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Past Questions — 2025
In a Type-C chopper (two quadrant), the power flow can be:
A)
Only from source to load
B)
Only from load to source
C)
Both directions
D)
Zero
A Type-C chopper is a two-quadrant chopper. In power electronics, choppers are classified based on the quadrants of the V-I (Voltage-Current) plane they operate in.
1. **Type-A (First Quadrant):** Power flows only from source to load (V > 0, I > 0).
2. **Type-B (Second Quadrant):** Power flows only from load to source (V > 0, I < 0).
3. **Type-C (Two Quadrant):** This chopper can operate in both the first and second quadrants. It allows the current to flow in both directions while the voltage polarity remains constant. Because current can flow from source to load (Quadrant I) and from load to source (Quadrant II), the power flow is bidirectional.
4. **Type-D (Four Quadrant):** Power flow is bidirectional and voltage polarity can also be reversed.
Since a Type-C chopper operates in two quadrants (specifically where current can be positive or negative), it enables regenerative braking (load to source) and motoring (source to load), meaning power can flow in both directions.
What is CMRR of Op-amp? List the ideal characteristics of Op-amp and write the assumptions made from ideal op-amp characteristics.
The control strategy for varying duty cycle a of a Chopper is:
A)
Constant frequency control
B)
Frequency modulation
C)
Current Limit Control
D)
All of the above
A chopper (DC-DC converter) varies the average output voltage by adjusting the duty cycle 'a' (where a = Ton / (Ton + Toff)). There are several control strategies to achieve this:
1. **Constant Frequency Control (Pulse Width Modulation - PWM):** The switching frequency is kept constant, and the duty cycle is varied by changing the 'on-time' (Ton). This is the most common method as it simplifies the design of output filters.
2. **Frequency Modulation (Variable Frequency Control):** The 'on-time' (Ton) is kept constant, and the duty cycle is varied by changing the switching frequency (varying Toff). This is used in specific applications where a constant pulse width is required.
3. **Current Limit Control (Hysteresis Control):** The duty cycle is varied based on the load current. The switch is turned on when the current hits a lower limit and turned off when it hits an upper limit. This results in a variable duty cycle and variable frequency based on the load conditions.
Since all three methods are valid strategies for varying the duty cycle to control the output of a chopper, the correct option is 'All of the above'.
The turn on method of thyristor is:
A)
Forward Voltage triggering
B)
Light triggering
C)
Temperature triggering
D)
All of the above
A thyristor (specifically an SCR - Silicon Controlled Rectifier) can be turned on (triggered) by various methods that cause the depletion region at the junctions to break down or inject sufficient charge carriers into the device.
1. **Forward Voltage Triggering (Option a):** When the anode-to-cathode voltage is increased beyond the forward breakover voltage (VBO), the device switches from the forward blocking state to the forward conduction state.
2. **Light Triggering (Option b):** In a Light Activated SCR (LASCR), light photons of sufficient energy hitting the gate region create electron-hole pairs, which trigger the device into conduction.
3. **Temperature Triggering (Option c):** As temperature increases, the leakage current increases. At a certain critical temperature, the leakage current becomes high enough to trigger the regenerative action of the internal PNPN structure, turning the device on.
Since all three methods are valid ways to initiate the turn-on process of a thyristor, the correct answer is 'All of the above'.
In a thyristor at which layer gate terminal is connected and why? Explain dv/dt triggering of the thyristor.
A __________ is a current controlled device whereas a __________ is a voltage controlled device.
A)
BJT, Power MOSFET
B)
Power MOSFET, BJT
C)
SCR, DIAC
D)
GTO, IGBT
To solve this question, we must analyze the control mechanism of the semiconductor devices mentioned in the options:
1. **BJT (Bipolar Junction Transistor):** A BJT is a current-controlled device. The current flowing through the collector-emitter junction is controlled by the current injected into the base terminal (Base Current, Ib). The relationship is defined by the current gain beta (β), where Ic = β * Ib.
2. **Power MOSFET (Metal-Oxide-Semiconductor Field-Effect Transistor):** A MOSFET is a voltage-controlled device. The conductivity of the channel between the drain and source is controlled by the voltage applied to the gate terminal (Gate-to-Source Voltage, Vgs). Because the gate is insulated by a layer of silicon dioxide, virtually no current flows into the gate, making it voltage-controlled.
Evaluating the options:
- **Option a (BJT, Power MOSFET):** Correct. BJT is current-controlled; Power MOSFET is voltage-controlled.
- **Option b (Power MOSFET, BJT):** Incorrect. This is the reverse of the correct order.
- **Option c (SCR, DIAC):** Incorrect. An SCR is triggered by a current pulse at the gate, but a DIAC is a bidirectional trigger diode that triggers based on a breakdown voltage, not a controlled 'device' in the same sense as a transistor.
- **Option d (GTO, IGBT):** Incorrect. While a GTO is current-controlled (for turn-off), an IGBT is voltage-controlled. However, the BJT/MOSFET pair is the classic textbook example of current-controlled vs. voltage-controlled devices.
Define: Pulse Amplitude modulation, Pulse Width modulation and Pulse Code modulation with suitable graph.
Which of the following is a basic gate?
A)
AND
B)
X-OR
C)
X-NOR
D)
NAND
In digital electronics, logic gates are categorized based on their complexity and function.
1. **Basic Gates**: These are the fundamental building blocks of digital circuits. There are three primary basic gates: AND, OR, and NOT. These gates perform the most elementary logical operations.
2. **Universal Gates**: These are gates that can be used to implement any other logic gate (including basic gates) without needing any other gate type. These are NAND and NOR. While NAND is a fundamental component in hardware manufacturing, it is classified as a 'Universal Gate' rather than a 'Basic Gate'.
3. **Derived/Special Purpose Gates**: These are constructed using combinations of basic gates to perform specific functions. Examples include XOR (Exclusive-OR) and XNOR (Exclusive-NOR).
Analyzing the options:
- (a) AND: A basic gate. (Correct)
- (b) X-OR: A derived gate. (Incorrect)
- (c) X-NOR: A derived gate. (Incorrect)
- (d) NAND: A universal gate. (Incorrect)
For 8085 Microprocessor the mnemonics for the operation of ‘Load 8-bit data in a register’ is:
A)
MVI Rd, Rs
B)
MVI reg., 8-bit
C)
Lxi Rp, 8- bit
D)
Out 8-bit
In the 8085 Microprocessor instruction set, loading an 8-bit immediate data value into a register is performed using the 'Move Immediate' (MVI) instruction.
1. **MVI reg, 8-bit data**: This instruction takes two bytes. The first byte is the opcode for MVI, and the second byte is the 8-bit data to be loaded into the specified register (e.g., MVI A, 05H). This matches the operation 'Load 8-bit data in a register'.
Analysis of other options:
- **Option a (MVI Rd, Rs)**: This is incorrect syntax. MVI requires an immediate value (data), not a source register (Rs). Moving data between registers is done using the 'MOV Rd, Rs' instruction.
- **Option c (Lxi Rp, 8-bit)**: LXI (Load Register Pair Immediate) is used to load a 16-bit constant into a register pair (like BC, DE, or HL), not a single 8-bit value into a single register.
- **Option d (Out 8-bit)**: The OUT instruction is used to send data from the Accumulator to an output port; it is not used to load data into a register.
The conversion of input power at one frequency to output power at a different frequency is seen in the operation of:
A)
Cycloconverter
B)
Phase Controlled rectifier
C)
Inverters
D)
ASCR
Step-by-step reasoning:
1. The question asks for a device that converts input power at one frequency to output power at a different frequency. This is the definition of a frequency converter.
2. A Cycloconverter is a power electronics circuit that converts AC power at one frequency to AC power at another frequency (usually lower) without an intermediate DC link. This directly matches the requirement.
3. Analyzing other options:
- Phase Controlled Rectifier: Converts AC to DC (Fixed frequency to 0 Hz). It is not a frequency-to-frequency converter.
- Inverters: Convert DC to AC. While they create a specific frequency, the input is DC (0 Hz), not a different AC frequency.
- ASCR (Asymmetric Thyristor Controlled Rectifier): Primarily used for AC voltage regulation and power factor correction, not for frequency conversion.
Therefore, the Cycloconverter is the only device listed that performs AC-to-AC frequency conversion.
A BJT has a collector current of 4 mA and base current of 20 mA. Determine beta:
A)
200
B)
400
C)
2000
D)
20
For a BJT, current gain beta (β) is defined as β = I_C / I_B. Here, I_C = 4 mA and I_B = 20 mA, so β = 4/20 = 0.2. However, since the given options are all integers and the standard exam intent is likely that the base current was meant to be 20 μA (not 20 mA), the intended calculation is β = 4 mA / 20 μA = 200. But among the provided options, 200 is option a, not listed as the correct key in the prompt. If we strictly use the stated values, none of the options match. If the question intended 4 mA and 20 μA, then option a (200) would be correct. The other options 20, 400, and 2000 do not match the beta formula for the intended or stated values.
In communication system, Modulation occurs in:
A)
Source
B)
Transmitter
C)
Receiver
D)
Destination
In a basic communication system, the process of modulation is the act of varying a property (amplitude, frequency, or phase) of a high-frequency carrier signal in accordance with the instantaneous value of a low-frequency message signal. This process is essential to allow the signal to be transmitted efficiently over a medium (like air or cable) without significant attenuation or interference.
1. **Source (a):** The source is where the original information (voice, data, image) is generated. It provides the baseband signal but does not perform the modulation process.
2. **Transmitter (b):** The transmitter's primary role is to prepare the signal for transmission. This includes amplification, filtering, and most importantly, modulation. The modulator is a key component located within the transmitter.
3. **Receiver (c):** The receiver performs the inverse operation of the transmitter. It receives the modulated signal and performs 'Demodulation' (or detection) to recover the original message.
4. **Destination (d):** The destination is the final point where the recovered information is delivered to the user.
The 8085 has __________ general purpose registers to store 8-bit data:
A)
six
B)
eight
C)
four
D)
twelve
The Intel 8085 microprocessor contains six general-purpose 8-bit registers. These are designated as B, C, D, E, H, and L. These registers can be used individually to store 8-bit data or used in pairs (BC, DE, and HL) to store 16-bit data (such as memory addresses). While the 8085 also has an Accumulator (A) and a Flag register, the Accumulator is considered a special-purpose register used for arithmetic and logical operations, and the Flag register stores the status of the CPU. Therefore, the count of general-purpose registers is exactly six.
In a sinusoidal PWM inverter with M = 0.8, Vdc = 300 V, peak output voltage is:
A)
150 V
B)
200 V
C)
240 V
D)
300 V
In a Sinusoidal Pulse Width Modulation (SPWM) inverter, the modulation index (M) is defined as the ratio of the peak value of the fundamental output voltage (Vout_peak) to the maximum possible peak output voltage that can be achieved in the linear modulation region. For a single-phase full-bridge inverter, the maximum peak output voltage (when M=1) is equal to the DC link voltage (Vdc).
The formula relating these parameters is:
Peak Output Voltage (Vout_peak) = M × Vdc
Given values:
Modulation Index (M) = 0.8
DC Link Voltage (Vdc) = 300 V
Calculation:
Vout_peak = 0.8 × 300 V
Vout_peak = 240 V
Therefore, the peak output voltage is 240 V. Option (c) is correct. Options (a), (b), and (d) are incorrect as they do not satisfy the mathematical relationship between the modulation index and the DC supply voltage.
A JK flip-flop toggles when:
A)
J = 0, K = 0
B)
J = 1, K = 0
C)
J = 0, K = 1
D)
J = 1, K = 1
A JK flip-flop is a universal flip-flop with four distinct operating states based on the inputs J and K:
1. Hold State (J=0, K=0): The output remains unchanged (Q_next = Q_prev).
2. Reset State (J=0, K=1): The output is forced to 0 (Q_next = 0).
3. Set State (J=1, K=0): The output is forced to 1 (Q_next = 1).
4. Toggle State (J=1, K=1): The output changes to its complement (Q_next = NOT Q_prev).
Therefore, the 'toggle' condition occurs specifically when both J and K inputs are high (1). This allows the JK flip-flop to act as a frequency divider or a counter element.
An FM signal with deviation of 50 kHz and modulating frequency of 10 kHz has a modulation index of:
A)
0.2
B)
2
C)
5
D)
10
To find the modulation index of a Frequency Modulated (FM) signal, we use the standard formula:
Modulation Index (β) = Frequency Deviation (Δf) / Modulating Frequency (fm)
Given data from the question:
1. Frequency Deviation (Δf) = 50 kHz
2. Modulating Frequency (fm) = 10 kHz
Calculation:
β = 50 kHz / 10 kHz
β = 5
Therefore, the modulation index is 5.
Analysis of options:
- Option a (0.2) is incorrect; it would result if the values were swapped (10/50).
- Option b (2) is incorrect.
- Option c (5) is correct based on the formula.
- Option d (10) is incorrect.
An amplifier in which the operating point and the input signal are such that the current in the output circuit flows all the time is called:
A)
Class A amplifier
B)
Class B amplifier
C)
Class AB amplifier
D)
Class C amplifier
The classification of power amplifiers is based on the conduction angle of the output current relative to the input signal cycle (360 degrees).
1. **Class A Amplifier**: In this configuration, the operating point (Q-point) is set exactly in the center of the load line. This ensures that the transistor remains in the active region for the entire cycle of the input signal. Consequently, current flows through the output circuit for the full 360° of the input cycle. This results in high linearity but low efficiency.
2. **Class B Amplifier**: The Q-point is set at the cutoff point. The transistor conducts for only half of the input cycle (180°). This leads to crossover distortion.
3. **Class AB Amplifier**: The Q-point is set slightly above the cutoff point. The transistor conducts for more than 180° but less than 360°. This is a compromise to eliminate crossover distortion while maintaining better efficiency than Class A.
4. **Class C Amplifier**: The Q-point is set below the cutoff point. The transistor conducts for a small fraction of the cycle (less than 180°), typically used in RF tuned amplifiers.
Since the question specifies that the current flows 'all the time' (full cycle), it refers to a Class A amplifier.
In a CB configuration current configuration factor is 0.9, if the emitter current is 1.8 mA. The value of IB is:
A)
1.17 mA
B)
0.18 mA
C)
13mA
D)
1.3 mA
1. Understanding the Terms: In a Common Base (CB) configuration, the 'current configuration factor' refers to the current gain, denoted as alpha (α). Alpha is the ratio of the collector current (Ic) to the emitter current (Ie).
2. Given Data:
- Current gain (α) = 0.9
- Emitter current (Ie) = 1.8 mA
3. Step-by-Step Calculation:
- First, find the Collector Current (Ic):
Ic = α × Ie
Ic = 0.9 × 1.8 mA = 1.62 mA
- Use the fundamental transistor current equation:
Ie = Ic + Ib
Therefore, Base Current (Ib) = Ie - Ic
- Substitute the values:
Ib = 1.8 mA - 1.62 mA
Ib = 0.18 mA
4. Verification of Options:
- Option (a) 1.17 mA: Incorrect.
- Option (b) 0.18 mA: Matches our calculation.
- Option (c) 13 mA: Incorrect.
- Option (d) 1.3 mA: Incorrect.
A Single-phase full bridge diode rectifier is supplied from 230 V, 50 Hz source. The load consists of R = 10 V and a large inductance so as to render the load current constant, the average value of output voltage is:
A)
20.704 V
B)
207.04 V
C)
10.352 V
D)
23 V
1. **Identify the Circuit**: The circuit is a Single-phase full bridge diode rectifier.
2. **Given Data**:
- Input Voltage ($V_{rms}$) = 230 V
- Frequency ($f$) = 50 Hz
- Load: R = 10 $\Omega$ (Note: The question says '10 V', but based on the context of 'R', it is clearly a typo for 10 $\Omega$) and a large inductance.
- The large inductance ensures the load current is constant (ripple-free), meaning the rectifier operates in continuous conduction mode.
3. **Formula for Average Output Voltage**: For a single-phase full-bridge rectifier, the average output voltage ($V_{dc}$) is given by the formula:
$V_{dc} = \frac{2V_m}{\pi}$
where $V_m$ is the peak value of the input AC voltage.
4. **Calculate Peak Voltage ($V_m$)**:
$V_m = V_{rms} \times \sqrt{2}$
$V_m = 230 \times 1.4142 \approx 325.27\text{ V}$
5. **Calculate Average Voltage ($V_{dc}$)**:
$V_{dc} = \frac{2 \times 325.27}{3.14159}$
$V_{dc} = \frac{650.54}{3.14159} \approx 207.07\text{ V}$
6. **Conclusion**: The calculated value matches option (b) 207.04 V (the slight difference is due to the precision of $\pi$ and $\sqrt{2}$ used in standard textbook calculations). The resistance value does not affect the average output voltage in a full-bridge rectifier when the current is constant/continuous.
The basic operation of microprocessor includes:
A)
Memory read
B)
I/O read
C)
I/O write
D)
All of the above
A microprocessor is the central processing unit (CPU) of a computer system, and its primary function is to execute instructions by interacting with memory and peripheral devices. The basic operations it performs include:
1. **Memory Read**: The microprocessor fetches an instruction or data from a specific memory address to be processed internally.
2. **I/O Read**: The microprocessor reads data from an input device (like a keyboard or sensor) via an I/O port.
3. **I/O Write**: The microprocessor sends data or control signals to an output device (like a display or actuator) via an I/O port.
Additionally, 'Memory Write' is another basic operation. Since options a, b, and c are all fundamental operations required for a microprocessor to function as a controller or processor, the correct choice is 'All of the above'.
Explain with diagram the principle of step-down chopper operation.
A 555 timer is used in astable mode with R1 = 2 kV, R2 = 5 kV, and C = 0.1 mF. Calculate the frequency of oscillation and duty cycle.
Define bus in 8085 microprocessor and explain address bus, data bus and control bus.
What are the interrupts of 8085 microprocessor? Explain the function of TRAP interrupt.
How does a PN junction diode work? Draw and explain V-I characteristics of PN diode with neat diagram.