobile ions form a space-charge region called the depletion region, depleted of mobile carriers.
4. The separated charges establish an electric field directed from the positive n-side ions towards the negative p-side ions.
5. The field drives carriers in a direction opposing their diffusion, until the two current contributions balance.

Why is the net current zero at equilibrium?

Drift is carrier motion caused by an electric field. The junction field moves minority electrons from p to n and minority holes from n to p. The resulting drift current opposes the diffusion current.

At equilibrium, drift and diffusion currents have equal magnitudes, so the net current is zero. This is a balance between processes, not an absence of all carrier movement. The ions producing the field remain bound within the material.

The barrier potential is the potential difference across the depletion region that opposes further majority-carrier diffusion. Denote its unbiased value by V₀. Potential difference means energy transferred per unit charge. The S.I. unit of potential difference is the volt, symbol V.

The p-region is on the left and the n-region on the right. Oppositely charged ions occupy the narrow junction region. The electric-field arrow points towards the p-side, and the potential curve rises towards the n-side.

See Fig. 14.11 in your NCERT textbook

How do forward and reverse bias change a junction diode?

Which battery connections produce each bias?

A semiconductor diode is a p-n junction provided with metallic contacts. It has two terminals. Bias means an externally applied voltage across the junction. The p-terminal is the anode and the n-terminal is the cathode.

For forward bias, connect the p-side to the battery's positive terminal and the n-side to its negative terminal. The applied field opposes the junction field, reducing the depletion width and barrier. More majority carriers can cross the junction as the forward voltage increases.

Let V denote the magnitude of the applied bias voltage. Let B denote the effective barrier potential, also measured in volts. In the elementary description, the forward barrier becomes B = V₀ − V. A small forward voltage reduces the barrier only slightly and produces a small current; a sufficiently increased forward voltage produces much greater current.

For reverse bias, connect the p-side to the negative terminal and the n-side to the positive terminal. The applied field reinforces the junction field. The depletion region widens, the barrier becomes B = V₀ + V, and majority-carrier diffusion is strongly suppressed.

Why does reverse current still flow?

Thermally generated minority carriers are swept across the depletion region. Their limited supply produces a small reverse saturation current. It is essentially voltage independent before breakdown, rather than exactly zero under all reverse-bias conditions.

Breakdown voltage is the reverse voltage at which current increases sharply. An external circuit must limit current below the rated value, otherwise overheating can destroy the junction. Excessive forward current can also damage a diode.

How are diode characteristics measured and used in numerical problems?

What does the current-voltage graph show?

A diode's current-voltage characteristic, also called its I-V or V-I characteristic, shows how current changes with applied voltage. A voltmeter measures voltage across the diode, while an ammeter measures current through it. A variable supply arrangement allows readings at different biases.

Forward current is measured conveniently in milliamperes, abbreviated mA; 1 mA equals 10⁻³ A. Reverse current before breakdown is much smaller and is measured in microamperes, abbreviated µA; 1 µA equals 10⁻⁶ A.

The threshold voltage or cut-in voltage marks the region beyond which forward current rises significantly. Typical values are about 0.2 V for germanium and about 0.7 V for silicon. These are approximate characteristic values, not exact switching voltages.

The circuits show a voltmeter across the diode and a current meter in series. The characteristic uses milliamperes for forward current and microamperes for reverse current, with a sharp reverse-current increase at breakdown.

See Fig. 14.16 in your NCERT textbook

How is dynamic resistance calculated?

Dynamic resistance, written r, is the ratio of a small voltage change to the corresponding current change near an operating point. Write these changes as ΔV and ΔI, where Δ means change. Then r = ΔV/ΔI.

The S.I. unit of resistance is the ohm, symbol Ω. Dynamic resistance uses nearby changes, whereas the static resistance at a point uses the voltage-to-current ratio there. Because the diode characteristic is curved, these two ratios are generally different.

Worked example 2. Near a forward current of 15 mA, a silicon diode characteristic gives 10 mA at 0.7 V and 20 mA at 0.8 V. Approximate that short segment as straight and calculate its dynamic resistance.

Answer: ΔV = 0.8 − 0.7 = 0.1 V and ΔI = 20 − 10 = 10 mA = 0.010 A. Therefore r = 0.1/0.010 = 10 Ω.

Worked example 3. At a reverse voltage of −10 V, a diode carries −1 µA. Calculate its static resistance using the magnitudes of voltage and current.

Answer: The current magnitude is 10⁻⁶ A. Resistance = 10/(10⁻⁶) = 1.0 × 10⁷ Ω. This large static resistance does not mean the reverse current is exactly zero.

How does a half-wave rectifier work?

What does each component do?

Rectification converts an alternating electrical input into a unidirectional output. An alternating current, abbreviated ac, reverses direction periodically. Direct current, abbreviated dc, flows in one direction; a rectifier's immediate output is pulsating rather than steady.

A half-wave rectifier uses a diode in series with a load. The load is the part of the circuit across which useful output is taken. Its resistance is labelled Rₗ. A transformer secondary supplies the alternating voltage to the diode-load circuit.

A transformer transfers electrical energy between its primary and secondary coils by electromagnetic induction. Here, the secondary provides the desired alternating voltage. The diode selects the conducting half-cycle, and the load resistor develops the output voltage when current passes through it.

1. During the half-cycle that makes the diode's p-side positive relative to its n-side, the diode becomes forward biased.
2. Current passes through the diode and load resistor, producing a load voltage.
3. During the opposite half-cycle, the diode becomes reverse biased and its small reverse current is neglected.
4. The load output is therefore approximately zero during this half-cycle; conduction resumes during the next forward-biased half-cycle.

Why is the output pulsating?

The load current retains one direction, but its magnitude rises and falls during each conducting interval. Between these intervals there is a gap. Thus rectification establishes direction without automatically producing constant voltage.

The diode's reverse breakdown voltage must be sufficiently greater than the peak reverse voltage imposed by the transformer secondary. This keeps ordinary rectification within the diode's intended operating region. The peak voltage is the largest voltage magnitude reached during a cycle.

The transformer secondary feeds one diode and a load resistor. The input graph alternates above and below its axis. The output graph retains positive half-cycle pulses, with zero-output intervals during the negative input half-cycles.

See Fig. 14.18 in your NCERT textbook

How does a centre-tapped full-wave rectifier use both half-cycles?

Why are two diodes needed?

A centre tap is a connection to the midpoint of a transformer secondary. A centre-tapped full-wave rectifier connects one diode to each end of that secondary. Their n-terminals meet at a common output point; the load connects between this point and the centre tap.

Label the secondary ends A and B and the diodes D₁ and D₂. These labels identify circuit components, not physical quantities. When A is positive relative to the centre tap, B is negative relative to it. Their roles reverse in the next half-cycle.

1. With A positive, D₁ is forward biased and supplies load current.
2. D₂ is then reverse biased and its current is neglected.
3. With B positive in the next half-cycle, D₂ conducts and D₁ is reverse biased.
4. Both conducting paths drive current through the load in the same direction, giving output during both input half-cycles.

Two diodes join the secondary ends to the top of the load. The centre tap connects to the load's lower end. The two input waveforms have opposite signs, while alternate output pulses are labelled as contributions from D₁ and D₂.

See Fig. 14.19 in your NCERT textbook

How do frequency and smoothing differ?

Frequency is the number of repeated cycles per second, and the S.I. unit of frequency is the hertz, symbol Hz. Let f be input frequency and F the frequency of output pulses. Half-wave rectification gives F = f; full-wave rectification gives F = 2f.

Derivation: Why does full-wave rectification double the output-pulse frequency?

Let $T$ be the input period in seconds and $f = 1/T$ its frequency in hertz. Consider the pulsating output before smoothing.

1. A half-wave rectifier conducts during one half-cycle and blocks the next. Successive output pulses are separated by one input period, so $T_{\mathrm{half}} = T$.
2. Taking the reciprocal of the output period gives $F_{\mathrm{half}} = \frac{1}{T} = f$.
3. In a full-wave rectifier, the two diodes conduct on alternate half-cycles, producing load pulses in the same direction. Successive pulses are separated by $T_{\mathrm{full}} = T/2$.
4. Therefore $F_{\mathrm{full}} = \frac{1}{T/2} = \frac{2}{T} = 2f$.

Result: $F_{\mathrm{half}} = f$ and $F_{\mathrm{full}} = 2f$. Full-wave rectification produces two output pulses per input cycle.

Worked example 4. An alternating input has frequency 50 Hz. Find the output-pulse frequencies of half-wave and full-wave rectifiers.

Answer: The half-wave circuit produces one pulse per input cycle, so its output frequency is 50 Hz. The full-wave circuit produces two pulses per input cycle, so its output frequency is 2 × 50 = 100 Hz.

A filter reduces the alternating variation, or ripple, in the rectified output. A capacitor, a component that stores charge, can be connected across the load. It charges as voltage rises and discharges through the load between peaks, making the output steadier.

How do LEDs, photodiodes and solar cells use light?

How does a light-emitting diode produce light?

A light-emitting diode, abbreviated LED, is a junction diode that emits light when forward biased. Electrons and holes injected across the junction recombine, releasing energy as light in suitable semiconductor materials. The device converts electrical energy into light energy.

The emitted light's colour depends on the semiconductor energy gap. Not every semiconductor junction is an efficient visible-light emitter. The current must be limited to the device's operating range, usually by a series resistor or a suitable current-control circuit.

LEDs are useful because they save electrical energy. They do not require combustion at the point of use to produce light. Their energy-saving role helps reduce the energy demand associated with lighting; this does not imply that manufacturing or electricity generation has no environmental impact.

How does a photodiode detect light?

A photodiode is a light-sensitive junction diode generally operated in reverse bias. Light reaching the junction can generate electron-hole pairs when its photon energy is sufficient. A photon is a quantum, or discrete packet, of light energy.

The junction field separates the generated carriers and produces a photocurrent, meaning current caused by illumination. Greater illumination increases the reverse photocurrent under otherwise unchanged conditions. A small reverse current can also exist without illumination; this is called dark current.

How does a solar cell supply electrical energy?

A solar cell converts incident solar energy into electrical energy through the photovoltaic effect, the production of voltage by illumination. Light generates electron-hole pairs, and the junction field separates them. An external load then provides a path for useful current.

Electrical power is the rate of transfer of electrical energy. The common feature of these devices is the junction and its charge carriers. The important difference is the intended action: emission, detection or energy conversion. A photodiode is used to sense illumination, while a solar cell is arranged to deliver electrical power from it.

How does a Zener diode regulate voltage?

What is special about its reverse characteristic?

A Zener diode is a specially designed, heavily doped junction diode intended to operate in reverse breakdown with controlled current. Heavy doping produces a thin depletion region. A strong electric field can then develop across this narrow region.

The Zener voltage, written U, is its specified reverse breakdown voltage. Below breakdown, reverse current is small. In the breakdown region, a substantial change in current produces only a small voltage change, so the voltage remains approximately constant.

A voltage regulator maintains an approximately constant output voltage despite changes in supply voltage or load demand within its operating limits. A Zener regulator uses the nearly constant voltage of the reverse-breakdown region for this purpose.

How is the regulating circuit connected?

Connect a series resistor between the positive supply terminal and the output node. Connect the Zener and load in parallel between that node and the negative return. The Zener cathode faces the positive output node, making it reverse biased.

Draw a supply feeding a series resistor, followed by a node that branches to the load and a reverse-biased Zener diode. Join both branches to the supply return. Mark output across the parallel branches and label the Zener cathode at the positive output node.

1. With the Zener in breakdown, the load voltage is approximately its Zener voltage.
2. If supply voltage rises while the load remains unchanged, more voltage appears across the series resistor and more current passes through it.
3. The additional current passes mainly through the Zener, keeping the load voltage nearly unchanged.
4. If supply voltage falls, Zener current falls; regulation continues provided sufficient breakdown current remains available.

If load current increases at an unchanged supply voltage, the Zener current decreases as more of the available current passes through the load. If load current decreases, more current can pass through the Zener. The load and Zener share the series-resistor current.

The series resistor limits current and takes up the excess supply voltage. Regulation fails if the supply cannot maintain breakdown, or if the load draws too much current. The Zener's current and power ratings must also be respected; breakdown operation is controlled, not unrestricted.

Glossary

• Energy band — A group of closely spaced allowed electron energy levels within a solid.
• Valence band — The energy band containing levels associated with electrons involved in atomic bonding.
• Conduction band — The band in which electrons can occupy available states and contribute to electrical conduction.
• Forbidden gap — The energy interval between allowed bands in which electron states are unavailable.
• Intrinsic semiconductor — A pure semiconductor with equal concentrations of thermally generated electrons and holes.
• Hole — An electron vacancy that behaves as a mobile carrier of effective positive charge.
• Doping — Deliberate addition of a small amount of suitable impurity to modify semiconductor carrier concentrations.
• Majority carrier — The more numerous type of mobile charge carrier in a doped semiconductor.
• Depletion region — The junction region depleted of mobile carriers and containing immobile charged impurity ions.
• Barrier potential — The junction potential difference that opposes further diffusion of majority charge carriers.
• Forward bias — A connection making the p-side positive relative to the n-side and reducing the junction barrier.
• Dynamic resistance — The ratio of a small voltage change to its corresponding current change near an operating point.
• Rectification — Conversion of an alternating electrical input into an output that retains one direction.
• Photocurrent — Electric current arising from charge carriers generated when light illuminates a suitable semiconductor.
• Zener regulation — Maintaining an approximately constant output voltage using a reverse-biased Zener diode in controlled breakdown.

Common errors and misconceptions

• Misconception: An n-type semiconductor has a net negative charge. Correct: It has electrons as majority carriers, but positive ionised donors balance their charge and the material remains neutral.
• Misconception: A hole is a moving proton. Correct: It is an electron vacancy; successive movements of bound electrons make the vacancy appear to move.
• Misconception: The depletion region contains no charge. Correct: It is depleted of mobile carriers but contains immobile charged donor and acceptor ions.
• Misconception: Reverse bias produces exactly zero current. Correct: A small minority-carrier current flows before breakdown, and reverse current increases sharply at breakdown.
• Misconception: A silicon diode switches on at an exact universal voltage. Correct: About 0.7 V is a typical threshold; the characteristic is continuous and current rises strongly through the threshold region.
• Misconception: A full-wave rectifier immediately provides steady voltage. Correct: Its output is unidirectional but pulsating; filtering makes it steadier.
• Misconception: LEDs, photodiodes and solar cells all need forward bias. Correct: LEDs use forward bias, photodiodes generally use reverse bias, and solar cells supply energy from illumination without an external bias source.
• Misconception: A Zener diode regulates every supply voltage. Correct: It regulates only while maintained in its intended breakdown range with adequate current and safe current and power limits.

Exam-style questions with model answers

Q1. Define an intrinsic semiconductor and state the relation between its electron and hole concentrations. [2 marks]

1. An intrinsic semiconductor is a pure semiconductor whose mobile electrons and holes arise from thermal generation.
2. Its electron concentration n and hole concentration p are equal: n = p = nᵢ, where nᵢ is the intrinsic carrier concentration.

Q2. Explain how pentavalent doping produces n-type silicon. Include the majority carriers, minority carriers and overall charge. [4 marks]

1. A pentavalent impurity has five valence electrons. Four participate in bonding with neighbouring silicon atoms, while the fifth remains weakly bound.
2. The fifth electron can become a conduction electron with a small energy supply, so electrons become the majority carriers.
3. Thermal generation still produces holes. These are minority carriers because their concentration is much smaller than the electron concentration.
4. The semiconductor remains electrically neutral overall because the extra mobile electrons are balanced by positive, immobile donor ions.

Q3. Explain the formation of the depletion region and barrier potential in an unbiased p-n junction, and explain why its equilibrium net current is zero. [5 marks]

1. Concentration differences cause electrons to diffuse from n to p and holes from p to n, with recombination near the boundary.
2. Loss of these mobile carriers exposes positive donor ions on the n-side and negative acceptor ions on the p-side.
3. The resulting depletion region contains immobile ions. Their electric field points from n to p and establishes a barrier opposing further majority-carrier diffusion.
4. This field produces minority-carrier drift in the opposite current direction to diffusion: electrons move from p to n and holes from n to p.
5. At equilibrium, drift and diffusion currents have equal magnitudes, giving zero net current even though both carrier processes continue.

Q4. A silicon diode carries 10 mA at 0.7 V and 20 mA at 0.8 V. Treat this short characteristic segment as straight. Calculate its dynamic resistance, showing both changes and converting current to amperes. [3 marks]

1. The voltage change ΔV is the difference between the two supplied voltage readings: 0.8 − 0.7 = 0.1 V.
2. The current change ΔI is 20 − 10 = 10 mA. Since 1 mA = 10⁻³ A, this is 0.010 A.
3. Dynamic resistance r is ΔV/ΔI, so r = 0.1/0.010 = 10 Ω for the stated straight-segment approximation.

Q5. Explain the connections and operation of a two-diode, centre-tapped full-wave rectifier. State whether its output is steady and give its output-pulse frequency for a 50 Hz input. [5 marks]

1. The diode p-terminals connect to opposite ends of the centre-tapped secondary. Their common n-terminal connects through the load to the centre tap.
2. When one secondary end is positive relative to the centre tap, its diode conducts while the other diode is reverse biased.
3. During the next half-cycle, the other diode conducts. Both paths drive load current in the same direction, using both halves of the input cycle.
4. The output remains pulsating rather than steady. A capacitor connected across the load can charge near peaks and discharge between them, reducing ripple.
5. Two output pulses occur per input cycle. For the given input frequency of 50 Hz, the output-pulse frequency is 2 × 50 = 100 Hz.

Q6. Distinguish an LED, a photodiode and a solar cell by their operating arrangements and functions. [3 marks]

1. An LED is forward biased and converts electrical energy into light through electron-hole recombination in a suitable semiconductor junction.
2. A photodiode is generally reverse biased. Incident light generates carriers and changes the reverse current, allowing the device to detect illumination.
3. A solar cell operates under illumination without an external bias source. Its junction separates light-generated carriers and supplies electrical energy to an external load.

Q7. Explain how a Zener regulator maintains its load voltage when the supply voltage rises. Describe its connections, the series resistor's role and one condition required for regulation. [5 marks]

1. The Zener is reverse biased across the load, with its cathode connected to the positive output node. Both receive current through a series resistor.
2. In reverse breakdown, a substantial current change causes only a small voltage change, keeping the load voltage approximately equal to the Zener voltage.
3. When supply voltage rises with the load unchanged, the extra voltage appears mainly across the series resistor, increasing the current through it.
4. The additional current passes mainly through the Zener branch. The resistor limits current and prevents an unrestricted rise through the diode.
5. Regulation requires operation within the intended breakdown range and safe current and power ratings. It cannot continue if the supply becomes insufficient to maintain breakdown.

Q8. Silicon contains 5 × 10²⁸ atoms m⁻³ and is doped with one part per million of pentavalent arsenic. At the stated temperature, nᵢ = 1.5 × 10¹⁶ m⁻³. Assume effective donor ionisation, donor-dominated electron concentration and thermal equilibrium. Calculate the electron and hole concentrations and identify the majority carriers. [4 marks]

1. One part per million is a fraction of 10⁻⁶, giving donor concentration N = (5 × 10²⁸) × 10⁻⁶ = 5 × 10²² m⁻³.
2. Under the given donor-dominated approximation, the electron concentration n is approximately N, hence n ≈ 5 × 10²² m⁻³.
3. Thermal equilibrium gives np = nᵢ², so the hole concentration p = (1.5 × 10¹⁶)²/(5 × 10²²) = 4.5 × 10⁹ m⁻³.
4. Electrons are the majority carriers because their concentration greatly exceeds the hole concentration. The material is n-type, with holes as minority carriers.

Key takeaways

• Energy-band occupation and forbidden-gap size explain the different electrical behaviour of conductors, insulators and semiconductors.
• Intrinsic semiconductors contain equal electron and hole concentrations, with both types contributing to conventional current.
• Donor doping produces n-type material, while acceptor doping produces p-type material; both remain electrically neutral overall.
• A junction's depletion region contains immobile ions whose field opposes diffusion and establishes the barrier potential.
• Forward bias reduces the barrier and depletion width; reverse bias increases both and leaves a small minority-carrier current.
• Half-wave rectification uses one input half-cycle, while centre-tapped full-wave rectification uses both and doubles the output-pulse frequency.
• LEDs emit light, photodiodes detect illumination, and solar cells supply electrical energy from incident light.
• A Zener regulator maintains approximately constant voltage in controlled reverse breakdown, subject to supply, load and device limits.

Test yourself

Why does a pure semiconductor behave as an insulator at absolute zero?

Its valence band is filled and its conduction band is empty, leaving no thermally generated mobile electron-hole pairs.

Which impurity type makes holes the majority carriers in silicon?

A trivalent acceptor impurity produces p-type silicon, with holes as majority carriers and electrons as minority carriers.

Why does donor doping not make the whole semiconductor negatively charged?

The additional mobile electrons are balanced by positive ionised donor atoms, so the semiconductor remains electrically neutral overall.

What distinguishes drift from diffusion?

Drift is carrier motion caused by an electric field; diffusion is carrier motion caused by a concentration difference.

Which battery connections forward bias a diode?

Connect the p-side to the positive terminal and the n-side to the negative terminal, reducing the barrier.

Does zero net current at an unbiased junction mean diffusion has stopped?

No. At equilibrium, diffusion and drift currents continue with equal magnitudes and opposite directions, giving zero net current.

Why does full-wave rectification produce twice the input frequency in its output pulses?

Each input cycle supplies two conducting half-cycles, and both produce load-current pulses in the same direction.

Why is a series resistor required in a simple Zener regulator?

It limits current and takes up excess supply voltage while the reverse-biased Zener maintains approximately constant load voltage.