PhysicsISC Class 12 · Chapter 14

Electronic Devices ISC Class 12: Notes, Circuits and Solved Questions

Updated 14 October 2026 · Fact-checked

Electronic Devices is Chapter 14 of ISC Class 12 Physics. It covers energy bands in conductors, insulators and semiconductors, intrinsic and extrinsic (p-type and n-type) semiconductors, the p-n junction with its depletion region and potential barrier, forward and reverse bias characteristics, half-wave and full-wave rectifiers, and the Zener diode, LED, photodiode and solar cell.

What Electronic Devices is about

This chapter (7 of the 70 theory marks) explains how semiconductors work and how the junction diode is used. After the February 2026 revision it no longer includes transistors or logic gates.

Energy bands: when atoms come together in a solid their energy levels split into bands of closely spaced levels separated by forbidden gaps. In a conductor the valence and conduction bands overlap (no gap); in an insulator the gap is large (above about 3 eV; diamond about 5.5 eV); in a semiconductor it is small (Si 1.1 eV, Ge 0.7 eV), so heating frees some electrons into the conduction band, leaving holes. That is why a semiconductor's resistance falls as its temperature rises.

Intrinsic (pure) semiconductors have equal numbers of electrons and holes. Doping gives extrinsic semiconductors: a pentavalent donor (P, As) gives n-type (majority electrons); a trivalent acceptor (B, Al, In) gives p-type (majority holes). Both remain electrically neutral.

At a p-n junction, diffusion of carriers leaves a depletion region of immobile ions with no mobile charges, and a potential barrier (about 0.7 V for Si, 0.3 V for Ge). Forward bias (p to +) lowers the barrier and narrows the depletion region, so a large current flows beyond the knee voltage; reverse bias widens it and only a tiny minority-carrier current flows until breakdown. You need the V-I characteristics with simple numericals, and the half-wave and full-wave (centre-tap) rectifiers with circuit diagrams and waveforms (the bridge rectifier is not included).

Special diodes: the Zener diode works in reverse breakdown and is used as a voltage regulator; the LED emits light when forward biased (and saves energy); the photodiode is used in reverse bias to detect light; the solar cell turns light into electrical energy without any bias.

Weightage: Electronic Devices is part of the unit Electronic Devices, which carries 7 of the 70 marks in the ISC Physics Paper 1 (Theory). Questions on it can appear in any section of the paper.Source: CISCE ISC Physics syllabus, examination year 2027, distribution of marks

Key concepts

Energy bands: conductors, insulators and semiconductors: An isolated atom has sharp energy levels (1s, 2s, 2p, …). When N atoms form a crystal, each level splits into N closely spaced levels: an energy band. The highest filled band is the valence band; the next, which can carry current, is the conduction band; between them lies the forbidden band (energy gap) $E_{g}$ . Conductors (metals such as copper): the bands overlap or the conduction band is partly filled; there is no forbidden gap, so many free electrons conduct at all temperatures. Insulators (diamond, $E_{g} \approx 5.5$ eV): a wide gap, so practically no electrons reach the conduction band. Semiconductors (Si 1.1 eV, Ge 0.7 eV): a narrow gap; at 0 K they insulate, but at room temperature some electrons gain enough thermal energy to cross, each leaving a hole. More cross as temperature rises, so resistance falls.; Worked example
What is the longest wavelength of light that can create an electron-hole pair in (a) silicon ( $E_{g}$ = 1.1 eV) and (b) germanium (0.7 eV)? (hc = 1240 eV nm)
Try it first, then show the solutionSolution
A photon must supply at least $E_{g}$ : $λ_{ma x} = \frac{1240}{E _{g}}$
(a) $\frac{1240}{1.1} = 1127 nm$ (b) $\frac{1240}{0.7} = 1771 nm$
Answer: (a) about 1130 nm (b) about 1770 nm, both in the near infrared
Intrinsic and extrinsic semiconductors: In a pure (intrinsic) semiconductor every electron freed to the conduction band leaves a hole (a missing electron in a bond, behaving like a positive charge) in the valence band, so $n_{e} = n_{h} = n_{i}$ . Conduction is by both. Doping adds a tiny amount of impurity (about 1 part per million) to make an extrinsic semiconductor. n-type: a pentavalent impurity (P, As, Sb) uses four electrons in bonds and donates the fifth: a donor. Electrons are majority carriers, holes minority. p-type: a trivalent impurity (B, Al, In, Ga) leaves one bond incomplete, creating a hole: an acceptor. Holes are majority carriers, electrons minority. Both types remain electrically neutral, because each impurity atom is itself neutral. Doping increases conductivity enormously.; Worked example
Silicon has $5 \times 1 0^{28}$ atoms per m³ and an intrinsic carrier concentration of $1.5 \times 1 0^{16} m^{- 3}$ . It is doped with arsenic at 1 part per million. Find the number of donor atoms per m³ and compare the majority-carrier density with the intrinsic value. What type is it?
Try it first, then show the solutionSolution
Donors $= 5 \times 1 0^{28} \times 1 0^{- 6} = 5 \times 1 0^{22} m^{- 3}$ ; each gives one free electron, so $n_{e} \approx 5 \times 1 0^{22} m^{- 3}$ .
Ratio $= \frac{5 \times 1 0 ^{22}}{1.5 \times 1 0 ^{16}} \approx 3.3 \times 1 0^{6}$
Arsenic is pentavalent, so this is n-type.
Answer: $5 \times 1 0^{22} m^{- 3}$ ; about 3 million times the intrinsic density; n-type
The p-n junction: depletion region and potential barrier: When p-type and n-type regions are formed in one crystal, electrons diffuse from n to p and holes from p to n, recombining near the junction. This leaves a thin layer of uncovered immobile ions: positive donor ions on the n side and negative acceptor ions on the p side. This layer, empty of mobile charge carriers, is the depletion region (about 1 µm wide). The ions set up an electric field from n to p, and the corresponding potential difference, the potential barrier (about 0.7 V for Si and 0.3 V for Ge), opposes further diffusion of majority carriers. In equilibrium the diffusion current balances the drift current of minority carriers pushed across by the field.; Worked example
Define (a) the depletion region and (b) the potential barrier of a p-n junction. Which way does the junction field point?
Try it first, then show the solutionSolution
(a) The region near the junction depleted of mobile charge carriers (electrons and holes), containing only immobile positive and negative ions.
(b) The potential difference across the depletion region, due to these ions, which opposes the diffusion of majority carriers across the junction.
The field points from the positive ions (n side) to the negative ions (p side).
Answer: Field from n to p; definitions as stated
Forward and reverse bias; V-I characteristics: Forward bias: p side connected to the positive terminal. The applied voltage opposes the barrier, so the barrier height and depletion width decrease. Little current flows until the applied voltage exceeds the knee (cut-in) voltage (about 0.7 V for Si, 0.3 V for Ge); then current rises steeply (mA), carried by majority carriers. Reverse bias: p side to the negative terminal. The barrier rises and the depletion region widens; only a tiny reverse saturation current (µA) of minority carriers flows, nearly independent of voltage, until breakdown, when it rises sharply. So the diode conducts well in one direction only. Resistance from the graph: static R = V/I; dynamic $r = \frac{Δ V}{Δ I}$ (small in forward bias).; Worked example
A silicon diode (0.7 V forward drop) is connected in series with a 430 Ω resistor across a 5 V supply. Find the current (a) when forward biased and (b) when the diode is reversed.
Try it first, then show the solutionSolution
(a) $I = \frac{5 - 0.7}{430} = \frac{4.3}{430} = 0.01 A = 10 m A$
(b) Reverse biased: only the tiny reverse saturation current (a few µA at most) flows; effectively zero.
Answer: (a) 10 mA (b) practically zero
Half-wave and full-wave rectifiers: A rectifier converts a.c. into (pulsating, unidirectional) d.c. Half-wave: one diode in series with the load across a transformer secondary. In the positive half-cycle the diode is forward biased and conducts; in the negative half it is reverse biased and blocks. The output is a series of positive half-waves; its frequency equals the input frequency (50 Hz for mains). Full-wave (centre-tap): a centre-tapped transformer and two diodes, $D_{1}$ and $D_{2}$ , feeding a common load. In one half-cycle $D_{1}$ conducts, in the next $D_{2}$ , and the current through the load is in the same direction both times. The output consists of every half-cycle made positive; the ripple frequency is twice the input (100 Hz). Full-wave is more efficient (uses both halves). (A capacitor filter can smooth the output; the four-diode bridge rectifier is not included in the syllabus.); Worked example
The secondary of a transformer gives 12 V (r.m.s.) at 50 Hz to a half-wave rectifier and, separately, to each half of a centre-tap full-wave rectifier. Find the peak output (Si diodes, 0.7 V drop) and the output pulse frequency in each case.
Try it first, then show the solutionSolution
Peak input $= 122 = 16.97 V$ ; peak output $\approx 16.97 - 0.7 = 16.3 V$ in both cases.
Half-wave: one pulse per cycle: 50 Hz. Full-wave: two pulses per cycle: 100 Hz.
Answer: About 16.3 V peak; 50 Hz (half-wave), 100 Hz (full-wave)
Zener diode as a voltage regulator: A Zener diode is a heavily doped p-n junction designed to operate in reverse breakdown. Its V-I curve shows a sharp breakdown at the Zener voltage $V_{Z}$ , after which the voltage stays almost constant over a wide range of current. Regulator: the Zener is connected in reverse bias in parallel with the load, with a series resistor $R_{S}$ between the unregulated supply and this pair. If the input voltage rises (or the load draws less current), the Zener current increases, the extra voltage drops across $R_{S}$ , and the load voltage stays at $V_{Z}$ . Current balance: $I_{S} = I_{Z} + I_{L}$ , with $I_{S} = \frac{V _{in} - V _{Z}}{R _{S}}$ and $I_{L} = \frac{V _{Z}}{R _{L}}$ .; Worked example
A 6 V Zener diode regulates a 15 V supply through a 300 Ω series resistor, feeding a 1 kΩ load. Find the current through the series resistor, the load and the Zener.
Try it first, then show the solutionSolution
$I_{S} = \frac{15 - 6}{300} = 0.03 A = 30 m A$
$I_{L} = \frac{6}{1000} = 6 m A$
$I_{Z} = 30 - 6 = 24 m A$
Answer: 30 mA; 6 mA; 24 mA
LED, photodiode and solar cell: Light-emitting diode (LED): a forward-biased p-n junction (of materials such as GaAs, GaAsP) in which electrons and holes recombine and release energy as light of photon energy about $E_{g}$ ; the band gap sets the colour ( $λ = h c / E_{g}$ ). LEDs need low voltage, switch fast, last long and convert far more of their energy to light than filament lamps, saving energy without atmospheric pollution or adding to global warming. Photodiode: operated in reverse bias; light of energy above $E_{g}$ creates electron-hole pairs in the depletion region, and the field sweeps them across, so the small reverse current increases with light intensity (used to detect light, e.g. in remote-control receivers). Solar cell: a p-n junction with a large area and no external bias; light creates electron-hole pairs, the junction field separates them, and an emf (about 0.5-0.6 V for Si) drives current through a load. Its V-I curve lies in the fourth quadrant.; Worked example
A red LED emits light of wavelength 653 nm. Estimate the band gap of its semiconductor. What colour would a material with a band gap of 2.0 eV emit?
Try it first, then show the solutionSolution
$E_{g} \approx \frac{1240}{653} = 1.9 e V$
For 2.0 eV: $λ = \frac{1240}{2.0} = 620 nm$ , orange-red.
Answer: About 1.9 eV; 620 nm (orange-red)

Physics in real life: Electronic Devices

Each concept, met in an everyday situation. Read the situation, try the problem, then check the solution. The one-line hook is what to remember.

Energy saving with LEDs

Suppose a family replaces a 60 W filament bulb with a 9 W LED lamp giving the same light, used 5 hours a day.

How much energy and money does this save in a 30-day month at ₹7 per unit?

Try it first, then show the solutionSolution

Power saved = 60 − 9 = 51 W; energy saved $= 51 \times 5 \times 30 = 7650 W h = 7.65 k W h$
Money saved $= 7.65 \times 7 = ₹53.55$

Answer: About 7.65 units and ₹54 a month for one lamp, with less heat and less pollution from power stations.

Memory hook: LED: same light, a sixth of the power.

Solar cells in series

Imagine a small rooftop solar panel made of 36 silicon cells connected in series, each giving 0.5 V and up to 3 A in sunlight.

What voltage and maximum power can the panel supply?

Try it first, then show the solutionSolution

In series the voltages add: $36 \times 0.5 = 18 V$
Current stays 3 A, so $P = 18 \times 3 = 54 W$

Answer: About 18 V and 54 W, enough to charge a 12 V battery.

Memory hook: Cells in series add volts, not amps.

Zener voltage regulation

Suppose a small circuit needs a steady 6 V, but the supply from a battery varies between 12 V and 18 V. A 6 V Zener with a 300 Ω series resistor feeds a load drawing 6 mA.

Find the Zener current at the lowest and highest supply voltages.

Try it first, then show the solutionSolution

At 12 V: $I_{S} = \frac{12 - 6}{300} = 20 m A$ , so $I_{Z} = 20 - 6 = 14 m A$
At 18 V: $I_{S} = \frac{18 - 6}{300} = 40 m A$ , so $I_{Z} = 34 m A$

Answer: 14 mA to 34 mA: the Zener absorbs the change and the load stays at 6 V.

Memory hook: The Zener soaks up the extra so the load never feels it.

Topics in this chapter

Electronic Devices formula sheet

Band gaps (typical): $E_{g} (S i) \approx 1.1 e V, E_{g} (G e) \approx 0.7 e V, E_{g} (C) \approx 5.5 e V$; Conductors: no gap.
Photon needed to create a pair: $\frac{h c}{λ} \geq E_{g}, λ_{ma x} = \frac{1240}{E _{g} ( e V )} nm$; Also gives LED wavelength.
Intrinsic semiconductor: $n_{e} = n_{h} = n_{i}$; Extrinsic: n-type $n_{e} ≫ n_{h}$ ; p-type $n_{h} ≫ n_{e}$ .
Barrier potential (typical): $V_{B} \approx 0.7 V (S i), 0.3 V (G e)$; Also the knee voltage in forward bias.
Diode in series with R: $I = \frac{V - V _{D}}{R}$; $V_{D}$ = forward drop of the diode.
Diode resistance: $R_{s t a t i c} = \frac{V}{I}, r_{d y nami c} = \frac{Δ V}{Δ I}$; Read from the V-I graph.
Rectifier output frequency: $f_{ha l f - w a v e} = f_{in}, f_{f u l l - w a v e} = 2 f_{in}$; Mains: 50 Hz and 100 Hz.
Zener regulator: $I_{S} = \frac{V _{in} - V _{Z}}{R _{S}}, I_{L} = \frac{V _{Z}}{R _{L}}, I_{Z} = I_{S} - I_{L}$; Load voltage stays at $V_{Z}$ .

Electronic Devices: diagrams to practise

Practise each one on paper: draw it from the checklist, then tick off every label.

Energy band diagrams

Distinguishing conductors, insulators and semiconductors; 'forbidden band is absent: identify the material'.

Draw

Three side-by-side sketches with energy on the vertical axis.
Conductor: valence and conduction bands overlapping. Insulator: valence band (filled), a wide forbidden gap (> 3 eV), empty conduction band.
Semiconductor: narrow gap (about 1 eV) with a few electrons in the conduction band and holes in the valence band.

Label

conduction band
valence band
forbidden gap $E_{g}$
electrons (dots), holes (circles)
energy axis

Watch out: No forbidden band = conductor; that one-word answer was often confused with insulator.

Diode V-I characteristic

'Draw the I-V characteristic of a junction diode in forward (and reverse) bias'.

Draw

Axes: V horizontal (forward to the right, reverse to the left), I vertical (forward current in mA up, reverse current in µA down).
Forward: nearly flat close to the axis up to the knee voltage (mark 0.7 V for Si), then rising steeply.
Reverse: a tiny, almost constant current just below the axis, then a sharp drop at the breakdown voltage.

Label

V (volts)
I (mA forward, µA reverse)
knee voltage
reverse saturation current
breakdown voltage

Watch out: Do not start the forward curve rising from the origin; mark the knee voltage and the different current scales.

Full-wave (centre-tap) rectifier with waveforms

'Draw a labelled diagram of a full-wave rectifier and show the input and output waveforms'.

Draw

A transformer with a centre-tapped secondary; ends A and B each connect to a diode ( $D_{1}$ , $D_{2}$ ); the diode outputs join at one end of the load $R_{L}$ ; the other end of $R_{L}$ goes to the centre tap.
Mark the direction of current through $R_{L}$ (the same in both half-cycles).
Below: input sine wave against time; output: every half-cycle above the axis (no negative parts).

Label

a.c. input
centre tap
$D_{1}$ , $D_{2}$
load $R_{L}$
input and output waveforms with axes
current direction

Watch out: Include the load resistor and label both waveforms; a single-diode circuit is half-wave, not full-wave.

Zener diode voltage regulator

'Draw a circuit showing a Zener diode as a voltage regulator and explain its working'.

Draw

Unregulated d.c. input; series resistor $R_{S}$ ; Zener diode connected in reverse bias (cathode to the positive line) across the output.
Load $R_{L}$ in parallel with the Zener.
Mark $I_{S}$ , $I_{Z}$ and $I_{L}$ , and $V_{Z}$ across the load.

Label

$V_{in}$
$R_{S}$
Zener (reverse biased)
$R_{L}$
$I_{S} = I_{Z} + I_{L}$
$V_{Z}$

Watch out: The Zener must be reverse biased; a forward-biased Zener just behaves as an ordinary diode.

How the board asks Electronic Devices

The question patterns to expect, and the method that works for each.

Question type	How it is asked	Method
One-mark facts	Majority carriers in p- or n-type; effect of temperature on a semiconductor's resistance; material with no forbidden band; function of a rectifier.	p-type: holes (not protons); resistance falls with temperature; no gap: conductor; rectifier: a.c. to d.c.
Definitions	Depletion region, potential barrier, forward bias, doping, donor and acceptor.	Use key words: immobile ions, no mobile carriers, opposes diffusion, p side to positive.
Characteristics and simple numericals	Draw forward/reverse V-I; current in a diode-resistor circuit; static and dynamic resistance.	Subtract the diode drop; read ΔV/ΔI from the graph.
Rectifiers (3-5 marks)	Circuit and waveforms of half-wave and full-wave rectifiers; output frequency; what happens if a diode fails.	Diagram with load, input and output graphs; full-wave gives twice the input frequency.
Special diodes	Zener regulator circuit and working; LED colour from band gap; why photodiodes are reverse biased; solar-cell working.	Tie each device to its bias: Zener reverse (breakdown), LED forward, photodiode reverse, solar cell none.
Case-based (Section D)	Components of a device such as a television or charger: p and n types, rectifier diagram, which diode regulates voltage.	Answer each part with the device name and one-line reason.

Tricks to remember Electronic Devices

Short memory aids. Say each one aloud once, then use it on a question the same day.

P is Positive holes, N is Negative electrons
p-type: majority holes (positive); n-type: majority electrons (negative). Both stay neutral overall.
Use it when: Majority-carrier questions (holes are not protons).
PENTA gives, TRI takes
Pentavalent = donor (gives an electron, n-type); trivalent = acceptor (takes an electron, p-type).
Use it when: Doping questions.
Forward: P to Plus
Forward bias: p side to the positive terminal; barrier lowers, depletion thins.
Use it when: Defining forward bias completely.
Full-wave doubles the beat
Half-wave output: 50 Hz; full-wave: 100 Hz.
Use it when: Rectifier frequency items.
Bias by device
LED forward; photodiode and Zener reverse; solar cell none.
Use it when: Special-diode questions.

Electronic Devices: board-level solved questions

Ordered from board level to HOTS. Cover the solution, write your own, then compare step by step.

Example 1: Majority carriers in n-type · 1 mark · MCQBoard level

In an n-type semiconductor:
(a) holes are the majority carriers and it is negatively charged
(b) electrons are the majority carriers and it is electrically neutral
(c) electrons are the majority carriers and it is negatively charged
(d) a trivalent impurity is added

Try it first, then show the solutionSolution

A pentavalent donor gives free electrons (majority), but each donor atom is neutral, so the crystal stays neutral.

Answer: (b)

Example 2: Semiconductors and temperature · 1 mark · Assertion-reasonBoard level

Assertion: The resistance of a semiconductor decreases as its temperature increases.
Reason: More electrons gain enough energy to cross the forbidden gap into the conduction band, creating more electron-hole pairs.
(a) Both Assertion and Reason are true, and Reason is the correct explanation for Assertion.
(b) Both Assertion and Reason are true, but Reason is not the correct explanation for Assertion.
(c) Assertion is true and Reason is false.
(d) Assertion is false and Reason is true.

Try it first, then show the solutionSolution

More charge carriers means higher conductivity, i.e. lower resistance. The reason explains the assertion.

Answer: (a)

Example 3: Diode and resistor in series · 2 marks · Short answerHarder board level

A silicon diode (forward drop 0.7 V) and a 430 Ω resistor are in series with a 5 V battery. Find the current when the diode is forward biased. What happens if the battery is reversed? [2]

Try it first, then show the solutionSolution

$I = \frac{5 - 0.7}{430} = 0.01 A = 10 m A$
Reversed: the diode is reverse biased; only a negligible reverse current (µA or less) flows.

Answer: 10 mA; practically zero when reversed

Example 4: Biasing a p-n junction · 3 marks · Short answerHarder board level

Explain how (a) forward bias and (b) reverse bias affect the depletion region and the potential barrier of a p-n junction. Sketch the V-I characteristic for forward bias, marking the knee voltage for silicon. [3]

Try it first, then show the solutionSolution

(a) Forward bias (p to +, n to −): the applied voltage opposes the junction field, so the barrier height falls and the depletion region narrows; majority carriers cross easily.
(b) Reverse bias: the applied voltage adds to the junction field; the barrier rises and the depletion region widens; only a tiny minority-carrier current flows.
Graph: current nearly zero up to about 0.7 V, then rising steeply; axes V (V) and I (mA).

Answer: Forward: thinner, lower; reverse: wider, higher; knee about 0.7 V

Example 5: Full-wave rectifier · 3 marks · Short answerHarder board level

Draw a labelled circuit of a full-wave rectifier using two junction diodes, explain its working, and state the output frequency for a 50 Hz input. [3]

Try it first, then show the solutionSolution

Circuit: centre-tapped transformer; ends to diodes $D_{1}$ and $D_{2}$ ; their outputs joined to one end of $R_{L}$ ; the other end to the centre tap.
In one half-cycle end A is positive: $D_{1}$ is forward biased and conducts through $R_{L}$ ; $D_{2}$ is reverse biased. In the next half-cycle B is positive: $D_{2}$ conducts, and current flows through $R_{L}$ in the same direction.
Both halves give output pulses in the same direction, so the output frequency is $2 \times 50 = 100 H z$ .

Answer: Unidirectional pulses at 100 Hz

Example 6: Zener diode as a voltage regulator · 5 marks · Long answerHarder board level

(a) Draw the V-I characteristic of a Zener diode and mark the Zener voltage. [1]
(b) Draw a circuit using a Zener diode as a voltage regulator and explain how it keeps the output steady when the input rises. [2]
(c) A 6 V Zener with a 300 Ω series resistor is fed from 15 V and supplies a 1 kΩ load. Find the Zener current. [2]

Try it first, then show the solutionSolution

(a) Forward part like an ordinary diode; in reverse, tiny current until $V_{Z}$ , then a sharp, almost vertical rise at constant voltage $V_{Z}$ .
(b) Series resistor $R_{S}$ from the input; Zener reverse biased in parallel with $R_{L}$ .
If $V_{in}$ rises, the current through $R_{S}$ rises; the Zener takes the extra current while its voltage stays $V_{Z}$ , so the extra voltage drops across $R_{S}$ and the load voltage is unchanged.
(c) $I_{S} = \frac{15 - 6}{300} = 30 m A$ ; $I_{L} = \frac{6}{1000} = 6 m A$
$I_{Z} = 30 - 6 = 24 m A$

Answer: (c) 24 mA

Example 7: Classifying materials by band gap · 3 marks · Short answerHOTS / competency

Materials X, Y and Z have energy gaps of 0 eV (bands overlap), 1.1 eV and 5.5 eV. (a) Classify each. (b) What is the longest wavelength that can free an electron into the conduction band of Y? (c) Why does Z look transparent to visible light? [3]

Try it first, then show the solutionSolution

(a) X: conductor; Y: semiconductor (silicon); Z: insulator (diamond).
(b) $λ_{ma x} = \frac{1240}{1.1} = 1127 nm$ .
(c) Visible photons have 1.8-3.1 eV, less than 5.5 eV, so they cannot raise electrons across the gap and are not absorbed.

Answer: (a) conductor, semiconductor, insulator (b) about 1130 nm (c) visible photons lack the energy to cross the gap

Example 8: Why a photodiode is reverse biased · 2 marks · Short answerHOTS / competency

Explain why a photodiode is operated in reverse bias rather than forward bias. [2]

Try it first, then show the solutionSolution

Light creates extra electron-hole pairs, which changes the number of minority carriers a lot (in proportion) but the majority carriers hardly at all.
In reverse bias the current is carried by minority carriers, so it changes noticeably with light intensity; in forward bias the large majority current would hide the change.

Answer: The fractional change in reverse (minority) current with light is large and easy to measure

Case-study practice (Section D style)

Read the situation, then answer every part. In the board paper this is a 5-mark question split into 1 + 1 + 1 + 2 marks.

Inside a phone charger

Suppose an older phone charger contains a step-down transformer with a centre-tapped secondary (12 V r.m.s. on each half), two silicon diodes and a load, fed from 50 Hz mains.

(i) · 1 mark
What is the function of the diodes in the charger?
Answer
Answer: To rectify: convert a.c. into unidirectional (d.c.) current.
Each diode conducts in only one direction, so current through the load always flows the same way.
(ii) · 1 mark
What type of rectifier is this, and what is the frequency of its output pulses?
Answer
Answer: Full-wave (centre-tap); 100 Hz
Two diodes conduct alternately, so there are two output pulses per input cycle.
(iii) · 1 mark
Estimate the peak output voltage.
Answer
Answer: About 16.3 V
$122 - 0.7 = 16.97 - 0.7$
(iv) · 2 marks
One diode burns out and becomes an open circuit. What is the output now? Sketch the waveform in words.
Answer
Answer: Half-wave output at 50 Hz.
Only the remaining diode conducts, in alternate half-cycles; the output is one positive pulse per cycle with gaps in between.

Holding a voltage steady

Suppose a hobby circuit needs a steady 6 V. The designer uses a 6 V Zener diode and a 300 Ω series resistor, fed from a supply that may vary from 12 V to 18 V. The load draws 6 mA.

(i) · 1 mark
In which bias is the Zener connected, and why?
Answer
Answer: Reverse bias.
Its useful property is the nearly constant voltage $V_{Z}$ in reverse breakdown.
(ii) · 1 mark
Find the Zener current when the supply is 12 V.
Answer
Answer: 14 mA
$I_{S} = 6/300 = 20 m A$ ; $I_{Z} = 20 - 6 = 14 m A$ .
(iii) · 1 mark
Find the Zener current when the supply is 18 V.
Answer
Answer: 34 mA
$I_{S} = 12/300 = 40 m A$ ; $I_{Z} = 34 m A$ .
(iv) · 2 marks
The load is changed to draw 25 mA at 12 V supply. Will regulation still work? Explain.
Answer
Answer: No.
$I_{S}$ is only 20 mA, less than the 25 mA the load needs; the Zener current would have to be negative, so it leaves breakdown and the output falls below 6 V.

Where the marks are

How a typical answer is split into step marks. Show every step: a right final answer alone rarely earns full marks.

With reference to a p-n junction, define (i) depletion region and (ii) potential barrier. (2 marks)

Step	Marks
(i) Region near the junction with no mobile charge carriers, containing only immobile ions	1
(ii) Potential difference across the junction (due to the immobile ions) that opposes diffusion of majority carriers	1

What is meant by forward biasing? Draw a labelled I-V characteristic for a forward-biased diode. (3 marks)

Step	Marks
p side connected to the positive terminal and n side to the negative terminal of the source	1
Axes labelled (V in volts, I in mA); curve stays near zero up to the knee voltage	1
Knee voltage marked (≈ 0.7 V for Si or 0.3 V for Ge) and steep rise beyond it	1

Explain the formation of the depletion layer, draw a full-wave rectifier circuit with input and output waveforms, and state why the output frequency is twice the input frequency. (5 marks)

Step	Marks
Diffusion of electrons and holes across the junction and recombination	1
Uncovered immobile ions form the depletion layer and barrier	1
Correct centre-tap circuit with two diodes and load	1
Input sine wave and full-wave output waveform with axes	1
Each half-cycle gives an output pulse, so two pulses per cycle	1

Where students lose marks in Electronic Devices

The slips examiners see most often, and what to write instead.

Mistake: Answering 'protons' for majority carriers in p-type
CISCE examiners' comments, 2024 and 2025: candidates chose protons for p-type, holes for n-type, or confused trivalent and pentavalent impurities.
Wrong working
p-type: majority carriers are protons.
Right working
p-type: holes (missing electrons); n-type: free electrons.
Do this: Remember p = positive holes, n = negative electrons; protons never move in a solid.
Usually costs: 1 mark.
Mistake: Incomplete definitions of depletion region and barrier
CISCE examiners' comments, 2024: candidates left out 'immobile ions' or 'absence of mobile charge carriers' and defined the potential barrier only as something that 'obstructs flow'.
Wrong working
Depletion region is the junction region; barrier stops current.
Right working
Depletion region: near the junction, no mobile carriers, only immobile ions. Barrier: p.d. across it opposing diffusion of majority carriers.
Do this: Learn both definitions with their key phrases.
Usually costs: 1 mark each.
Mistake: Forward-bias graph rising from the origin
CISCE examiners' comments, 2024: forward-bias I-V graphs started rising from the origin, had swapped or unlabelled axes, no knee voltage marked, or the wrong shape.
Wrong working
A straight line through the origin.
Right working
Flat up to the knee (0.7 V Si, 0.3 V Ge), then rising steeply; axes labelled V and I with units.
Do this: Always mark the knee voltage.
Usually costs: 1-2 marks.
Mistake: Half-wave circuit drawn for a full-wave question
CISCE examiners' comments, 2025: candidates drew a half-wave instead of a full-wave rectifier, omitted the load resistor, left the circuit incomplete or mislabelled the input and output waveforms; some named the wrong diode as a voltage regulator.
Wrong working
One diode and no load; output drawn as a full sine wave.
Right working
Centre-tapped transformer, two diodes, load $R_{L}$ ; output: all half-cycles on one side. Regulator: Zener diode.
Do this: Check the number of diodes and include $R_{L}$ every time.
Usually costs: 1-3 marks.
Mistake: Confusing conductors with insulators on band diagrams
CISCE examiners' comments, 2025: asked to identify a material whose band diagram has no forbidden band, candidates confused conductor with insulator; others said a semiconductor's resistance rises with temperature.
Wrong working
No forbidden band: insulator. Semiconductor resistance ∝ temperature.
Right working
No forbidden gap: conductor (bands overlap). A semiconductor's resistance falls as temperature rises.
Do this: Gap size decides: none = conductor, small = semiconductor, large = insulator.
Usually costs: 1 mark.
Mistake: Confusing a rectifier with a transformer
CISCE examiners' comments, 2024: asked for the function of a rectifier, candidates gave the function of a transformer or drew waveforms instead of stating it.
Wrong working
A rectifier steps down the voltage.
Right working
A rectifier converts alternating current into direct (unidirectional) current.
Do this: Transformer changes a.c. voltage; rectifier changes a.c. to d.c.
Usually costs: 1 mark.

Practice questions on Electronic Devices

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

Q1 · 1 mark · MCQBoard level
The energy gap of an insulator is typically:
(a) zero
(b) about 0.7 eV
(c) about 1.1 eV
(d) more than 3 eV
Hint
Diamond ≈ 5.5 eV.
Answer
(d)
Q2 · 1 mark · MCQBoard level
In forward bias, the width of the depletion region of a p-n junction:
(a) increases
(b) decreases
(c) stays the same
(d) becomes infinite
Hint
The barrier is lowered.
Answer
(b)
Q3 · 1 mark · MCQBoard level
A Zener diode used as a voltage regulator works in:
(a) forward bias below the knee
(b) forward bias above the knee
(c) reverse bias below breakdown
(d) reverse breakdown
Hint
Constant voltage region.
Answer
(d)
Q4 · 1 mark · Very short answerBoard level
What is the function of a rectifier?
Hint
a.c. to …
Answer
It converts alternating current into direct (unidirectional) current.
Q5 · 2 marks · Short answerBoard level
Distinguish between n-type and p-type semiconductors (impurity used and majority carriers).
Hint
Donor versus acceptor.
Answer
n-type: pentavalent (donor) impurity such as P or As; majority carriers electrons. p-type: trivalent (acceptor) impurity such as B or Al; majority carriers holes.
Q6 · 2 marks · Short answerHarder board level
An LED emits light of wavelength 620 nm. Find the band gap of its material in eV.
Hint
$E_{g} = 1240/ λ$ .
Answer
$E_{g} = 1240/620 = 2.0$ eV
Q7 · 2 marks · Short answerHarder board level
The input to a rectifier is 50 Hz a.c. State the frequency of the output pulses for (a) a half-wave and (b) a full-wave rectifier, and explain the difference.
Hint
Pulses per cycle.
Answer
(a) 50 Hz: one pulse per cycle. (b) 100 Hz: both half-cycles produce pulses in the same direction.
Q8 · 2 marks · Short answerHOTS / competency
Explain how a solar cell produces an emf without any external bias. In which quadrant does its V-I characteristic lie?
Hint
Built-in junction field.
Answer
Light creates electron-hole pairs near the junction; the built-in field sweeps electrons to the n side and holes to the p side, building up a p.d. that drives current through a load. Its V-I curve lies in the fourth quadrant (current flows opposite to forward current while V is positive).
Q9 · 3 marks · Short answerHarder board level
Silicon ( $5 \times 1 0^{28}$ atoms m⁻³, $n_{i} = 1.5 \times 1 0^{16} m^{- 3}$ ) is doped with 1 ppm of arsenic. Find the electron density and the ratio to the intrinsic value. Name the majority and minority carriers.
Hint
One electron per donor.
Answer
$n_{e} \approx 5 \times 1 0^{22} m^{- 3}$ ; ratio about $3.3 \times 1 0^{6}$ . Majority: electrons; minority: holes (n-type).
Q10 · 3 marks · Short answerHarder board level
A 6 V Zener with a 300 Ω series resistor is fed from 15 V. Find the Zener current when the load is (a) 1 kΩ and (b) 600 Ω.
Hint
$I_{Z} = I_{S} - I_{L}$ .
Answer
$I_{S} = 30$ mA in both. (a) $I_{L} = 6$ mA, $I_{Z} = 24$ mA (b) $I_{L} = 10$ mA, $I_{Z} = 20$ mA.
Q11 · 3 marks · Short answerHarder board level
Explain the formation of the depletion region and potential barrier at a p-n junction, including the roles of diffusion and drift.
Hint
Majority carriers diffuse; the field drives minority carriers.
Answer
Electrons diffuse from n to p and holes from p to n, recombining near the junction and leaving immobile donor (+) and acceptor (−) ions: the depletion region. These ions create a field (n to p) and a barrier potential that opposes further diffusion. The same field drifts minority carriers across; equilibrium is reached when diffusion and drift currents are equal.
Q12 · 5 marks · Long answerHarder board level
(a) Draw energy band diagrams for a conductor, an insulator and a semiconductor. [2]
(b) Distinguish between intrinsic and extrinsic semiconductors. [2]
(c) Find the longest wavelength that can produce electron-hole pairs in germanium ( $E_{g}$ = 0.7 eV). [1]
Hint
Gap sizes.
Answer
(a) Conductor: overlapping bands; insulator: wide gap (> 3 eV); semiconductor: narrow gap (≈ 1 eV). (b) Intrinsic: pure, $n_{e} = n_{h}$ , low conductivity. Extrinsic: doped, one type of carrier in majority, much higher conductivity; n-type or p-type. (c) $λ = 1240/0.7 = 1771$ nm.
Q13 · 2 marks · Short answerHarder board level
From a diode's forward characteristic, the current is 20 mA at 0.8 V and rises from 10 mA to 30 mA as V goes from 0.75 V to 0.85 V. Find the static resistance at 0.8 V and the dynamic resistance in this range.
Hint
V/I and ΔV/ΔI.
Answer
Static: 0.8/0.02 = 40 Ω. Dynamic: 0.10/0.020 = 5 Ω.

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1.What is a forbidden energy gap?Show answer
The range of energies between the valence and conduction bands that electrons cannot have.
2.What is doping?Show answer
Adding a small, controlled amount of impurity to a pure semiconductor to increase its conductivity.
3.Majority carriers in p-type and n-type?Show answer
Holes in p-type; electrons in n-type.
4.Barrier potential of Si and Ge?Show answer
About 0.7 V and 0.3 V.
5.What is forward bias?Show answer
p side connected to the positive terminal and n side to the negative terminal.
6.Output frequency of a full-wave rectifier on 50 Hz mains?Show answer
100 Hz.
7.Use of a Zener diode?Show answer
As a voltage regulator (in reverse breakdown).
8.Why are LEDs preferred to filament lamps?Show answer
They use much less power for the same light, last longer and switch faster, saving energy and reducing pollution.

Best order to study it

Learn energy bands and classify materials by gap size.
Learn intrinsic semiconductors, holes, and n-type and p-type doping.
Understand the p-n junction: diffusion, depletion region, barrier.
Learn forward and reverse bias and draw the V-I characteristic; practise simple diode numericals.
Draw half-wave and full-wave rectifiers with waveforms.
Finish with the Zener regulator, LED, photodiode and solar cell.

Quick revision checklist

The night before the exam: can you do each of these without looking?

Can you draw band diagrams for a conductor, an insulator and a semiconductor?
Can you explain why a semiconductor's resistance falls when heated?
Can you state the impurity and majority carriers for n-type and p-type?
Can you define depletion region and potential barrier with key words?
Can you draw the forward and reverse V-I curve with the knee marked?
Can you draw a full-wave rectifier with input and output waveforms?
Can you draw a Zener regulator and calculate $I_{Z}$ ?
Can you state the bias and use of the LED, photodiode and solar cell?

Electronic Devices: frequently asked questions

Are transistors and logic gates in the ISC 2027 syllabus?

No. The junction transistor, transistor as an amplifier and logic gates were removed in the February 2026 revision. This chapter now covers semiconductors, the junction diode, rectifiers and special diodes.

Is the bridge rectifier required?

No. The syllabus says the bridge rectifier of four diodes is not included. Learn the half-wave rectifier and the two-diode (centre-tap) full-wave rectifier.

Do I need numericals on diodes?

Yes, simple ones: current through a diode and resistor, static and dynamic resistance from a characteristic, rectifier output frequency and Zener regulator currents.

Why are LEDs described as environmentally friendly?

They convert a much larger share of electrical energy into light, so they need less power from power stations for the same lighting, which reduces fuel burnt, atmospheric pollution and contribution to global warming. The syllabus specifically mentions this.

What is the difference between a photodiode and a solar cell?

Both are p-n junctions lit by light. A photodiode is reverse biased by a battery and used to detect light (its current depends on intensity). A solar cell has no bias and a large area; it generates an emf and supplies power.

How many marks does this chapter carry?

Electronic Devices carries 7 of the 70 theory marks; a recent paper set a 5-mark case-based question on diodes and the full-wave rectifier.

Based on: CISCE ISC 2027 Physics syllabus (revised, uploaded February 2026); CISCE ISC 2027 Physics specimen Paper 1 (Theory) question paper and answer key; CISCE Analysis of Pupil Performance, ISC Physics, 2025; CISCE Analysis of Pupil Performance, ISC Physics, 2024