## Semiconductors

**Question 1. In a semiconductor at room temperature,**

- The valence band is completely filled
- The valence band is partially empty and the conduction band is partially filled
- The conduction band is completely empty
- The valence band is partially filled and the conduction band is partially filled

**Answer:** 2. The valence band is partially empty and the conduction band is partially filled

At room temperature in semiconductors, the electrons from the covalent bond get sufficient thermal energy so that they get detached from the bond and are able to overcome the forbidden energy gap. Thus, the valence band is partially empty (or ‘holes’ are created) and the conduction band is partially filled. At the absolute zero temperature (0 K), the conduction band in a semiconductor is completely empty and thus behaves as a perfect insulator.

**Question 2. In a p-type semiconductor, the majority carriers of current are**

- Protons
- Holes
- Neutrons
- Electrons

**Answer:** 2. Holes

When a trivalent impurity, such as boron (B) or gallium (Ga), is added to an intrinsic (or pure) semiconductor, such as silicon (Si) or germanium (Ge), it is said to be a p-type semiconductor. Each boron atom has three valence electrons and each silicon atom has four.

The three valence electrons of each. boron atoms form three covalent bonds with the three neighboring atoms. In the fourth covalent bond, one electron from a silicon atom is present and the bond is incomplete with the shortage of one electron.

This missing electron is called a hole. These holes act as the majority charge carriers in the conduction of electricity in a p-type semiconductor.

**Question 3. The depletion layer consists of**

- Electrons
- Protons
- Mobile ions
- Immobile ions

**Answer:** 4. Immobile ions

A depletion layer is formed instantaneously across a p-n junction due to the diffusion of electrons and holes. The n-side has an excess of electrons, which diffuse through the p-side. Similarly, holes diffuse from the p-side to the n-side. During the diffusion, the majority of charge carriers (free electrons in the n-type and holes in the p-type) are gone due to recombinations. This constitutes the depletion layer containing only immobile ions.

**Question 4. At absolute zero, Si acts as**

- A conducting metal
- A nonmetal
- An insulator
- None of these

**Answer:** 3. An insulator

At absolute zero, all the electrons are tied to their atoms in the valence band and no thermal energy is available to detach the electrons. This means that the conduction band is completely empty (no free electrons). Hence, the semiconducting material (Si) acts as an insulator.

**Question 5. When an n-type semiconductor is heated**,

- The number of electrons increases, while that of holes decreases
- The number of holes increases, while that of electrons decreases
- The number of electrons and holes increases equally
- The number of electrons and holes remains the same

**Answer:** 3. The numbers of electrons and holes increase equally

When an n-type extrinsic semiconductor is heated, the thermal energy is sufficient for the electron to break the bond and become free. Simultaneously, a hole is also created. The numbers of electrons and holes remain equal and increase equally with time.

**Question 6. The depletion layer in a p-n junction region is caused by**

- Drift of holes
- Diffusion of charge carriers
- Migration of impurity ions
- Drift of electrons

**Answer:** 2. Diffusion of charge carriers

In a p-n junction, the depletion layer is caused by the diffusion of charge carriers and the consequent recombinations.

**Question 7. In a junction diode, the holes are nothing but**

- Protons
- Neutrons
- Missing of electrons
- Extra electrons

**Answer:** 3. Missing of electrons

Holes and electrons are the two types of charge carriers responsible for currents in a semiconductor. A hole is nothing but the absence of an electron. Although a hole is not a physical particle like an electron, it can ‘shift’ from one atom to another in a semiconductor.

**Question 8. To obtain a p-type germanium semiconductor, it must be doped with**

- Indium
- Arsenic
- Antimony
- Phosphorus

**Answer:** 1. Indium

When an intrinsic tetravalent semiconductor (such as germanium) is doped with a trivalent impurity (such as indium) atom, the impurity atom leaves an empty space in the bond structure and the intrinsic semiconductor becomes a p-type semiconductor.

**Question 9. When arsenic is added as an impurity to silicon, the resulting material is**

- An n-type conductor
- An n-type semiconductor
- A p-type semiconductor
- None of these

**Answer:** 2. An n-type semiconductor

Arsenic is a pentavalent impurity, which when doped with a silicon semiconductor leaves one free electron. Such free electrons (negative particles) constitute currents and thus the semiconductor is called an n-type semiconductor.

**Question 10. Which one of the following statements is false?**

- Pure Si doped with trivalent impurities gives a p-type semiconductor.
- The majority of carriers in an n-type semiconductor are holes.
- The minority carriers in a p-type semiconductor are electrons.
- The resistance of an intrinsic semiconductor decreases with an increase in temperature.

**Answer:** 2. The majority of carriers in an n-type semiconductor are holes.

In a p-type semiconductor, silicon is doped with a trivalent impurity atom leaving behind a hole in the covalent bonds. Here, the majority of carriers are holes and not electrons

**Question 11. The device that can act as a complete electronic circuit is an**

- Junction diode
- Integrated circuit
- Zener diode
- Junction transition

**Answer:** 2. Integrated circuit

An integrated circuit (IC), also called a chip or a microchip, is a set of electronic circuits on one small flat piece of a semiconductor material (normally silicon)

**Question 12. The application of a forward bias to a p-n junction**

- Widens the depletion zone
- Increases the potential difference across the depletion zone
- Increases the number of donors on the n-side
- Decreases the electric field in the depletion zone

**Answer:** 4. Decreases the electric field in the depletion zone

The electric field across a p-n junction is reduced which thus supports the flow of electric current).

**Question 13. Which of the following elements produces an n-type semiconductor when added as an impurity into silicon?**

- B
- P
- Al
- Mg

**Answer: **2. P

For the growth of an n-type semiconductor, silicon (a tetravalent semiconductor) has to be doped with a pentavalent impurity, such as phosphorus.

**Question 14. A semiconductor device is connected in series with a battery and a resistor. A current is found to pass through the circuit. If the polarity of the battery is reversed, the current drops to almost zero. The device may be**

- A p-type semiconductor
- An intrinsic semiconductor
- A p-n junction diode
- An n-type semiconductor

**Answer:** 3. A p-n junction diode

A p-n junction diode conducts fairly well in its forward bias. Reversing the polarity of the cell brings the diode in the reverse bias, which restricts the current. Hence, the device is a p-n junction diode.

**Question 15. In forward-biasing a p-n junction diode,**

- The positive terminal of the battery is connected to the p-side and the depletion region becomes thick
- The positive terminal of the battery is connected to the n-side and the depletion layer becomes thin
- The positive terminal of the battery is connected to the n-side and the depletion region becomes thick
- The positive terminal of the battery is connected to the p-side and the depletion region becomes thin.

**Answer:** 4. The positive terminal of the battery is connected to the p-side and the depletion region becomes thin.

In a p-n junction diode with forward bias, the depletion layer becomes thin.

**Question 16. A p-n junction photodiode is made of a material having a band gap of 2.0 eV. The minimum frequency of radiation that can be absorbed by the material is nearly**

- 1 x 10
^{14}Hz - 20 x 10
^{14}Hz - 5 x 10
^{14}Hz - 10 x 10
^{14}Hz

**Answer:** 3. 5 x 10^{14} Hz

The incident photon (radiation) must have an energy of 2.0 eV.

Now, E = hv

∴ \(\mathrm{v}=\frac{E}{h}=\frac{2.0 \mathrm{eV}}{6.67 \times 10^{-34} \mathrm{Js}}=\frac{2.0 \times 1.6 \times 10^{-19} \mathrm{~J}}{6.67 \times 10^{-34} \mathrm{~J} \mathrm{~s}}\)

= 4.8 x 10^{14} Hz ≈ 5.0 x 10^{14} Hz.

**Question 17. A p-n junction photodiode is fabricated from a semiconductor having a band gap of 2.5 eV. It can detect a signal of wavelength**

- 6000 Å
- 6000 nm
- 4000 Å
- 4000 nm

**Answer:** 3. 4000 Å

Energy of photon = \(E=\frac{h c}{\lambda} \Rightarrow \lambda=\frac{h c}{E}\)

Given that band gap = E = 2.5 eV and he =1242 eV nm.

∴ \(\lambda=\frac{h c}{E}=\frac{1242 \mathrm{eV} \mathrm{nm}}{2.5 \mathrm{eV}}=496.8 \mathrm{~nm}=4968 Å\)

Hence, the required wavelength will be 4000 Å

**Question 18. In a p-n junction photocell, the value of the photoelectromotive force produced by a beam of monochromatic light is proportional to the**

- Barrier voltage at the p-n junction
- The intensity of the light falling on the cell
- Frequency of the light falling on the cell
- Voltage applied at the p-n junction

**Answer:** 2. Intensity of the light falling on the cell

In a p-n junction photocell, the photoelectromotive force produced by incident monochromatic radiation is proportional to the intensity of the incident radiation.

**Question 19. Which of the following bonds produces a solid that reflects light in the visible region whose electrical conductivity decreases with temperature and has a high melting point?**

- Metallic bonding
- Ionic bonding
- van der Waals bonding
- Covalent bonding

**Answer:** 1. Metallic bonding

For solids with metallic bonding, there is a reflection in the visible range. They also have a high melting point, and the electrical conductivity decreases with an increase in temperature.

**Question 20. The carbon, silicon, and germanium atoms have four valence electrons each. Their valence and conduction bands are separated by energy-band gaps represented by EGC, EGSi, and EGGe respectively. Which of the following relationships is true in their cases?**

- EG
_{C}> EG_{Si} - EG
_{C}– EG_{Si} - EG
_{C}< EG_{Ge} - EG
_{C}< EG_{Si}

**Answer:** 1. EG_{C} > EG_{Si}

The energy-band gap in carbon is 5.5 eV, that in silicon is 1,14 eV, and that in germanium is 0.67 eV. Thus, EG_{C} > EG_{Si}.

**Question 21. A reverse bias applied across a p-n junction diode**

- Lowers the potential barrier
- Raises the potential barrier
- Increases the majority-carrier current
- Increases the minority-carrier current

**Answer:** 2. Raises the potential barrier

A p-n junction diode operating in a reverse bias has its potential barrier raised across the depletion layer.

**Question 22. A p-n junction diode can be used as**

- A capacitor
- An amplifier
- A rectifier
- A regulator

**Answer:** 3. A rectifier

A p-n junction conducts only in a forward bias and offers high resistance in a reverse bias. Hence, it is used as a rectifier.

**Question 23. The cause of the potential barrier in a p-n junction diode is the**

- Depletion of negative charges near the junction
- The concentration of positive charges near the junction
- Depletion of positive charges near the junction
- The concentration of positive and negative charges near the junction

**Answer:** 4. Concentration of positive and negative charges near the junction

During the formation of a junction diode, holes from the p-region diffuse into the n-region, and electrons from the n-region diffuse into the p-region.

In the course of this diffusion process, a thin layer at the junction becomes free from any charge carriers and is called the depletion layer. There is a potential gradient in the depletion layer – negative on the p-side and positive on the n-side. This is called the potential barrier.

**Question 24. A piece of copper and another of germanium are cooled from room temperature to 80 K. Then,**

- The resistance of each will increase
- The resistance of copper will decrease
- The resistance of copper will decrease, while that of germanium will increase
- The resistance of copper will increase, while that of germanium will decrease

**Answer:** 3. The resistance of copper will decrease, while that of germanium will increase

With a decrease in temperature, the resistance of copper (a metallic conductor) decreases due to a decrease in thermal agitation. Germanium (a semiconductor) has its resistance increased, as the thermal energy is insufficient to free the electrons from the covalent bonds.

**Question 25. In a forward bias, the width of the potential barrier in a p-n junction diode**

- Increases
- Decreases
- Remains constant
- First increases, then decreases

**Answer:** 2. Decreases

As more electrons and holes flow in the depletion region in a forward bias, the number of positive and negative ions is reduced, which causes the width of the potential barrier to narrow.

**Question 26. The barrier potential of a p-n junction depends on**

- The type of the semiconductor material,
- The amount of doping,
- Temperature.

**Which of the following combinations is correct?**

- (1) and (2) only
- (1) only
- (2) and (3) only
- (1), (2), and (3)

**Answer:** 4. (1), (2) and (3)

The potential barrier of a p-n junction depends on

- The type of the material (Si: 1.14 V but Ge: 0.67 V),
- The amount of doping, and
- The temperature.

**Question 27. For an n-type semiconductor, which of the following statements is true?**

- Electrons are majority carriers and trivalent atoms are dopants.
- Electrons are minority carriers and pentavalent atoms are dopants.
- Holes are minority carriers and pentavalent atoms are dopants.
- Holes are majority carriers and pentavalent atoms are dopants.

**Answer:** 3. Holes are minority carriers and pentavalent atoms are dopants.

Doping an intrinsic tetravalent semiconductor (Si or Ge) by a pentavalent impurity brings out a drastic increase in conductivity due to electrons as the majority charge carriers and holes as the minority charge carriers.

**Question 28. If a small amount of antimony is added to a germanium crystal,**

- The antimony atoms become acceptor atoms
- There will be more free electrons than holes in the semiconductor
- The resistance of the germanium crystal increases
- The germanium crystal becomes a p-type semiconductor

**Answer:** 2. There will be more free electrons than holes in the semiconductor

On adding a pentavalent impurity, such as antimony, to germanium, the intrinsic semiconductor becomes an n-type extrinsic semiconductor. Hence, there will be more free electrons than holes.

**Question 29. Choose only the false statement from the following.**

- Substances with energy gaps of the order of 10 eV are insulators.
- The conductivity of a semiconductor increases with an increase in temperature.
- In conductors, the valence and conduction bands may overlap.
- The resistivity of a semiconductor increases with an increase in temperature.

**Answer:** 4. The resistivity of a semiconductor increases with an increase in temperature.

With an increase in the temperature of a semiconductor, the electrons forming covalent bonds get sufficient thermal energy to get detached. Hence, the conductivity increases and the resistivity decreases. Thus, an increase in resistivity is a false consequence.

**Question 30. In a semiconducting material, the mobilities of electrons and holes are μ _{e} and μ_{h} respectively. Which of the following is true?**

- \(\mu_{\mathrm{e}}>\mu_{\mathrm{h}}\)
- \(\mu_{\mathrm{e}}<\mu_{\mathrm{h}}\)
- \(\mu_e=\mu_h\)
- \(\mu_{\mathrm{e}}<0 \text { but } \mu_{\mathrm{h}}>0\)

**Answer:** 1. \(\mu_{\mathrm{e}}>\mu_{\mathrm{h}}\)

The mobility (μ) of an electron or a hole is defined as the drift velocity per unit electric field (μ = v_{d}/E). The effective mass of an electron is smaller than that of a hole. Hence, the mobility of an electron is higher than that of a hole (μ_{e} > μ_{h}).

**Question 31. When there is a saturation current in a diode, what is its resistance?**

- Zero
- Infinity
- Periodic
- Data insufficient

**Answer:** 2. Infinity

The dynamic resistance is defined as the slope of the V-I characteristic curve representing the equation r = ΔV/ΔI. At saturation, ΔI = 0 but

∴ \(\Delta V \neq 0, \text { so } r=\frac{\Delta V}{0}=\infty\)

**Question 32. The temperature-dependence of resistivity, ρ = f(T), of a semiconductor, is represented by**

**Answer:** 3.

As the temperature of a semiconductor is increased, the electrons in the valence band gain sufficient energy to escape the pull of their atoms. Consequently, a conduction of electrons results. This decreases the resistivity nonlinearly, as shown in the option (3).

**Question 33. The peak voltage in the output of a half-wave diode rectifier fed with a sinusoidal signal without any filter is 10 V. The DC component of the output voltage is**

- 10√2 V
- 10 V
- \(\frac{10}{\pi} \mathrm{V}\)
- \(\frac{20}{\pi} \mathrm{V}\)

**Answer:** 3. \(\frac{10}{\pi} \mathrm{V}\)

The input AC voltage is V = (10 V)sin ωt. The output is rectified to be V = (10 V)sinωt (only for the positive half cycle, as shown). The average output is

⇒ \(V_{\mathrm{av}}=\frac{1}{T} \int V_0 \sin \omega t d t\)

⇒ \(\frac{1}{T}\left[V_0 \int_0^{T / 2} \sin \omega t d t\right]\)

⇒ \(\frac{V_0}{T}\left[\frac{\cos \omega t}{\omega}\right]_{T / 2}^0\)

∴ \(\frac{V_0}{\pi}=\frac{10 \mathrm{~V}}{\pi}\)

**Question 34. If a full-wave rectifier circuit is operating from 50-Hz mains, the fundamental frequency of the ripple will be**

- 25 Hz
- 100 Hz
- 70.7 Hz
- 50 Hz

**Answer:** 2. 100 Hz

In a full-wave rectifier circuit, the output has current and voltage signals for both positive and negative halves of the input sinusoidal cycles. Hence, the frequency of the output will be double that of the input, i.e., 2 x 50 Hz = 100 Hz.

**Question 35. For conduction through a p-n junction diode, which of the following is correct?**

- A high potential on the n-side and a low potential on the p-side
- A high potential on the p-side and a low potential on the n-side
- The same potential on both the p- and n-sides
- Undetermined

**Answer:** 2. A high potential on the p-side and a low potential on the n-side

Conduction occurs in a p-n junction when it is in a forward bias, i.e., when the p-side is connected to its positive terminal (higher potential) and the n-side to its negative terminal (lower potential).

**Question 36. In the energy-band diagram of a material shown in the adjoining figure., the empty circles and filled circles represent holes and electrons respectively. The material is**

- An insulator
- A metal
- An n-type semiconductor
- A p-type semiconductor

**Answer:** 4. A p-type semiconductor

In a p-type semiconductor, holes are the majority carriers (in the valence band) and electrons are the minority carriers, as shown in the energy-band diagram. Hence, the given diagram corresponds to a p-type semiconductor.

**Question 37. In the given circuit, the voltage across the load is 12 V. The current in the Zener diode varies from 0 A to 50 mA. What is the maximum wattage of the diode?**

- 12 W
- 6 W
- 0.6 W
- 1.2 W

**Answer:** 3. 0.6 W

The maximum current through the Zener diode is I_{z} = 50 mA = 0.05 A, and the voltage across the load is V =12 V.

∴ the maximum, wattage of the diode will be

I_{z}V = (0.05 A)(12 V) = 0.6 W.

**Question 38. When a p-n junction diode is reverse-biased, the resistance measured by an ohmmeter will be**

- High
- Zero
- Infinity
- Low

**Answer:** 1. High

In the reverse bias, the obstruction to the current through a p-n junction diode is very high. So, the ohmmeter will measure a high resistance.

**Question 39. If the internal resistance of the cell is negligible then the current flowing through the circuit is**

- 0.06 A
- 0.02 A
- 0.08 A
- 0.1 A

**Answer:** 4. 0.1 A

The diode D_{1} is forward-biased, while D_{2} is reverse-biased. Thus, D_{2} offers a high resistance and hence it is nonconducting. The current through the circuit is

∴ \(I=\frac{5 \mathrm{~V}}{20 \Omega+30 \Omega}=\frac{5 \mathrm{~V}}{50 \Omega}=0.1 \mathrm{~A}\)

**Question 40. In the given figure, a diode D is connected to an external resistor of resistance R = 100 Ω and a battery of emf 3.5 V. If the barrier potential developed across the diode is 0.5 V, the current in the circuit will be**

- 20 mA
- 35 mA
- 30 mA
- 40 mA

**Answer:** 3. 30 mA

The potential drop across the resistor R is

V = 3.5V – 0.5 V = 3.0 V.

∴ the current through the circuit is

⇒ \(I=\frac{V}{R}=\frac{3.0 \mathrm{~V}}{100 \Omega}=0.03 \mathrm{~A}=30 \mathrm{~mA}\)

**Question 41. Two ideal diodes are connected to a battery as shown in the circuit. The current supplied by the battery is**

- 0.75 A
- 0.25 A
- 0.5 A
- zero

**Answer:** 3. 0.5 A

In the given circuit, D_{1} is in a forward bias (conducting), whereas D_{2} is in a reverse bias (nonconducting). Hence, the effective resistance is R =10Ω and the current delivered by the cell is

∴ \(I=\frac{5 \mathrm{~V}}{10 \Omega}=0.5 \mathrm{~A}\)

**Question 42. In the given diagram, the input is across the terminals A and C, and the output is across B and D. Then, the output is**

- Zero
- The same as the input
- Half-wave-rectified
- Full-wave-rectified

**Answer:** 4. Full-wave-rectified

The given circuit containing four p-n junction diodes forms a bridgelike arrangement, as shown in the circuit diagram, which gives a full-wave rectified output, as shown in the graph given below.

An AC voltage source is applied across AC, which gives a DC output across BD. As shown source in the waveforms, in a positive half cycle across AC, the diodes D_{1} and D_{4} are conducting, and the current through the load is along BD. Similarly, in the negative half cycle, the diodes D_{2} and D_{3} are conducting, with the current through R from B to D.

**Question 43. The given circuit has two ideal diodes connected as shown in the figure. The current flowing through the resistor of resistance R _{1} will be**

- 2.5 A
- 10.0 A
- 1.43 A
- 4.13 A

**Answer:** 1. 2.5 A

In the given diagram, D_{1} is nonconducting.

So, the current through the cell is

∴ \(I=\frac{10 \mathrm{~V}}{R_1+R_3}=\frac{10 \mathrm{~V}}{4 \Omega}=2.5 \mathrm{~A}\)

**Question 44. Consider the junction diode as ideal. The value of the current flowing through AB is**

- 10
^{-2 }A - 10
^{-3}A - 10
^{-4}A - 0 A

**Answer:** 1. 10^{-2 }A

Potential difference = V_{A}– V_{B} = 4 V- (-6 V) =10 V.

∴ Current = I = \(\frac{V_{\mathrm{A}}-V_{\mathrm{B}}}{R}=\frac{10 \mathrm{~V}}{1000 \Omega}=0.01 \mathrm{~A}=10^{-2} \mathrm{~A}\)

**Question 45. Which one of the following represents a forward-biased diode?**

**Answer:** 1.

A p-n junction diode is in its forward bias when the p-side is at a higher potential compared to the n-side, as shown in option (1).

**Question 46. The diode used in the circuit shown in the figure has a constant voltage drop of 0.5 V at all currents and a maximum power rating of 100 mW. What should be the value of the resistance R connected in series L with the diode for obtaining the maximum current?**

- 6:76 Ω
- 20 Ω.
- 5 Ω
- 5.6 Ω

**Answer:** 3. 5 Ω

For the diode, the voltage drop is V_{D} = 0.5 V, and its maximum power rating is P_{max} = 100 mW = 100 x 10^{-3} W.

∴ diode resistance = \(R_{\mathrm{D}}=\frac{V_{\mathrm{D}}^2}{P_{\max }}=\frac{(0.5 \mathrm{~V})^2}{10^{-1} \mathrm{~W}}=2.5 \Omega\).

∴ the current through the diode is

⇒ \(I_D=\frac{V_{\mathrm{D}}}{R_{\mathrm{D}}}=\frac{0.5 \mathrm{~V}}{2.5 \Omega}=0.2 \mathrm{~A}\)

Since this 0.2 A current flows through the circuit, the total resistance is

⇒ \(R_{\mathrm{T}}=\frac{V}{I_{\mathrm{D}}}=\frac{1.5 \mathrm{~V}}{0.2 \mathrm{~A}}=7.5 \Omega\)

the resistance of the series resistor is

R = R_{T} – R_{D} = 7.5Ω – 2.5Ω = 5.0Ω.

**Question 47. If in a p-n junction, a Square input signal of 10 V is applied as shown in the figure, the output signal across R _{L} will be**

**Answer:** 2.

The p-n junction diode is conducting in a forward bias. In the given case, it is for V = 0 V to V = 5 V only.

So, the output across R will be 0-5 V, as shown in the option (2).

**Question 48. In the case of forward-biasing a p-n junction diode, which of the following figures correctly depicts the direction of the flow of carriers?**

**Answer:** 2.

When a p-n junction is formed, a diffusion of charge carriers across the junction takes place. The charge carriers undergo recombinations with their opposite charges. This is indicated by the direction of the flow of carriers, as shown in option (2).

**Question 49. Of the diodes shown in the following diagrams, which one is reverse-biased?**

**Answer:** 3.

In option (3), the p-side is at a zero potential (i.e., earth-connected), while the n-side is at +5 V. This indicates a reverse bias.

**Question 50. For a given circuit with an ideal p-n junction diode (D), which of the following statements is correct?**

- In a forward bias, the voltage across R is V.
- In a reverse bias, the voltage across R is V.
- In a forward bias, the voltage across R is 2V.
- In a reverse bias, the voltage across R is 2V.

**Answer:** 1. In a forward bias, the voltage across R is V.

An ideal diode has effectively zero resistance in its forward bias. So, the potential drop across R is V.

**Question 51. The application of a forward bias in a p-n junction**

- Increases the number of donors on the n-side
- Increases the electricity Held in the depletion zone
- Increases the potential difference across the depletion zone
- Widens the depletion zone

**Answer:** 1. Increases the number of donors on the n-side

In a forward-biased p-n junction diode, there is an increase in the number of donors on the n-side and a decrease in the potential barrier.

**Question 52. A Zener diode is used for**

- Producing oscillations in an oscillator
- Amplification
- Stabilization
- Rectification

**Answer:** 3. Stabilization

A Zener diode is a highly doped p-n junction device used for voltage stabilization.

**Question 53. A Zener diode is specified as having a breakdown voltage of 9.1 V with a maximum power dissipation of 364 mW. What is the maximum current that the diode can handle?**

- 40 mA
- 60 mA
- 50 mA
- 45 mA

**Answer:** 1. 40 mA

Power = voltage x current.

∴ \(I_{\max }=\frac{\text { power }}{\text { voltage }}=\frac{364 \times 10^{-3} \mathrm{~W}}{9.1 \mathrm{~V}}=40 \times 10^{-3} \mathrm{~A}=40 \mathrm{~mA}\)

**Question 54. What is the direction of the electric field (in the depletion layer) of a p-n junction diode?**

- From the p-side to the n-side
- From the n-side to the p-side
- Randomly oriented
- No electric field

**Answer:** 2. From the n-side to the p-side

Across the p-n junction, the electric field \(\overrightarrow{E_{\mathrm{b}}}\) is directed from the n-side to the p-side.

**Question 55. A Zener diode having a breakdown voltage equal to 15 V is used in a voltage regulator circuit. The current through the diode is**

- 5 mA
- 10 mA
- 15 mA
- 20 mA

**Answer:** 1. 5 mA

The p.d. across the l-kΩ resistor is V_{z} =15 V.

∴ Current = \(I^{\prime}=\frac{15 \mathrm{~V}}{1 \mathrm{k} \Omega}=15 \mathrm{~mA}\)

The p.d. across the 25-Ω resistor is

20 V-15 V = 5 V

∴ Current = I = \(\frac{5 \mathrm{~V}}{250 \Omega}=20 \mathrm{~mA} .\)

Hence, the current through the Zener is

I_{z} = I-I’ = 20 mA – 15 mA = 5 mA.

**Question 56. The given graph represents the V-I characteristics of a semiconductor device. Which of the following statements is correct?**

- It is the V-I characteristic of a solar cell where point A represents the open-circuit voltage and point B is the short-circuit current.
- It is the V-I characteristics of a solar cell, and the points A and B represent the open-circuit voltage and current respectively.
- It is the V—I characteristics of a photodiode, and the points A and B represent the open-circuit voltage and current respectively.
- It is the V-I characteristics of an LED, and the points A and B represent the open-circuit voltage and the short-circuit current respectively.

**Answer:** 1. It is the V-I characteristic of a solar cell where point A represents the open-circuit voltage and point B is the short-circuit current.

The given graph represents the V-I characteristic curve of a solar cell with V_{oc} as the open-circuit voltage and I_{sc} as the short-circuit current.

**Question 57. In an unbiased p-n junction, holes diffuse from the p-region to the n-region because of the**

- Attraction of the free electrons of the n-region
- Higher hole concentration in the p-region than the n-region
- Higher concentration of electrons in the n-region than in the p-region
- Potential difference across the p-n junction

**Answer:** 2. Higher hole concentration in the p-region than the n-region

Diffusion takes place from the higher concentration to the lower concentration. Hence, in an unbiased p-n junction, holes diffuse from the p-side to the n-side as the p-region has a higher hole concentration.

**Question 58. For a p-type semiconductor, which of the following statements is true?**

- Electrons are majority carriers and trivalent atoms are dopants.
- Holes are majority carriers and trivalent atoms are dopants.
- Holes are majority carriers and pentavalent atoms are dopants.
- Electrons are majority carriers and pentavalent atoms are dopants.

**Answer:** 2. Holes are majority carriers and trivalent atoms are dopants.

In a p-type semiconductor, an intrinsic (tetravalent Ge or Si) semiconductor is doped with a trivalent impurity. This creates vacancies of valence electrons called holes, which act as the majority charge carriers.

**Question 59. The reverse-breakdown voltage of a Zener diode is 5.6 V in the given circuit. The current I _{z} through the Zener is**

- 7 mA
- 17 mA
- 10 mA
- 15 mA

**Answer:** 3. 10 mA

Given that I_{z} = 5.6 V.

Hence, the voltage across R_{1} is 9 V- 5.6 V = 3.4 V.

∴ main current = I = \(\frac{3.4 \mathrm{~V}}{200 \Omega}\)

= 17 mA.

The current through R_{z} is

∴ \(I_2=\frac{V_{\mathrm{Z}}}{R_{\mathrm{Z}}}=\frac{5.6 \mathrm{~V}}{800 \Omega}=7 \mathrm{~mA}\)

∴ I_{z} =I – I_{2} = 17 mA – 7 mA = 10 mA.

**Question 60. At 0.3 V and 0.7 V, the Ge and Si diodes become conducting respectively, as shown in the given figure. If the ends of the Ge diode are reversed, the change in the potential V _{0} will be**

- 0.2 V
- 0.6 V
- 0.4 V
- 0.8 V

**Answer:** 3. 0.4 V

When the Ge and Si diodes are both forward-biased, a current passes through Ge with a potential drop of 0.3 V.

Hence, the initial output voltage is V_{0} =12 V- 0.3 V =11.7 V.

When Ge is reverse-biased, a current flows through Si alone with a potential drop of 0.7 V. Hence, the output voltage is

V_{0} =1 2 V – 0.7 V =11.3 V.

∴ the change in the output voltage is

V_{0 }– V_{0}‘ = 11.7 V – 11.3 V = 0.4 V

**Question 61. The current through the tire Zener diode in the given circuit will be**

- 14 mA
- 5 nrA
- 9 mA
- Zero

**Answer:** 3. 9 mA

The potential drop across the 5-kΩ resistor is

V =120 V – V_{z} =120 V- 50 V = 70 V.

the main current through the battery is \(I=\frac{70 \mathrm{~V}}{5 \mathrm{k} \Omega}=14 \mathrm{~mA}\)

The current through the 10-kΩ resistor is

∴ [latexI_0=\frac{50 \mathrm{~V}}{10 \mathrm{k} \Omega}=5 \mathrm{~mA}][/latex).

the current through the Zener diode is

I_{z} = I – I_{0} = 14 mA – 5 mA = 9 mA

**Question 62. In the given circuit, the current through the Zener diode is close to**

- 6.0 mA
- 6.7 mA
- 4.0 mA
- Zero

A**nswer:** 4. Zero

When the Zener voltage (Vz) reaches 10 V, the potential drop across the R_{1} resistor is 12 V – 10 V = 2 V.

∴ main current = \(=\frac{2 \mathrm{~V}}{500 \Omega}=4 \mathrm{~mA}\)

The Zener diode with two R_{2} resistors makes a parallel combination.

The current through each R_{2} resistor (=1500 Ω.) is

∴ \(I_2=\frac{V_{\mathrm{Z}}}{R_2 / 2}=\frac{10 \mathrm{~V}}{750 \Omega}=13.3 \mathrm{~mA}\)

Since l_{2} cannot be greater than the main current, the Zener diode will not reach its breakdown voltage. So, I_{z} = 0.

**Question 63. The adjoining figure represents a voltage-regulator circuit using a Zener diode. The breakdown voltage of the Zener diode is 6 V and the load resistance of the circuit is R _{L}= 4 kΩ. The series resistance
of the circuit is R_{i }**

**= 1 kΩ. If the battery voltage VB varies from 8 V to 16 V, what are the minimum and maximum values of the current through the Zener diode?**

- 0.5 mA and 8.5 mA
- 1.5 mA and 8.5 mA
- 0.5 mA and 6 mA
- 1 mA and 8.5 mA

**Answer:** 1. 0.5 mA and 8.5 mA

Given that breakdown voltage = V_{z} = 6 V.

With V_{B} = 8 V, the potential drop across R_{L} is 8V – 6V = 2V.

∴ main current = \(I=\frac{2 \mathrm{~V}}{R_{\mathrm{i}}}=\frac{2 \mathrm{~V}}{1 \mathrm{k} \Omega}=2 \mathrm{~mA}\)

The current through R_{L} is

⇒ \(I_1=\frac{V_{\mathrm{Z}}}{R_{\mathrm{L}}}=\frac{6 \mathrm{~V}}{4 \mathrm{k} \Omega}=1.5 \mathrm{~mA}\)

∴ I_{z} = I- I_{1} = 2 mA-1.5 mA = 0.5 mA.

With V_{B} = 16 V

⇒ \(I=\frac{16 \mathrm{~V}-6 \mathrm{~V}}{1 \mathrm{k} \Omega}=10 \mathrm{~mA}\)

The current through R_{L} is I_{1}=1.5 mA.

So, I_{z} = I – I_{2} = 10 mA – 1.5 mA = 8.5 mA.

**Question 64. The adjoining figure shows a DC-voltage-regulator circuit with a Zener diode of breakdown voltage 6 V. If the unregulated input voltage varies between 10 V and 16 V, what is the maximum Zener current?**

- 7.2 mA
- 3.5 mA
- 1.5 mA
- 2.5 mA

**Answer:** 2. 3.5 mA

Given that Zener breakdown voltage = V_{z} = 6 V.

The Zener current will be maximum when the supply voltage is maximum (= 16 V).

The voltage across R_{s} is 16 V- 6 V = 10 V.

The main current (through R_{s}) is \(I_{\mathrm{s}}=\frac{10 \mathrm{~V}}{2 \mathrm{k} \Omega}=5 \mathrm{~mA}\)

But \(I_{\mathrm{L}}=\frac{V_{\mathrm{Z}}}{R_{\mathrm{L}}}=\frac{6 \mathrm{~V}}{4 \mathrm{k} \Omega}=1.5 \mathrm{~mA}\)

∴ The maximum current through the zener is

I_{z} = I_{s}– I_{L} = 5 mA-1.5 mA = 3.5 mA.

**Question 65. A diode has a potential drop of 0.5 V in its forward bias. The maximum current that can flow through the diode is 10 mA. Find the resistance R to be connected in series with the diode so that the setup can be connected across a battery of 1.5 V.**

- 50 Ω
- 25 Ω
- 100 Ω
- 20 Ω

**Answer:** 3. 100 Ω

The potential drop across the resistor is

V = IR = 1.5 V- 0.5 V = 1.0 V.

Given that I = 10 mA = 10^{-2} A.

the required resistance is

⇒ \(R=\frac{1.0 \mathrm{~V}}{10^{-2} \mathrm{~A}}=100 \Omega\)

**Question 66. If a semiconductor photodiode can detect a photon with a maximum wavelength of 400 nm then its band-gap energy is**

- 1.1 eV
- 3.1 eV
- 2.0 eV
- 1.5 eV

**Answer:** 2. 3.1 eV

The band gap is equal to the energy (hv) of the photon.

∴ bad gap = \(\Delta E=\frac{h c}{\lambda}=\frac{1240 \mathrm{eV} \mathrm{nm}}{400 \mathrm{~nm}}=3.1 \mathrm{eV}\).

**Question 67. The increase in the depletion region in a p-n junction diode is due to**

- Its reverse bias
- Its forward bias
- Both its forward and reverse biases
- An increase in the forward current

**Answer:** 1. Its reverse bias

In a reverse bias, the potential barrier becomes higher, so the width of the depletion layer is increased and the junction offers a large resistance.

**Question 68. The solid materials which have negative temperature coefficients of resistance are**

- Insulators only
- Semiconductors only
- Insulators and semiconductors
- Metals only

**Answer:** 3. Insulators and semiconductors

The resistance and resistivity of metallic conductors increase with increasing temperature, so they have a positive coefficient of resistance. In (intrinsic) semiconductors, the resistivity decreases with increasing temperature. The same is the case with insulators, in which electrons absorb energy at a higher temperature to contribute to conduction. Hence, both semiconductors and insulators have a negative temperature coefficient of resistance.