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Past Questions
The efficiency of a class-C power amplifier is typically:
A)
25–50%
B)
50–70%
C)
70–85%
D)
>85%
Power amplifiers are classified based on their conduction angle (the portion of the input cycle for which the transistor conducts).
1. **Class A:** Conducts for the full 360° of the cycle. It has the highest linearity but the lowest efficiency (maximum theoretical efficiency is 25% for series-fed and 50% for transformer-coupled).
2. **Class B:** Conducts for 180° of the cycle. It is more efficient than Class A, with a maximum theoretical efficiency of 78.5%.
3. **Class AB:** Conducts for slightly more than 180° to reduce crossover distortion. Its efficiency falls between Class A and Class B.
4. **Class C:** Conducts for less than 180° (typically much less). Because the transistor is 'off' for a large portion of the cycle, it draws very little average current from the power supply compared to the output power delivered to the load (which is usually a tuned LC circuit). This results in very high efficiency, typically ranging from 80% to 90% or even higher, often exceeding 85% in practical high-frequency applications.
Therefore, among the given options, '>85%' is the most accurate description of Class C amplifier efficiency.
Draw the circuit diagram of a Dual Slope A/D converter and explain the operation of the conversion process. Prove that in Dual Slope A/D converter the output of the counter is proportional to the analog voltage Va.
In a differential amplifier, the common-mode gain should be:
A)
High
B)
Low
C)
Equal to differential gain
D)
Zero
A differential amplifier is designed to amplify the difference between two input signals (differential mode) while rejecting signals that are common to both inputs (common-mode), such as noise or interference.
1. **Differential Gain (Ad):** This is the gain applied to the difference between the two inputs. For a useful amplifier, this should be high.
2. **Common-Mode Gain (Ac):** This is the gain applied to signals that appear identically on both inputs. In an ideal scenario, the amplifier should completely ignore these signals.
3. **Common-Mode Rejection Ratio (CMRR):** This is defined as the ratio of differential gain to common-mode gain (CMRR = Ad / Ac). To maximize the CMRR (which indicates a better quality amplifier), the common-mode gain (Ac) must be as small as possible.
Therefore, for optimal performance and noise rejection, the common-mode gain should be low. While 'Zero' (option d) is the ideal theoretical value, in practical engineering and circuit design, the goal is to keep it 'Low'. In the context of standard multiple-choice questions for electronics, 'Low' is the correct descriptive requirement for a functional differential amplifier.
The radiation resistance of an antenna is 80Ω and loss-resistance is 10Ω. What is its directivity if the power gain is 20?
How is a J-K flip-flop made to toggle?
A)
J = 0, K = 0
B)
J = 1, K = 0
C)
J = 0, K = 1
D)
J = 1, K = 1
A J-K flip-flop is a universal flip-flop that overcomes the 'invalid state' problem of the S-R flip-flop. Its behavior is determined by the inputs J and K as follows:
1. **Hold State (J=0, K=0):** The output remains unchanged (Q_next = Q_present). This corresponds to option (a).
2. **Reset State (J=0, K=1):** The output is forced to 0 (Q_next = 0). This corresponds to option (c).
3. **Set State (J=1, K=0):** The output is forced to 1 (Q_next = 1). This corresponds to option (b).
4. **Toggle State (J=1, K=1):** The output switches to the complement of its current state (Q_next = NOT Q_present). If the output was 0, it becomes 1; if it was 1, it becomes 0. This is known as the 'toggle' condition.
Therefore, to make a J-K flip-flop toggle, both inputs J and K must be set to high (1).
The unit of S-parameters is:
A)
Ohms
B)
Siemens
C)
Dimensionless
D)
Decibels
S-parameters, or Scattering parameters, are used in microwave engineering to describe the electrical behavior of linear electrical networks. They are defined as the ratio of the reflected or transmitted voltage wave amplitude to the incident voltage wave amplitude. Mathematically, an S-parameter is expressed as: S = (b / a), where 'a' is the amplitude of the incident wave and 'b' is the amplitude of the reflected or transmitted wave. Since it is a ratio of two similar quantities (voltage waves), the units cancel out, making S-parameters dimensionless.
Analysis of other options:
- a. Ohms: This is the unit of impedance (Z-parameters), not S-parameters.
- b. Siemens: This is the unit of admittance (Y-parameters), not S-parameters.
- d. Decibels: While S-parameters are often converted to decibels (dB) for easier representation of gain or loss (e.g., Return Loss), the S-parameter itself is a complex coefficient and is fundamentally dimensionless.
Which of the following amplifier class have the highest linearity and lowest distortion?
A)
Class A
B)
Class B
C)
Class C
D)
Class B push-pull
To determine which amplifier class has the highest linearity and lowest distortion, we analyze the conduction angle and biasing of the active device (transistor) for each class:
1. **Class A**: In a Class A amplifier, the transistor is biased such that it conducts for the full cycle (360°) of the input signal. Because the operating point (Q-point) is placed in the center of the load line, the signal remains within the linear region of the transistor's characteristic curve. This results in the highest linearity and the lowest harmonic distortion among all classes.
2. **Class B**: The transistor conducts for only half of the input cycle (180°). This leads to 'crossover distortion' at the point where the signal switches from the positive half-cycle to the negative half-cycle, significantly increasing distortion compared to Class A.
3. **Class C**: The transistor conducts for less than half of the cycle (less than 180°). This class is highly non-linear and produces massive distortion, making it unsuitable for audio amplification; it is primarily used in tuned RF amplifiers where a LC tank circuit filters out the harmonics.
4. **Class B push-pull**: While this configuration improves efficiency and reduces some distortion compared to a single Class B stage, it still suffers from crossover distortion unless modified into Class AB. It cannot match the inherent linearity of a pure Class A amplifier.
Conclusion: Class A provides the most faithful reproduction of the input waveform, ensuring the highest linearity and lowest distortion.
Draw and explain the working of a square wave generator.
The main advantage of negative feedback in amplifiers is:
A)
Increased gain
B)
Increased bandwidth
C)
Increased distortion
D)
Increased input impedance only
Negative feedback occurs when a portion of the output signal is fed back to the input with a 180-degree phase shift. This process involves a trade-off: while it reduces the overall gain of the amplifier, it provides several critical advantages.
1. **Bandwidth**: The gain-bandwidth product of an amplifier remains constant. Since negative feedback reduces the gain, the bandwidth (the range of frequencies over which the amplifier operates effectively) increases proportionally. This is the primary advantage listed in the options.
2. **Gain Stability**: It makes the gain less dependent on temperature variations or component tolerances.
3. **Distortion**: It reduces non-linear distortion and noise, making option (c) incorrect.
4. **Impedance**: While negative feedback can modify input and output impedances, it does not 'only' increase input impedance; the effect depends on the topology (series or shunt) used for feedback.
5. **Gain Reduction**: Negative feedback explicitly decreases the gain, making option (a) incorrect.
The dominant mode in a circular waveguide is:
A)
TE 11
B)
TM 01
C)
TE 01
D)
TM 11
In waveguide theory, the 'dominant mode' is defined as the mode with the lowest cutoff frequency. For a circular waveguide of radius 'a', the cutoff frequency for any mode is given by the formula: fc = (chi_mn / 2πa) * c, where chi_mn is the root of the Bessel function (for TM modes) or the root of the derivative of the Bessel function (for TE modes).
1. For TE modes, the smallest root is chi'_11 ≈ 1.841 (for TE11).
2. For TM modes, the smallest root is chi_01 ≈ 2.405 (for TM01).
3. The TE01 mode has a root chi'_01 ≈ 3.832.
Comparing these values, the TE11 mode has the smallest root (1.841), which results in the lowest cutoff frequency. Therefore, TE11 is the dominant mode in a circular waveguide.
- Option b (TM 01) is the dominant TM mode, but its cutoff frequency is higher than TE 11.
- Option c (TE 01) has a significantly higher cutoff frequency than TE 11.
- Option d (TM 11) has a higher cutoff frequency than both TE 11 and TM 01.
Define gain margin and phase margin. What is their significance in stability analysis?
What should be the nature of bandwidth for a good control system?
A)
Small
B)
Medium
C)
Large
D)
All of the mentioned
In control systems, bandwidth is defined as the range of frequencies over which the system tracks the input signal with acceptable accuracy (typically where the magnitude is within -3dB of the DC gain).
1. **Speed of Response**: There is a direct relationship between bandwidth and the rise time (speed) of a system. A larger bandwidth implies a faster response and a shorter rise time, allowing the system to respond quickly to changes in the input.
2. **Tracking Ability**: A system with a larger bandwidth can track higher-frequency input signals without significant attenuation or phase lag.
3. **Trade-off (Noise)**: While a very large bandwidth can make a system more susceptible to high-frequency noise, in the context of general 'performance' and 'speed' for a good control system, a larger bandwidth is desired to ensure agility and fast transient response.
Therefore, for a system to be considered 'good' in terms of responsiveness and tracking capability, a large bandwidth is preferred.
The excess-3 code for 597 is given by:
A)
100011001010
B)
100010100111
C)
010110010111
D)
010110101101
To convert a decimal number to Excess-3 code, follow these steps:
1. Separate the decimal number into its individual digits: 5, 9, and 7.
2. Add 3 to each individual digit:
- For 5: 5 + 3 = 8
- For 9: 9 + 3 = 12
- For 7: 7 + 3 = 10
3. Convert each of these resulting sums into their 4-bit binary equivalents:
- 8 in binary is 1000
- 12 in binary is 1100
- 10 in binary is 1010
4. Concatenate the binary groups together: 1000 1100 1010.
Comparing this result to the options:
- Option (a) 100011001010 matches our calculation.
- Option (b) 100010100111 is incorrect.
- Option (c) 010110010111 is incorrect.
- Option (d) 010110101101 is incorrect.
State and explain Gauss’s Law.
Which op-amp circuit produces an output proportional to the time integral of the input?
A)
Differentiator
B)
Integrator
C)
Comparator
D)
Summing amplifier
An Op-Amp Integrator is a circuit where the output voltage is proportional to the integral of the input voltage over time. In a typical inverting integrator, a resistor is placed at the input and a capacitor is placed in the feedback loop. The output voltage is given by the formula: Vout = -(1/RC) ∫ Vin dt.
Analysis of other options:
- a. Differentiator: This circuit produces an output proportional to the rate of change (derivative) of the input voltage (Vout ∝ dVin/dt).
- c. Comparator: This circuit compares two input voltages and outputs a high or low signal based on which input is larger; it does not perform integration.
- d. Summing amplifier: This circuit produces an output that is the weighted sum of multiple input voltages.
Sketch the root locus for a system with open-loop transfer function G(s)H(s)= K/s(s+4).
Explain the operation of a class-AB amplifier with the help of suitable diagram.
The number of select lines for a 16-to-1 multiplexer is:
A)
2
B)
4
C)
8
D)
16
A multiplexer (MUX) is a combinational circuit that selects one of many input signals and forwards the selected input into a single output line. The relationship between the number of input lines (n) and the number of select lines (m) is given by the formula: 2^m = n. In this question, the multiplexer is a 16-to-1 MUX, meaning there are n = 16 input lines. Substituting this into the formula: 2^m = 16. Since 16 = 2^4, it follows that m = 4. Therefore, 4 select lines are required to uniquely address each of the 16 input lines.
Analysis of options:
- Option (a) 2: 2^2 = 4 inputs (Incorrect).
- Option (b) 4: 2^4 = 16 inputs (Correct).
- Option (c) 8: 2^8 = 256 inputs (Incorrect).
- Option (d) 16: 2^16 = 65,536 inputs (Incorrect).
The Poynting vector represents:
A)
Electric field strength
B)
Magnetic flux density
C)
Power flow per unit area
D)
Energy density
The Poynting vector, denoted by $\mathbf{S}$, is defined as the cross product of the electric field intensity $\mathbf{E}$ and the magnetic field intensity $\mathbf{H}$: $\mathbf{S} = \mathbf{E} \times \mathbf{H}$.
1. **Physical Significance**: The Poynting vector represents the directional energy flux (the rate of energy transfer per unit area) of an electromagnetic field. In simpler terms, it describes the power flow per unit area (measured in Watts per square meter, $\text{W/m}^2$).
2. **Analysis of Options**:
- **Option a (Electric field strength)**: This is represented by $\mathbf{E}$, not the Poynting vector.
- **Option b (Magnetic flux density)**: This is represented by $\mathbf{B}$, not the Poynting vector.
- **Option c (Power flow per unit area)**: This is the correct definition of the magnitude and direction of the Poynting vector.
- **Option d (Energy density)**: Energy density is the energy stored per unit volume, given by $w = \frac{1}{2}(\epsilon E^2 + \mu H^2)$, which is a scalar quantity, whereas the Poynting vector is a vector representing flow.
A magnetron is commonly used in:
A)
Radar transmitters
B)
Satellite receivers
C)
Mobile phones
D)
Optical fibers
A magnetron is a high-power vacuum tube that functions as a cross-field microwave generator. It uses the interaction between a magnetic field and an electric field to generate high-frequency electromagnetic waves (microwaves). Due to its ability to produce very high peak power outputs, it is the primary source of microwave energy used in radar transmitters and microwave ovens.
Analysis of other options:
- b. Satellite receivers: These typically use Low Noise Amplifiers (LNAs) and mixers, not high-power generators like magnetrons, as they are designed to receive weak signals, not transmit high-power pulses.
- c. Mobile phones: These use solid-state power amplifiers (SSPAs) and integrated circuits (RFICs) based on CMOS or GaAs technology for efficiency and miniaturization.
- d. Optical fibers: These transmit data using light (photons) via lasers or LEDs, not microwave radiation generated by vacuum tubes.
Calculate the emf when a coil of 100 turns is subjected to a flux rate of 0.3 tesla/sec.
A)
3
B)
30
C)
-30
D)
-300
To calculate the electromotive force (emf) induced in a coil, we use Faraday's Law of Electromagnetic Induction.
1. **Given Data:**
- Number of turns (N) = 100
- Rate of change of flux (dΦ/dt) = 0.3 Tesla/sec (Note: The question uses 'tesla/sec', which refers to the rate of change of magnetic flux density per unit area, effectively representing the rate of change of flux per turn, dΦ/dt).
2. **Formula:**
The magnitude of the induced emf (ε) is given by:
|ε| = N * (dΦ/dt)
3. **Calculation:**
|ε| = 100 turns * 0.3 Wb/s (or T·m²/s)
|ε| = 30 Volts
4. **Analysis of Options:**
- Option (a) 3: Incorrect calculation.
- Option (b) 30: Correct magnitude of the induced emf.
- Option (c) -30: While Lenz's Law states the emf opposes the change (introducing a negative sign), in standard multiple-choice technical exams, if the magnitude is provided as a positive value and the direction is not specified, the positive magnitude is typically selected. However, if the question specifically asks for the vector/directional value, -30 could be argued. In the context of this specific exam's marking scheme for this question, the magnitude 30 is the intended answer.
- Option (d) -300: Incorrect calculation.
Therefore, the correct magnitude is 30 V.
A lossless coaxial cable has L = 250 nH/m and C = 100 pF/m. Calculate its characteristic impedance and velocity of propagation.
Standing wave ratio is defined as the
A)
Ratio of voltage maxima to voltage minima
B)
Ratio of current maxima to current minima
C)
Product of voltage maxima and voltage minima
D)
Product of current maxima and current minima
In a transmission line, when a load is not perfectly matched to the characteristic impedance of the line, a portion of the incident wave is reflected back toward the source. The interference between the incident wave and the reflected wave creates a 'standing wave' pattern along the line. The Voltage Standing Wave Ratio (VSWR or SWR) is a measure of this mismatch. It is mathematically defined as the ratio of the maximum voltage amplitude (Vmax) to the minimum voltage amplitude (Vmin) found along the line: VSWR = Vmax / Vmin. Because the current standing wave ratio is identical to the voltage standing wave ratio (Imax / Imin = Vmax / Vmin), option 'a' is the standard definition. Options 'c' and 'd' are incorrect because SWR is a ratio, not a product. While option 'b' is numerically equivalent, the primary and standard definition of SWR in electrical engineering textbooks is based on voltage magnitudes.
Divergence theorem is based on:
A)
Gauss law
B)
Stoke’s law
C)
Ampere law
D)
Lenz law
The Divergence Theorem (also known as Gauss's Theorem) relates the flux of a vector field through a closed surface to the divergence of the field within the volume enclosed by that surface. Mathematically, it is expressed as: ∬S (V · n) dS = ∭V (∇ · V) dV. In electromagnetics, this theorem is the fundamental mathematical basis for Gauss's Law, which states that the total electric flux out of a closed surface is equal to the charge enclosed divided by the permittivity of free space.
Why other options are incorrect:
- b. Stoke's law: This is based on the Curl theorem, relating a surface integral of the curl of a vector field to a line integral around the boundary curve.
- c. Ampere law: While Ampere's law can be converted using the Divergence theorem in certain contexts, it is fundamentally a manifestation of the Curl theorem (relating current to the circulation of the magnetic field).
- d. Lenz law: This is a physical law regarding the direction of induced electromotive force (EMF) and is not a mathematical theorem of vector calculus.
What happens when a P-type semiconductor is joined with an N-type semiconductor?
A)
They repel each other
B)
They form an insulator
C)
They create a diode
D)
They have no interaction
When a P-type semiconductor (rich in holes) is joined with an N-type semiconductor (rich in electrons), a PN junction is formed. At the interface, electrons from the N-side diffuse into the P-side and holes from the P-side diffuse into the N-side, creating a 'depletion region'. This structure allows current to flow easily in one direction (forward bias) while blocking it in the opposite direction (reverse bias). This one-way conduction property is the fundamental characteristic of a diode.
Why other options are incorrect:
- (a) They do not repel each other; they are physically joined to form a crystal lattice.
- (b) While the depletion region has high resistance, the overall device is a semiconductor, not an insulator, as it can conduct under specific bias conditions.
- (d) There is a significant interaction involving charge carrier diffusion and the creation of a built-in potential barrier.