Electronics Engineering 1 - Part 1

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    1/34 Questions

  • It is a process when the reverse voltage applied to zener diode exceeds the selected reverse breakdown voltage that occurs in the depletion layer and the current through the diode increases to the maximum circuit value, which is usually limited by a series resistor.

    • Avalanche breakdown
    • Regulation
    • Stabilization
    • Reverse breakdown
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About This Quiz

Explore the fundamentals of electronics engineering in this engaging quiz. Covering topics from zener diodes and avalanche breakdowns to semiconductor properties and impurities, this quiz assesses key concepts vital for understanding electrical circuits and components.

Electronics Engineering 1 - Part 1 - Quiz
Questions and Answers
  • 2. 

    The most basic passive components in electrical or electronic circuits are

    • Resistors

    • Capacitors

    • Inductors

    • Diodes

    • Batteries

    Correct Answer
    A. Resistors
    Explanation
    Resistors are the most basic passive components in electrical or electronic circuits because they oppose the flow of electric current. They are used to control the amount of current flowing in a circuit and to limit the voltage across components. Resistors are made of materials with high resistance, which means they have a low conductivity. They are commonly used to regulate current, divide voltage, and provide load in circuits. Unlike active components like diodes or batteries, resistors do not require a power source and do not amplify or generate signals.

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  • 3. 

    Semiconductors ability to conduct electricity can be greatly improved by adding certain impurities to the crystalline structure thereby, producing more free electrons than holes or vice versa.

    • The statement is TRUE.

    • The statement is FALSE.

    • The statement is sometimes TRUE.

    • Not applicable

    Correct Answer
    A. The statement is TRUE.
    Explanation
    Adding certain impurities to the crystalline structure of semiconductors can indeed improve their ability to conduct electricity. This process is known as doping, and it involves introducing atoms of a different element into the crystal lattice of the semiconductor material. These impurity atoms either provide extra electrons (n-type doping) or create vacancies for electrons (p-type doping), thus increasing the number of charge carriers and enhancing the conductivity of the material. Therefore, the given statement is true.

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  • 4. 

    These atoms allow some of the electrons to bond with its neighbouring silicon atoms leaving one "free electron" to move about when an electrical voltage is applied and are also called "donors".

    • Pentavalent impurities

    • Crystals

    • Trivalent impurities

    • P-type semiconductor

    Correct Answer
    A. Pentavalent impurities
    Explanation
    Pentavalent impurities are atoms that have five valence electrons, such as phosphorus or arsenic. When these impurities are added to a crystal lattice, they can bond with neighboring silicon atoms, leaving one extra electron that is not involved in any bonding. This extra electron is called a "free electron" and can move freely within the lattice when an electrical voltage is applied. This behavior of pentavalent impurities makes the crystal a p-type semiconductor, as it has an excess of positive charge carriers (holes) due to the presence of the free electrons.

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  • 5. 

    These are negatively charged and there are a large number of holes with a small number of free electrons in relation to the number of holes.

    • Pentavalent impurities

    • Crystals

    • Trivalent impurities

    • N-type semiconductor

    Correct Answer
    A. Trivalent impurities
    Explanation
    Trivalent impurities are the correct answer because they introduce excess or additional valence electrons in a crystal lattice. These additional electrons are able to move freely, creating a surplus of negative charge. At the same time, the trivalent impurities also create holes, or vacancies, where electrons could be present. However, the number of free electrons is much smaller compared to the number of holes. This imbalance between the number of holes and free electrons is characteristic of a p-type semiconductor.

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  • 6. 

    In a semiconductor crystal, the atoms are held together by

    • The interaction of valence electrons

    • Forces of attraction

    • Covalent bonds

    • Answers (A), (B), and (C)

    Correct Answer
    A. Answers (A), (B), and (C)
    Explanation
    The correct answer is (A), (B), and (C). In a semiconductor crystal, the atoms are held together by the interaction of valence electrons, forces of attraction, and covalent bonds. Valence electrons are the outermost electrons of an atom and they play a crucial role in bonding. The forces of attraction between atoms are responsible for keeping them close together. Covalent bonds occur when atoms share electrons, creating a strong bond. All three of these factors contribute to the stability and structure of a semiconductor crystal.

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  • 7. 

    When the N and P-type semiconductor materials are first brought together some of the free electrons move across the junction to fill up the holes in the P-type material producing _________________.

    • Free electrons

    • Positive ions

    • Negative ions

    • Pn junction

    Correct Answer
    A. Negative ions
    Explanation
    When the N and P-type semiconductor materials are first brought together, some of the free electrons from the N-type material move across the junction to fill up the holes in the P-type material. This movement of electrons creates an excess of negative charges in the P-type material, resulting in the formation of negative ions.

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  • 8. 

    A widely used application of the power diode is in the conversion of alternating voltages (AC) to direct voltages (DC), called _____________.

    • Diffusion

    • Rectification

    • Amplification

    • Modification

    Correct Answer
    A. Rectification
    Explanation
    The power diode is commonly used in the process of converting alternating voltages (AC) to direct voltages (DC), which is known as rectification. Rectification involves the conversion of the alternating current to a unidirectional flow of current, allowing for the production of a direct voltage. This process is essential in various applications, such as power supplies and electronic devices, where a steady and constant flow of DC voltage is required.

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  • 9. 

    In a half-wave rectifier circuit, during each "positive" half cycle of the AC sine wave, the diode is ______________.

    • Reverse bias

    • Forward bias

    • Unidirectional

    • Ideal

    Correct Answer
    A. Forward bias
    Explanation
    During each "positive" half cycle of the AC sine wave, the diode in a half-wave rectifier circuit is forward biased. This means that the voltage at the anode of the diode is higher than the voltage at the cathode, allowing current to flow through the diode. As a result, only the positive half of the AC waveform is allowed to pass through the circuit, while the negative half is blocked.

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  • 10. 

    One of the ways of reducing the ripple or voltage variations on a direct DC voltage is by _______________.

    • Connecting capacitors across the load resistance

    • Connecting resistor across the diode

    • Connecting a transformer to the circuit

    • Connect a diode across the load resistance

    Correct Answer
    A. Connecting capacitors across the load resistance
    Explanation
    Connecting capacitors across the load resistance helps to reduce the ripple or voltage variations on a direct DC voltage. Capacitors act as energy storage devices and can smooth out the voltage by storing charge when the voltage is high and releasing it when the voltage is low. This helps to stabilize the voltage and reduce any fluctuations or variations in the DC output.

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  • 11. 

    The main advantage of a full-wave bridge rectifier compared to half-wave is that ___________.

    • It has a smaller AC ripple value for a given load

    • It has a smaller reservoir or smoothing capacitor than

    • It produces fundamental frequency of the ripple voltage twice that of the AC supply frequency

    • All of the above

    • (A) and (B)

    Correct Answer
    A. (A) and (B)
    Explanation
    The main advantage of a full-wave bridge rectifier compared to half-wave is that it has a smaller AC ripple value for a given load and it has a smaller reservoir or smoothing capacitor than a half-wave rectifier. This means that the output of the full-wave rectifier will have less variation or fluctuations in voltage, resulting in a smoother and more stable DC output. Additionally, the smaller capacitor requirement makes the full-wave bridge rectifier more cost-effective and compact. Therefore, both options (A) and (B) are correct.

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  • 12. 

    If the load resistance of a capacitor-filtered full-wave rectifier is reduced, the ripple voltage _________.

    • Decreases

    • Increases

    • Is not affected

    • Has a different frequency

    Correct Answer
    A. Increases
    Explanation
    When the load resistance of a capacitor-filtered full-wave rectifier is reduced, the ripple voltage increases. This is because a lower load resistance allows more current to flow through the circuit, resulting in a larger voltage drop across the resistor. As a result, the ripple voltage, which is the difference between the maximum and minimum voltage levels, becomes larger. Therefore, reducing the load resistance increases the ripple voltage.

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  • 13. 

    Valence electrons are

    • In the closest orbit to the nucleus

    • In the most distant orbit from the nucleus

    • In various orbits around the nucleus

    • Not associated with a particular atom

    Correct Answer
    A. In the most distant orbit from the nucleus
    Explanation
    Valence electrons are located in the most distant orbit from the nucleus. These electrons are involved in chemical bonding and determine the reactivity of an atom. They are the outermost electrons and are responsible for forming bonds with other atoms.

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  • 14. 

    A positive ion is formed when ________.

    • there are more holes than electrons in the outer orbit

    • Two atoms bond together

    • An atom gains an extra valence electron

    • A valence electron breaks away from the atom

    Correct Answer
    A. A valence electron breaks away from the atom
    Explanation
    A positive ion is formed when a valence electron breaks away from the atom. When an atom loses one or more valence electrons, it becomes positively charged because the number of protons in the nucleus is now greater than the number of electrons orbiting the nucleus. This creates an imbalance of positive and negative charges, resulting in the formation of a positive ion.

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  • 15. 

    A technique where no external voltage potential is applied to the PN-junction.

    • Reversed bias

    • Forward bias

    • Zero bias

    • Avalance breakdown

    Correct Answer
    A. Zero bias
    Explanation
    Zero bias refers to a technique where no external voltage potential is applied to the PN-junction. In this state, the PN-junction operates without any bias, meaning that there is no forward or reverse voltage applied. As a result, no current flows through the junction, and the depletion region is not altered. This allows for a more accurate measurement of the intrinsic properties of the PN-junction, such as the width of the depletion region and the diffusion of charge carriers.

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  • 16. 

    The voltage potential is connected negative to the P-type material and positive to the N-type material across the diode which has the effect of increasing the PN-junction width.

    • Zero bias

    • Reverse bias

    • Forward bias

    • Depletion layer

    Correct Answer
    A. Reverse bias
    Explanation
    When a diode is in reverse bias, the voltage potential is connected negative to the P-type material and positive to the N-type material. This causes the depletion layer, which is the region around the PN-junction where there are no majority charge carriers, to widen. As a result, the reverse bias reduces the flow of current through the diode, making it act as an insulator.

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  • 17. 

    Semiconductors contain two types of mobile charge carriers.

    • TRUE

    • FALSE

    • The statement is TRUE depending on the type of material.

    • The statement is FALSE when fully charged with electrons.

    Correct Answer
    A. TRUE
    Explanation
    Semiconductors do indeed contain two types of mobile charge carriers. These carriers are electrons, which have a negative charge, and holes, which can be thought of as the absence of an electron and have a positive charge. The behavior and conductivity of semiconductors depend on the movement and interaction of these charge carriers. Therefore, the statement is true.

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  • 18. 

    A natural potential barrier is developed across a PN-junction which is about 0.7v for Silicon diodes and about 0.3v for Germanium diodes when __________.

    • Zero bias

    • Forward bias

    • Reverse bias

    • Connected with capacitor

    Correct Answer
    A. Zero bias
    Explanation
    When a PN-junction is in zero bias, it means that no external voltage is applied across the junction. In this state, the natural potential barrier is developed across the junction due to the difference in doping concentrations between the P and N regions. For silicon diodes, this barrier is about 0.7V, while for germanium diodes, it is about 0.3V. This potential barrier prevents current flow in the absence of an applied voltage.

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  • 19. 

    Which of the following is not included in the group?

    • Boron

    • Antimony

    • Indium

    • Gallium

    • Aluminum

    Correct Answer
    A. Antimony
    Explanation
    The question asks for the element that is not included in the group. The group consists of Boron, Indium, Gallium, and Aluminum. Antimony is not included in this group, making it the correct answer.

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  • 20. 

    Which of the following is not included in the group?

    • Phosphorus

    • Arsenic

    • Antimony

    • Aluminum

    • Bismuth

    Correct Answer
    A. Aluminum
    Explanation
    The given options consist of elements from the periodic table. Phosphorus, Arsenic, Antimony, and Bismuth are all metalloids or nonmetals. However, Aluminum is a metal and does not belong to the same group as the other elements mentioned.

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  • 21. 

    The energy band in which free electrons exist is the _________.

    • First band

    • Second band

    • Conduction band

    • Valence band

    Correct Answer
    A. Conduction band
    Explanation
    The conduction band is the energy band in which free electrons exist. In this band, electrons have enough energy to move freely and conduct electricity. The valence band, on the other hand, is the energy band in which electrons are tightly bound to atoms and do not contribute to electrical conductivity. Therefore, the correct answer is conduction band.

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  • 22. 

    Each atom  in a silicon crystal has _________.

    • Four valence electrons

    • Four conduction electrons

    • Eight valence electrons, four of its own and four shared

    • No valence electrons because all are shared with other atoms

    Correct Answer
    A. Eight valence electrons, four of its own and four shared
    Explanation
    Each atom in a silicon crystal has eight valence electrons, four of its own and four shared. This is because silicon is in Group 14 of the periodic table, which means it has four valence electrons in its outermost energy level. These electrons can form covalent bonds with neighboring silicon atoms, resulting in a crystal lattice structure. In this structure, each silicon atom shares one electron with each of its four neighboring atoms, giving a total of four shared electrons. Therefore, each silicon atom effectively has eight valence electrons, four of its own and four shared.

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  • 23. 

    The current in a semiconductor is produced by _________.

    • Electron only

    • Holes only

    • Negative ions

    • Both electrons and holes

    Correct Answer
    A. Both electrons and holes
    Explanation
    In a semiconductor, the current is produced by both electrons and holes. Electrons are negatively charged particles that carry negative charge, while holes are essentially the absence of an electron and carry positive charge. The movement of both electrons and holes contributes to the flow of current in a semiconductor material. This is because electrons can move freely in the conduction band, while holes can move in the valence band. Therefore, both electrons and holes play a role in generating the current in a semiconductor.

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  • 24. 

    The purpose of pentavalent impurity is to ________.

    • Reduce the conductivity of silicon

    • Increase the number of holes

    • Increase the number of free electrons

    • Create minority carriers

    Correct Answer
    A. Increase the number of free electrons
    Explanation
    Pentavalent impurities, such as phosphorus or arsenic, have five valence electrons which are greater than the four valence electrons of silicon. When these impurities are added to silicon, they introduce extra electrons into the crystal lattice, creating more free electrons. This increases the conductivity of silicon as the free electrons are able to move more easily through the material, making it a better conductor. Therefore, the purpose of pentavalent impurities is to increase the number of free electrons in silicon.

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  • 25. 

    The term bias means ___________.

    • The ratio of majority carriers to minority carriers

    • The amount of current across the diode

    • A dc voltage is applied to control the operation of a device

    • Neither (A), (B), nor (C)

    Correct Answer
    A. A dc voltage is applied to control the operation of a device
    Explanation
    The term bias refers to the application of a direct current (dc) voltage to control the operation of a device. This bias voltage is used to establish the desired operating conditions and characteristics of the device, such as its current flow or voltage levels. It is a crucial aspect in controlling the behavior and performance of electronic components and circuits.

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  • 26. 

    To forward-bias a diode, _____________________.

    • An external voltage is applied that is positive at the anode and negative at the cathode

    • An external voltage is applied that is negative at the anode and positive at the cathode

    • An external voltage is applied that is positive at the p region and negative at the n region

    • Answers (A) and (C)

    • Answers (B) and (C)

    Correct Answer
    A. Answers (A) and (C)
    Explanation
    To forward-bias a diode, an external voltage is applied that is positive at the anode and negative at the cathode. Additionally, an external voltage is applied that is positive at the p region and negative at the n region.

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  • 27. 

    Although current is blocked in reverse bias, ______________.

    • There is some current due to majority carriers

    • There is a very small current due to minority carriers

    • There is an avalanche current

    • There is a very small current due to battery

    • There is some current due to free electrons

    Correct Answer
    A. There is a very small current due to minority carriers
    Explanation
    In a reverse biased diode, the majority carriers (which are the electrons in an N-type material and the holes in a P-type material) are pushed away from the junction and towards the depletion region, creating a barrier for current flow. However, a very small current can still flow due to the minority carriers (holes in the N-type material and electrons in the P-type material) that are present in the material. These minority carriers can overcome the barrier and contribute to a small current flow.

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  • 28. 

    When a voltmeter is placed across a forward-biased diode, it will read a voltage approximately equal to ______________.

    • The bias battery voltage

    • 0 V

    • The diode barrier potential

    • The total circuit voltage

    Correct Answer
    A. The diode barrier potential
    Explanation
    When a voltmeter is placed across a forward-biased diode, it will read a voltage approximately equal to the diode barrier potential. This is because the diode barrier potential is the voltage required to overcome the potential barrier at the junction of the diode, allowing current to flow through the diode. Therefore, when the diode is forward-biased, the voltmeter will measure a voltage close to the diode barrier potential.

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  • 29. 

    A silicon diode is in series with a 1.0kW resistor and a 5 V battery. If the anode is connected to the positve battery terminal, the cathode voltage with respect to the negative battery terminal is ___________.

    • 0.7V

    • 0.3V

    • 5.3V

    • 4.3V

    Correct Answer
    A. 4.3V
    Explanation
    When a silicon diode is in series with a resistor and a battery, the voltage drop across the diode is typically around 0.7V. This means that the cathode voltage, which is the voltage at the end of the diode connected to the resistor, will be 0.7V less than the battery voltage. In this case, the battery voltage is 5V, so the cathode voltage with respect to the negative battery terminal will be 5V - 0.7V = 4.3V.

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  • 30. 

    The postive lead of an ohmmeter is connected to the anode of a diode and the negative lead is connected to the cathode. The diode is _____________.

    • Reversed-bias

    • Open

    • Forward-bias

    • Faulty

    • Answers (B) and (D)

    Correct Answer
    A. Forward-bias
    Explanation
    When the positive lead of an ohmmeter is connected to the anode of a diode and the negative lead is connected to the cathode, it creates a forward-bias condition. In forward-bias, the diode allows current to flow easily from the anode to the cathode, as it is designed to conduct electricity in this direction. This indicates that the diode is functioning correctly and is not faulty. Therefore, the correct answer is forward-bias.

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  • 31. 

    If one of the diodes in a bridge full-wave rectifier opens, the output is _______.

    • 0V

    • One-fourth the amplitude of the input voltage

    • A half-wave rectified voltage

    • A 120 Hz voltage

    Correct Answer
    A. A half-wave rectified voltage
    Explanation
    If one of the diodes in a bridge full-wave rectifier opens, it means that the current can only flow in one direction through the circuit. This results in only half of the input voltage being rectified, while the other half is blocked. Therefore, the output will be a half-wave rectified voltage, where only the positive or negative half of the input waveform is present in the output.

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  • 32. 

    A 10V peak-to-peak sinusoidal voltage is applied across a silicon diode and series resistor. The maximum voltage across the diode is ________.

    • 9.3V

    • 10V

    • 5

    • 4.3V

    • 0.7V

    Correct Answer
    A. 4.3V
    Explanation
    When a sinusoidal voltage is applied across a diode and series resistor, the diode will only conduct when the voltage across it exceeds the forward voltage drop (typically around 0.7V for a silicon diode). In this case, the peak-to-peak voltage is 10V, so the maximum voltage across the diode will be 10V - 0.7V = 9.3V. Therefore, the correct answer is 9.3V.

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  • 33. 

    What is the input rms voltage, when the average voltage value of a half-wave rectifier is 35V?

    • 110V

    • 49.5V

    • 78V

    • 220V

    Correct Answer
    A. 78V
    Explanation
    The average voltage value of a half-wave rectifier is equal to the RMS voltage divided by π. To find the RMS voltage, we can multiply the average voltage by π. Therefore, if the average voltage is 35V, the RMS voltage would be 35V * π ≈ 109.96V. The closest option to this value is 110V, which is the correct answer.

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  • 34. 

    What is the secondary rms voltage of a bridge rectifier when the peak-to-peak voltage is 40V (use second approximation)?

    • 19.8V

    • 27.29V

    • 29.27V

    • 13.15V

    • 15.13V

    Correct Answer
    A. 15.13V
    Explanation
    The secondary rms voltage of a bridge rectifier can be calculated using the formula Vrms = Vpp / 2√2. In this case, the peak-to-peak voltage is given as 40V. Using the formula, we can calculate the secondary rms voltage as 40V / 2√2 = 15.13V. Therefore, the correct answer is 15.13V.

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  • Mar 21, 2023
    Quiz Edited by
    ProProfs Editorial Team
  • Jun 09, 2010
    Quiz Created by
    Bievangelista

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