← Labour Officer
Past Questions
Explain working of DC chopper.
FM is preferred over AM because:
A)
Requires less bandwidth
B)
Less affected by noise
C)
Simpler circuits
D)
Lower power consumption
Frequency Modulation (FM) is preferred over Amplitude Modulation (AM) primarily due to its superior noise immunity. Most electrical noise (such as atmospheric interference or industrial sparks) manifests as amplitude spikes. In AM, the information is carried in the amplitude of the carrier wave, meaning the receiver cannot distinguish between the signal and the noise. In FM, the information is carried in the frequency variations, and the amplitude remains constant. FM receivers use 'limiters' to clip off amplitude variations (noise) without losing the frequency-encoded information, resulting in a much clearer signal.
Analysis of other options:
- Option (a) is incorrect: FM actually requires significantly more bandwidth than AM (e.g., an FM channel is typically 200 kHz, whereas an AM channel is about 10 kHz).
- Option (c) is incorrect: FM circuits (modulators and demodulators/PLLs) are generally more complex than the simple envelope detectors used in AM.
- Option (d) is incorrect: While FM transmitters can be more power-efficient in terms of the carrier, the primary reason for its preference in communication quality is noise rejection, not power consumption.
If a BJT has: IC = 4mA, IB = 0.1mA, Emitter current IE is:
A)
4.1 mA
B)
3.9 mA
C)
4.5 mA
D)
5 mA
Option A is correct: 4.1 mA.
For a half wave bridge inverter, the output voltage:
A)
Vo = – Vs/2 for 0 < t < T/2
B)
Vo = – Vs/2 for T/2< t
C)
Vo = – Vs for 0 < t < T/2
D)
Vo = Vs/2 for T/2< t < T
In a half-bridge inverter, the DC source (Vs) is split into two equal capacitors, each providing a voltage of Vs/2. The output voltage (Vo) is measured across the load connected between the midpoint of the capacitor bank and the switching nodes.
1. During the first half-cycle (0 < t < T/2), one switch (e.g., S1) is turned ON. This connects the load to the negative terminal of the capacitor bank, resulting in an output voltage of Vo = -Vs/2.
2. During the second half-cycle (T/2 < t < T), the other switch (e.g., S2) is turned ON. This connects the load to the positive terminal of the capacitor bank, resulting in an output voltage of Vo = +Vs/2.
Evaluating the options:
- Option (a) correctly describes the first half-cycle: Vo = -Vs/2 for 0 < t < T/2.
- Option (b) is incorrect because for T/2 < t < T, the voltage should be +Vs/2.
- Option (c) is incorrect because the magnitude in a half-bridge inverter is Vs/2, not Vs (Vs is the magnitude for a full-bridge inverter).
- Option (d) is partially correct in magnitude, but the standard convention for the first half-cycle is negative and the second is positive; however, compared to the standard definition of a half-bridge operation, option (a) is the definitive characteristic of the initial state.
A microprocessor can be interfaced with which type of memory?
A)
ROM
B)
RAM
C)
Both ROM and RAM
D)
None
A microprocessor requires two primary types of memory to function effectively:
1. **ROM (Read-Only Memory):** This is non-volatile memory used to store the 'firmware' or the boot-up program (BIOS/Monitor program). Since a microprocessor cannot perform any action without instructions, the ROM provides the initial set of instructions that the CPU executes upon reset.
2. **RAM (Random Access Memory):** This is volatile memory used to store temporary data, variables, and the stack during program execution. It allows the microprocessor to read and write data quickly while the system is running.
In the context of the 8085 Microprocessor, the address bus and data bus are designed to interface with both types of memory chips. The control signals (RD' and WR') allow the processor to read from both ROM and RAM, but write only to RAM. Therefore, a complete microprocessor system must be interfaced with both to be functional.
The opcode of an instruction specifies:
A)
Operation to be performed
B)
Memory location
C)
Input data
D)
Output data
In the context of the 8085 Microprocessor and computer architecture, an instruction consists of two main parts: the Opcode (Operation Code) and the Operand.
1. **Opcode (Operation Code):** This is the part of the instruction that tells the microprocessor exactly what task to perform. For example, in the instruction 'MOV A, B', the 'MOV' part is the opcode, which tells the CPU to move data from one register to another.
2. **Operand:** This is the data on which the operation is performed or the address of the memory location where the data is stored. In 'MOV A, B', 'A' and 'B' are the operands.
Evaluating the options:
- **Option (a) is correct** because the opcode specifically defines the operation (e.g., ADD, SUB, MOV, JMP).
- **Option (b) is incorrect** because the memory location is specified by the operand (addressing mode).
- **Option (c) is incorrect** because input data is the operand.
- **Option (d) is incorrect** because output data is the result of the operation, not the specification of the instruction itself.
A half-controlled bridge rectifier consists of:
A)
Four SCRs
B)
Two SCRs and two diodes
C)
Four diodes
D)
One SCR and one diode
A bridge rectifier is a circuit that converts AC to DC. Depending on the components used, it can be categorized as follows:
1. Fully Uncontrolled Rectifier: Uses four diodes. The output voltage cannot be controlled.
2. Fully Controlled Rectifier: Uses four SCRs (Thyristors). The output voltage can be precisely controlled by adjusting the firing angle.
3. Half-Controlled (Semi-Controlled) Rectifier: This is a hybrid configuration. It replaces two of the diodes in a full bridge with two SCRs. Typically, the top two components are SCRs and the bottom two are diodes (or vice versa). This allows for partial control of the output voltage while reducing the cost and complexity of the triggering circuitry compared to a fully controlled bridge.
Therefore, a half-controlled bridge rectifier consists of two SCRs and two diodes.
Pulse Width Modulation (PWM) is used to:
A)
Control output voltage of inverters
B)
Rectify AC
C)
Convert AC to DC
D)
Reduce ripple in DC
Pulse Width Modulation (PWM) is a technique used in power electronics to control the average voltage delivered to a load. In the context of inverters (which convert DC to AC), PWM is used to vary the width of the pulses of the DC voltage applied to the load. By adjusting the duty cycle (the ratio of 'on' time to the total period), the fundamental component of the output AC voltage can be precisely controlled, and the harmonic content can be reduced.
Analysis of other options:
- Option b (Rectify AC): This is the function of a rectifier (using diodes or thyristors), not PWM.
- Option c (Convert AC to DC): This is the definition of rectification, which is the opposite of what an inverter does.
- Option d (Reduce ripple in DC): Ripple reduction is achieved using filters (capacitors or inductors), not PWM. While PWM is used in DC-DC converters (Buck/Boost), its primary purpose there is voltage regulation, and the ripple is actually a byproduct of the switching process that must be filtered out.
XOR gate output is HIGH when:
A)
Both inputs are LOW
B)
Both inputs are HIGH
C)
Only one input is HIGH
D)
Inputs are equal
An XOR (Exclusive OR) gate is a digital logic gate that implements an exclusive disjunction. The output of an XOR gate is HIGH (1) if and only if the inputs are different.
Let's analyze the Truth Table for a 2-input XOR gate:
1. Input A=0, B=0: Output = 0 (Both are LOW)
2. Input A=0, B=1: Output = 1 (Only one is HIGH)
3. Input A=1, B=0: Output = 1 (Only one is HIGH)
4. Input A=1, B=1: Output = 0 (Both are HIGH)
Evaluating the options:
- Option (a) 'Both inputs are LOW' results in a LOW output.
- Option (b) 'Both inputs are HIGH' results in a LOW output.
- Option (c) 'Only one input is HIGH' results in a HIGH output, which matches the XOR logic.
- Option (d) 'Inputs are equal' (both 0 or both 1) results in a LOW output (this is actually the logic for an XNOR gate).
Carrier communication requires:
A)
Modulation
B)
Demodulation
C)
Transmission over channel
D)
All of the above
Carrier communication (wireless or wired high-frequency communication) is a complete system process that involves several critical stages to successfully move information from a source to a destination.
1. **Modulation (Option a):** Baseband signals (like voice or data) have low frequencies and cannot be transmitted efficiently over long distances. Modulation is the process of superimposing the message signal onto a high-frequency 'carrier wave' to allow for efficient radiation via antennas and to avoid signal interference.
2. **Transmission over channel (Option c):** Once modulated, the signal must be physically transported from the transmitter to the receiver through a medium (the channel), such as free space (air/vacuum), coaxial cables, or optical fibers.
3. **Demodulation (Option b):** At the receiving end, the high-frequency carrier is no longer needed. Demodulation is the process of extracting the original baseband message signal from the received modulated carrier wave.
Since a functional carrier communication system fundamentally requires the process of preparing the signal (modulation), sending it (transmission), and recovering it (demodulation), all three options are essential components of the system.
In an AM system, the maximum and minimum envelope voltages are 10 V and 4 V respectively. The modulation index is approximately:
A)
0.25
B)
0.43
C)
0.67
D)
0.80
Option C is correct: 0.67.
The value of anode current required to maintain the conduction of an SCR even though the gate signal is removed is called as the:
A)
Latching Current
B)
Holding Current
C)
Switching Current
D)
All of these
In a Silicon Controlled Rectifier (SCR), there are two critical current levels related to the transition between the OFF state and the ON state:
1. **Latching Current (I_L):** This is the minimum anode current required to keep the SCR in the ON state immediately after the gate trigger signal is removed. If the anode current does not reach the latching level before the gate pulse ends, the SCR will return to the OFF state. Therefore, it is associated with the 'turn-on' process.
2. **Holding Current (I_H):** This is the minimum anode current below which the SCR will stop conducting and return to the OFF state. It is associated with the 'turn-off' process. Once the SCR is already conducting, if the current drops below the holding current, the device turns off.
Since the question specifically asks for the current required to maintain conduction *even though the gate signal is removed*, it refers to the Latching Current.
Comparison: Latching Current > Holding Current.
Explain interrupt priority in microprocessor.
Explain the working principle and V–I characteristics of a Zener diode. Describe how it is used as a voltage regulator in power supply circuits.
In a step-up chopper circuit, if Vs is the source voltage and α is duty cycle, then the output voltage is:
A)
Vs/ (1 + α)
B)
Vs (1 + α)
C)
Vs (1 - α)
D)
Vs/ (1 - α)
A step-up chopper (also known as a Boost Converter) is used to produce an output voltage higher than the input source voltage.
1. **Operation Phase 1 (Switch ON):** When the switch is closed for a time period 'T_on', the inductor stores energy from the source (Vs). The diode is reverse-biased, and no current flows to the load.
2. **Operation Phase 2 (Switch OFF):** When the switch is open for a time period 'T_off', the inductor voltage adds to the source voltage, and the combined voltage is delivered to the load through the diode.
**Mathematical Derivation:**
- Let T be the total switching period.
- Duty cycle (α) = T_on / T.
- The average voltage across the inductor over one complete cycle is zero (steady-state assumption).
- Average Voltage = (Vs * T_on) + (Vs - Vo) * T_off = 0
- Vs * T_on + Vs * T_off - Vo * T_off = 0
- Vs * (T_on + T_off) = Vo * T_off
- Vs * T = Vo * (T - T_on)
- Vo = Vs * [T / (T - T_on)]
- Divide numerator and denominator by T: Vo = Vs / [1 - (T_on/T)]
- Since α = T_on/T, the formula becomes: Vo = Vs / (1 - α).
**Analysis of Options:**
- Option (a) Vs/(1 + α) is incorrect.
- Option (b) Vs(1 + α) is incorrect.
- Option (c) Vs(1 - α) is the formula for a step-down (Buck) chopper.
- Option (d) Vs/(1 - α) is the correct formula for a step-up (Boost) chopper.
Memory interfacing with a microprocessor requires:
A)
Address bus, data bus, control signals
B)
Only data bus
C)
Only power supply
D)
Only clock signal
Memory interfacing is the process of connecting the microprocessor to memory chips so that the CPU can read from or write to specific memory locations. To achieve this, three primary components are required:
1. Address Bus: The microprocessor uses the address bus to specify the unique location (address) in memory that it wants to access.
2. Data Bus: This is the bidirectional path used to transfer the actual data between the microprocessor and the memory chip.
3. Control Signals: These signals (such as RD (Read) and WR (Write) in the 8085) tell the memory whether the CPU wants to read data from the memory or write data into it.
Without any one of these three, the microprocessor cannot communicate with the memory. Options b, c, and d are incorrect because they list only a single component, which is insufficient for the complete operation of memory interfacing.
Explain basic block diagram of digital communication system.
Memory-mapped I/O means:
A)
I/O devices share the address space with memory
B)
Separate address space for I/O
C)
Only RAM is used
D)
Only ROM is used
In microprocessor architecture, there are two primary methods for interfacing I/O devices: Memory-Mapped I/O and I/O-Mapped I/O (Isolated I/O).
1. **Memory-Mapped I/O**: In this configuration, the microprocessor treats I/O devices as if they were memory locations. The I/O devices are assigned specific addresses within the same address space used by the RAM and ROM. Consequently, the processor uses the same instructions (like MOV or LDA in 8085) to access both memory and I/O devices. The trade-off is that a portion of the available memory address space is 'consumed' by the I/O devices.
2. **I/O-Mapped I/O (Isolated I/O)**: This uses a separate address space for I/O devices, distinct from the memory address space. Special instructions (like IN and OUT in 8085) are used to communicate with these devices, and the processor uses a control signal (like IO/M) to distinguish between a memory access and an I/O access.
Analysis of options:
- Option (a) is correct as it defines the core concept of memory-mapping.
- Option (b) describes 'Isolated I/O' or 'I/O-Mapped I/O'.
- Option (c) and (d) are incorrect as they refer to types of memory, not the method of interfacing I/O devices.
An op-amp used for mathematical integration is called:
A)
Differentiator
B)
Oscillator
C)
Comparator
D)
Integrator
An operational amplifier (op-amp) circuit designed to perform the mathematical operation of integration is specifically called an Integrator. In a typical inverting integrator circuit, a resistor is placed at the input and a capacitor is placed in the feedback loop. The output voltage is proportional to the integral of the input voltage over time.
Analysis of other options:
- a. Differentiator: This circuit performs the opposite operation (differentiation), where the output is proportional to the rate of change of the input voltage. It uses a capacitor at the input and a resistor in the feedback loop.
- b. Oscillator: This is a circuit that generates a periodic waveform (like a sine or square wave) without an external input signal; it is not a mathematical integration tool.
- c. Comparator: This circuit compares two input voltages and outputs a digital signal (high or low) based on which input is larger; it does not perform integration.
If the firing angle becomes negative, then the rectifier begins to work as
A)
A rectifier
B)
An inverter
C)
A chopper
D)
A regulator
In power electronics, specifically regarding controlled rectifiers (like those using Thyristors or GTOs), the firing angle (α) determines the point in the AC cycle where the device begins to conduct.
1. **Rectification Mode (0° < α < 180°):** The device converts AC to DC. The average output voltage is positive, and power flows from the AC source to the DC load.
2. **Inversion Mode (180° < α < 360°, or effectively negative firing angles in certain control contexts):** When the firing angle is shifted such that the device conducts during the negative half of the supply voltage cycle (or when the load voltage is higher than the source voltage and the firing occurs at a negative angle relative to the natural commutation point), the average output voltage becomes negative.
3. **Power Flow:** In this state, the power flow reverses; energy is transferred from the DC side (e.g., a battery or a DC motor acting as a generator) back to the AC source. This process of converting DC back to AC is the definition of an inverter.
Why other options are incorrect:
- **a. A rectifier:** This is the standard operation for positive firing angles (0 to 180 degrees).
- **c. A chopper:** A chopper is a DC-to-DC converter; it does not involve firing angles relative to an AC sine wave.
- **d. A regulator:** While a rectifier can be used for regulation, the specific act of shifting the firing angle into the negative region changes the fundamental nature of the power conversion (direction of flow), turning it into an inverter.
Describe sampling, quantization, and encoding.
A full-adder has:
A)
Two inputs and one output
B)
Three inputs and two outputs
C)
Two inputs and two outputs
D)
One input and one output
A Full-Adder is a combinational digital circuit that performs the addition of three binary bits.
1. **Inputs**: It takes three inputs: two significant bits to be added (A and B) and a carry-in bit (Cin) coming from a previous less-significant stage of addition. Total = 3 inputs.
2. **Outputs**: It produces two outputs: the Sum (S) and the Carry-out (Cout). Total = 2 outputs.
Comparison with other options:
- Option (a) describes a Half-Adder's inputs but not its outputs (a Half-Adder has 2 inputs and 2 outputs).
- Option (c) describes a Half-Adder (2 inputs: A, B; 2 outputs: Sum, Carry).
- Option (d) is incorrect as no standard adder operates with a single input and output.
Draw and explain the V-I characteristics of SCR.
Explain ideal op-amp characteristics.
Explain the formation of a PN junction and describe the concept of depletion region and barrier potential. Discuss how forward and reverse bias affect the PN junction characteristics.