Class 12Physics · Electronic DevicesFull chapter

Semiconductor Electronics

The whole chapter in one place — read it, then test yourself. Clear notes, formula sheet, a practice quiz, and worked NCERT solutions & PYQs.

Energy Bands, Intrinsic and Extrinsic Semiconductors

Quick answer Classifies solids as conductors, insulators and semiconductors using energy-band theory, then explains how doping creates n-type and p-type semiconductors.

When a very large number of atoms come together to form a solid, the sharply defined energy levels of an isolated atom split and merge into continuous energy bands. The highest band that is normally filled with electrons is the valence band; the next higher band, where electrons can move freely and carry current, is the conduction band. Between them lies the forbidden energy gap, Eg, a range of electron energies that cannot exist in the crystal.

Solids are classified by the size of this gap. In metals the valence and conduction bands overlap (or the conduction band is partly filled), so electrons conduct freely even at low temperature. In insulators, Eg is large (typically above 3 eV; diamond has Eg ≈ 5.4 eV), so essentially no electron is thermally excited across it at room temperature. Semiconductors have a small gap (silicon ≈ 1.1 eV, germanium ≈ 0.7 eV) — small enough that some electrons gain enough thermal energy at room temperature to jump into the conduction band, leaving behind vacant bonds called holes in the valence band. A hole behaves as a mobile, effectively positive charge carrier.

In a chemically pure crystal, called an intrinsic semiconductor, every thermally excited electron leaves exactly one hole behind, so the electron concentration ne equals the hole concentration nh, both equal to the intrinsic concentration ni. Conductivity of an intrinsic semiconductor is low and rises rapidly with temperature, unlike a metal whose resistance rises with temperature.

Useful semiconductor behaviour comes from doping — deliberately adding a controlled trace of impurity atoms. Adding a pentavalent impurity (phosphorus, arsenic, antimony) to silicon or germanium gives an n-type semiconductor: four of the five valence electrons of each impurity atom form covalent bonds with neighbouring Si/Ge atoms, and the fifth electron is very loosely bound, occupying a donor energy level just below the conduction band. This electron is easily donated to the conduction band, so electrons become the majority carriers and holes the minority carriers. Adding a trivalent impurity (boron, aluminium, indium) instead creates a p-type semiconductor: the impurity atom can complete only three covalent bonds, leaving one bond incomplete, which readily accepts an electron from a neighbouring atom. This creates an acceptor energy level just above the valence band and generates a mobile hole, so holes become the majority carriers and electrons the minority carriers. In both cases the crystal remains electrically neutral overall — the extra carriers are mobile, but the impurity ions left behind are fixed in the lattice.

Even in a doped (extrinsic) semiconductor, the product of electron and hole concentrations remains fixed at a given temperature: nenh = ni2, known as the law of mass action.

Worked Example: Silicon is doped with a donor concentration Nd = 1×1020 m-3, with ni = 1.5×1016 m-3 at 300 K. Since Nd ≫ ni, almost every donor atom is ionised, so ne ≈ Nd = 1×1020 m-3. Using the law of mass action, the hole concentration is nh = ni2/ne = (1.5×1016)2 / (1×1020) = 2.25×1012 m-3. Electrons outnumber holes by nearly eight orders of magnitude, confirming this is a strongly n-type sample.

Energy Gap Eg = Ec − Ev eV · Forbidden gap between the bottom of the conduction band (Ec) and top of the valence band (Ev); Eg(Si) ≈ 1.1 eV, Eg(Ge) ≈ 0.7 eV, Eg(insulator) > 3 eV
Law of Mass Action ni² = nₑ × nₕ Product of electron and hole concentrations equals the square of the intrinsic concentration, for intrinsic or doped material at a given temperature
Conductivity of a Semiconductor σ = e(nₑμₑ + nₕμₕ) S m⁻¹ · μₑ, μₕ are electron and hole mobilities; e is the electronic charge
n-type Approximation nₑ ≈ Nd (when Nd ≫ ni) Electron concentration approximately equals donor concentration when donor doping dominates
p-type Approximation nₕ ≈ Na (when Na ≫ ni) Hole concentration approximately equals acceptor concentration when acceptor doping dominates
Remember
  • Solids are classified by the forbidden energy gap Eg between the valence and conduction bands: metals (no gap/overlap), semiconductors (small Eg ~1 eV), insulators (large Eg > 3 eV)
  • Intrinsic (pure) semiconductor: thermally generated electron-hole pairs give ne = nh = ni, and conductivity rises with temperature
  • Doping with a pentavalent impurity (P, As, Sb) gives an n-type semiconductor with electrons as majority carriers
  • Doping with a trivalent impurity (B, Al, In) gives a p-type semiconductor with holes as majority carriers
  • The law of mass action, nenh = ni^2, holds for both intrinsic and doped semiconductors at a given temperature
  • The semiconductor crystal remains overall electrically neutral even though majority and minority carrier concentrations differ hugely

p-n Junction Diode: Formation and V-I Characteristics

Quick answer Explains how a p-n junction forms a depletion region and potential barrier, and how forward and reverse bias change the current through the diode.

A p-n junction is formed when a p-type and an n-type region are created within a single continuous semiconductor crystal (commonly by diffusing acceptor and donor impurities from opposite faces). Near the junction, the sharp concentration difference causes majority carriers to diffuse across it: electrons diffuse from the n-side into the p-side and holes diffuse from the p-side into the n-side. As they cross, electrons and holes recombine near the junction, leaving behind fixed, uncompensated ion cores — positive donor ions on the n-side and negative acceptor ions on the p-side. This region, stripped of mobile charge carriers, is the depletion region.

The exposed ion charges set up an internal electric field pointing from the n-side to the p-side, which opposes further diffusion of majority carriers (though it assists a small drift of minority carriers across the junction in the opposite sense). At equilibrium, diffusion and drift currents balance, and the field corresponds to an internal potential barrier V0 (about 0.7 V for silicon, 0.3 V for germanium at room temperature).

When an external voltage is applied, the diode's behaviour depends on the direction (bias):

  • Forward bias: the p-side is connected to the positive terminal and the n-side to the negative terminal. The applied field opposes the internal field, so the potential barrier is lowered and the depletion region narrows. Once the applied voltage approaches V0 (the knee/threshold voltage), majority carriers flood across the junction and the current rises very rapidly (approximately exponentially) with small further increases in voltage.
  • Reverse bias: the p-side is connected to the negative terminal and the n-side to the positive terminal. The applied field adds to the internal field, so the barrier height increases and the depletion region widens, blocking majority-carrier flow almost completely. Only a very small, nearly voltage-independent reverse saturation current I0 flows, carried by the few thermally generated minority carriers. If the reverse voltage is increased beyond a critical breakdown voltage, the current rises sharply (used deliberately in Zener diodes).

Because the I-V relationship is strongly non-linear, a diode does not have a single fixed resistance; instead we define a dynamic (AC) resistance at an operating point as the ratio of a small change in voltage to the resulting small change in current.

Worked Example: In forward bias, the current through a diode rises from 2 mA to 7 mA when the voltage across it changes by 50 mV. The dynamic resistance is rd = ΔV/ΔI = 0.050 V / (5×10-3 A) = 10 Ω, a typical low value for a forward-biased silicon diode operating well above the knee voltage. In reverse bias, by contrast, a similar voltage change produces almost no change in the (already tiny) current, so the dynamic resistance there is enormously larger (of order MΩ), reflecting the diode's near-ideal blocking action.

Dynamic Resistance (Forward Bias) r_d = ΔV_f / ΔI_f Ω · Slope of the forward I-V curve at the operating point; typically tens of ohms for a Si diode
Dynamic Resistance (Reverse Bias) r_d = ΔV_r / ΔI_r Ω · Very large (order of MΩ) because the reverse current stays close to I0 until breakdown
Barrier (Knee) Potential V0(Si) ≈ 0.7 V, V0(Ge) ≈ 0.3 V V · Approximate forward voltage at which current begins rising sharply
Remember
  • Diffusion of majority carriers across the junction creates a depletion region and leaves behind fixed ion charges
  • The fixed ion charges set up an internal electric field and a built-in potential barrier V0 (~0.7 V for Si, ~0.3 V for Ge)
  • Forward bias lowers the barrier and narrows the depletion region, letting current rise rapidly once the knee voltage is crossed
  • Reverse bias raises the barrier and widens the depletion region, allowing only a small, nearly constant reverse saturation current I0
  • A diode's resistance is not constant; dynamic resistance rd = ΔV/ΔI is used at a chosen operating point
  • Beyond the reverse breakdown voltage, current rises sharply — this behaviour is exploited by Zener diodes

Diode as a Rectifier

Quick answer Shows how the one-way conduction property of a diode is used to convert AC into pulsating DC in half-wave and full-wave rectifier circuits.

Because a p-n junction diode conducts appreciably only when forward biased and blocks current when reverse biased, it acts like a one-way valve for current. This property is used to convert alternating current (AC), which reverses direction every half cycle, into a unidirectional (though pulsating) output — a process called rectification.

In a half-wave rectifier, a single diode is connected in series with the AC source and the load. During the half-cycle that forward-biases the diode, it conducts and current flows through the load, reproducing (approximately) that half of the input waveform. During the other half-cycle, the diode is reverse biased and blocks conduction, so the output is zero. The result is a series of one-directional pulses occurring at the same frequency as the input, with long "dead" gaps in between — a poor approximation to steady DC.

A full-wave rectifier overcomes this using either a centre-tapped transformer with two diodes, or four diodes in a bridge arrangement, so that current is driven through the load in the same direction during both half-cycles of the input. Whichever diode is forward biased in a given half-cycle conducts, and the two conduction paths are arranged so the load current never reverses. Consequently the output consists of a pulse for every half-cycle of the input, i.e. the ripple (pulsation) frequency of a full-wave rectifier is twice the input frequency, compared to the half-wave rectifier's ripple frequency, which equals the input frequency. This makes the full-wave output smoother, gives it a higher average (DC) value for the same peak input, and makes the residual ripple easier to remove with a capacitor filter connected across the load.

Worked Example: A step-down transformer feeds a peak voltage Vm = 15 V to a full-wave bridge rectifier operating from 50 Hz mains. (a) The ripple frequency of the output is twice the input frequency: 2 × 50 Hz = 100 Hz. (b) Assuming ideal diodes (no forward voltage drop), the average (DC) output voltage is Vdc = 2Vm/π = (2 × 15)/3.1416 ≈ 9.55 V, compared to only Vm/π ≈ 4.77 V that a half-wave rectifier would give for the same peak input.

Half-wave Ripple Frequency f_ripple = f_input Hz · Output pulsates once per input cycle since only one half-cycle is used
Full-wave Ripple Frequency f_ripple = 2 × f_input Hz · Both half-cycles are used, giving twice the pulsation rate and easier smoothing
Average DC Output – Half-wave V_dc = Vm / π V · Vm is the peak value of the AC input; ideal diode assumed
Average DC Output – Full-wave V_dc = 2Vm / π V · Twice the half-wave average since both halves of the input contribute
Remember
  • A diode conducts only when forward biased, so it can convert AC into pulsating, unidirectional current (rectification)
  • A half-wave rectifier uses one diode and passes only alternate half-cycles of the input
  • A full-wave rectifier (centre-tap or bridge, 2 or 4 diodes) uses both half-cycles, giving a higher average output
  • Ripple frequency: half-wave = input frequency f; full-wave = 2f
  • A capacitor filter across the load smooths the pulsating output closer to steady DC
  • For the same peak input, a full-wave rectifier gives double the average DC output of a half-wave rectifier

Special Purpose p-n Junction Diodes

Quick answer Covers the Zener diode as a voltage regulator and the optoelectronic junction devices: photodiode, LED and solar cell.

Ordinary junction diodes are designed to avoid reverse breakdown, but some diodes are built specifically to exploit it or to interact with light.

A Zener diode is a heavily doped p-n junction diode designed to operate safely and repeatedly in reverse breakdown. Because of the heavy doping, the depletion region is very thin, and at a well-defined reverse voltage VZ the junction breaks down (via Zener tunnelling for lower voltages, or avalanche multiplication for higher voltages), after which the voltage across the diode stays almost constant over a wide range of current. This makes it ideal as a voltage regulator: the Zener is connected in reverse bias, in series with a resistor Rs across the unregulated input, with the load connected in parallel with the Zener. Any increase in input voltage (or decrease in load current) is absorbed as extra current through the Zener rather than extra voltage across the load, so the load voltage remains close to VZ.

A photodiode is a p-n junction operated in reverse bias, enclosed so that light can fall on the junction. Photons with energy at least equal to the band gap generate electron-hole pairs; the reverse-bias field sweeps these apart before they can recombine, producing a photocurrent that increases almost linearly with the intensity of incident light. Reverse bias is used (rather than zero or forward bias) because it gives a faster response and a current that tracks light intensity more linearly.

A light-emitting diode (LED) is a heavily doped p-n junction, usually of a direct-band-gap compound semiconductor (such as GaAs or GaP-based alloys), operated in forward bias. Injected electrons and holes recombine near the junction, releasing the recombination energy as photons instead of heat; the emitted wavelength is set by the material's band gap. LEDs are compact, low-voltage, low-power, fast-switching and need no warm-up time, unlike incandescent bulbs.

A solar cell is also a p-n junction, but requires no external bias at all: when light falls on it, electron-hole pairs generated near the junction are separated by the built-in field itself, so the device generates an emf directly (the photovoltaic effect) and can deliver power to an external circuit. Its performance is described by the open-circuit voltage Voc, short-circuit current Isc, and the maximum-power point (Vm, Im) on its I-V characteristic. The fill factor compares the actual maximum power obtainable to the (unattainable) ideal product VocIsc, and the efficiency compares the maximum electrical power output to the incident optical power.

Worked Example: An LED is fabricated from a compound semiconductor with band gap Eg = 1.9 eV. Using hc ≈ 1240 eV·nm, the emitted wavelength is λ = hc/Eg = 1240/1.9 ≈ 652.6 nm, which falls in the red region of the visible spectrum — consistent with typical red LEDs.

LED Emission Wavelength λ = hc / Eg ≈ 1240 / Eg(eV) nm · Photon energy released on recombination approximately equals the band gap energy; hc ≈ 1240 eV·nm
Fill Factor FF = (Vm × Im) / (Voc × Isc) Ratio of the actual maximum power to the product of open-circuit voltage and short-circuit current; typically 0.7–0.8 for a good cell
Solar Cell Efficiency η = Pm / Pin = (Vm × Im) / Pin Pin is the incident optical power on the cell
Zener Series Resistor (Design) Rs = (Vin − Vz) / (IL + Iz) Ω · Current-limiting resistor needed in a Zener voltage-regulator circuit
Remember
  • A Zener diode is heavily doped and deliberately operated in reverse breakdown, where its voltage VZ stays nearly constant over a wide current range
  • A series resistor Rs limits current through a Zener regulator while the load voltage is held close to VZ
  • A photodiode is operated in reverse bias; its photocurrent is proportional to incident light intensity
  • An LED is a forward-biased, direct-band-gap junction that emits light on electron-hole recombination; emitted wavelength depends on Eg
  • A solar cell needs no external bias — it generates its own emf via the photovoltaic effect when illuminated
  • Solar cell performance is summarised by Voc, Isc, the maximum power point, fill factor, and efficiency

Junction Transistor: Structure, Action and Characteristics

Quick answer Describes the three-region structure of a bipolar transistor, the conditions needed for transistor action, and its input/output characteristic curves.

A bipolar junction transistor consists of three alternately doped regions forming two back-to-back p-n junctions: the emitter (heavily doped, to supply a large number of majority carriers), the base (made very thin and lightly doped), and the collector (moderately doped and physically larger, to collect carriers and dissipate heat). An npn or a pnp arrangement is possible; the npn transistor is more commonly used because electrons (which move faster than holes) form the main current.

For normal (active-region) operation, the emitter-base junction is forward biased and the collector-base junction is reverse biased. Forward bias at the emitter-base junction injects a large number of majority carriers from the emitter into the base. Because the base is thin and lightly doped, only a small fraction of these recombine there (contributing to the small base current); the vast majority diffuse across the narrow base and are swept into the collector by the reverse-biased collector junction's field. This is transistor action: a small base current controls a much larger collector current.

By Kirchhoff's current law applied to the transistor, the emitter current splits into the base and collector currents: IE = IB + IC. Two current-gain factors describe this: the common-base current gain αdc = IC/IE (always slightly less than 1, since IC is slightly less than IE), and the common-emitter current gain βdc = IC/IB (much greater than 1, since IB is a very small fraction of IE). The two are related by β = α/(1−α).

A transistor's behaviour is summarised by two characteristic curves (for a given configuration, commonly common-emitter): the input characteristic (IB vs VBE at constant VCE), which resembles the forward characteristic of a diode, and the output characteristic (IC vs VCE for different fixed values of IB), which shows three regions: cutoff (both junctions reverse biased, IC ≈ 0), the active region (IC nearly constant for a given IB, rising only slightly with VCE, and closely proportional to IB), and saturation (both junctions forward biased, VCE small, IC large and limited mainly by the external circuit).

Worked Example: A transistor operates with IE = 5 mA and IB = 0.1 mA. Then IC = IE − IB = 5 − 0.1 = 4.9 mA. The common-base gain is α = IC/IE = 4.9/5 = 0.98, and the common-emitter gain is β = IC/IB = 4.9/0.1 = 49. Check: β = α/(1−α) = 0.98/0.02 = 49, which matches.

Emitter Current Relation IE = IB + IC Total current entering the emitter splits between the base and collector terminals
DC Current Gain (Common Base) αdc = IC / IE Always slightly less than 1, typically 0.95–0.99
DC Current Gain (Common Emitter) βdc = IC / IB Much greater than 1, typically 20–500; measures current amplification
α–β Relation β = α / (1 − α) ; α = β / (1 + β) Interconversion between common-base and common-emitter current gains
Remember
  • A transistor has three regions: heavily doped emitter, thin lightly doped base, moderately doped larger collector
  • Active-region operation needs the emitter-base junction forward biased and the collector-base junction reverse biased
  • The thin, lightly doped base lets most carriers injected from the emitter reach the collector instead of recombining
  • IE = IB + IC always holds; the common-emitter gain beta is much larger than the common-base gain alpha
  • beta = alpha/(1-alpha) links the common-base and common-emitter current gains
  • The output characteristic shows three regions: cutoff, active, and saturation

Transistor as a Switch and CE Amplifier

Quick answer Explains how the cutoff and saturation regions let a transistor act as a switch, and how the active region lets a common-emitter circuit act as a voltage amplifier.

The two extreme regions of the output characteristic give a transistor its role as an electronic switch. When the base-emitter voltage is below the threshold needed to forward bias the emitter junction, the transistor sits in cutoff: both junctions are effectively reverse biased, negligible collector current flows, and the output (collector) voltage stays close to the supply voltage ("high"/OFF state for current flow, but output logic-high). Driving enough base current instead pushes the transistor into saturation, where both junctions are forward biased, the collector-emitter voltage drops to a small value, and collector current is limited mainly by the external load resistor ("ON" state, output logic-low). Switching rapidly between these two states, rather than operating anywhere in between, is the basis of transistor use in digital circuits.

Operating instead in the active region, with a fixed DC bias point (Q-point) chosen roughly in the middle of the usable range, the transistor can amplify small AC signals. In the common-emitter (CE) configuration, the input AC signal is applied between base and emitter, and the amplified output is taken across a load resistor RC connected between collector and the supply. A small AC change in base current ΔIB produces a much larger change in collector current ΔIC = βacΔIB, where βac = ΔIC/ΔIB (measured at constant VCE) is the AC current gain. This changing collector current develops a changing voltage across RC, giving voltage amplification; combined with the current gain, the CE configuration also gives substantial power gain, which is why it is the most widely used amplifier configuration. A characteristic feature of the CE amplifier is that the output voltage is 180° out of phase with (i.e. inverted relative to) the input voltage.

Worked Example: A CE amplifier has βac = 80, collector load RC = 4 kΩ, and AC input (base) resistance Ri = 800 Ω. If the input AC voltage changes by ΔVi = 5 mV: the base current change is ΔIB = ΔVi/Ri = 5 mV / 800 Ω = 6.25 μA; the collector current change is ΔIC = βacΔIB = 80 × 6.25 μA = 500 μA = 0.5 mA; the voltage gain is Av = βac(RC/Ri) = 80 × (4000/800) = 80 × 5 = 400; and the resulting output voltage change is ΔVo = AvΔVi = 400 × 5 mV = 2 V (inverted in phase relative to the input).

AC Current Gain βac = ΔIC / ΔIB Measured at constant VCE from the output characteristics
Voltage Gain (CE Amplifier) Av = ΔVo / ΔVi = βac × (RC / Ri) Ri is the AC input (base-emitter) resistance, RC is the collector load resistance
Input Resistance ri = ΔVBE / ΔIB Ω · Slope of the input characteristic at the operating point
Power Gain Ap = Av × βac Approximate overall power amplification of a CE amplifier
Remember
  • In cutoff both junctions are reverse biased and IC ≈ 0; in saturation both are forward biased and VCE is small — the basis of transistor switching
  • Active-region operation with a fixed Q-point allows the transistor to amplify small AC signals
  • The CE configuration gives both current gain and voltage gain, so it also gives high power gain and is the most common amplifier configuration
  • A small change in base current produces a much larger change in collector current, amplified further into a voltage swing across RC
  • The CE amplifier output is 180 degrees out of phase with the input
  • Voltage gain Av = betaac x (RC/Ri) depends on the AC current gain and the ratio of load to input resistance

The formula sheet

Every formula in this chapter, in one place — screenshot it before your exam.

Eg = Ec − Ev
Energy GapeV
ni² = nₑ × nₕ
Law of Mass Action
σ = e(nₑμₑ + nₕμₕ)
Conductivity of a SemiconductorS m⁻¹
nₑ ≈ Nd (when Nd ≫ ni)
n-type Approximation
nₕ ≈ Na (when Na ≫ ni)
p-type Approximation
r_d = ΔV_f / ΔI_f
Dynamic Resistance (Forward Bias)Ω
r_d = ΔV_r / ΔI_r
Dynamic Resistance (Reverse Bias)Ω
V0(Si) ≈ 0.7 V, V0(Ge) ≈ 0.3 V
Barrier (Knee) PotentialV
f_ripple = f_input
Half-wave Ripple FrequencyHz
f_ripple = 2 × f_input
Full-wave Ripple FrequencyHz
V_dc = Vm / π
Average DC Output – Half-waveV
V_dc = 2Vm / π
Average DC Output – Full-waveV
λ = hc / Eg ≈ 1240 / Eg(eV)
LED Emission Wavelengthnm
FF = (Vm × Im) / (Voc × Isc)
Fill Factor
η = Pm / Pin = (Vm × Im) / Pin
Solar Cell Efficiency
Rs = (Vin − Vz) / (IL + Iz)
Zener Series Resistor (Design)Ω
IE = IB + IC
Emitter Current Relation
αdc = IC / IE
DC Current Gain (Common Base)
βdc = IC / IB
DC Current Gain (Common Emitter)
β = α / (1 − α) ; α = β / (1 + β)
α–β Relation
βac = ΔIC / ΔIB
AC Current Gain
Av = ΔVo / ΔVi = βac × (RC / Ri)
Voltage Gain (CE Amplifier)
ri = ΔVBE / ΔIB
Input ResistanceΩ
Ap = Av × βac
Power Gain

Test yourself

Tap an answer to check it instantly — you'll see why it's right, and what to revise if it isn't.

0 correct · 0/12 answered
Q1 Energy bands easy

Which of the following best explains why silicon (Eg ≈ 1.1 eV) behaves as a semiconductor while diamond (Eg ≈ 5.4 eV) behaves as an insulator at room temperature?

Q2 Extrinsic semiconductors easy

A pure germanium crystal is doped with a small amount of arsenic (a pentavalent element). What type of semiconductor is formed and what are the majority carriers?

Q3 p-n junction easy

In which biasing condition of a p-n junction diode does the width of the depletion region increase?

Q4 Mass-action law medium

The intrinsic carrier concentration of a semiconductor sample is 1.2×10¹⁶ m⁻³. On doping, the electron concentration rises to 4.8×10¹⁶ m⁻³. What is the new hole concentration?

Q5 Diode dynamic resistance medium

The forward current through a diode changes from 2 mA to 7 mA when the voltage across it changes by 50 mV. What is the dynamic resistance of the diode in this range?

Q6 Rectifiers medium

A full-wave rectifier is fed with a 50 Hz AC input. What is the frequency of the ripples (pulsations) in its output?

Q7 Zener diode medium

Why is a Zener diode operated in the reverse breakdown region when used as a voltage regulator?

Q8 Photodiode medium

A photodiode used as a light sensor is normally operated in which mode, and why?

Q9 Transistor current relations hard

A transistor has a common-emitter current gain β = 99. If the base current is IB = 20 μA, what are the collector current IC and emitter current IE?

Q10 Solar cell hard

A solar cell has Voc = 0.6 V and Isc = 100 mA. At its maximum power point, Vm = 0.45 V and Im = 80 mA. What is the fill factor of this cell?

Q11 CE amplifier hard

In a CE amplifier, RC = 5 kΩ, the AC input (base) resistance Ri = 500 Ω, and βac = 100. What is the voltage gain of the amplifier?

Q12 LED hard

An LED is made from a compound semiconductor with a band gap of 2.0 eV. Approximately what wavelength of light does it emit? (Use hc ≈ 1240 eV·nm)

NCERT solutions & previous-year questions

Step-by-step model answers — tap a question to reveal the full solution.

NCERT questions 6

1 In an n-type silicon sample, the fifth (donor) electron contributed by a pentavalent impurity atom needs only a very small amount of energy to become a free conduction electron. Explain why this is so, and state where the donor energy level lies relative to the conduction band.Donor energy levels

When a pentavalent impurity atom (such as phosphorus or arsenic) replaces a silicon atom in the crystal lattice, four of its five valence electrons form normal covalent bonds with the four neighbouring silicon atoms, exactly as a silicon atom's own electrons would. The fifth electron cannot take part in any covalent bond; it remains attached to its parent (donor) ion only by the weak, screened Coulomb attraction inside the dielectric medium of the crystal.

Because this attraction is weak, only a very small amount of energy (a few hundredths of an eV, much smaller than the band gap Eg ≈ 1.1 eV of silicon) is required to detach this electron and set it free to move through the crystal as a conduction electron. In energy-band language, this means the donor energy level ED lies just below the bottom of the conduction band Ec, separated from it by only this small ionisation energy.

Consequently, at room temperature, where the thermal energy kT (≈ 0.026 eV) is comparable to or larger than this small donor ionisation energy, essentially all the donor atoms get ionised and contribute a free electron each. This is why the majority carrier (electron) concentration in an n-type semiconductor is very nearly equal to the donor doping concentration.

2 Pure silicon at 300 K has an intrinsic carrier concentration ni = 1.5×10¹⁶ m⁻³. A silicon sample is doped simultaneously with 5×10²² m⁻³ of arsenic (donor) atoms and 5×10²⁰ m⁻³ of indium (acceptor) atoms. Calculate the equilibrium electron and hole concentrations, and state whether the resulting material is n-type or p-type.Compensated doping

Given: Nd (arsenic, donor) = 5×1022 m-3, Na (indium, acceptor) = 5×1020 m-3, ni = 1.5×1016 m-3.

Step 1: Net donor concentration. Since both donor and acceptor impurities are present, the net effective donor concentration is Nd − Na = 5×1022 − 5×1020 = 4.95×1022 m-3.

Step 2: Electron concentration. Because Nd − Na ≫ ni, almost all of this net donor concentration appears as free electrons: ne ≈ Nd − Na = 4.95×1022 m-3.

Step 3: Hole concentration. Using the law of mass action, nh = ni2/ne = (1.5×1016)2 / (4.95×1022) = 2.25×1032 / 4.95×10224.5×109 m-3.

Conclusion: Since ne (≈ 5×1022 m-3) is enormously larger than nh (≈ 4.5×109 m-3), electrons are the majority carriers, so the doped sample is an n-type semiconductor.

3 An intrinsic semiconductor has a band gap Eg = 1.2 eV. Assuming its intrinsic carrier concentration (and hence its conductivity) varies with absolute temperature as ni ∝ exp(−Eg/2kBT), find the ratio of its electrical conductivity at 600 K to that at 300 K. (Take kB = 8.6×10⁻⁵ eV/K.)Temperature dependence of conductivity

Since conductivity of an intrinsic semiconductor is proportional to the carrier concentration ni, and ni ∝ exp(−Eg/2kBT), the ratio of conductivities is:

σ(600)/σ(300) = exp[ (−Eg/2kB)(1/600) ] / exp[ (−Eg/2kB)(1/300) ] = exp[ (Eg/2kB)(1/300 − 1/600) ]

Step 1: Eg/(2kB) = 1.2 / (2 × 8.6×10-5) = 1.2 / (1.72×10-4) ≈ 6977 K.

Step 2: (1/300 − 1/600) = 1/600 = 1.667×10-3 K-1.

Step 3: Exponent = 6977 × 1.667×10-3 ≈ 11.63.

Step 4: σ(600)/σ(300) = e11.631.1×105.

This enormous ratio (roughly a hundred-thousand-fold increase) illustrates why semiconductor conductivity is so strongly temperature-dependent — in sharp contrast to a metal, whose resistance only rises slowly and linearly with temperature.

4 In a p-n junction diode, the forward current can be expressed as I = I0[exp(eV/2kBT) − 1], where I0 is the reverse saturation current. For a certain diode, I0 = 5×10⁻¹² A and T = 300 K (take kB = 8.6×10⁻⁵ eV/K). Find: (a) the forward current at V = 0.6 V, (b) the increase in current when V is raised to 0.7 V, (c) the dynamic resistance between these two voltages, and (d) how the current changes when a reverse bias is changed from 1 V to 2 V.Diode I-V equation

Given: I0 = 5×10-12 A, T = 300 K, kB = 8.6×10-5 eV/K, so 2kBT = 2 × 8.6×10-5 × 300 = 0.0516 eV (using V in volts and e = 1, the exponent is simply V/0.0516).

(a) At V = 0.6 V: exponent = 0.6/0.0516 = 11.63. exp(11.63) ≈ 1.12×105. So I ≈ I0 × 1.12×105 = 5×10-12 × 1.12×1055.6×10-7 A (0.56 μA) (the “−1” in the formula is negligible here).

(b) At V = 0.7 V: exponent = 0.7/0.0516 = 13.57. exp(13.57) ≈ 7.79×105. So I ≈ 5×10-12 × 7.79×105 ≈ 3.9×10-6 A (3.9 μA). The increase in current is ΔI ≈ 3.9 μA − 0.56 μA ≈ 3.34 μA.

(c) Dynamic resistance: rd = ΔV/ΔI = 0.1 V / (3.34×10-6 A) ≈ 3.0×104 Ω.

(d) Reverse bias 1 V to 2 V: For reverse bias, V is negative, so the exponential term exp(eV/2kBT) becomes vanishingly small, and I ≈ −I0 = −5×10-12 A regardless of exactly how large the reverse voltage is (1 V or 2 V). So the reverse current stays essentially unchanged at about 5 pA — the diode's reverse (dynamic) resistance in this range is extremely high, demonstrating its near-ideal current-blocking behaviour under reverse bias.

5 Explain why the base region of a transistor is made thin and lightly doped, the emitter is heavily doped, and the collector region is made physically larger than the emitter.Transistor structure

Heavily doped emitter: The emitter's job is to inject as many majority carriers as possible into the base when the emitter-base junction is forward biased. Heavy doping ensures a large supply of majority carriers, giving high emitter injection efficiency — most of the junction current consists of carriers entering the base from the emitter, not the other way round.

Thin, lightly doped base: The base must let almost all the carriers injected from the emitter pass through to the collector rather than recombining within it. Making the base physically thin reduces the distance (and hence time) available for recombination, and light doping keeps the number of base majority carriers small, further reducing the recombination rate. Both effects push the common-base current gain α (= IC/IE) close to 1, which in turn makes the common-emitter current gain β = α/(1−α) very large — this large β is exactly what gives a transistor its current-amplifying ability.

Larger, moderately doped collector: The collector must efficiently collect the carriers diffusing across from the base (so it is made physically larger in area than the emitter to “catch” the spreading carrier flow) and must also withstand the reverse bias voltage applied to the collector-base junction without breaking down; moderate doping (lighter than the emitter) gives the collector-base junction a higher reverse breakdown voltage. The larger area also helps the collector dissipate the heat generated as carriers are collected, since it usually handles the largest power dissipation in the transistor.

6 An AC input of frequency 50 Hz is applied first to a half-wave rectifier and then to a full-wave rectifier. (a) What is the ripple frequency of the output in each case? (b) Which type of rectifier is generally preferred in practice, and why?Rectifier comparison

(a) Ripple frequency:

  • Half-wave rectifier: Only one half-cycle of the input is used per full input cycle, so the output pulsates once per input cycle. Ripple frequency = input frequency = 50 Hz.
  • Full-wave rectifier: Both half-cycles of the input are converted into output pulses of the same polarity, so the output pulsates twice per input cycle. Ripple frequency = 2 × input frequency = 100 Hz.

(b) Preferred rectifier: The full-wave rectifier is generally preferred because: (i) it uses both half-cycles of the input, giving a higher average (DC) output voltage for the same peak input than a half-wave rectifier; (ii) its higher ripple frequency (100 Hz vs 50 Hz for 50 Hz mains) is easier to filter out with a comparatively smaller smoothing capacitor, giving a steadier DC output; and (iii) it makes more efficient use of the transformer secondary winding, since current flows through the load during both halves of every input cycle rather than only alternate halves.

Previous-year board questions 4

Q1 With the help of a circuit description, explain the working of a full-wave rectifier using a centre-tapped transformer and two p-n junction diodes. Describe the waveforms of the input and output voltages. 2023 3 marks

Circuit: A transformer with a centre-tapped secondary winding is used. The two ends of the secondary are connected to the anodes of two diodes D1 and D2; the cathodes of both diodes are joined together and connected through the load resistor RL back to the centre tap of the secondary, which is taken as the reference (ground) point. The output voltage is developed across RL.

Working: During one half-cycle of the AC input, the upper end of the secondary is positive relative to the centre tap. This forward biases D1 (which conducts) while the lower end being negative reverse biases D2 (which is cut off). Current flows through D1 and RL. During the next half-cycle, the polarity reverses: the lower end becomes positive relative to the centre tap, forward biasing D2 (which now conducts) while D1 is reverse biased and cut off. Current now flows through D2 and RL.

In both half-cycles, current through RL flows in the same direction (from the common cathode point, through RL, back to the centre tap), because the two diodes alternately conduct for opposite half-cycles but always deliver current the same way through the load.

Waveforms: The input voltage across the secondary is a symmetric sine wave, alternating between positive and negative half-cycles. The output voltage across RL consists of a series of positive humps (each shaped like a rectified half-sine-wave), one for the positive half-cycle (via D1) and one for the negative half-cycle (via D2), so that a pulse appears for every half-cycle of the input. As a result, the output ripple frequency is twice the input frequency, and the output never goes negative — it is a unidirectional, pulsating voltage.

Q2 In a common-emitter transistor amplifier, βac = 50, the collector load resistance RC = 2 kΩ, and the AC input (base) resistance Ri = 1 kΩ. If the input AC voltage changes by ΔVi = 10 mV, calculate (a) the change in base current ΔIB, (b) the change in collector current ΔIC, (c) the voltage gain of the amplifier, and (d) the resulting change in output voltage ΔVo. 2022 4 marks

Given: βac = 50, RC = 2 kΩ = 2000 Ω, Ri = 1 kΩ = 1000 Ω, ΔVi = 10 mV = 0.01 V.

(a) Change in base current: ΔIB = ΔVi / Ri = 0.01 V / 1000 Ω = 1×10-5 A = 10 μA.

(b) Change in collector current: ΔIC = βac × ΔIB = 50 × 10 μA = 500 μA = 0.5 mA.

(c) Voltage gain: Av = βac × (RC/Ri) = 50 × (2000/1000) = 50 × 2 = 100.

(d) Change in output voltage: ΔVo = Av × ΔVi = 100 × 10 mV = 1000 mV = 1 V (this output swing is 180° out of phase with the input, as expected for a CE amplifier).

Q3 Explain the formation of the depletion region and the potential barrier in a p-n junction diode. How do the width of the depletion region and the barrier height change when the junction is (i) forward biased and (ii) reverse biased? 2024 3 marks

Formation of the depletion region and potential barrier: When a p-region and an n-region are joined to form a p-n junction, the large concentration gradient at the junction causes majority carriers to diffuse across it: electrons diffuse from the n-side into the p-side, and holes diffuse from the p-side into the n-side. As these carriers cross and recombine near the junction, they leave behind fixed, uncompensated ion charges: positively charged donor ions on the n-side and negatively charged acceptor ions on the p-side, in a narrow region on either side of the junction that becomes depleted of mobile charge carriers — the depletion region. The exposed ion charges create an internal electric field directed from the n-side to the p-side, which opposes further diffusion of majority carriers. At equilibrium this field corresponds to a built-in potential barrier V0 (about 0.7 V for silicon, 0.3 V for germanium), across which diffusion current (majority carriers) is exactly balanced by a small drift current (minority carriers pulled the other way by the same field).

(i) Forward bias (p-side connected to positive terminal, n-side to negative terminal): the applied field opposes the internal field, so the net field at the junction weakens. This lowers the potential barrier and narrows the depletion region, allowing majority carriers to cross the junction easily; the forward current rises rapidly, approximately exponentially, once the applied voltage approaches V0.

(ii) Reverse bias (p-side connected to negative terminal, n-side to positive terminal): the applied field reinforces the internal field, so the net field strengthens. This raises the potential barrier and widens the depletion region, making it even harder for majority carriers to cross; only a small, nearly constant reverse saturation current I0 flows, due to minority carriers, until reverse breakdown occurs at sufficiently high reverse voltage.

Q4 A Zener diode with breakdown voltage VZ = 6 V is to regulate the voltage across a load drawing a constant current IL = 10 mA, from an unregulated supply that varies between 10 V and 16 V. If the minimum current needed to keep the Zener diode in breakdown is Iz(min) = 5 mA, calculate the series resistor Rs required, and verify that the Zener current does not fall below Iz(min) even at the lowest input voltage. 2021 4 marks

Given: VZ = 6 V, IL = 10 mA = 0.010 A, Vin(min) = 10 V, Vin(max) = 16 V, Iz(min) = 5 mA = 0.005 A.

Step 1: Design Rs using the worst case (lowest input voltage). At the lowest input voltage, the total current through Rs must still be enough to supply the load current AND keep at least Iz(min) flowing through the Zener:

Rs = (Vin(min) − VZ) / (IL + Iz(min)) = (10 − 6) / (0.010 + 0.005) = 4 / 0.015 ≈ 266.7 Ω (choose Rs ≈ 267 Ω, or a standard value such as 270 Ω).

Step 2: Verify at Vin = 10 V (lowest input): Total current through Rs = (Vin − VZ)/Rs = (10 − 6)/267 ≈ 0.01498 A = 14.98 mA. Zener current = total current − load current = 14.98 − 10 = 4.98 mA ≈ 5 mA, which just meets the minimum required Zener current — confirming the design is correct at the worst case.

Step 3: Check at Vin = 16 V (highest input): Total current = (16 − 6)/267 ≈ 37.5 mA. Zener current = 37.5 − 10 = 27.5 mA (comfortably above Iz(min), and assumed within the Zener's rated maximum current). Thus with Rs ≈ 267 Ω, the load voltage stays regulated at 6 V across the full 10–16 V input range, with the Zener current always at or above 5 mA.

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