Analog Electronics MCQ Quiz - Objective Question with Answer for Analog Electronics - Download Free PDF

Concept:

The given circuit is an op-amp differentiator. For an ideal differentiator with input \( v_i(t) \), the output is given by:

\( v_o(t) = -RC \cdot \frac{d}{dt}v_i(t) \)

Given that \( RC = 1 \), the equation simplifies to:

\( v_o(t) = -\frac{d}{dt}v_i(t) \)

Given:

The input signal \( v_i(t) \) is a triangular waveform.

• During the rising edge (slope = +1): \( \frac{d}{dt}v_i(t) = 1 \Rightarrow v_o(t) = -1 \)
• During the falling edge (slope = –1): \( \frac{d}{dt}v_i(t) = -1 \Rightarrow v_o(t) = +1 \)

Thus, the output waveform will be a square wave alternating between +1 and –1, switching at every point where the slope of \( v_i(t) \) changes.

Answer:

Option 2: Square wave output alternating between +1 and –1 (in response to triangular wave input)

Three-Stage Cascade Amplifier and Overall Noise Figure Calculation

Problem Statement: In a three-stage cascade amplifier, each stage has a gain of 10 dB and a noise figure of 10 dB. We are tasked with calculating the overall noise figure of this cascade system.

Solution:

To calculate the overall noise figure (F_overall) of a cascade amplifier, we use Friis's formula for noise figure in cascaded systems:

Friis's Formula:

F_overall = F₁ + ^{(F₂ - 1)}/_G1 + ^{(F₃ - 1)}/_{G1 × G2} + ...

Where:

• F_n = Noise figure of the nth stage (in linear scale).
• G_n = Gain of the nth stage (in linear scale).

Step 1: Convert Gains and Noise Figures from dB to Linear Scale

The given gain (G) and noise figure (F) for each stage are 10 dB:

• G = 10 dB → G (linear) = 10^G(dB)/10 = 10^10/10 = 10.
• F = 10 dB → F (linear) = 10^F(dB)/10 = 10^10/10 = 10.

Step 2: Apply Friis's Formula

Since there are three stages, Friis's formula for the overall noise figure becomes:

F_overall = F₁ + ^{(F₂ - 1)}/_G1 + ^{(F₃ - 1)}/_{G1 × G2}

Substitute the values for F₁, F₂, F₃, G₁, and G₂:

• F₁ = 10 (linear).
• F₂ = 10 (linear).
• F₃ = 10 (linear).
• G₁ = G₂ = 10 (linear).

Thus:

F_overall = 10 + ^{(10 - 1)}/₁₀ + ^{(10 - 1)}/_{10 × 10}

F_overall = 10 + 0.9 + 0.09

F_overall = 10.99

Final Answer: The overall noise figure of the three-stage cascade amplifier is 10.99 (linear scale).

Explanation:

• An amplifier can be converted into an oscillator by doing some change in the amplifier circuit as:
• Connect the output of the amplifier to the input by a positive feedback circuit.
• The phase-shifted the output by 180° and feed this phase shift output to the input via a feedback circuit.
• An arrangement of the RC tuned circuit is connected as a load to the amplifier.
• Oscillator circuit block diagram.

Explanation:

Step-by-Step Solution:

Step 1: Analyze the Sine Wave Input

The sine wave is given by the equation:

V(t) = A × sin(ωt)

Where:

• A: Amplitude of the sine wave = 2V
• ω: Angular frequency of the sine wave

The sine wave oscillates between -2V and +2V, crossing the reference voltage (1V) twice per cycle.

Step 2: Determine When the Sine Wave Exceeds the Reference Voltage

To find the time intervals during which the sine wave is greater than the reference voltage (1V), we solve the equation:

A × sin(ωt) > V_ref

Substituting the values:

2 × sin(ωt) > 1

Dividing through by 2:

sin(ωt) > 0.5

From trigonometric principles, sin(ωt) > 0.5 occurs between two angles:

ωt = π/6 and ωt = 5π/6

Therefore, the sine wave exceeds the reference voltage for the duration:

Δt = (5π/6 - π/6)/ω = (4π/6)/ω = (2π/3)/ω

Step 3: Calculate the Duty Cycle

The total time period of the sine wave is:

T = 2π/ω

The duty cycle is the ratio of the time the output is high (Δt) to the total time period (T):

Duty Cycle = Δt / T

Substituting the values:

Duty Cycle = [(2π/3)/ω] / [2π/ω]

Simplifying:

Duty Cycle = (2π/3) / (2π) = 1/3

Expressing this as a decimal:

Duty Cycle = 0.3333

Final Answer: The duty cycle of the output signal is 0.3333.

Hence the correct answer is 33.33 %

Explanation:

Schmitt trigger circuit:

V0 = + Vsat (or) -Vsat

\({V_{ref}} = \frac{{{R_2}}}{{{R_1} + {R_2}}}{V_0}\)

\({R_{01}} = {R_1}||{R_2} = \frac{{{R_1}{R_2}}}{{{R_1} + {R_2}}}\)

VUT is the upper threshold point if the output is +ve reference

VLT = Lower threshold point if the output voltage is –ve reference

\({V_{UT}} = \frac{{{R_2}}}{{{R_1} + {R_2}}}\left( { + {V_{sat}}} \right)\)

\({V_{LT}} = \frac{{{R_2}}}{{{R_1} + {R_2}}}\left( { - {V_{sat}}} \right)\)

Calculation:

On comparing this circuit with the question provided, we get:

In the Schmitt trigger

Vi = Vp-p, R1 = 10 kΩ, R2 = 5 kΩ , Vref = 0 V

Sinusoidal saturation voltage = ± 5 V (Vsat)

\({V_{LT}} = \frac{{{R_2}}}{{{R_1} + {R_2}}}\left( { - {V_{sat}}} \right)\) = 5 x (-5)/15  = -1.667 V

Concept:

The efficiency of a rectifier is defined as the ratio of dc output power to input power.

The efficiency of a half-wave rectifier will be:

\(\eta = \frac{{{P_{dc}}}}{{{P_{ac}}}}\)

\(\eta= \frac{{\frac{{V_{dc}^2}}{{{R_L}}}}}{{\frac{{V_{rms}^2}}{{{R_L}}}}} \)

V_DC = DC or average output voltage

R_L = Load Resistance

For a half-wave rectifier, the output DC voltage or the average voltage is given by:

\(V_{DC}=\frac{V_m}{\pi}\)

Also, the RMS voltage for a half-wave rectifier is given by:

\(V_{rms}=\frac{V_m}{2}\)

Calculation:

The efficiency for a half-wave rectifier will be:

\(\eta= \frac{{{{\left( {\frac{{{V_m}}}{\pi }} \right)}^2}}}{{{{\left( {\frac{{{V_m}}}{{2 }}} \right)}^2}}} = 40.6\;\% \)

For Half wave rectifier maximum efficiency = 40.6%

Note: For Full wave rectifier maximum efficiency = 81.2%

Concept:

For a transistor, the base current, the emitter current, and the collector current are related as:

IE = IB + IC

where IC = β IB

β = Current gain of the transistor

Typical base-emitter voltages, VBE for both NPN and PNP transistors are as follows:

• If the transistor is made up of a silicon material, the base-emitter voltage VBE will be 0.7 V.
• If the transistor is made up of a germanium material, the base-emitter voltage VBE will be 0.3 V.
   

Application:

From the given figure, Apply KVL

10 - I_B × R_B - V_BE = 0

Let us assume V_BE = 0.7 V

10 - I_B (1 × 10⁶) - 0.7 = 0

I_B = 9.3 μA

We know that,

I_C = β I_B

Where,

I_C  & I_B = collector current and base current

Therefore,

I_C = 100 × 9.3 μA

= 930 μA

= 0.93 mA

≈ 1 mA

• A diode is an electronic device allowing current to move through it only in one direction.
• Current flow is permitted when the diode is forwaforward-biased
• Current flow is prohibited when the diode is reversed-biased.
• The direction of the arrow represents the direction of conventional current flow when the diode is forward biased

• In the figure given above, the symbol represents the circuit symbol of a semiconductor junction diode.
• The ‘P’ side of the diode is always positive terminal and is designated as anode for forward bias.
• Another side that is negative is designated as cathode and is the ‘N’ side of diode.

Concept: 

The working of the Zener diode is explained in the below figures.

Calculation:

Given,

Zener voltage Vz = 10 V

E_in = 6 V ⇒ E_in < V_z

Hence zener will be reverse biased and get open-circuited.

Output voltage E₀ = 0 V

Varactor diode:

• It is represented by a symbol of diode terminated in the variable capacitor as shown below:

• Varactor diode refers to the variable Capacitor diode, which means the capacitance of the diode varies linearly with the applied voltage when it is reversed biased.
• The junction capacitance across a reverse bias pn junction is given by

​           \(C=\frac{A\epsilon}{W}\)

• As the reverse bias voltage increases, the depletion region width increases resulting in a decrease in the junction capacitance.
•  Varactor diodes are used in electronic tuning systems to eliminate the need for moving parts
• Varactor [also called voltacap, varicap, voltage-variable capacitor diode, variable reactance diode, or tuning diode] diodes are the semiconductor, voltage-dependent, variable capacitors
• Varactors are used as voltage-controlled capacitors and it operated in a reverse-biased state

BJT Amplifier:

• Transistors biasing is done to keep stable DC operating conditions needed for its functioning as an amplifier.
• A properly biased transistor must have it's Q-point (DC operating parameters like IC and VCE) at the center of saturation mode and cut-off mode i.e. active mode.
• In the active mode of transistor operation, the emitter-base junction is forward biased and the collector-base junction is reverse biased.​
• In the cut-off mode of transistor operation, the emitter-base junction is reverse biased and the collector-base junction is reverse biased.​

​

Different modes of BJT operations are:

Early Effect:

• A large collector base reverse bias is the reason behind the early effect manifested by BJTs.
• As reverse biasing of the collector to base junction increases, the depletion region penetrates more into the base, as the base is lightly doped.
• This reduces the effective base width and hence the concentration gradient in the base increases.
• This reduction in the effective base width causes less recombination of carriers in the base region which results in an increase in collector current. This is known as the Early effect.
• The decrease in base width causes ß to increase and hence collector current increases with collector voltage rather than staying constant.
• The slope introduced by the Early effect is almost linear with IC and the common-emitter characteristics extrapolate to an intersection with the voltage axis VA, called the Early voltage.

This is explained with the help of the following VCE (Reverse voltage) vs IC (Collector current) curve:

• A limiter circuit is also known as a clipper circuit.
• A clipper is a device that removes either the positive half (top half) or negative half (bottom half), or both positive and negative halves of the input AC signal.
• The clipping (removal) of the input AC signal is done in such a way that the remaining part of the input AC signal will not be distorted
• In the below circuit diagram, the positive half cycles are removed by using the series positive clipper.

​Note: A Clamper circuit can be defined as the circuit that consists of a diode, a resistor, and a capacitor that shifts the waveform to the desired DC level without changing the actual appearance of the applied signal.

Concept:

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

Where β = common-emitter current gain

α = Common base current gain

Calculation:

Common base current gain = α = 0.98

\(\beta = \frac{{0.98}}{{1 - 0.98}} = 49\)

Note: \(\alpha = \frac{{{I_C}}}{{{I_E}}}\) & \(\beta = \frac{{{I_C}}}{{{I_B}}}\)

Where I_C = Collector current

I_E = Emitter current