1Which of the following feedback types is primarily used in the design of oscillators?
A.Negative feedback
B.Positive feedback
C.Zero feedback
D.Degenerative feedback
Correct Answer: Positive feedback
Explanation:Positive feedback is used in oscillators to sustain oscillations, as it feeds a portion of the output signal back to the input in phase with the input signal.
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2What is the primary reason for using negative feedback in amplifier circuits?
A.To increase the overall voltage gain
B.To induce oscillations
C.To stabilize the gain and improve linearity
D.To decrease the bandwidth
Correct Answer: To stabilize the gain and improve linearity
Explanation:Negative feedback intentionally reduces the gain to achieve higher stability, lower distortion, increased bandwidth, and reduced sensitivity to component variations.
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3If an amplifier has an open-loop gain of and a feedback factor of , which expression represents the loop gain?
A.
B.
C.
D.
Correct Answer:
Explanation:The loop gain of a feedback amplifier is the product of the open-loop gain and the feedback factor , denoted as .
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4According to the Barkhausen criterion, what condition must be met for an electronic circuit to sustain oscillations?
A. and phase shift is or
B. and phase shift is
C. and phase shift is
D. and phase shift is
Correct Answer: and phase shift is or
Explanation:The Barkhausen criterion states that for sustained oscillations, the loop gain magnitude must be exactly 1, and the total phase shift around the loop must be an integer multiple of .
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5What is the formula for the closed-loop gain of an amplifier with negative feedback?
A.
B.
C.
D.
Correct Answer:
Explanation:For negative feedback, the closed-loop gain is reduced by a factor of compared to the open-loop gain .
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6How does negative feedback affect the bandwidth of an amplifier?
A.It decreases the bandwidth by
B.It increases the bandwidth by a factor of
C.It leaves the bandwidth unchanged
D.It reduces the bandwidth to zero
Correct Answer: It increases the bandwidth by a factor of
Explanation:Negative feedback increases the bandwidth of an amplifier by the same factor by which it reduces the midband gain, keeping the gain-bandwidth product roughly constant.
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7What is the effect of negative feedback on non-linear distortion generated within the amplifier itself?
A.It increases distortion by
B.It eliminates all distortion completely
C.It reduces the distortion by a factor of
D.It shifts the distortion to a higher frequency range
Correct Answer: It reduces the distortion by a factor of
Explanation:Negative feedback reduces the internally generated non-linear distortion by the desensitivity factor .
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8In a voltage-series feedback amplifier, how are the input and output impedances affected?
Explanation:Series mixing at the input increases the input impedance, while voltage sampling (shunt connection) at the output decreases the output impedance.
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9Which feedback topology is ideal for a transconductance amplifier?
A.Voltage-Series
B.Current-Series
C.Voltage-Shunt
D.Current-Shunt
Correct Answer: Current-Series
Explanation:A transconductance amplifier takes a voltage input and provides a current output. Therefore, it requires series mixing at the input and series sampling at the output (Current-Series feedback).
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10What is the effect of a voltage-shunt feedback configuration on input impedance and output impedance ?
A. increases, increases
B. decreases, decreases
C. increases, decreases
D. decreases, increases
Correct Answer: decreases, decreases
Explanation:Shunt mixing at the input decreases the input impedance, and voltage sampling (shunt) at the output decreases the output impedance.
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11How does negative feedback affect the noise internally generated within an amplifier stage?
A.It amplifies the internal noise
B.It reduces the internal noise relative to the signal
C.It has no effect on internal noise
D.It changes the noise to a different frequency spectrum
Correct Answer: It reduces the internal noise relative to the signal
Explanation:Negative feedback reduces internally generated noise by the factor , thereby improving the Signal-to-Noise Ratio (SNR) for noise originating inside the loop.
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12If an amplifier has an open-loop gain of $1000$ and a feedback factor of $0.009$, what is the approximate closed-loop gain?
A.10
B.100
C.90
D.1000
Correct Answer: 100
Explanation:The closed loop gain is .
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13The sensitivity of an amplifier's gain with respect to changes in its internal parameters is given by which formula in the presence of negative feedback?
A.
B.
C.
D.
Correct Answer:
Explanation:Sensitivity is reduced in a negative feedback amplifier, and the formula is .
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14Current-shunt feedback is most suitable for which type of ideal amplifier?
A.Voltage amplifier
B.Current amplifier
C.Transconductance amplifier
D.Transresistance amplifier
Correct Answer: Current amplifier
Explanation:A current amplifier operates best with low input impedance and high output impedance. Current-shunt feedback provides shunt mixing (lowers ) and current sampling (increases ).
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15In an RC-coupled BJT amplifier, what is the primary purpose of the coupling capacitors?
A.To block DC from the input signal and between stages
B.To increase the high-frequency gain
C.To provide positive feedback
D.To bypass the emitter resistor
Correct Answer: To block DC from the input signal and between stages
Explanation:Coupling capacitors allow AC signals to pass from one stage to another while blocking DC, maintaining the independent DC bias conditions of each stage.
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16What component is responsible for the decrease in gain at low frequencies in an RC-coupled BJT amplifier?
A.Transistor junction capacitances
B.Stray wiring capacitances
C.Coupling and bypass capacitors
D.Load resistors
Correct Answer: Coupling and bypass capacitors
Explanation:At low frequencies, the reactances of the coupling and bypass capacitors become significant, dropping voltage and thereby reducing the overall gain of the amplifier.
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17Which of the following causes the gain of an RC-coupled amplifier to drop at high frequencies?
A.Coupling capacitors
B.Bypass capacitors
C.Inter-electrode and stray capacitances
D.Emitter resistors
Correct Answer: Inter-electrode and stray capacitances
Explanation:At high frequencies, the parasitic junction capacitances of the BJT and stray wiring capacitances exhibit low reactance, shunting the signal to ground and reducing gain.
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18In a common-emitter RC-coupled amplifier, what is the phase shift between the input and output signals in the midband frequency range?
A.
B.
C.
D.
Correct Answer:
Explanation:A single-stage Common-Emitter (CE) amplifier inherently introduces a phase inversion, which is a phase shift.
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19What is the function of the emitter bypass capacitor () in an RC-coupled BJT amplifier?
A.To stabilize the DC bias
B.To prevent AC signal degeneration and maximize voltage gain
C.To filter out high-frequency noise
D.To couple the output to the next stage
Correct Answer: To prevent AC signal degeneration and maximize voltage gain
Explanation:The bypass capacitor provides a low-impedance path for the AC signal around the emitter resistor, preventing negative feedback (degeneration) at signal frequencies and thereby increasing the AC gain.
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20If three amplifier stages with gains of $10$, $20$, and $30$ are cascaded, what is the overall voltage gain of the system (assuming no loading effects)?
A.60
B.6000
C.50
D.20
Correct Answer: 6000
Explanation:The overall gain of cascaded amplifier stages is the product of their individual gains: .
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21When amplifier stages are cascaded, how are their individual decibel (dB) gains combined to find the total gain?
A.They are multiplied
B.They are divided
C.They are added
D.They are subtracted
Correct Answer: They are added
Explanation:In the logarithmic decibel scale, the overall gain of cascaded stages is the sum of the individual gains:
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22What happens to the overall bandwidth of a system when multiple identical amplifier stages are cascaded?
A.The overall bandwidth increases
B.The overall bandwidth remains the same
C.The overall bandwidth decreases
D.The bandwidth becomes infinite
Correct Answer: The overall bandwidth decreases
Explanation:Cascading identical stages reduces the overall bandwidth due to the cumulative attenuation at the cutoff frequencies, a phenomenon known as bandwidth shrinkage.
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23For identical cascaded amplifier stages, each with a bandwidth , the overall bandwidth is approximately given by which formula?
A.
B.
C.
D.
Correct Answer:
Explanation:The overall bandwidth for synchronous cascaded stages is calculated using the shrinkage factor .
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24In a cascaded amplifier system, what is the 'loading effect'?
A.The increase in gain due to adding a load resistor
B.The reduction in a stage's gain because the next stage draws current from it
C.The shifting of the Q-point due to temperature changes
D.The generation of high-frequency oscillations
Correct Answer: The reduction in a stage's gain because the next stage draws current from it
Explanation:Loading effect occurs when the finite input impedance of a subsequent stage acts as a load on the previous stage, drawing current and thereby reducing the effective voltage gain of the previous stage.
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25Which of the following is the primary objective of a power amplifier?
A.To provide maximum voltage gain
B.To operate with the smallest possible input signal
C.To deliver maximum power to a low-impedance load
D.To maximize the input impedance of the system
Correct Answer: To deliver maximum power to a low-impedance load
Explanation:Unlike voltage amplifiers, power amplifiers are designed to handle large signals and deliver significant current and power to low-impedance loads like speakers or motors.
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26Which matching component is frequently used at the output of a power amplifier to transfer maximum power to a low-impedance load?
A.Coupling capacitor
B.Zener diode
C.Step-down transformer
D.Inductor
Correct Answer: Step-down transformer
Explanation:A step-down transformer is often used in power amplifiers for impedance matching, transforming the low load impedance to a higher value seen by the transistor, enabling maximum power transfer.
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27Power amplifiers are classified as large-signal amplifiers. What does this imply about the transistor's operation?
A.The transistor operates exclusively in the cutoff region
B.The transistor's operating point swings over a large portion of the load line
C.The transistor only handles high-frequency signals
D.The transistor parameters remain perfectly linear throughout operation
Correct Answer: The transistor's operating point swings over a large portion of the load line
Explanation:Large-signal operation means the input signal is large enough to drive the transistor over a significant portion of its active region, often leading to non-linearities and distortion.
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28What is the conduction angle for a Class A power amplifier?
A.
B.
C.Between and
D.
Correct Answer:
Explanation:In a Class A amplifier, the transistor is biased such that it conducts current during the entire cycle () of the input signal.
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29Where is the quiescent (Q) point located on the AC load line for an ideal Class A power amplifier?
A.At the cutoff point
B.At the saturation point
C.Exactly at the center of the active region
D.Slightly above the cutoff point
Correct Answer: Exactly at the center of the active region
Explanation:To allow maximum symmetrical swing of the output signal without clipping, the Q-point is placed at the center of the AC load line in a Class A amplifier.
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30What is the maximum theoretical efficiency of a series-fed (resistive load) Class A power amplifier?
A.
B.
C.
D.
Correct Answer:
Explanation:A series-fed Class A amplifier has a maximum theoretical conversion efficiency of due to the continuous power dissipation in the transistor and the series load resistor.
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31By using transformer coupling in a Class A amplifier, the maximum theoretical efficiency can be increased to:
A.
B.
C.
D.
Correct Answer:
Explanation:Transformer coupling eliminates DC power loss in the load resistor, raising the maximum theoretical efficiency of a Class A amplifier to .
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32What is the conduction angle for a single transistor in a Class B power amplifier?
A.Less than
B.
C.Between and
D.
Correct Answer:
Explanation:In a Class B amplifier, the transistor is biased at cutoff, so it conducts for exactly half of the input signal cycle ().
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33What is the maximum theoretical efficiency of a Class B push-pull power amplifier?
A.
B.
C.
D.
Correct Answer:
Explanation:The maximum theoretical efficiency of a Class B push-pull amplifier is or approximately .
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34Which type of distortion is most prominently associated with Class B push-pull amplifiers?
A.Harmonic distortion
B.Intermodulation distortion
C.Crossover distortion
D.Phase distortion
Correct Answer: Crossover distortion
Explanation:Crossover distortion occurs in Class B amplifiers because the transistors do not begin to conduct until the input signal exceeds their base-emitter turn-on voltage (approx. 0.7V), creating a 'dead zone' where the signal crosses zero.
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35How many transistors are typically required to amplify a full sine wave in a Class B amplifier circuit?
A.One
B.Two
C.Three
D.Four
Correct Answer: Two
Explanation:A Class B amplifier requires a push-pull configuration with at least two transistors (one for the positive half-cycle and one for the negative half-cycle) to reproduce the full sine wave.
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36What is the power dissipation of a Class B amplifier when there is no input signal (zero signal condition)?
A.Maximum
B.Half of maximum
C.Ideally zero
D.Equal to the load power
Correct Answer: Ideally zero
Explanation:Because Class B transistors are biased at cutoff, they draw virtually no current when there is no input signal, resulting in zero ideal quiescent power dissipation.
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37What is the primary motivation for using a Class AB power amplifier instead of a Class B amplifier?
A.To achieve higher efficiency than Class B
B.To eliminate crossover distortion
C.To reduce the number of components
D.To increase the voltage gain
Correct Answer: To eliminate crossover distortion
Explanation:Class AB amplifiers apply a small forward bias to the transistors so they conduct slightly even when the input signal is zero. This eliminates the dead zone and fixes crossover distortion.
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38What is the conduction angle for a transistor operating in Class AB mode?
A.Exactly
B.Between and
C.Exactly
D.Less than
Correct Answer: Between and
Explanation:In Class AB, the transistor conducts for slightly more than half a cycle, so its conduction angle is greater than but strictly less than .
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39Where is the Q-point located on the load line for a Class AB amplifier?
A.Exactly in the middle of the active region
B.Slightly above the cutoff point
C.Deep in the cutoff region
D.At the saturation point
Correct Answer: Slightly above the cutoff point
Explanation:The Q-point in Class AB is set slightly above the cutoff point. This small quiescent current ensures the transistor is already 'on' when the signal crosses zero.
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40Which components are commonly used in the biasing network of a Class AB push-pull amplifier to provide temperature compensation?
A.Inductors
B.Capacitors
C.Diodes
D.Transformers
Correct Answer: Diodes
Explanation:Diodes are often used in the biasing network of Class AB amplifiers because their thermal characteristics match those of the BJT base-emitter junctions, stabilizing the quiescent current against temperature variations.
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41What is the primary function of a voltage regulator in an electronic circuit?
A.To step up DC voltage
B.To convert AC voltage to DC voltage
C.To maintain a constant output DC voltage despite variations in input voltage or load current
D.To filter out high-frequency noise from an AC signal
Correct Answer: To maintain a constant output DC voltage despite variations in input voltage or load current
Explanation:A voltage regulator is designed to automatically maintain a constant voltage level at its output, regardless of changes to the load current or the input supply voltage.
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42Which semiconductor device is most fundamentally used as a simple voltage regulator?
A.Light Emitting Diode (LED)
B.Schottky Diode
C.Zener Diode
D.Tunnel Diode
Correct Answer: Zener Diode
Explanation:A Zener diode, when reverse-biased into its breakdown region, maintains a nearly constant voltage across its terminals, making it a basic voltage regulator.
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43Which term describes a regulator's ability to maintain a constant output voltage when the input (line) voltage changes?
A.Load regulation
B.Line regulation
C.Thermal regulation
D.Ripple rejection
Correct Answer: Line regulation
Explanation:Line regulation is a measure of how well the voltage regulator maintains its output voltage despite variations in the input (unregulated) line voltage.
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44What is 'Load Regulation' in the context of power supplies?
A.The change in output voltage for a given change in load current
B.The change in output current for a given change in input voltage
C.The maximum power a load can dissipate
D.The variation of input voltage as the load changes
Correct Answer: The change in output voltage for a given change in load current
Explanation:Load regulation specifies how much the output voltage changes when the load current varies from zero (no-load) to its maximum rated value (full-load).
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45In a series voltage regulator, the control element (usually a transistor) is placed in what relation to the load?
A.In parallel with the load
B.In series with the load
C.In parallel with the input supply
D.Between the ground and the negative terminal
Correct Answer: In series with the load
Explanation:In a series regulator, the control transistor is in series with the load. It acts as a variable resistor to drop the excess input voltage and maintain a constant voltage across the load.
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46What is the role of an Operational Amplifier (Op-Amp) when used in a linear voltage regulator circuit?
A.It acts as a high-power switch
B.It generates the unregulated input voltage
C.It acts as an error amplifier to compare the output voltage with a reference voltage
D.It provides the primary load current
Correct Answer: It acts as an error amplifier to compare the output voltage with a reference voltage
Explanation:The Op-Amp functions as an error amplifier. It compares a fraction of the output voltage to a stable reference voltage (like a Zener) and adjusts the series pass transistor accordingly.
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47In a shunt voltage regulator, how does the control element maintain constant output voltage?
A.By varying its resistance in series with the load
B.By shutting down the input supply during overvoltage
C.By shunting (diverting) variable amounts of current away from the load
D.By increasing the input voltage dynamically
Correct Answer: By shunting (diverting) variable amounts of current away from the load
Explanation:A shunt regulator is placed in parallel with the load. It regulates the output voltage by sinking more or less current to ground, thus controlling the voltage drop across a series resistor.
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48What does the term 'dropout voltage' refer to in a linear voltage regulator?
A.The voltage at which the regulator shuts down completely
B.The minimum difference between input and output voltage required to maintain regulation
C.The voltage drop across the load
D.The absolute maximum input voltage allowed
Correct Answer: The minimum difference between input and output voltage required to maintain regulation
Explanation:Dropout voltage is the minimum input-to-output voltage differential required by the regulator to function properly and maintain a stable output voltage.
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49Which of the following ICs is a popular standard fixed positive voltage regulator?
A.LM7905
B.LM317
C.LM7805
D.NE555
Correct Answer: LM7805
Explanation:The 78xx series (like the LM7805) provides a fixed positive regulated voltage (5V in the case of 7805). The 79xx series is for negative voltages, and LM317 is an adjustable regulator.
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50What advantage does a series regulator have over a basic Zener shunt regulator?
A.It requires fewer components
B.It dissipates less power when the load is disconnected
C.It naturally limits output current to zero
D.It can only step up voltages
Correct Answer: It dissipates less power when the load is disconnected
Explanation:In a basic shunt regulator, maximum current flows through the shunt element when no load is connected, wasting power. A series regulator only draws the current demanded by the load plus a small quiescent current, making it more efficient under light or no-load conditions.