Electronics Devices And Circuit 1(TTA)
INSTRUCTIONS-1. NUMBER OF QUESTIONS 25
2. HAS A TIME LIMIT OF 15 MINUTES
3. HAS A PASS MARKS OF 30%
4. QUESTIONS PER PAGE 1
5. EACH QUESTIONS HAS 1 MARKS
6. NEGATIVE MARKING FOR EACH QUESTIONS IS 0.257. WILL ALLOW TO YOU GO BACK ,SKIP AND CHANGE YOUR ANSWERS
8. WILL ALLOW TO YOU PRINT OUT YOUR RESULT AND CERTIFICATE
9. WILL ALLOW TO YOU PRINT OUT YOURS RESPONSE SHEET WITH
CORRECT ANSWER KEY AND EXPLANATION.
- 1.
.A long specimen of p-type semiconductor material:
- A.
Is positively charged
- B.
Is electrically neutral
- C.
Has an electric filed directed along its length
- D.
None of the above
Correct Answer
B. Is electrically neutralExplanation
A long specimen of p-type semiconductor material is electrically neutral because it has equal numbers of positive charges (holes) and negative charges (electrons). In a p-type semiconductor, impurities are added to the material, creating excess holes which are positively charged. However, the material also has an equal number of negatively charged electrons. As a result, the positive and negative charges balance each other out, making the material electrically neutral.Rate this question:
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- 2.
N-type semiconductors are:
- A.
Negatively charged
- B.
Produced when Indium is added as an impurity to Germanium
- C.
Produced when Phosphorous is added as an impurity to Silicon
- D.
None of the above
Correct Answer
C. Produced when pHospHorous is added as an impurity to SiliconExplanation
n-type semiconductors are produced when Phosphorous is added as an impurity to Silicon. This is because Phosphorous has five valence electrons, while Silicon has four. When Phosphorous is added to Silicon, the extra electron becomes a free electron that is able to move freely within the crystal lattice. This creates an excess of negatively charged electrons, giving the semiconductor an overall negative charge. Therefore, the correct answer is "Produced when Phosphorous is added as an impurity to Silicon."Rate this question:
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- 3.
The probability that an electron in a metal occupies the Fermi-level, at any temperature (>0 K) is:
- A.
0
- B.
1
- C.
0.5
- D.
None of the above None of the above none of these
Correct Answer
C. 0.5Explanation
The probability that an electron in a metal occupies the Fermi-level at any temperature (>0 K) is not 0 or 1. The Fermi level represents the highest energy level that is occupied by electrons at absolute zero temperature. As the temperature increases, some electrons may gain enough energy to occupy higher energy levels above the Fermi level. Therefore, the probability that an electron occupies the Fermi level at any temperature (>0 K) is not definite and can be considered as 0.5.Rate this question:
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- 4.
.Measurement of Hall coefficient enables the determination of:
- A.
Mobility of charge carriers
- B.
Type of conductivity and concentration of charge carriers
- C.
Temperature coefficient and thermal conductivity
- D.
None of these
Correct Answer
B. Type of conductivity and concentration of charge carriersExplanation
The measurement of Hall coefficient allows for the determination of the type of conductivity and concentration of charge carriers in a material. The Hall coefficient is a parameter that describes the behavior of charge carriers in a magnetic field. By measuring this coefficient, one can determine whether the material has positive or negative charge carriers, as well as their concentration. Therefore, the correct answer is "Type of conductivity and concentration of charge carriers."Rate this question:
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- 5.
If the energy gap of a semiconductor is 1.1 e V it would be:
- A.
Opaque to the visible light
- B.
Transparent to the visible light
- C.
Transparent to the ultraviolet radiation
- D.
None of the above
Correct Answer
A. Opaque to the visible lightExplanation
A semiconductor with an energy gap of 1.1 eV would be opaque to visible light because the energy of visible light photons (approximately 1.65 - 3.1 eV) is higher than the energy gap. This means that visible light photons do not have enough energy to excite electrons across the energy gap, resulting in the semiconductor being unable to transmit or transmit only a small portion of visible light.Rate this question:
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- 6.
In an intrinsic semiconductor, the mobility of electrons in the conduction band is:
- A.
Less than the mobility of holes in the valence band
- B.
Zero
- C.
Greater than the mobility of holes in the valence band
- D.
None of the above
Correct Answer
C. Greater than the mobility of holes in the valence bandExplanation
In an intrinsic semiconductor, the mobility of electrons in the conduction band is greater than the mobility of holes in the valence band. This is because electrons are negatively charged and therefore have a higher mobility compared to positively charged holes. Intrinsic semiconductors have an equal number of electrons and holes, but due to their different mobilities, electrons are more likely to move and contribute to the conduction of electric current.Rate this question:
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- 7.
The Hall Effect voltage in intrinsic silicon is:
- A.
Positive
- B.
Zero
- C.
Negative
- D.
None of the above
Correct Answer
C. NegativeExplanation
Intrinsic silicon is a pure form of silicon with no impurities. In this case, the Hall Effect voltage refers to the voltage generated when an electric current passes through the silicon in the presence of a magnetic field. In intrinsic silicon, the charge carriers are both electrons and holes. When a magnetic field is applied, the electrons and holes will experience a force perpendicular to their motion, resulting in a separation of charge carriers. This separation creates a voltage difference, with the negative charge carriers (electrons) accumulating on one side and the positive charge carriers (holes) accumulating on the other side. Therefore, the Hall Effect voltage in intrinsic silicon is negative.Rate this question:
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- 8.
The Hall coefficient of an intrinsic semiconductor is
- A.
Positive under all conditions
- B.
Negative under all conditions
- C.
Zero under all conditions
- D.
None of the above
Correct Answer
B. Negative under all conditionsExplanation
The Hall coefficient of an intrinsic semiconductor is negative under all conditions. The Hall coefficient is a measure of the sign and magnitude of the charge carriers in a material. In an intrinsic semiconductor, the charge carriers are electrons and holes. The negative Hall coefficient indicates that the majority charge carriers in an intrinsic semiconductor are electrons.Rate this question:
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- 9.
The electron and hole concentrations in a intrinsic semiconductor are ni and pi respectively. When doped with a p-type material, these change to n and p, respectively. Then
- A.
n + p = ni + pi
- B.
n + ni = p + pi
- C.
np = nipi
- D.
None of the above
Correct Answer
C. np = nipiExplanation
The correct answer is "np = nipi". In an intrinsic semiconductor, the electron concentration (n) and hole concentration (p) are equal to the intrinsic carrier concentrations (ni and pi, respectively). When doped with a p-type material, the electron concentration increases to n and the hole concentration decreases to p. The product of the electron and hole concentrations (np) remains constant and is equal to the product of the intrinsic carrier concentrations (nipi). This relationship is known as the law of mass action and holds true in doped semiconductors.Rate this question:
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- 10.
If the temperature of an extrinsic semiconductor is increased so that the intrinsic carrier concentration is doubled, then
- A.
The majority carrier density doubles
- B.
The minority carrier density doubles
- C.
Both majority and minority carrier densities double
- D.
None of the above
Correct Answer
C. Both majority and minority carrier densities doubleExplanation
When the temperature of an extrinsic semiconductor is increased, the intrinsic carrier concentration increases. This increase in intrinsic carrier concentration leads to an increase in both majority and minority carrier densities. The majority carriers are the majority charge carriers in the semiconductor, while the minority carriers are the minority charge carriers. Therefore, when the intrinsic carrier concentration is doubled, both the majority and minority carrier densities double as well.Rate this question:
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- 11.
At room temperature, the current in an intrinsic semiconductor is due to
- A.
Holes
- B.
Electrons
- C.
Holes and electrons
- D.
None of the above
Correct Answer
C. Holes and electronsExplanation
In an intrinsic semiconductor at room temperature, the current is due to the movement of both holes and electrons. In an intrinsic semiconductor, the number of electrons and holes is equal, and they are generated by thermal excitation. The movement of both charge carriers contributes to the overall current flow in the material. Therefore, the correct answer is "Holes and electrons."Rate this question:
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- 12.
Measurement of Hall coefficient in a semiconductor provides information on the
- A.
Sign and mass of charge carriers
- B.
Mass and concentration of charge carriers
- C.
Sign of charge carriers alone
- D.
Sign and concentration of charge carriers
Correct Answer
D. Sign and concentration of charge carriersExplanation
The measurement of Hall coefficient in a semiconductor provides information on the sign and concentration of charge carriers. The Hall coefficient is a parameter that describes the behavior of charge carriers in a magnetic field. By measuring the Hall coefficient, we can determine whether the charge carriers are positive or negative (sign) and also obtain information about their concentration, which is the number of charge carriers per unit volume.Rate this question:
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- 13.
.In a p-n junction diode: identify the false statement
- A.
The depletion capacitance increases with increase in the reverse-bias
- B.
The depletion capacitance decreases with increase in the reverse-bias
- C.
The diffusion capacitance increases with increase in the forward-bias
- D.
The diffusion capacitance is much higher than the depletion capacitance when it is forward-biased
Correct Answer
C. The diffusion capacitance increases with increase in the forward-biasExplanation
The diffusion capacitance is not affected by the forward-bias. It remains constant regardless of the biasing condition.Rate this question:
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- 14.
The width of the depletion region is
- A.
None of the above none of these
- B.
Directly proportional to doping
- C.
Inversely proportional to doping
- D.
Independent of doping
Correct Answer
C. Inversely proportional to dopingExplanation
The width of the depletion region in a semiconductor is inversely proportional to the doping level. This means that as the doping level increases, the width of the depletion region decreases, and vice versa. The depletion region is a region near the junction of two differently doped regions in a semiconductor device. It is an area depleted of majority charge carriers, creating a barrier for the flow of current. Higher doping levels introduce more charge carriers, reducing the width of the depletion region and allowing for easier current flow. Conversely, lower doping levels result in a wider depletion region and higher resistance to current flow.Rate this question:
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- 15.
The Fermi energy in p-n junction at thermal equilibrium is
- A.
Proportional to distance
- B.
Directly increases with the temperature
- C.
Invariant with respect to distance
- D.
None of the above
Correct Answer
C. Invariant with respect to distanceExplanation
The Fermi energy in a p-n junction at thermal equilibrium is invariant with respect to distance. This means that regardless of the position within the junction, the Fermi energy remains constant.Rate this question:
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- 16.
Gold is often diffused into silicon p-n junction devices to
- A.
Is proportional to the square of the recombination rate
- B.
Is proportional to the cube of the recombination rate
- C.
Make silicon a direct gap semiconductor
- D.
None of the above
Correct Answer
D. None of the aboveExplanation
Gold is often diffused into silicon p-n junction devices to modify the electrical properties of the semiconductor material. It does not have a direct relation to the recombination rate or make silicon a direct gap semiconductor. Therefore, the correct answer is "None of the above".Rate this question:
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- 17.
18.In a forward-biased photo diode with increase in incident light intensity, the diode current
- A.
Increases
- B.
Remains constant
- C.
Decreases
- D.
None of the above
Correct Answer
A. IncreasesExplanation
In a forward-biased photo diode, the diode current is directly proportional to the incident light intensity. This means that as the incident light intensity increases, more electrons are excited and flow through the diode, resulting in an increase in diode current. Therefore, the correct answer is "Increases".Rate this question:
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- 18.
The Varactor diode is
- A.
Voltage-dependent resistance
- B.
Voltage-dependent capacitance
- C.
Voltage-dependent inductor
- D.
None of the above
Correct Answer
B. Voltage-dependent capacitanceExplanation
A varactor diode is a type of diode that exhibits a voltage-dependent capacitance. This means that the capacitance of the diode can be varied by changing the applied voltage across it. As the voltage increases, the capacitance decreases, and vice versa. This property makes varactor diodes useful in applications such as voltage-controlled oscillators, frequency multipliers, and voltage-controlled filters.Rate this question:
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- 19.
The electric field in abrupt p-n junction is
- A.
Linear function of distance
- B.
Parabolic function of distance
- C.
Independent of distance
- D.
None of the above NONE OF THESE
Correct Answer
A. Linear function of distanceExplanation
The electric field in an abrupt p-n junction is a linear function of distance. This means that the electric field strength increases or decreases uniformly as the distance from the junction increases. In an abrupt p-n junction, there is a sharp transition between the p-type and n-type regions, and the electric field is constant within these regions. However, at the junction itself, there is a linear change in the electric field due to the difference in charge carriers and the resulting electric potential.Rate this question:
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- 20.
A p-n junction, which is produced by recrystallization on a base crystal, from a liquid phase of one or more components and the semiconductor is called
- A.
Doped junction
- B.
Alloy junction
- C.
Fused junction
- D.
None of the above
Correct Answer
C. Fused junctionExplanation
A fused junction refers to a p-n junction that is created by recrystallization on a base crystal from a liquid phase of one or more components. In this process, the semiconductor material is melted and then solidified to form the junction. This method allows for precise control over the doping levels and composition of the junction, making it suitable for specific applications in semiconductor devices. The other options, doped junction and alloy junction, do not accurately describe the process of creating a junction through recrystallization from a liquid phase. Therefore, the correct answer is fused junction.Rate this question:
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- 21.
.Each diode of full wave center-taped rectifier conducts for
- A.
450 only
- B.
1800 only
- C.
3600 complete period
- D.
2700 only
Correct Answer
B. 1800 onlyExplanation
In a full wave center-taped rectifier, each diode conducts only during the positive half cycle of the input AC signal. This means that the diodes conduct for half of the complete period of the input signal. Since the input signal has a complete period of 3600, each diode conducts for 1800 only.Rate this question:
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- 22.
.Bulk resistance of a diode is
- A.
The sum of resistance values of n-material and p-material
- B.
The sum of half the resistance value of n-material and p-material
- C.
Equivalent resistance of the resistance value of p- and n-material is parallel
- D.
None of the above
Correct Answer
A. The sum of resistance values of n-material and p-materialExplanation
The bulk resistance of a diode is the sum of the resistance values of the n-material and p-material. This is because the diode is made up of these two materials, and their individual resistances add up to determine the overall resistance of the diode.Rate this question:
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- 23.
.In a diode circuit, the point where the diode starts conducting is known as
- A.
Cut-in point
- B.
Cut-out point
- C.
Knee point
- D.
Cut-off point
Correct Answer
A. Cut-in pointExplanation
The point where the diode starts conducting is known as the cut-in point. This is the voltage threshold at which the diode allows current to flow in the forward direction. Below this voltage, the diode is in a non-conducting state and does not allow current to pass through. The cut-in point is an important characteristic of a diode as it determines when it starts to function as a rectifier or switch in a circuit.Rate this question:
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- 24.
A Zener diode should hav
- A.
Heavily doped p- and n-regions
- B.
Lightly doped p- and n-regions
- C.
Narrow depletion region
- D.
Both Heavily doped p- and n-regions AND Narrow depletion region
Correct Answer
D. Both Heavily doped p- and n-regions AND Narrow depletion regionExplanation
A Zener diode should have both heavily doped p- and n-regions and a narrow depletion region. The heavily doped p- and n-regions allow for a high concentration of charge carriers, which enables the diode to conduct current in both forward and reverse bias. The narrow depletion region is important because it allows for a high electric field to be established across the diode, which is necessary for the Zener effect to occur. The Zener effect is the phenomenon where the diode can maintain a constant voltage drop across it, even with large changes in current.Rate this question:
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- 25.
When a diode is forward-biased, the recombination of free electron and holes may produce
- A.
Heat
- B.
LighT
- C.
Radiation
- D.
All of the above
Correct Answer
D. All of the aboveExplanation
When a diode is forward-biased, the recombination of free electrons and holes may produce heat, light, and radiation. This is because when the diode is forward-biased, the p-n junction allows the flow of current in the forward direction. During this process, free electrons and holes combine, releasing energy in the form of heat. Additionally, depending on the energy level of the recombination, photons of light may be emitted, resulting in the production of light. Furthermore, in certain cases, the recombination process may also produce radiation. Therefore, all of the above options are possible outcomes when a diode is forward-biased.Rate this question:
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