## WBCHSE Class 12 Physics Semiconductor Electronics Multiple Choice Questions

**Question 1. In an n-type silicon, which of Ibe following statements is true?**

- Electrons are majority carriers and trivalent atoms are the dopants
- Electrons are minority carriers and pentavalent atoms are dopants
- Holes are minority carriers and pentavalent atoms dopants (oi holes are majority carriers and trivalent atoms are dopants
- Holes are majority carriers and trivalent atoms are dopants

**Answer: **3. Holes are minority carriers and pentavalent atoms dopants (oi holes are majority carriers and trivalent atoms are dopants

**Question 2. In an unbiased p-n junction, boles diffuse from the p-region to the n -region because**

- Free electrons In the n -region attract them f v they move across the junction by the potential difference
- They move across the junction by the potential difference
- Hole concentration In p -the region is higher as compared To the n-region
- All the above

**Answer:** 3. Hole concentration in p – the region is more as compared To n-region

**Question 3. Carbon, silicon, and germanium have four valence electrons each.’ These are characterisedi>y valence and conduction bands separated by energy band gap equal to (£g)c, (£g)s. and (£g)Ge> Which of the following statements is true?**

- (E
_{g})S_{i}<(E_{g})_{Ge}E<(E_{g})_{C} - (E
_{g})_{C}<(E_{g})_{Ge}<(E_{g})S_{i} - (E
_{g})_{C}>(E_{g})S_{i}>(E_{g})_{Ge} - (E
_{g})_{C}= (E_{g})S_{i}= (E_{g})_{Ge}

**Answer:** 3.(E_{g})_{C}>(E_{g})S_{i}>(E_{g})_{Ge}

**Read and Learn More Class 12 Physics Multiple Choice Questions**

**Question 4. In V _{0} is the potential barrier across a p-n junction, when no battery is connected across the junction**

- 1 and 3 both correspond to the forward bias of the junction
- 3 corresponds to the forward bias of the junction and 1 corresponds to the reverse bias of the junction
- 1 corresponds to forward bias and 3 corresponds to reverse bias of junction
- 3 and 1 both correspond to the reverse bias of the junction

**Answer:** 2. 3 corresponds to the forward bias of the junction and 1 corresponds to the reverse bias of the junction

**Question 5. Assuming the diodes to be ideal**

- D
_{1}is forward-biased and D_{2}is reverse biased, hence current flows from A to B - D
_{2}is forward biased and D_{1}is reverse biased, hence n current flows from B to A and vice versa - D
_{1}and D_{2}are both forward-biased and hence current flows from A to B - D
_{1}and D_{2}, are both reverse biased and hence no current flows from A to B and vice versa

**Answer:** 2. D_{2} is forward biased and D_{1} is reverse biased, hence n current flows from B to A and vice versa

**Question 6. A 220 V AC supply is connected between points A and B will be the potential difference V across the capacitor**

- 220V
- 110V
- 0V
- 222√2V

**Answer:** 4. 222√2V

**Question 7. In the circuit, if the diode forward voltage drop is 0.3 V, the voltage** **difference between A and B is**

- 1.3 V0
- 2.3 V
- 0
- 0.5 V

**Answer:** 2. 2.3 V

**Question 8. The output of the given circuit in**

- Would be zero
- Would be like a half-wave rectifier with positive cycles in the output
- Would be like a half-wave rectifier with negative cycles in the output
- Would be like that of a full-wave rectifier

**Answer:** 3. Would be like a half-wave rectifier with negative cycles in the output

**Question 9. Consider an n-p-n transistor with its base-emitter junction forward-biased and collector-base junction reverse-biased. Which of the following statements is true?**

- Electrons cross over from emitter to collector
- Holes move from base to collector
- Electrons move from emitter to base
- Electrons from the emitter move out of the base without going to the collector

**Answer:** 1 And 3

**Question 10. The breakdown in a reverse-biased p-n junction is more likely to occur due to the**

- The large velocity of the minority charge carriers if the doping concentration is small
- The large velocity of the minority carriers if the doping concentration is large
- The strong electric field in a depletion region if the doping concentration is small
- Strong electric field in the depletion region if the doping concentration is large

**Answer:** 1 And 4

**Question 11. In an n-p-n transistor circuit, the collector current is 10 mA. If 95% of the electrons emitted reach the collector, which of the following statements is true?**

- The emitter current will be 8 mA
- The emitter current will be 10.53 mA
- The base current will be 0.53 mA
- The base current will be 2 mA

**Answer:** 2 And 3

**Question 12. If the potential difference is applied, the value of the electric current**

- Becomes infinite for an insulator kept at OK
- Becomes zero for a semiconductor kept at 0 K
- Becomes finite for a metal kept at 0 K
- Becomes infinite in a forward-biased p-n junction diode kept at 300 K

**Answer:** 3. Becomes finite for a metal kept at 0 K

**Question 13. Two pieces of copper and germanium are cooled from room temperature to 77 K. As a result,**

- Resistance of both the pieces will increase
- Resistance of both pieces will decrease
- Resistance of copper will decrease but that of germanium will increase
- Resistance of copper will increase but that of germanium will decrease

**Answer:** 3. Resistance of copper will decrease but that of germanium will increase

**Question 14. For Intrinsic semiconductors, the energy gap of the forbidden zone Is approximately**

- 0.01 eV
- 0.1 eV
- 1 eV
- 10 eV

**Answer:** 3. 1 eV

**Question 15. In a semiconducting material, the mobilities of electrons and holes are pe and ph respectively. Which of the following is time?**

- μ
_{e}>μ_{h} - μ
_{e}<μ_{h} - μ
_{e}= μ_{h} - μ
_{e}>< 0 , μ_{h}>0

**Answer:** 1.μ_{e}>μ_{h}

**Question 16. In some substances, the charge can flow at ordinary temperatures, but not at very low temperatures. These are called**

- Conductors
- Insulators
- Semiconductors
- Dielectric

**Answer:** 3. Semiconductors

**Question 17. If a small amount of antimony is added to the germanium crystal**

- It becomes a P-type semiconductor
- The antimony becomes an acceptor atom
- There will be more free electrons than holes in the semiconductor
- Its resistance is increased

**Answer:** 3. There will be more free electrons than holes in the semiconductor

**Question 18. Pure S _{i} at 500K has an equal number of electron (nf) hole (n_{h}) concentrations of 1.5 × 10^{16}.m^{-3} Doping by indium increases n_{h} to 4.5 × 10^{22}.m^{-3}. The doped semiconductor is of concentration**

- p-type basing electron concentration n
_{e}= 5 × 10^{9}.m^{-3} - n-type having electron concentration n
_{e}= 5 × 10^{22}.m^{-3} - p-type having electron concentration n
_{e}= 2. 5 × 10^{10}.m^{-3} - n-type having electron concentration n
_{e}= 2.5 × 10^{23}m^{-3}

**Answer: **1. p-type basing electron concentration n_{e}= 5 × 10^{9}.m^{-3}

**Question 19. A rectifier converts**

- Mechanical energy in electrical
- Liglit energy into electrical
- ac to dc
- dc to ac

**Answer:** 3. ac to dc

**Question 20. If a p-n junction diode is not connected in a circuit, then**

- The potential is the same everywhere
- The potential of p -end is more than n -end
- An electric field acts from the N-end junction
- An electric field acts from the p -end to n -end at the junction

**Answer:** 3. An electric field acts from the N-end junction

**Question 21. In the circuit, The forward bias resistances of both the diodes are 50, and reverse bias resistances are infinite. If the battery voltage is 6 V, the current through 100fl resistance in unit A is**

- Zero
- 0.02
- 0.03
- 0.036

**Answer:** 2. 0.02

**Question 22. Two identical p-n junctions are connected in series with a battery in three different ways. The potential drop across the two p-n junctions will be equal**

- In the circuits 1 and 2
- In the circuits 2 and 3
- In the circuits 3 and 1
- In the circuit 1 only

**Answer:** 2. In the circuits 2 and 3

**Question 23. When a p-n diode is reverse-biased, then **

- No current flows 20 V 15V 1kft
- The depletion region is increased
- The depletion region is reduced
- The height of the potential barrier is reduced

**Answer:** 2. The depletion region is increased

**Question 24. The current through the given circuit **

- \(\frac{3}{10}\) A
- \(\frac{1}{10}\) A
- \(\frac{3}{50}\) A
- \(\frac{3}{10}\) A

**Answer:** 3. \(\frac{3}{50}\) A

**Question 25. The impurity atoms that should be added to germanium to make it n-type of semiconductor is**

- Iodine
- Indium
- Arsenic
- Aluminium

**Answer:** 3. Arsenic

**Question 26. The circuit has two oppositely connected ideal diodes in parallel. What is the current in the circuit**

- 1.33A
- 1.71A
- 2.00A
- 2.31A

**Answer:** 3. 2.00A

**Question 27. When a p-n junction is reverse-biased, then**

- No current flows
- The depletion layer is increased
- The depletion layer is decreased
- The height of potential barriers is reduced

**Answer:** 3. The depletion layer is increased

**Question 28. A Zener diode having a breakdown voltage equal to 15Y is used in a voltage regulator circuit. The current through the diode is**

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

**Answer:** 5mA

**Question 29. If the rate of dropping in the emitter, base, and collector of a transistor is De, Db, and De respectively, then**

- D
_{e}= D_{b}= D_{c} - D
_{e}< D_{b}= D_{c} - D
_{e}> D_{b}> D_{c} - D
_{e}> D_{c}> D_{b}

**Answer:** 4. D_{e} > D_{c} > D_{b}

**Question 30. In the active region of a transistor, the biasing at the emitter-base junction and collector-base junction are respectively**

- Forward, Forward
- Forward, Reverse
- Reverse, Forward
- Reverse, Reverse

**Answer:** 2. Forward, Reverse

**Question 31. A transistor is used as**

- A rectifier
- An amplifier
- An oscillator
- A source of electrons and holes

**Answer: ** 2. An amplifier

**Question 32. In an n-p-n transistor circuit. collector current Is 10 mA. If 90% A of electrons from the emitter enter into the collector then**

- Emitter current = 11.11 mA
- Emitter current
- = 9mA
- Base current = 2.1 mA
- Base current = 0.9 mA

**Answer:** 1. Emitter current = 11.11 mA

**Question 33. For a transistor, if ΔI _{C}/ΔI_{E} = 0.96, then the current gain = β is**

- 6
- 12
- 24
- 48

**Answer:** 3. 24

**Question 34. A transistor is operated in a common-emitter configuration at Vc = 2V such that a change in the base current from 100 A to 300 A produces a change in the collector current from 10 mA to 20 mA. the current gain is**

- 50
- 75
- 100
- 25

**Answer:** 1. 50

**Question 35. If the open-loop gain feedback ratio of a positive feedback oscillator is A are r respectively, then the closed-loop gain A _{f }is**

- 1
- <A
- <A
- Infinity

**Answer:** 4. Infinity

**Question 36. The depletion region of the p-n junction completely vanishes, if.**

- A suitable forward bias is applied
- a suitable reverse bias is applied
- a light of suitable frequency is made incident from outside
- The Zener effect takes place

**Answer: ** 1,3,4

**Question 37. If a pentavalent element is doped in an intrinsic semiconductor, then**

- a p-type semiconductor is formed
- The effective energy gap of the forbidden zone decreases
- The energy states of free electrons in the pentavalent element come nearer to the conduction band
- The energy states of free electrons in the pentavalent element come down near the valence band

**Answer:** 2,3

**Question 38. If a p-n junction is reverse-biased, then**

- A potential barrier is created at the junction
- The effective resistance reaches an infinite value
- A small current is obtained due to the diffusion of the majority of carrier electrons and holes
- Despite increasing the reverse bias, the reverse current remains almost unchanged

**Answer: ** 1,2,3

**Question 39. In case of a transfer**

- The emitter region is heavily doped
- In CE configuration, if the base-emitter and the collector-emitter both junctions are forward biased, then the transistor acts in the saturation region
- To use as an amplifier, it is operated in the active region
- To use as a switch, it is operated in either cut region or saturation region

**Answer:** 1,2,3,4

**Question 40. In case of a Zener diode,**

- To use as a voltage regulator, it is kept in reverse bias.
- Power consumption in reverse bias is higher than that of an ordinary semiconductor diode
- On forward bias, _ it behaves as an ordinary semiconductor diode is
- If Zeiifer voltage is applied in reverse bias, then a high potential barrier is created at its junction

**Answer:** 1,2,3

**Question 41. In a feedback oscillator**

- Effective amplification of the amplifier used in this oscillator is infinity
- The output AC signal frequency is only determined by the feedback ratio
- The output AC signal frequency is controlled by a tank circuit in the feedback device
- The output AC signal is obtained without any externally applied input

**Answer:** 1,3,4

**Question 42. In an n-type semiconductor, If n and p am dm WHvmtratlon of electrons and holes respectively and n( Is die concentration, of electron-hole pairs In a pure semi doctor, then**

- n = p
- n>p-
- n<p
- np = n²
_{1}

**Answer: ** 1,3,4

**Question 43. A common-emitter DC circuit of the n-p-n transistor is shown. V _{BB} and V_{CC} are applied emf of the external source at the base end and collector end respectively. Generally, if we need to amplify an AC signal, it is to be applied at the base end. Hence base end is called the input end. R_{B} and I_{B} are input resistance and input DC respectively.**

⇒ In most of the cases, input resistance R_{B} is induced due to the internal resistance of the transistor. The amplified signal of the transistor is obtained at the collector end. Thus the col¬lector end is called the output end. Here, R_{L} is external load resistance and I_{C} is the output DC

The terminal voltage across RL, V_{0} = I_{C}R_{L }is the output voltage; V_{i} = I_{B}R_{B} is the input voltage. In the case of only DC biasing of the transistor, the current gain iff = voltage gain β =\(\frac{I_C}{I_B}\), power gain =. \(=\frac{P_o}{P_i}\)

**In a given circuit, V _{CC} – 8V, the terminal voltage across load resistance =0.8V, load resistance = 800 Ω, the internal resistance of the transistor = 200 Ω, and β_{dc} – 250.**

** 1. The value of base current (in μA)**

- 0.8
- 2.5
- 4.0
- 8.0

**Answer: ** 3. 4.0

** 2. The value of collector current (in mA)**

- 1.0
- 2.5
- 8.0
- 10.0

**Answer: ** 1. 1.0

**3. The collector-emitter voltage (in volt)**

- 2.5
- 7.2
- 8.0
- 8.8

**Answer:** 2. 7.2

**4. Voltage gain**

- 250
- 1000
- 2000
- 4000

**Answer: ** 2. 1000

**5. Power gain**

- 250
- 2500
- 25000
- 250000

**Answer:** 4. 250000

**Question 44. To maintain a constant potential difference across a regulative load resistance RL, it is not connected directly with an unregulated voltage source Vf. ****Hence a reverse biased Zener diode may be connected parallel to the load resistance. If the range of the current through the load resistance in which the value ofthe terminal voltage is well maintained within maximum value VQ and minimum value V, then percentage load regulation**

r = \(\frac{V_0-V}{V_0} \times 100\)

Each Zener diode has a voltage rating and a watt rating. According to this, the current through the Zener diode has

a specific value above which the Zener diode will be burnt Hence to bypass this damage, an additional resistance R is to be connected in the circuit

**In a given circuit, the maximum voltage of the external unregulated source is 15V, the rating of the Zener diode is 6 V-0.25 W and the maximum and minimum voltages across the load resistance are 5.95 V and 5.90 V respectively.**

**1. Maximum safe current through the Zener diode is approximately (in mA)**

- 1.5
- 9.0
- 24.0
- 41.67

**Answer:** 4. 41.67

**2. The minimum value of the resistance R is approximately**

- 360
- 216
- 166
- 83

**Answer:** 2. 216

**3. Percentage load regulation**

- 0.84
- 4.2
- 5.95
- 8.4

**Answer:** 1. 0. 84

**Question 45. The impurity atom which when added to germanium makes it an n-type semiconductor**

- Boron
- Indium
- Arsenic
- Aluminium

**Answer:** 3. Arsenic

**Question 46. What will happenifthe amount of reverse biasing a p-n junction diode is gradually increased?**

- The thickness of the depletion region will increase
- The flow of current due to majority carriers will increase
- The thickness of the depletion region will decrease
- The flow of current due to majority carriers will decrease

**Answer:** 1. Thickness of the depletion region will increase

**Question 47. For a transistor if β = 100, then α will be**

- 0.99
- 1.01
- 100
- 0.0

**Answer:** 1. 0.99

β = \(\frac{\alpha}{1-\alpha}\)

Or, \(\frac{\beta}{1}=\frac{\alpha}{1-\alpha}\)

= \(\frac{\beta}{1+\beta}=\frac{\alpha}{(1-\alpha)+\alpha}\)

∴ α = \(\frac{\beta}{1+\beta}=\frac{100}{1+100}\)

= \(\frac{100}{101}\)

= 0.99

**Question 48. The lengths, radii, and specific resistances of two conducting wires are each in the ratio of 1: 3. If the resistance of the thinner wire is 10Ω, then the resistance of the other wire will be**

- 40 Ω
- 20 Ω
- 10 Ω
- 5 Ω

**Answer: **3. 10Ω

We know, R = \(\rho \frac{l}{A}=\frac{\rho l}{\pi r^2}\)

∴ \(\frac{R_1}{R_2}=\frac{\rho_1}{\rho_2} \cdot \frac{l_1}{l_2} \cdot\left(\frac{r_2}{r_1}\right)^2\)

= \(=\frac{1}{3} \times \frac{1}{3} \cdot\left(\frac{3}{1}\right)^2\)

= 1

R_{2}= R_{1 }= _{ }10

**Question 49. Innormal transistor operation**

- Emitter-base junction and collector-base junction are both in reverse bias
- Emitter-base junction is in forward bias and collector base junction is in reverse bias
- Emitter-base junction and collector-base junction both are in forward bias
- Emitter-base junction is in reverse bias and the collector-base junction is in forward bias

**Answer:** 2. Emitter-base junction is in forward bias and the collector-base junction is in reverse bias

In the common-base (CB) mode of a transistor, the emitterbase junction is kept at forward bias keeping the base grounded and the collector-base junction is kept at reverse bias to get the proper circuit action.

**Question 50. In a transistor output characteristics commonly used in the common-emitter configuration, the base current I _{B}, the collector current I_{C} and the collector-emitter voltage V_{CE} have values of the following orders of the magnitude in the active region**

- I
_{B}and I_{C}are both in μA and V_{CE}in V - I
_{B}is in μA, IQ in mA, and V_{CE}in V - I
_{B}is in mA, I_{C}in μA, and V_{CE}in mV - I
_{B}is in mA, I_{C}is in mA and V_{CE }in mV

**Answer:** 2. I_{B} is in μA, IQ in mA, and V_{CE} in V

**Question 51. In the circuit shown assume the diode to be ideal. When Vf increases from2Vto 6V, the changes in the current is(In mA)**

- Zero
- 20
- 80/3
- 40

**Answer:** 2. 20

When V_{i}. increases from 2V to 3V, the diode remains in reverse bias, so current I = 0

When V_{i} increases from 3V to 6V, the current increases From 0 to; so,I = \(\frac{(6-3) V}{150 \Omega}\)= 0.02A

= 20mA

**Question 52. If the band gap between the valence band and conduction band in a material is 5.0 eV, then the material**

- Semiconductor
- Good conductors
- Superconductor
- Insulator

**Answer:** 4. Insulator

**Question 53. Assume that each diode. 1.70 has a forward bias resistance of 50Ω and an infinite reverse bias resistance. The current through the resistance 150Ω is**

- 0.66 A
- 0.05 A
- Zero
- 0.04 A

**Answer:** 4. 0.04 A

The uppermost diode in the circuit is forward-biased. But the diode beneath it is reverse biased, hence would not conduct any current

The current through the closed circuit

= \(\frac{10 \mathrm{~V}}{(50+50+50) \Omega}\)

= \(\frac{10}{250}\)

= 0.04A

**Question 54. In the case of a bipolar transistor β = 45. The potential drop T across the collector resistance of 1 kΩ is 5 V. The base current is approximately**

- 22 μA
- 55 μA
- 11μA
- 45 μA

**Answer:** 3. 11 μA

Here β = 45

∴ \(\frac{I_C}{I_B}\) = 45

Now, I_{C }× 1 × 10³ = 5 or, I_{C} = 5 × 10^{-3}

∴ \(I_B=\frac{I_C}{45}\)

= \(\frac{5 \times 10^{-3}}{45}\)

= 0.111 × 10^{-3}

= 111 μA

**Question 55. A zener diode having breakdown voltage 5.6 Vis connected in reverse bias with a battery ofemf10 V and a resistance of 100 fl in series. The current flowing through the more zen is**

- 88 mA
- 4.4 mA
- 0.88 mA
- 44 mA

**Answer:** 4. 44 mA

Her, I= \(\frac{10-5.6}{100}\)

= 0.044 A

= 44 × 10^{-3}

= 44 mA

**Question 56. When a semiconductor device is connected in series with a battery and a resistance, a current is found to flow in the circuit. If, however, the polarity of the battery is reversed, practically no current flows in the circuit. The device may be**

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

**Answer:** 4. A p-n junction

**Question 57. What will be the current flowing through the 6kil resistor in the circuit shown, where the breakdown voltage of the Zener is 6 V?**

- \(\frac{2}{3}\)mA
- 1 mA
- 10 mA
- \(\frac{3}{2}\)mA

**Answer:** 1. \(\frac{2}{3}\)mA

As the breakdown voltage of Zener is 6v, the potential drop in the 4 kΩ resistor is 6V.

The potential drop in 6 kΩ resistor is 4 V

∴ The current flowing through the 6kΩ resistor

= \(\frac{4}{6 \times 10^3}\) A

= \(\frac{2}{3}\)mA

**Question 58. The forward-biased diode connection is**

**Answer:** 1

In the case of forward bias of a p-n junction diode, the potential ofp -side is greater than n -side.

**Question 59. Identify the semiconductor devices whose characteristics are given below, in the order (1), (2), (3), (4)**

- Simple diode, zener diode, solar cell, light-dependent resistance
- Zener diode, simple diode, light-dependent resistance, solar cell
- Solar cell, light-dependent resistance, zener diode, simple diode
- Zener diode, solar cell, simple diode, light-dependent resistance

**Answer:** 1. Simple diode, zener diode, solar cell, light-dependent resistance

**Question 60. For a common emitter configuration,if a and /? have their usual meanings, the incorrect relationship between cr and is**

- \(\frac{1}{\alpha}=\frac{1}{\beta}+1\)
- \(\alpha=\frac{\beta}{1-\beta}\)
- \(\alpha=\frac{\beta}{1+\beta}\)
- \(\alpha=\frac{\beta^2}{1+\beta^2}\)

**Answer:** 3. \(\alpha=\frac{\beta}{1+\beta}\)

**Question 61. In a common-emitter amplifier circuit using an n-p-n transistor, the phase difference between the input and the output voltages will be**

- 45°
- 90°
- 135°
- 180°

**Answer:** 4. 180v

**Question 62. The reading of the ammeter for a silicon diode in the given circuit is**

- 11.5 mA
- 13.5 mA
- 0
- 15mA

**Answer:** 1. 11.5 mA

For silicon diode, the stopping potential is 0.7 V

I = \(\frac{V-V_{\text {diode }}}{R}\)

= \(\frac{3-0.7}{200} \times 1000\)

= 11.5 mA

**Question 63. The barrier potential of a p-n junction depends on **

1. Type of semiconductor material

2. The amount of doping and

3. Temperature. Which one of the following is correct?

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

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

The barrier potential V of-n junction diode,

- Is proportional to temperature T increases with the Increase of density of the donor atom and acceptor atom In the two parts of the diode.
- Decreases with the increase of number density (nÿ) of the thermal electron and hole of the pure semiconductor. This
- H depends on the material of the semiconductor.

**Question 64. The given graph represents V-I characteristics for a semiconductor device. Which ofthe following statements Is correct?**

- It Is a V-I characteristic for solar cells where point A represents open circuit voltage and point B short circuit current
- It is for a solar cell and points A and B represent open circuit voltage and current respectively
- It Is for a photodiode and points A and B represent open circuit voltage and current respectively
- It is for LED and points A and B represent open circuit voltage and short circuit current respectively

**Answer:** 1. It Is a V-I characteristic for solar cells where point A represents open circuit voltage and point B short circuit current

**Question 65. If p-n junction, a square input signal of 10 V is applied, as shown,**

**Then the output across RL will be**

**Answer:** 4

Only positive voltage can pass through a p-n junction diode

**Question 66. Consider the junction diode as ideal. The value of current flowing through AB Is**

- 10
^{-2}A - 10
^{-1}A - 10
^{-3}A - 0

**Answer:** 1. 10^{-3} A

The p-n junction diode is forward-biased

I = \(\frac{V}{R}=\frac{4-(-6)}{1000}\)

= 10^{-2} A

**Question 67. A n-p-n transistor is connected to a common emitter configuration in a given amplifier. A load resistance of 800 ft is connected in the collector circuit and the voltage drop across it is 0.8 V. If the current amplification factor is 0.96 and the input resistance of the circuit is 192 ft, the voltage gain and the power gain of the amplifier will respectively be**

- 3.69, 3.84
- 4,4
- 4,369
- 4,3.84

**Answer:** 4. 4,3.84

Here β = 0.96

R_{L} = 800

V_{0} = 0.8V

R_{B} = 192Ω

∴ Voltage grain = \(\beta \frac{R_L}{R_n}=0.96 \times \frac{800}{192}\)

= 4

∴ Power gain = \(\beta^2 \frac{R_L}{R_B}=(0.96)^2 \times \frac{800}{192}\)

= 3.84

**Question 68. Two sides of a semiconductor germanium crystals A and B are doped with arsenic and indium, respectively. They are connected to a battery. **

**The correct graph between current and voltage for the arrangement is**

**Answer:** 1.

Arsenic is a pentavalent sample and indium is a trivalent sample. As side A is doped with arsenic is n-type and as side B is doped with indium it is p-type. Therefore, the circuit is reverse-biased. But the given electric cell is a fixed source. If it is considered as an ideal source then there will be no change in the electromotive force.

**Question 69. A common-emitter amplifier circuit is shown in the figure below. For the transistor used in the circuit the current amplification factor, β _{dc} = 100. Other parameters are mentioned.**

We find that

- V
_{BE}= . + 18.2 V, V_{BC}= -3.45 V, and the amplifier is working - V
_{BE}= + 18.5 V, V_{BC}= +2.85 V, and the amplifier is not working - V
_{BE }=. + 120.7 V, V_{BC}= + 3.75 V= and amplifier Is not working - V
_{BE}= . + 21.5 V, V_{BC}= -2.75 V, and the amplifier is working

**Answer:** 3. V_{BE }=. + 120.7 V, V_{BC} = + 3.75 V= and amplifier Is not working

V_{CE }= V_{CC }– I_{C }R_{C }= 24- (1.5 × 10^{-3}) × (4.7 × 10³)

= 16.95 V

V_{BE }V_{CC }– I_{B }R_{B }= \(V_{C C}-\frac{I_C}{\beta} R_B\)

= 24 – \(\frac{1.5 \times 10^{-3}}{100} \times\left(220 \times 10^9\right)\)

= 20.7 V

The p – side of the n-p-n transistor i.e the base connected with +20 .7V, means it is kept in forward.

Again, the n-type collector is connected with +16.95 V, which means it is kept in reverse bias.

∴ This transistor is in the active region. So it acts as an amplifier

V_{BC }= V_{BC }+ V_{EC} = V_{BE }– V_{CE}

20.7-16.95 = 3.75 V

**Question 70. In the circuit shown in the figure, the input voltage V. is 20 V, VB£ = 0 and VCE – 0. The values of IB, Ic, and /? are given by**

- I
_{B}= 20μA, I_{C}= 5 mA, β = 250 - I
_{B}= 25μA, I_{C}= 5 mA, β = 200 - I
_{B}= 40 μA, I_{C}= 5 mA, β = 250 - I
_{B}= 40μM, I_{C}= 5 mA, β = 125

**Answer:** 4. I_{B} = 40μM, Ic = 5 mA, β = 125

Applying KVL in loop ABEKD we get,

500 × l0³ I_{B} + V_{BE}-20 = 0

Or, \(=\frac{20}{5 \times 10^5}\)

= 40 μA

Applying KVL in loop FGHKEBC we get

⇒ \(4 \times 10^3 I_C+V_{C E}-20\) = 0

⇒ \(I_C=\frac{20}{4 \times 10^3}\)

= 5 mA

Or, I_{C}= βI_{B}

Or, β = \(\frac{I_C}{I_B}=\frac{5 \times 10^{-3}}{40 \times 10^{-6}}\)

= 125

**Question 71. Inap-w junction diode, changeintemperaturedue to heating**

- Does not affect the resistance of the p-n junction
- Affects only forward resistance
- Affects only reverse resistance
- Affects the overall V-I characteristics of the p-n junction

**Answer:** 4. Affects the overall V-I characteristics of the p-n junction

The number density of electrons and holes of a p-n junction diode increases due to heating. So, the characteristics graph of the p-n junction diode will change totally