Semiconductor Electronics

What to remember

• In a pure semiconductor the number of electrons equals the number of holes (nₑ = n_h = nᵢ). Doping with a pentavalent atom gives an n-type semiconductor, and doping with a trivalent atom gives a p-type semiconductor. In every case nₑ × n_h = nᵢ².
• A p-n junction has a depletion region and a barrier potential (about 0.7 V for silicon, 0.3 V for germanium). It conducts well in forward bias and gives only a tiny current in reverse bias, so it works as a rectifier.
• The LED is forward biased, the photodiode is reverse biased, the solar cell works with no external bias, and the Zener diode works in reverse breakdown as a voltage regulator.
• A half-wave rectifier gives an output of the same frequency as the input. A full-wave rectifier gives an output of twice the input frequency. NAND and NOR are the universal gates.

Conductors, insulators and semiconductors: energy bands

In a solid the closely spaced atoms make the energy levels of the outer electrons spread into bands.

• Valence band: the band of energies of the valence electrons, which is filled or nearly filled at low temperature.
• Conduction band: the band above the valence band, in which electrons can move freely and conduct.
• Energy gap (band gap) E_g: the gap between the top of the valence band and the bottom of the conduction band.

Class	Energy band picture	Band gap	Resistivity (about)
Metal (conductor)	Conduction and valence bands overlap, or the conduction band is partly filled	Nil or very small	10⁻² to 10⁻⁸ Ω m
Semiconductor	Small gap	Less than about 3 eV; Ge 0.67 eV, Si 1.1 eV	10⁻⁵ to 10⁶ Ω m
Insulator	Large gap	More than about 3 eV	10¹¹ to 10¹⁹ Ω m

• As the temperature rises, some electrons gain energy ≥ E_g and jump to the conduction band. So the conductivity of a semiconductor increases and its resistance decreases with a rise in temperature. In a metal the resistance increases with temperature.

Intrinsic semiconductors, electrons and holes

An intrinsic semiconductor is a pure semiconductor, such as pure silicon (Si) or germanium (Ge). Each atom has four valence electrons and forms four covalent bonds with its neighbours.

• When a bond is broken by thermal energy, an electron is freed. The place it leaves is a vacancy, called a hole. A hole acts as a positive charge carrier of charge +e.
• Electrons and holes are created in pairs. So nₑ = n_h = nᵢ, where nᵢ is the intrinsic carrier concentration.
• The total current is the sum of the electron current and the hole current. In an electric field electrons move towards the positive terminal and holes towards the negative terminal.
• nᵢ increases with temperature. For silicon at room temperature nᵢ is about 1.5 × 10¹⁶ m⁻³.

Extrinsic semiconductors: n-type and p-type

Adding a small amount of impurity (a dopant, about 1 part in 10⁶) to a pure semiconductor raises its conductivity greatly. This is doping, and the doped crystal is an extrinsic semiconductor. The dopant ions are fixed and the crystal stays electrically neutral.

Feature	n-type	p-type
Dopant	Pentavalent (donor): arsenic, phosphorus, antimony	Trivalent (acceptor): boron, aluminium, gallium, indium
Dopant valence electrons	5	3
Effect	Each dopant atom gives one extra free electron	Each dopant atom makes one hole
Majority carriers	Electrons	Holes
Minority carriers	Holes	Electrons
Relation	nₑ ≫ n_h	n_h ≫ nₑ
Overall charge	Neutral	Neutral

• Law of mass action: nₑ × n_h = nᵢ² for any semiconductor at a given temperature. If doping raises nₑ, then n_h falls.

p-n junction

A p-n junction is formed when a p-type and an n-type crystal are joined as one crystal.

• 1. Diffusion: electrons diffuse from the n side (high electron concentration) to the p side. Holes diffuse from the p side to the n side. This gives the diffusion current.
• 2. The electrons that cross leave fixed positive donor ions behind on the n side. The holes that cross leave fixed negative acceptor ions behind on the p side.
• 3. A region around the junction becomes empty of mobile carriers. This is the depletion region. Its width is of the order of 10⁻⁶ m.
• 4. The fixed ions set up an electric field from the n side towards the p side. It is the barrier potential (about 0.7 V for Si and 0.3 V for Ge). The field opposes the diffusion.
• 5. The field drives minority carriers across the junction. This drift current is opposite to the diffusion current. At equilibrium the two are equal and the net current is zero.

For a barrier potential V across a depletion region of width d, the field is about E = V/d. For 0.6 V across 10⁻⁶ m, E = 6 × 10⁵ V m⁻¹.

Semiconductor diode: forward and reverse bias

A p-n junction with two metal contacts is a diode. Its symbol is an arrow pointing from the p side to the n side with a bar at the n side. The arrow shows the direction of conventional current in forward bias.

Feature	Forward bias	Reverse bias
Connection	p to positive terminal, n to negative terminal	p to negative terminal, n to positive terminal
Applied field	Opposes the barrier field	Adds to the barrier field
Depletion region	Narrows	Widens
Barrier height	Decreases	Increases
Current	Large (mA); majority carriers; rises steeply once the knee voltage is crossed	Very small (μA); minority carriers; nearly constant
Resistance	Low (a few ohms near the knee)	Very high

V–I characteristic, in words.

• In forward bias the current is almost zero until the voltage crosses a threshold (knee or cut-in voltage). It is about 0.7 V for Si and 0.3 V for Ge. Above it the current rises rapidly.
• In reverse bias a small constant current (reverse saturation current, in μA) flows, nearly independent of the voltage.
• At a large enough reverse voltage (breakdown voltage) the reverse current rises suddenly, and an ordinary diode can be damaged.

Rectifier: half-wave and full-wave

A rectifier converts alternating voltage into a unidirectional (pulsating) voltage. It uses the one-way conduction of a diode.

Half-wave rectifier. One diode is connected in series with the load resistor R_L and the AC source (through a transformer).

• During the positive half-cycle the diode is forward biased and conducts. The output across R_L has the shape of that half-cycle.
• During the negative half-cycle the diode is reverse biased. No current flows, and the output is zero.
• The output frequency equals the input frequency f (50 pulses per second for a 50 Hz input).

Full-wave rectifier. Two diodes are used with a centre-tapped transformer. The two halves of the secondary winding give voltages in opposite phase.

• During one half-cycle diode D₁ conducts and D₂ is reverse biased. During the next half-cycle D₂ conducts and D₁ is reverse biased.
• The current through the load flows in the same direction in both half-cycles.
• The output frequency is 2f (100 pulses per second for a 50 Hz input).

The full-wave rectifier uses two diodes, conducts in both half-cycles, and has a smaller ripple than the half-wave rectifier.

Filter and ripple. The rectifier output is pulsating DC. Its AC part is called ripple. A capacitor in parallel with the load acts as a filter: it charges when the voltage rises and discharges through the load when the voltage falls. A larger capacitance gives less ripple.

Special-purpose diodes

LED (light-emitting diode). A heavily doped p-n junction, made from a material such as GaAs, GaAsP or GaP. It is forward biased.

• When electrons and holes recombine near the junction, energy is emitted as light. The photon energy is nearly equal to the band gap, so the colour depends on E_g.
• It works in forward bias. Its I–V curve is like that of an ordinary diode, with a knee voltage that is larger (about 1.5 V to 2.5 V) than that of a Si diode.
• Wavelength: λ (nm) = 1240 / E (eV), using hc = 1240 eV nm. An LED for visible light needs E_g of about 1.8 eV to 3 eV.

Photodiode. A p-n junction that is operated in reverse bias and is exposed to light through a window.

• A photon of energy hν ≥ E_g creates electron-hole pairs in the depletion region. These raise the reverse current.
• The reverse current increases with the intensity of the light. So it is used to detect light signals and to measure light intensity.

Solar cell. A p-n junction that converts light into electrical power, with no external bias.

• Photons produce electron-hole pairs near the junction. The junction field separates them: electrons go to the n side and holes to the p side. The p side becomes positive and the n side negative, so an emf appears.
• The I–V curve is drawn in the fourth quadrant, where the voltage is positive and the current is negative (the cell is a source).

Zener diode. A heavily doped p-n junction designed to work in reverse breakdown.

• Its I–V curve in forward bias is like a normal diode. In reverse bias the current is very small until the breakdown (Zener) voltage V_Z. Then the current rises sharply while the voltage across the diode stays almost constant at V_Z.
• Voltage regulator: the Zener is connected in reverse bias in parallel with the load, with a series resistor R_s. If the input or the load changes, the extra current passes through the Zener and the load voltage remains at V_Z.
• Current relations: I (through R_s) = (V_in − V_Z) / R_s; I_load = V_Z / R_L; I_Z = I − I_load.

Device	Bias	Working principle	Use
LED	Forward	Recombination emits light	Indicators, displays
Photodiode	Reverse	Light increases the reverse current	Light detection
Solar cell	None	Light creates an emf at the junction	Power from sunlight
Zener diode	Reverse (breakdown)	Constant voltage at V_Z	Voltage regulator

Logic gates

Logic gates are digital circuits that follow a Boolean rule. The input and output are 0 (low) or 1 (high).

Gate	Symbol in words	Expression	Rule
OR	A shield-like shape (curved back) with two inputs and one output	Y = A + B	Output 1 if any input is 1
AND	A flat-backed, D-shaped body with two inputs	Y = A · B	Output 1 only if all inputs are 1
NOT	A triangle with a small circle at its output; one input	Y = A̅	Output is the opposite of the input
NAND	AND shape with a small circle at the output	Y = (A · B)̅	Output 0 only if all inputs are 1
NOR	OR shape with a small circle at the output	Y = (A + B)̅	Output 1 only if all inputs are 0

Truth table:

A	B	OR	AND	NAND	NOR
0	0	0	0	1	1
0	1	1	0	1	0
1	0	1	0	1	0
1	1	1	1	0	0

NOT gate: input 0 gives output 1, and input 1 gives output 0.

Simple combinations.

• AND followed by NOT = NAND. OR followed by NOT = NOR.
• A NAND gate with its two inputs joined together acts as a NOT gate, because Y = (A · A)̅ = A̅.
• NAND and NOR gates are called universal gates, because any other gate can be built from them.

Worked numericals

Example 1. Law of mass action. In a silicon sample nᵢ = 1.5 × 10¹⁶ m⁻³. It is doped so that nₑ = 4.5 × 10²² m⁻³. Find n_h.

n_h = nᵢ² / nₑ = (1.5 × 10¹⁶)² / (4.5 × 10²²) = 2.25 × 10³² / 4.5 × 10²² = 5 × 10⁹ m⁻³. The sample is n-type.

Example 2. Diode in series with a resistor. A silicon diode (knee voltage 0.7 V) is forward biased in series with a 430 Ω resistor and a 5 V battery. Find the current.

The voltage across the resistor = 5 − 0.7 = 4.3 V. I = 4.3 / 430 = 0.01 A = 10 mA.

Example 3. Zener regulator. A 12 V supply is connected through a 100 Ω resistor to a 6 V Zener diode in parallel with a 200 Ω load.

Current in the series resistor = (12 − 6) / 100 = 0.06 A = 60 mA. Load current = 6 / 200 = 0.03 A = 30 mA. Zener current = 60 − 30 = 30 mA. The Zener power = 6 × 0.03 = 0.18 W.

Example 4. LED wavelength. An LED has a band gap of 2.48 eV. Use hc = 1240 eV nm.

λ = 1240 / 2.48 = 500 nm. The light is green.

Formula summary

Constants: e = 1.6 × 10⁻¹⁹ C; h = 6.63 × 10⁻³⁴ J s; c = 3 × 10⁸ m s⁻¹; 1 eV = 1.6 × 10⁻¹⁹ J; hc = 1240 eV nm.

Quantity	Formula	Notes
Intrinsic carriers	nₑ = n_h = nᵢ	pure semiconductor
Mass action	nₑ n_h = nᵢ²	any semiconductor, fixed temperature
Barrier field	E = V/d	V in volt, d in m
Photon energy	E = hc/λ, so λ (nm) = 1240 / E (eV)	wavelength to band gap
Condition for pair creation	hν ≥ E_g	photodiode, solar cell
Output frequency	half-wave f; full-wave 2f	f is the input frequency
Zener current	I_Z = (V_in − V_Z)/R_s − V_Z/R_L	Zener regulator
NOT, AND, OR	A̅, A · B, A + B	Boolean expressions
NAND, NOR	(A · B)̅, (A + B)̅	universal gates

NEET traps

• 1. In a p-type semiconductor the holes are the majority carriers, but the crystal is still electrically neutral. It is not positively charged.
• 2. n-type uses pentavalent dopants (donors); p-type uses trivalent dopants (acceptors).
• 3. The minority carriers decide the reverse current. The majority carriers decide the forward current.
• 4. In forward bias the depletion region narrows; in reverse bias it widens.
• 5. The knee voltage is about 0.7 V for Si and 0.3 V for Ge. Do not swap them.
• 6. The resistance of a semiconductor falls when the temperature rises; the resistance of a metal rises.
• 7. The full-wave rectifier doubles the output frequency; the half-wave rectifier does not change it.
• 8. A photodiode works in reverse bias, an LED in forward bias, and a solar cell with no bias.
• 9. A Zener diode is used in reverse breakdown, with a series resistor, and its voltage stays nearly constant.
• 10. A NAND gate gives 0 only when all inputs are 1; a NOR gate gives 1 only when all inputs are 0. An OR gate followed by NOT is NOR, not NAND.

One-liners

• 1. Band gap of silicon is 1.1 eV and that of germanium is 0.67 eV.
• 2. In a metal the valence and conduction bands overlap.
• 3. An insulator has a band gap greater than about 3 eV.
• 4. In a pure semiconductor, nₑ = n_h = nᵢ.
• 5. A pentavalent dopant makes an n-type semiconductor; a trivalent dopant makes a p-type semiconductor.
• 6. The product nₑ n_h equals nᵢ² for every semiconductor.
• 7. The depletion region has no mobile carriers and has a width of about 10⁻⁶ m.
• 8. The barrier potential is about 0.7 V for Si and 0.3 V for Ge.
• 9. Forward bias lowers the barrier; reverse bias raises it.
• 10. A half-wave rectifier gives 50 Hz output for a 50 Hz input; a full-wave rectifier gives 100 Hz.
• 11. A capacitor connected across the load acts as a filter to reduce ripple.
• 12. An LED is forward biased and its light colour depends on its band gap.
• 13. A photodiode is reverse biased and its current increases with the light intensity.
• 14. A Zener diode keeps the voltage constant in reverse breakdown.
• 15. NAND and NOR gates are universal gates.

Practice questions

1. Which of the following has an energy gap of about 1.1 eV?

   1. A metal such as copper
   2. Germanium
   3. Silicon
   4. An insulator such as diamond
   Answer

   C. Silicon

   Si has E_g = 1.1 eV and Ge has 0.67 eV. A metal has no gap and an insulator has more than about 3 eV.

2. A silicon sample has an intrinsic carrier concentration nᵢ = 2 × 10¹⁶ m⁻³. It is doped so that the electron concentration is 1 × 10²² m⁻³. The hole concentration is

   1. 4 × 10¹⁰ m⁻³
   2. 4 × 10⁵⁴ m⁻³
   3. 2 × 10⁻⁶ m⁻³
   4. 2 × 10¹⁰ m⁻³
   Answer

   A. 4 × 10¹⁰ m⁻³

   n_h = nᵢ²/nₑ = (2 × 10¹⁶)² / 10²² = 4 × 10³² / 10²² = 4 × 10¹⁰ m⁻³. Check: nₑ × n_h = 10²² × 4 × 10¹⁰ = 4 × 10³² = nᵢ².

3. A p-type semiconductor is obtained by doping pure silicon with

   1. phosphorus
   2. boron
   3. antimony
   4. arsenic
   Answer

   B. boron

   Boron is trivalent (an acceptor). Phosphorus, arsenic and antimony are pentavalent and give n-type.

4. In a p-type semiconductor the majority charge carriers are

   1. electrons
   2. positive ions that can move
   3. negative ions that can move
   4. holes
   Answer

   D. holes

   Trivalent doping makes holes the majority carriers. The ions are fixed in the lattice.

5. The barrier potential across a germanium p-n junction is about

   1. 0.7 V
   2. 0.67 V
   3. 1.1 V
   4. 0.3 V
   Answer

   D. 0.3 V

   About 0.3 V for Ge and 0.7 V for Si. (0.67 eV is the band gap of Ge.)

6. A silicon diode (knee voltage 0.7 V) is forward biased in series with a 860 Ω resistor and a 5 V cell. The current is

   1. 5 mA
   2. 8.1 mA
   3. 4.3 mA
   4. 5.8 mA
   Answer

   A. 5 mA

   Voltage across R = 5 − 0.7 = 4.3 V. I = 4.3 / 860 = 5 × 10⁻³ A = 5 mA. Check: 5 mA × 860 Ω = 4.3 V.

7. A full-wave rectifier is fed with an AC input of frequency 50 Hz. The frequency of the output pulses is

   1. 200 Hz
   2. 50 Hz
   3. 100 Hz
   4. 25 Hz
   Answer

   C. 100 Hz

   The output frequency of a full-wave rectifier is 2f = 100 Hz.

8. A depletion region of width 2 × 10⁻⁶ m has a potential difference of 0.6 V across it. The average electric field in the region is

   1. 1.2 × 10⁶ V m⁻¹
   2. 3 × 10⁵ V m⁻¹
   3. 1.2 × 10⁻⁶ V m⁻¹
   4. 3 × 10⁻⁶ V m⁻¹
   Answer

   B. 3 × 10⁵ V m⁻¹

   E = V/d = 0.6 / (2 × 10⁻⁶) = 3 × 10⁵ V m⁻¹. Check: 3 × 10⁵ × 2 × 10⁻⁶ = 0.6 V.

9. An LED has a band gap of 2.0 eV. Taking hc = 1240 eV nm, the wavelength of the light it emits is about

   1. 620 nm
   2. 1240 nm
   3. 62 nm
   4. 310 nm
   Answer

   A. 620 nm

   λ = hc/E = 1240 / 2.0 = 620 nm. Check: 1240 / 620 = 2.0 eV.

10. Light of wavelength 400 nm falls on a semiconductor. Taking hc = 1240 eV nm, the energy of each photon is

    1. 4.96 eV
    2. 1.55 eV
    3. 3.1 eV
    4. 0.31 eV
    Answer

    C. 3.1 eV

    E = 1240 / 400 = 3.1 eV. Check: 3.1 eV × 400 nm = 1240 eV nm.

11. A 15 V supply feeds a Zener diode of breakdown voltage 9 V through a 250 Ω series resistor. A 600 Ω load is connected across the Zener diode. The current through the Zener diode is

    1. 15 mA
    2. 39 mA
    3. 24 mA
    4. 9 mA
    Answer

    D. 9 mA

    Total current = (15 − 9)/250 = 24 mA. Load current = 9/600 = 15 mA. I_Z = 24 − 15 = 9 mA.

12. A photon of wavelength 620 nm is absorbed in a semiconductor. Taking hc = 1240 eV nm, the largest band gap across which this photon can create an electron-hole pair is about

    1. 1.2 eV
    2. 2.0 eV
    3. 0.5 eV
    4. 620 eV
    Answer

    B. 2.0 eV

    E = 1240 / 620 = 2.0 eV. A gap of 2.0 eV or less can be bridged.

13. A NAND gate has its two inputs joined together to a single input A. The circuit behaves as

    1. a NOT gate
    2. an AND gate
    3. a NOR gate
    4. an OR gate
    Answer

    A. a NOT gate

    Y = (A · A)̅ = A̅. When A = 0 the output is 1 and when A = 1 the output is 0.

14. A germanium diode (knee voltage 0.3 V) is forward biased in series with a 470 Ω resistor and a 5 V cell. The current is

    1. 9.1 mA
    2. 10.6 mA
    3. 10 mA
    4. 11.3 mA
    Answer

    C. 10 mA

    Voltage across R = 5 − 0.3 = 4.7 V. I = 4.7/470 = 0.01 A = 10 mA. Check: 10 mA × 470 Ω = 4.7 V.

15. When an ordinary p-n junction is forward biased, its depletion region

    1. disappears completely and stays so at any voltage
    2. becomes wider
    3. remains unchanged
    4. becomes narrower
    Answer

    D. becomes narrower

    The forward bias opposes the barrier field, so the barrier height falls and the depletion region narrows.

16. Which logic gate gives output 1 only when both of its inputs are 0?

    1. NAND
    2. NOR
    3. OR
    4. AND
    Answer

    B. NOR

    NOR(0, 0) = NOT(0 + 0) = 1; for every other input the output is 0. NAND(0, 0) = 1 too, but NAND is also 1 for (0,1) and (1,0).

17. Given below are two statements: Statement I: In a pure semiconductor the number of electrons equals the number of holes. Statement II: In an n-type semiconductor the product of the electron and hole concentrations is equal to nᵢ².

    1. Both Statement I and Statement II are correct
    2. Both Statement I and Statement II are incorrect
    3. Statement I is correct but Statement II is incorrect
    4. Statement I is incorrect but Statement II is correct
    Answer

    A. Both Statement I and Statement II are correct

    Electrons and holes are created in pairs. The mass-action relation nₑ n_h = nᵢ² holds for doped samples as well.

18. Given below are two statements: Statement I: The resistance of a semiconductor increases when its temperature rises. Statement II: A p-type semiconductor is made by doping with a pentavalent impurity.

    1. Both Statement I and Statement II are correct
    2. Both Statement I and Statement II are incorrect
    3. Statement I is correct but Statement II is incorrect
    4. Statement I is incorrect but Statement II is correct
    Answer

    B. Both Statement I and Statement II are incorrect

    The resistance of a semiconductor falls with temperature, and p-type needs a trivalent impurity. Both are incorrect.

19. Given below are two statements: Statement I: The reverse current of a p-n junction is due to minority carriers. Statement II: Forward bias increases the height of the potential barrier.

    1. Both Statement I and Statement II are correct
    2. Both Statement I and Statement II are incorrect
    3. Statement I is correct but Statement II is incorrect
    4. Statement I is incorrect but Statement II is correct
    Answer

    C. Statement I is correct but Statement II is incorrect

    Reverse current is carried by minority carriers. Forward bias lowers the barrier, so Statement II is incorrect.

20. Which of the following statements about rectifiers are correct? 1. The output frequency of a half-wave rectifier equals the input frequency. 2. A centre-tap full-wave rectifier uses two diodes. 3. The output frequency of a full-wave rectifier equals the input frequency.

    1. 2 and 3 only
    2. 1 and 3 only
    3. 1 and 2 only
    4. 1, 2 and 3
    Answer

    C. 1 and 2 only

    Statements 1 and 2 are correct. A full-wave rectifier gives 2f, so 3 is wrong.

21. Given below are two statements: Statement I: A solar cell is operated with an external forward bias. Statement II: A photodiode is operated in reverse bias.

    1. Both Statement I and Statement II are correct
    2. Both Statement I and Statement II are incorrect
    3. Statement I is correct but Statement II is incorrect
    4. Statement I is incorrect but Statement II is correct
    Answer

    D. Statement I is incorrect but Statement II is correct

    A solar cell works with no external bias, so Statement I is incorrect. A photodiode is reverse biased, so Statement II is correct.

22. Given below are two statements: Statement I: A Zener diode used as a voltage regulator is connected in reverse bias. Statement II: Once in breakdown, the voltage across a Zener diode stays almost constant while the current changes.

    1. Both Statement I and Statement II are correct
    2. Both Statement I and Statement II are incorrect
    3. Statement I is correct but Statement II is incorrect
    4. Statement I is incorrect but Statement II is correct
    Answer

    A. Both Statement I and Statement II are correct

    Both are correct. This constant voltage V_Z at the load is what makes the regulator work.

23. Which of the following statements about energy bands are correct? 1. In a metal the conduction and valence bands overlap, or the conduction band is partly filled. 2. The band gap of germanium is smaller than that of silicon. 3. An insulator has a band gap of less than 1 eV.

    1. 1 only
    2. 1 and 2 only
    3. 2 and 3 only
    4. 1, 2 and 3
    Answer

    B. 1 and 2 only

    1 and 2 are correct (Ge 0.67 eV, Si 1.1 eV). An insulator has a gap greater than about 3 eV, so 3 is wrong.

24. Given below are two statements: Statement I: An OR gate gives output 1 for inputs A = 1 and B = 0. Statement II: A NAND gate gives output 0 for inputs A = 0 and B = 0.

    1. Both Statement I and Statement II are correct
    2. Both Statement I and Statement II are incorrect
    3. Statement I is correct but Statement II is incorrect
    4. Statement I is incorrect but Statement II is correct
    Answer

    C. Statement I is correct but Statement II is incorrect

    OR(1, 0) = 1. NAND(0, 0) = NOT(0 · 0) = 1, so Statement II is incorrect.

25. Given below are two statements: Statement I: An LED with a band gap of 2.48 eV emits light of wavelength 500 nm (hc = 1240 eV nm). Statement II: Light of longer wavelength has photons of greater energy.

    1. Both Statement I and Statement II are correct
    2. Both Statement I and Statement II are incorrect
    3. Statement I is correct but Statement II is incorrect
    4. Statement I is incorrect but Statement II is correct
    Answer

    C. Statement I is correct but Statement II is incorrect

    λ = 1240 / 2.48 = 500 nm. Since E = hc/λ, longer wavelength means less energy, so Statement II is incorrect.

26. Given below are two statements: Statement I: The depletion region of a p-n junction contains a large number of mobile electrons and holes. Statement II: The barrier potential of a silicon junction is about 0.7 V.

    1. Both Statement I and Statement II are correct
    2. Both Statement I and Statement II are incorrect
    3. Statement I is correct but Statement II is incorrect
    4. Statement I is incorrect but Statement II is correct
    Answer

    D. Statement I is incorrect but Statement II is correct

    The depletion region is empty of mobile carriers, so Statement I is incorrect. Statement II is correct.

27. Assertion (A): The resistance of a semiconductor decreases as its temperature increases. Reason (R): More electron-hole pairs are created at a higher temperature.

    1. Both A and R are true and R is the correct explanation of A
    2. Both A and R are true but R is not the correct explanation of A
    3. A is true but R is false
    4. A is false but R is true
    Answer

    A. Both A and R are true and R is the correct explanation of A

    More carriers mean greater conductivity, hence lower resistance.

28. Assertion (A): A half-wave rectifier gives an output of the same frequency as the input. Reason (R): A half-wave rectifier uses a single diode.

    1. Both A and R are true and R is the correct explanation of A
    2. Both A and R are true but R is not the correct explanation of A
    3. A is true but R is false
    4. A is false but R is true
    Answer

    B. Both A and R are true but R is not the correct explanation of A

    Both are true. The output has one pulse per input cycle because only one half-cycle is conducted; the number of diodes is not the explanation.

29. Assertion (A): The NAND gate is a universal gate. Reason (R): A NAND gate gives output 1 only when all its inputs are 1.

    1. Both A and R are true and R is the correct explanation of A
    2. Both A and R are true but R is not the correct explanation of A
    3. A is true but R is false
    4. A is false but R is true
    Answer

    C. A is true but R is false

    Any gate can be built from NAND gates. A NAND gate gives 0, not 1, when all inputs are 1, so R is false.

30. Assertion (A): The width of the depletion region decreases in reverse bias. Reason (R): Reverse bias increases the height of the potential barrier.

    1. Both A and R are true and R is the correct explanation of A
    2. Both A and R are true but R is not the correct explanation of A
    3. A is true but R is false
    4. A is false but R is true
    Answer

    D. A is false but R is true

    Reverse bias widens the depletion region and raises the barrier. A is false and R is true.

31. Assertion (A): A Zener diode is used as a voltage regulator. Reason (R): In breakdown the voltage across the Zener diode stays nearly constant over a wide range of current.

    1. Both A and R are true and R is the correct explanation of A
    2. Both A and R are true but R is not the correct explanation of A
    3. A is true but R is false
    4. A is false but R is true
    Answer

    A. Both A and R are true and R is the correct explanation of A

    The nearly constant V_Z keeps the load voltage steady.

32. Assertion (A): A solar cell is operated with a forward-bias battery connected across it. Reason (R): A solar cell gives an emf because light creates electron-hole pairs at the junction, which the junction field separates.

    1. Both A and R are true and R is the correct explanation of A
    2. Both A and R are true but R is not the correct explanation of A
    3. A is true but R is false
    4. A is false but R is true
    Answer

    D. A is false but R is true

    A solar cell works without an external bias; A is false. R is the true description of its working.

33. Match List I with List II. List I: a. LED b. Photodiode c. Solar cell d. Zener diode. List II: i. Forward bias ii. Reverse bias iii. No external bias iv. Reverse breakdown

    1. a-iii, b-ii, c-i, d-iv
    2. a-i, b-ii, c-iii, d-iv
    3. a-ii, b-i, c-iii, d-iv
    4. a-i, b-ii, c-iv, d-iii
    Answer

    B. a-i, b-ii, c-iii, d-iv

    LED is forward biased, the photodiode reverse biased, the solar cell has no external bias, and the Zener diode works in reverse breakdown.

34. Match List I with List II. List I: a. Majority carriers in n-type b. Majority carriers in p-type c. Intrinsic semiconductor d. Acceptor impurity. List II: i. Electrons ii. Holes iii. Equal numbers of electrons and holes iv. Trivalent atom

    1. a-iii, b-ii, c-i, d-iv
    2. a-i, b-ii, c-iii, d-iv
    3. a-ii, b-i, c-iii, d-iv
    4. a-i, b-ii, c-iv, d-iii
    Answer

    B. a-i, b-ii, c-iii, d-iv

    n-type: electrons; p-type: holes; intrinsic: nₑ = n_h; an acceptor is a trivalent atom.

35. Match List I with List II. List I (gate): a. OR b. AND c. NAND d. NOR. List II (expression): i. (A + B)̅ ii. A · B iii. A + B iv. (A · B)̅

    1. a-iv, b-ii, c-iii, d-i
    2. a-ii, b-iii, c-iv, d-i
    3. a-iii, b-ii, c-i, d-iv
    4. a-iii, b-ii, c-iv, d-i
    Answer

    D. a-iii, b-ii, c-iv, d-i

    OR is A + B, AND is A · B, NAND is (A · B)̅, NOR is (A + B)̅.

36. Match List I with List II. List I: a. Output frequency of a half-wave rectifier for a 50 Hz input b. Output frequency of a full-wave rectifier for a 50 Hz input c. Number of diodes in a centre-tap full-wave rectifier d. Number of diodes in a half-wave rectifier. List II: i. 100 Hz ii. 1 iii. 50 Hz iv. 2

    1. a-iii, b-i, c-iv, d-ii
    2. a-i, b-iii, c-iv, d-ii
    3. a-iv, b-iii, c-i, d-ii
    4. a-iii, b-i, c-ii, d-iv
    Answer

    A. a-iii, b-i, c-iv, d-ii

    Half-wave: f = 50 Hz with one diode. Full-wave: 2f = 100 Hz with two diodes.

37. Match List I with List II. List I: a. Band gap of silicon b. Band gap of germanium c. Barrier potential of silicon junction d. Barrier potential of germanium junction. List II: i. 0.3 V ii. 0.67 eV iii. 1.1 eV iv. 0.7 V

    1. a-ii, b-iii, c-iv, d-i
    2. a-iv, b-ii, c-iii, d-i
    3. a-iii, b-ii, c-iv, d-i
    4. a-iii, b-ii, c-i, d-iv
    Answer

    C. a-iii, b-ii, c-iv, d-i

    Si: 1.1 eV and 0.7 V. Ge: 0.67 eV and 0.3 V.

38. Match List I with List II. List I: a. Metal b. Semiconductor c. Insulator d. Forward-biased p-n junction. List II: i. Band gap greater than about 3 eV ii. Conduction and valence bands overlap iii. Small band gap, less than about 3 eV iv. Depletion region narrows

    1. a-ii, b-iii, c-i, d-iv
    2. a-iii, b-ii, c-i, d-iv
    3. a-ii, b-i, c-iii, d-iv
    4. a-ii, b-iii, c-iv, d-i
    Answer

    A. a-ii, b-iii, c-i, d-iv

    Metal: overlapping bands; semiconductor: small gap; insulator: large gap; forward bias narrows the depletion region.

39. A signal A goes to one input of an OR gate. A signal B goes through a NOT gate to the other input of the same OR gate. For which input values (A, B) is the output Y equal to 0?

    1. A = 1, B = 0
    2. A = 0, B = 1
    3. A = 1, B = 1
    4. A = 0, B = 0
    Answer

    B. A = 0, B = 1

    Y = A + B̅. Y = 0 needs A = 0 and B̅ = 0, so B = 1. Row check: (0,0) gives 0 + 1 = 1; (0,1) gives 0 + 0 = 0; (1,0) gives 1; (1,1) gives 1.

40. The output of a NOR gate is fed into a NOT gate. The combination behaves as a single

    1. NOT gate
    2. AND gate
    3. NAND gate
    4. OR gate
    Answer

    D. OR gate

    NOT(NOR(A, B)) = A + B. Row check: (0,0) → NOR 1 → 0; (0,1) → 0 → 1; (1,0) → 0 → 1; (1,1) → 0 → 1. This is the OR truth table.

41. A Zener diode of breakdown voltage 8 V is connected in parallel with a 400 Ω load. The combination is connected to an 18 V source through a 200 Ω series resistor. The power dissipated in the Zener diode is

    1. 0.9 W
    2. 0.16 W
    3. 0.4 W
    4. 0.24 W
    Answer

    D. 0.24 W

    Total current = (18 − 8)/200 = 50 mA. Load current = 8/400 = 20 mA. I_Z = 30 mA. P = 8 × 0.03 = 0.24 W.

42. A photodiode has a band gap of 1.24 eV. Taking hc = 1240 eV nm, it can detect light of wavelength up to about

    1. 500 nm
    2. 1000 nm
    3. 1.0 nm
    4. 2000 nm
    Answer

    B. 1000 nm

    Pair creation needs hν ≥ E_g, so λ ≤ hc/E_g = 1240 / 1.24 = 1000 nm.

43. A student connects a much larger capacitor across the load resistor of a rectifier. The ripple in the output voltage will

    1. increase
    2. become 100% and the output will be zero
    3. decrease
    4. remain exactly the same
    Answer

    C. decrease

    A larger capacitor discharges more slowly through the load, so the output is smoother and the ripple falls.

44. An n-type silicon crystal has an electron concentration of 5 × 10²² m⁻³. The intrinsic concentration of silicon is 1.5 × 10¹⁶ m⁻³. The hole concentration is

    1. 4.5 × 10⁹ m⁻³
    2. 7.5 × 10³⁸ m⁻³
    3. 1.5 × 10¹⁶ m⁻³
    4. 3 × 10⁻⁷ m⁻³
    Answer

    A. 4.5 × 10⁹ m⁻³

    n_h = nᵢ²/nₑ = 2.25 × 10³² / 5 × 10²² = 4.5 × 10⁹ m⁻³. The holes are the minority carriers.

45. A full-wave rectifier is fed with a 50 Hz AC supply. The time interval between two successive output pulses is

    1. 5 ms
    2. 20 ms
    3. 100 ms
    4. 10 ms
    Answer

    D. 10 ms

    Output frequency = 100 Hz. T = 1/100 = 0.01 s = 10 ms. Check: the input period is 20 ms and two pulses occur in each period.