Analog Electronics — Part 1 MCQs with Explanations

Option A: Current controlled Current source A current-controlled current source (CCCS) means the output current is proportional to the input current. However, the operation of a JFET does not involve control of current by another current. Therefore, this option is incorrect . Option B: Current controlled Voltage source A current-controlled voltage source (CCVS) implies the output voltage is controlled by the input current. Since the JFET is voltage-controlled, its output voltage is not directly dependent on the current. Hence, this option is incorrect . Option C: Voltage controlled Voltage source A voltage-controlled voltage source (VCVS) suggests that the output voltage is influenced by the input voltage. Although the JFET is voltage-controlled, its output behavior is better described as controlling current, not voltage. Thus, this option is also incorrect . Option D: Voltage controlled Current source A voltage-controlled current source (VCCS) means the output current is controlled by the input voltage. In a JFET, the gate-to-source voltage controls the current flowing through the drain-source channel. This aligns with the operation of the JFET, making this option correct . Conclusion: The action of a JFET in its equivalent circuit can best be represented as a Voltage controlled Current source (Option D).

Option A: 99% An efficiency of 99% is highly unlikely for a Common Base (CB) amplifier. Such a high efficiency is typically not achievable in practical amplifier circuits due to inherent losses in the components and design. Therefore, this option is incorrect . Option B: 85% An efficiency of 85% is possible for certain types of amplifiers like Class C amplifiers, which are specifically designed for high efficiency. However, the CB amplifier operates in Class A mode, which inherently limits its maximum efficiency. Hence, this option is incorrect . Option C: 50% For a CB amplifier operating in Class A mode, the theoretical maximum efficiency is 50%. This is due to the symmetrical operation of the circuit, where half the power is dissipated as heat in the transistor. Therefore, this option is correct . Option D: 25% An efficiency of 25% is too low for a CB amplifier operating in Class A mode. While practical losses can reduce efficiency, it typically does not drop this low under normal operating conditions. Hence, this option is incorrect . Conclusion: The maximum efficiency of a CB amplifier could be 50% (Option C).

Option A: The edge of the depletion region on the p-side At the edge of the depletion region on the p-side, the electric field is not maximum. The field strength gradually reduces as you move away from the junction towards the edges of the depletion region. Therefore, this option is incorrect . Option B: The edge of the depletion region on the n-side Similarly, at the edge of the depletion region on the n-side, the electric field is not at its peak. The field strength diminishes as you approach the edges of the depletion region. Thus, this option is incorrect . Option C: The p + n junction In a p + n junction diode under reverse bias, the electric field is maximum at the junction itself. This is because the maximum charge separation occurs at the junction, resulting in the highest electric field strength. Hence, this option is correct . Option D: The center of the depletion region on the n-side The center of the depletion region on the n-side does not experience the maximum electric field. As you move away from the junction, the electric field strength decreases, making this option incorrect . Conclusion: In a p + n junction diode under reverse bias, the magnitude of the electric field is maximum at the p + n junction (Option C).

Option A: Resistor between source and load A resistor between the source and load allows both AC and DC components of the signal to pass. It does not block the DC component, which is not suitable when the goal is to prevent DC return. Therefore, this option is incorrect . Option B: Inductor between source and load An inductor primarily opposes changes in current and allows DC to pass through while blocking high-frequency AC signals. Since it does not block DC, it cannot prevent a DC return. Hence, this option is incorrect . Option C: Capacitor between source and load A capacitor blocks DC components and allows AC signals to pass through due to its property of impedance, which is inversely proportional to frequency. This makes it the ideal component to prevent DC return between the source and load. Therefore, this option is correct . Option D: Either A or B Neither a resistor nor an inductor can effectively prevent DC return as they allow DC to pass through. Thus, this option is incorrect . Conclusion: To prevent a DC return between the source and load, it is necessary to use a capacitor (Option C).

A constant current signal across a parallel RLC circuit gives an output of 1.4v at the signal frequencies of 3.89kHz and 4.1kHz. This indicates that the circuit is operating near its resonant frequency. At resonance, the impedance of the parallel RLC circuit is maximum, leading to a maximum voltage across it for a given current. In this case, the output voltage is 1.4v at frequencies close to resonance. As we move further away from the resonant frequency, the impedance decreases, and consequently, the output voltage also decreases. Therefore, at 4kHz, which is closer to the resonant frequency than 3.89kHz or 4.1kHz, the output voltage will be **slightly higher** than 1.4v. Among the given options, Option B: 2v is the closest to this expected value.

$$\eqalign{ & {I_{\text{C}}} = \frac{{{I_{{\text{CO}}}}}}{{1 - \alpha }} + \frac{\alpha }{{1 - \alpha }} \times {I_\beta } \cr & = \frac{6}{{1 - 0.98}} + \frac{{0.98}}{{1 - 0.98}} \times 100 \cr & = 5.2\,{\text{mA}} \cr} $$

Explanation of each option: Option A: Junction capacitance is associated with a p-n junction. This capacitance arises due to the separation of charge carriers across the junction, creating an electric field that contributes to capacitance. Option B: Charge storage capacitance is also associated with a p-n junction. It refers to the capacitance that stores charge at the junction when it is forward biased, as the charge carriers accumulate at the junction. Option C: Depletion capacitance is related to a p-n junction as well. It occurs due to the depletion region that forms at the junction when it is reverse biased, and this region acts like a capacitor. Option D: Channel length modulation is not associated with a p-n junction. This phenomenon is related to field-effect transistors (FETs), where the effective length of the channel is modulated by the drain-to-source voltage, and does not occur in p-n junctions. Conclusion: The correct answer is Option D: Channel length modulation because it is not associated with a p-n junction.

Explanation of each option: Option A: CC-CB is incorrect because this configuration consists of a common collector (CC) stage followed by a common base (CB) stage. While this can be used in some amplifiers, it is not a typical configuration for a cascade amplifier. Option B: CE-CB is incorrect because this configuration consists of a common emitter (CE) stage followed by a common base (CB) stage. While this may be used in certain applications, it is not the standard configuration for a cascade amplifier. Option C: CB-CC is incorrect because this configuration starts with a common base (CB) stage followed by a common collector (CC) stage. This is also not a common arrangement for cascade amplifiers. Option D: CE-CC is correct because the typical cascade amplifier configuration uses a common emitter (CE) stage followed by a common collector (CC) stage. The CE stage provides high voltage gain, while the CC stage provides current gain and impedance matching, making this a common choice for multistage amplifiers. Conclusion: The correct answer is Option D: CE-CC because this is the standard multistage configuration for cascade amplifiers.

Option A: Reduces gain is correct because negative feedback in an amplifier reduces the overall gain of the system. This reduction in gain, however, leads to improved linearity, reduced distortion, and better stability in the amplifier's operation. Option B: Increase frequency & phase distortion is incorrect because negative feedback actually helps reduce both frequency and phase distortion. It enhances the fidelity of the amplifier by minimizing these distortions. Option C: Reduces bandwidth is incorrect because negative feedback typically increases the bandwidth of an amplifier. It makes the amplifier more stable across a wider range of frequencies. Option D: Increases noise is incorrect because negative feedback reduces the overall noise in an amplifier. By stabilizing the gain and reducing distortion, it typically lowers the noise figure of the amplifier. Conclusion: The correct answer is Option A: Reduces gain because negative feedback decreases the amplifier's gain but improves its overall performance in terms of linearity, stability, and distortion reduction.

The correct answer is Option A: \( \frac{1}{11} \) kΩ Explanation: The input resistance of a current shunt negative feedback amplifier is given by the formula: \[ R_{in(f)} = \frac{R_{in}}{1 + A\beta} \] where, \( R_{in} \) = Input resistance without feedback = 1 kΩ \( A \) = Open-loop voltage gain = 50 \( \beta \) = Feedback factor = 0.2 Substituting the values: \[ R_{in(f)} = \frac{1}{1 + (50 \times 0.2)} \] \[ R_{in(f)} = \frac{1}{1 + 10} = \frac{1}{11} \text{ kΩ} \] Thus, the input resistance of the current shunt negative feedback amplifier is \( \frac{1}{11} \) kΩ , which corresponds to Option A .

Option A: A unity gain non-inverting amplifier is correct because the most commonly used amplifier in sample & hold circuits is a unity gain non-inverting amplifier. This configuration provides a stable output with no amplification (gain = 1), which is ideal for sample & hold applications where the input signal is maintained without distortion. Option B: A unity gain inverting amplifier is incorrect because inverting amplifiers flip the signal, meaning the output is inverted with respect to the input. This is not typically desired in sample & hold circuits, where the signal should remain in phase with the input. Option C: An inverting amplifier with a gain of 10 is incorrect because amplifying the signal in this way would distort the original input, which is undesirable in sample & hold circuits where the purpose is to maintain the exact value of the sampled signal. Option D: An inverting amplifier with a gain of 100 is incorrect because, like Option C, amplifying the signal with a high gain (such as 100) would lead to a distorted output, which is not suitable for sample & hold circuits. Conclusion: The correct answer is Option A: A unity gain non-inverting amplifier because this configuration is commonly used in sample & hold circuits due to its ability to maintain the input signal without inversion or amplification.

Option A: Voltage shunt feedback is incorrect because voltage shunt feedback involves taking a portion of the output voltage and feeding it back in parallel with the input. This is not what an unbypassed resistor in a common emitter configuration does. Option B: Current series feedback is incorrect because current series feedback involves taking a portion of the output current and feeding it back in series with the input. This is not characteristic of an unbypassed resistor in a common emitter amplifier. Option C: Negative voltage feedback is correct because the unbypassed resistor in a common emitter amplifier provides negative voltage feedback. This resistor, typically placed in the emitter leg of the transistor, causes a decrease in the gain by reducing the overall amplification and stabilizing the output, leading to better linearity and reduced distortion. Option D: Positive current feedback is incorrect because positive current feedback would increase the overall gain and instability, which is not the role of an unbypassed resistor in a common emitter configuration. Conclusion: The correct answer is Option C: Negative voltage feedback because the unbypassed resistor in a common emitter amplifier provides negative voltage feedback, which stabilizes the gain and reduces distortion in the amplifier.

Option A: $$g_m r_o$$ is incorrect because this expression represents the voltage gain in a transistor, not the current gain. The term $$r_o$$ is the output resistance, and multiplying $$g_m$$ (transconductance) with $$r_o$$ gives the voltage gain, not the current gain. Option B: $$\frac{g_m}{r_o}$$ is incorrect because this expression represents the inverse relationship between transconductance and output resistance, which is related to the voltage gain of the amplifier, not the current gain. Option C: $$g_m r_\pi$$ is correct because this is the standard expression for the current gain of a BJT in terms of the transconductance ($$g_m$$) and the base-emitter resistance ($$r_\pi$$). The current gain is the product of these two parameters. Option D: $$\frac{g_m}{r_\pi}$$ is incorrect because this expression does not correctly represent the current gain of a BJT. Instead, it gives a ratio that relates to the voltage gain and other characteristics of the transistor, but not the current gain. Conclusion: The correct answer is Option C: $$g_m r_\pi$$ because the current gain of a BJT is given by the product of transconductance ($$g_m$$) and base-emitter resistance ($$r_\pi$$).

Option A: Transistor capacitances is correct because at high frequencies, the capacitances present in a bipolar transistor (such as base-emitter and base-collector junction capacitances) start to have a significant effect. These capacitances limit the current gain, causing a drop in the transistor's performance at high frequencies. The capacitive effects become more dominant as frequency increases, leading to a reduction in the current gain. Option B: High current effects in the base is incorrect because high current effects in the base generally affect the transistor's thermal stability and the possibility of transistor saturation. While they can affect the performance, they are not the primary reason for the drop in current gain at high frequencies. Option C: Parasitic inductive elements is incorrect because while parasitic inductive elements may have some impact on the high-frequency behavior of a transistor, they do not directly cause the current gain to drop. The primary limiting factor at high frequencies is the capacitance, not inductance. Option D: The early effect is incorrect because the Early effect refers to the variation of the transistor's collector current due to changes in the collector-base voltage. While the Early effect can influence the transistor's operation, it is not the main cause for the reduction in current gain at high frequencies. Conclusion: The correct answer is Option A: Transistor capacitances because at high frequencies, the transistor's junction capacitances play a significant role in limiting the current gain, leading to a reduction in performance.

Option A: Increase in both input & output impedances is incorrect because voltage series (series-shunt) feedback does not increase both input and output impedances. While it does increase the input impedance, it decreases the output impedance. Therefore, this option does not correctly describe the effect of series-shunt feedback. Option B: Decreases in both input & output impedances is incorrect because voltage series feedback actually increases the input impedance and decreases the output impedance. Hence, this option is not applicable. Option C: Increase in input impedance & decreases in output impedance is correct because when voltage series feedback (also known as series-shunt feedback) is applied, it increases the input impedance and decreases the output impedance. This makes the amplifier more stable and improves its performance in certain applications. Option D: Decrease in input impedance & increase in output impedance is incorrect because voltage series feedback increases the input impedance, not decreases it, and it also decreases the output impedance, not increases it. Therefore, this option does not describe the behavior of voltage series feedback. Conclusion: The correct answer is Option C: Increase in input impedance & decreases in output impedance because voltage series (series-shunt) feedback increases the input impedance while decreasing the output impedance, improving amplifier performance.

Explanation: The MOSFET switch in its on state behaves like a resistor . When the MOSFET is on, the conducting channel is established between the drain and source terminals, and it allows current to flow through it with minimal resistance, making it act as a low-value resistor. Therefore, the correct answer is Option A: Resistor. Why the other options are incorrect: Option B: Inductor – An inductor stores energy in a magnetic field when current passes through it. A MOSFET in the on state does not function in this manner. Option C: Capacitor – A capacitor stores energy in an electric field and releases it when the circuit conditions change. This is not the behavior of a MOSFET in the on state. Option D: Battery – A battery provides a constant voltage or current in a circuit, whereas a MOSFET switch in its on state allows current to flow with minimal resistance, similar to a resistor, not a voltage source.

The direction of induced electromotive force (e.m.f.) can be found by Lenz's law . Lenz's law states that the direction of the induced electromotive force (e.m.f.) in a circuit is such that it opposes the change in magnetic flux that causes it. In other words, when there is a change in magnetic flux passing through a coil or a conductor, an e.m.f. is induced in the conductor in a direction that produces a magnetic field opposing the change in the original magnetic field. This law helps to understand the direction of the induced current and the polarity of the induced voltage when a magnetic field is changing in a conductor, coil, or any closed-loop circuit. It is a fundamental principle in electromagnetic induction and is used to analyze and predict the behavior of inductors and transformers in electrical and electronic circuits.

In an iron cored coil , the presence of the iron core significantly increases the inductance due to the higher magnetic permeability of iron compared to air. When the iron core is removed , the coil becomes an air cored coil , which has much lower magnetic permeability. As a result, the inductance of the coil will decrease .

A crack in the magnetic path of an inductor creates a gap in the core, disrupting the continuous magnetic flux. This disruption increases the reluctance of the magnetic circuit, which reduces the overall inductance of the inductor. As a result, the inductance will be reduced rather than remaining unchanged or increasing.

The correct answer is Option A: gate-source voltage . Let's break down why: gm (transconductance) of a MOSFET tells us how much the drain current (Id) changes for a small change in the gate-source voltage (Vgs). Think of it like this: If you slightly adjust the voltage on the gate (Vgs), how much does the current flowing through the MOSFET from drain to source (Id) change? Option A: gate-source voltage (Vgs) - This is the primary way to control gm. A higher Vgs generally leads to a higher gm. Option B: drain-source voltage (Vds) - While Vds *does* influence the drain current, and therefore indirectly *can* affect gm, it's not the primary control. gm is *defined* by the relationship between gate voltage and drain current. Vds mainly affects the *saturation* region of operation, but the *control* is still largely through the gate. Option C: drain current (Id) - gm and Id are related, but Id doesn't directly *control* gm. Instead, gm is a *property* that describes how Id responds to changes in Vgs. Option D: gate current (Ig) - In a MOSFET, the gate current is ideally very, very small (close to zero). Therefore, it does not control gm. The gate is insulated. In simpler terms: The gate-source voltage (Vgs) is like the accelerator pedal in a car . It directly controls how much current (Id) flows through the MOSFET. gm is a measure of how sensitive the current flow is to changes in the accelerator pedal.

The correct answer is Option A: When it is given as input to a monostable multivibrator . Let's break down why: Monostable Multivibrator: A monostable multivibrator has one stable state . When triggered by an input pulse (even an irregular one) , it goes into a temporary unstable state for a fixed period of time . This fixed period is determined by the circuit's components (resistors and capacitors) , not the shape of the input pulse. So, regardless of how the input pulse looks, the output will be a pulse with a definite, predictable width . Why other options are incorrect: Bistable Multivibrator: Requires a trigger to flip between two stable states, it doesn't create a pulse of definite width from an irregular input. Schmitt-trigger: Converts an analog input signal to a digital output signal with hysteresis, but it doesn't necessarily create a pulse of definite width. Pulse Transformer: Primarily used for voltage or current transformation of pulses, not for shaping them into definite widths. It basically transfers the input pulse, with some modifications to its amplitude and polarity. In simple terms: Imagine the monostable multivibrator as a timer. You can start the timer with any kind of signal(irregular), but the timer will always run for the same amount of time (definite width).

fT = 1.47 x 1010 Hz & fβ = 1.64 x 108 Hz is the correct answer to “An npn BJT has g m = 38 mA/v, C µ = 10 -14 F, C π = 4 × 10 -13 F and DC current gain β 0 = 90. For this transistor f T & f β are”. This question is standard in competitive examination syllabi for Analog Electronics.

0.5 is the correct answer to “An n-channel JFET has $${I_{{\text{DSS}}}} = 2\,{\text{mA}}$$ and V p = -4 v. Its transconductance g m = in (mA/V) for an applied gate to source voltage V GS = -2 v is”. This question is standard in competitive examination syllabi for Analog Electronics.

Internal Capacitance of the device is the correct answer to “Generally, the gain of a transistor amplifier falls at high frequencies due to the”. This question is standard in competitive examination syllabi for Analog Electronics.

Voltage controlled capacitor is the correct answer to “MOSFET can be used as a”. This question is standard in competitive examination syllabi for Analog Electronics.

Affects the common mode gain Ac is the correct answer to “A change in the value of the emitter resistance Re in a differential amplifier”. This question is standard in competitive examination syllabi for Analog Electronics.

Q-factor of the tuned output circuit is the correct answer to “The bandwidth of an RF tuned amplifier is dependent on”. This question is standard in competitive examination syllabi for Analog Electronics.

Both Temperature & β is the correct answer to “Introducing a resistor in the emitter of a common amplifier stabilizes the dc operating point against variations in”. This question is standard in competitive examination syllabi for Analog Electronics.

High CMRR is the correct answer to “Most of the linear ICs are based on the two-transistor differential amplifier because of its”. This question is standard in competitive examination syllabi for Analog Electronics.

Cut off point is the correct answer to “The horizontal intercept of dc load line is the same as ideal”. This question is standard in competitive examination syllabi for Analog Electronics.

100 KHz is the correct answer to “A 741-type OP-amp has a gain-bandwidth product of 1MHz. A non-inverting amplifier using this OP-amp & having a voltage gain of 20 db will exhibit -3 db bandwidth of”. This question is standard in competitive examination syllabi for Analog Electronics.

CMRR is the correct answer to “A differential amplifier is invariably used in the input stage of all OP-amps. This is dome basically to provide the OP-amps with a very high”. This question is standard in competitive examination syllabi for Analog Electronics.

Large collector-base reverse bias is the correct answer to “The early effect in a bipolar junction transistor is caused by”. This question is standard in competitive examination syllabi for Analog Electronics.

Temperature is the correct answer to “In a transistor leakage current mainly depends on”. This question is standard in competitive examination syllabi for Analog Electronics.

Greater than +100 is the correct answer to “Asource follower using an FET usually has a voltage gain which is”. This question is standard in competitive examination syllabi for Analog Electronics.

98dB is the correct answer to “Three identical amplifiers with each one having a voltage gain of 50,input resistance of 1KΩ & output resistance of 250, are cascaded. The open circuit voltage gain of combined amplifier is”. This question is standard in competitive examination syllabi for Analog Electronics.

25 Ω & 20 % is the correct answer to “A dc power supply has no-load voltage of 30v,and a full-load voltage of 25v at full-load current of 1A.Its output resistance & load regulation ,respectively are”. This question is standard in competitive examination syllabi for Analog Electronics.

The amplitude of output is proportional to rate of change of input is the correct answer to “In an OP-amp differentiator”. This question is standard in competitive examination syllabi for Analog Electronics.

Overcome a crossover distortion is the correct answer to “Class AB operation is often used in power large signal) amplifiers in order to”. This question is standard in competitive examination syllabi for Analog Electronics.

Common emitter is the correct answer to “Which of the following amplifier circuit using junction transistor has the best gain”. This question is standard in competitive examination syllabi for Analog Electronics.