← MES Jr. Grade
Past Questions
Which of the following 3-phase AC to DC converter requires neutral point connection?
A)
3-phase semi-conduction
B)
3-phase full converter
C)
3-phase half-wave converter
D)
3-phase full converter with diodes
In a 3-phase half-wave converter (also known as a 3-pulse converter), only one thyristor or diode is conducted at a time. The current flows from the AC supply line through the switching device to the load, and then returns to the source. For the circuit to be complete, the return path must be connected to the neutral point of the 3-phase supply (star connection). Without a neutral connection, the current cannot return to the source, and the converter will not function.
In contrast, 3-phase full converters (6-pulse converters), whether they use thyristors or diodes, are bridge configurations. In these circuits, the current flows from one phase and returns through another phase (line-to-line conduction). Therefore, they do not require a neutral point connection to operate.
- Option (a) 'semi-conduction' is not a standard term for a specific converter topology in this context.
- Option (b) and (d) are full-bridge converters which are 3-wire systems and do not require a neutral.
Output of assembler in machine codes is referred to as
A)
Object program
B)
Source program
C)
Macro-instruction
D)
Symbolic addressing
An assembler is a utility program that translates assembly language (a low-level, human-readable symbolic language) into machine code (binary 0s and 1s) that the CPU can execute directly.
1. **Source Program**: This is the original code written by the programmer in assembly language. It is the input to the assembler, not the output.
2. **Object Program**: This is the output produced by the assembler. It consists of the machine code instructions and data that are ready to be linked and loaded into memory for execution.
3. **Macro-instruction**: A macro is a sequence of instructions assigned a name; it is a tool used within the source program to simplify repetitive coding, not the final output of the assembly process.
4. **Symbolic addressing**: This refers to the use of labels or names (symbols) instead of absolute memory addresses in the source code. The assembler resolves these symbols into actual addresses during the translation process to create the object program.
Therefore, the output of the assembler in machine codes is the Object program.
In a class B amplifier, the output current flows for
A)
Less than half input
B)
More than half input
C)
Half input
D)
One complete cycle
In a Class B amplifier, the biasing point (Q-point) is set exactly at the cutoff point of the transistor's characteristic curve. Because the transistor is biased at cutoff, it only conducts when the input signal swings positive (for an NPN transistor), allowing current to flow for only one half-cycle of the input signal. Specifically, the conduction angle is 180 degrees. To achieve a full output waveform without distortion, Class B amplifiers are typically used in 'push-pull' configurations where two transistors work in tandem—one handling the positive half-cycle and the other handling the negative half-cycle. Therefore, a single transistor in a Class B amplifier conducts for exactly half of the input cycle.
Which of the following is not a method of turning ON an SCR?
A)
Gate triggering
B)
Temperature triggering
C)
Light triggering
D)
Emitter triggering
An SCR (Silicon Controlled Rectifier) is a four-layer thyristor that can be turned ON (triggered) using several methods.
1. **Gate Triggering (Option a):** This is the most common method where a positive voltage is applied to the gate terminal, injecting holes into the P-layer, which reduces the depletion region and allows the device to conduct.
2. **Temperature Triggering (Option b):** At very high temperatures, the intrinsic carrier concentration increases, reducing the forward breakover voltage. If the temperature is high enough, the SCR can turn ON even without a gate signal.
3. **Light Triggering (Option c):** In a Light Activated SCR (LASCR), photons of light are directed at the gate region, creating electron-hole pairs that trigger the device into conduction.
**Emitter triggering (Option d)** is not a recognized method for turning on an SCR. While the SCR is composed of P-N junctions (which are similar to the emitter-base junctions in a BJT), 'Emitter triggering' is not a standard terminology or a physical method used to trigger a thyristor. Therefore, it is the correct answer as it is NOT a method of turning ON an SCR.
The action of JFET in its equivalent circuit can be represented as
A)
Current controlled current source
B)
Current controlled voltage source
C)
Voltage controlled current source
D)
Voltage controlled voltage source
A Junction Field Effect Transistor (JFET) is a three-terminal semiconductor device where the current flowing from the drain to the source (Id) is controlled by the voltage applied to the gate-source junction (Vgs).
1. **Mechanism**: In a JFET, the gate-source voltage (Vgs) modifies the width of the depletion region within the channel. As Vgs becomes more negative (for an n-channel JFET), the depletion region widens, narrowing the conductive channel and reducing the drain current (Id).
2. **Mathematical Representation**: This relationship is governed by Shockley's equation: Id = Idss [1 - (Vgs/Vp)]^2. This shows that the output current (Id) is a function of the input voltage (Vgs).
3. **Equivalent Circuit**: In small-signal analysis, the JFET is modeled as a voltage-controlled current source (VCCS). The output current is represented as Id = gm * Vgs, where 'gm' is the transconductance of the device.
Comparing this to the options:
- Option (a) and (b) are incorrect because the input is a voltage, not a current.
- Option (d) is incorrect because the output of a JFET is a current (Id), not a voltage.
The simplified form of logic function Y = (A+B)(A+B) is
A)
A+B
B)
AB
C)
A+B
D)
AB+AB
The given logic function is Y = (A + B)(A + B).
Step 1: According to the Idempotent Law of Boolean Algebra, any variable or expression operated on by itself using AND or OR results in the original variable or expression.
Specifically, the Idempotent Law states: X · X = X.
Step 2: In this problem, let the expression (A + B) be represented by a single variable X.
Therefore, Y = X · X.
Step 3: Applying the Idempotent Law (X · X = X), we get Y = X.
Step 4: Substituting (A + B) back for X, we get Y = A + B.
Alternative Algebraic Expansion:
Y = (A + B)(A + B)
= A·A + A·B + B·A + B·B (Distributive Law)
= A + AB + AB + B (Since A·A = A and B·B = B)
= A(1 + B) + B(1 + A) (Factoring)
= A(1) + B(1) (Since 1 + X = 1)
= A + B.
Comparing this result with the options:
Option (a) A+B is correct.
Option (b) AB is incorrect as it represents an AND operation.
Option (c) A+B is identical to (a).
Option (d) AB+AB simplifies to AB, which is incorrect.
Which of the following logic families is well suited for high speed operations?
A)
TTL
B)
MOS
C)
ECL
D)
CMOS
The question asks for the logic family best suited for high-speed operations.
1. **ECL (Emitter-Coupled Logic):** This is a non-saturating logic family. In most other logic families (like TTL), transistors are driven into saturation, which leads to a 'storage time' delay when switching from an ON state to an OFF state. ECL avoids saturation by operating transistors in the active region, which drastically reduces switching time and makes it the fastest logic family available. It is widely used in high-performance computing and RF applications.
2. **TTL (Transistor-Transistor Logic):** While faster than standard CMOS in early iterations, TTL transistors operate in saturation, making them slower than ECL.
3. **MOS (Metal-Oxide-Semiconductor) and CMOS (Complementary MOS):** These families are primarily valued for their extremely low power consumption and high packing density. While modern CMOS is very fast, in the context of fundamental logic family comparisons in engineering exams, ECL is recognized as the gold standard for raw speed due to its differential amplifier configuration and lack of saturation.
Therefore, ECL is the correct choice for high-speed operations.
How many I/O ports can be accessed by direct method?
A)
8
B)
256
C)
32K
D)
64K
In the context of the 8085 and 8086 microprocessors (which are the standard for these engineering exams), I/O ports are accessed using two methods: Direct Addressing (also known as I/O Mapped I/O) and Indirect Addressing.
1. Direct I/O Addressing: In this method, the address of the I/O port is specified directly within the instruction (e.g., IN 05H or OUT 05H). Since the instruction uses an 8-bit address to identify the port, the total number of unique port addresses that can be accessed is 2^8 = 256 (ranging from 00H to FFH).
2. Indirect I/O Addressing: In this method, the port address is stored in a register (like the DX register in 8086). Because the register is 16-bit, it can access up to 2^16 = 65,536 (64K) ports.
Since the question specifically asks for the 'direct method', the limit is determined by the 8-bit address field, which is 256 ports.
In a double side band (DSB) full carrier AM transmission system, if modulation index is doubled, then ratio of the total sideband power to the carrier power increases by a factor of
A)
1
B)
2
C)
3
D)
4
Step-by-step derivation:
1. In a standard Double Sideband Full Carrier (DSB-FC) AM system, the total power is the sum of the carrier power and the sideband power.
2. The formula for Carrier Power is: Pc = (V_c^2) / (2R).
3. The formula for Total Sideband Power (sum of Upper Sideband and Lower Sideband) is: P_sb = (m^2 * Pc) / 2, where 'm' is the modulation index.
4. The ratio of the total sideband power to the carrier power is given by: Ratio = P_sb / Pc = m^2 / 2.
5. Let the initial modulation index be m1. The initial ratio is R1 = (m1^2) / 2.
6. According to the problem, the modulation index is doubled. So, m2 = 2 * m1.
7. The new ratio is R2 = (m2^2) / 2 = ((2 * m1)^2) / 2 = (4 * m1^2) / 2.
8. To find the factor by which the ratio increases, we divide R2 by R1: Factor = R2 / R1 = [(4 * m1^2) / 2] / [(m1^2) / 2] = 4.
Therefore, if the modulation index is doubled, the ratio of sideband power to carrier power increases by a factor of 4.
For a full adder, determine the output for the input A= 1, B= 0, and carry in=1
A)
Sum = 0 with carry of 0
B)
Sum = 0 with carry of 1
C)
Sum = 1 with carry of 0
D)
Sum = 1 with carry of 1
A Full Adder is a combinational circuit that performs the addition of three binary bits: two significant bits (A and B) and a carry bit (Cin) from a previous stage.
1. **Input Values**: A = 1, B = 0, Cin = 1.
2. **Sum Calculation**: The Sum (S) is calculated using the XOR operation: S = A ⊕ B ⊕ Cin.
- Step 1: A ⊕ B = 1 ⊕ 0 = 1
- Step 2: (A ⊕ B) ⊕ Cin = 1 ⊕ 1 = 0
Therefore, Sum = 0.
3. **Carry Calculation**: The Carry-out (Cout) is calculated using the formula: Cout = (A AND B) OR (Cin AND (A ⊕ B)).
- Step 1: A AND B = 1 AND 0 = 0
- Step 2: Cin AND (A ⊕ B) = 1 AND (1 ⊕ 0) = 1 AND 1 = 1
- Step 3: 0 OR 1 = 1
Therefore, Carry = 1.
Alternatively, using simple binary addition: 1 + 0 + 1 = 2 in decimal, which is '10' in binary. The rightmost bit (0) is the Sum and the leftmost bit (1) is the Carry. Thus, Sum = 0 and Carry = 1.
The current gain of a bipolar transistor drops at high frequencies because of
A)
Transistor capacitance
B)
High current effects in the base
C)
Parasitic inductive elements
D)
The early effect
In a Bipolar Junction Transistor (BJT), the current gain decreases at high frequencies primarily due to internal junction capacitances. There are two main types of capacitance involved: 1) Junction Capacitances (Diffusion and Depletion capacitance) such as Cbe (base-emitter) and Cbc (base-collector). 2) The Miller Effect, which amplifies the effect of the collector-base capacitance in common-emitter configurations. At high frequencies, these capacitances provide low-impedance paths (shunts) for the signal, causing the signal to bypass the active region of the transistor and effectively reducing the current gain.
Why other options are incorrect:
- b. High current effects: These typically relate to saturation or thermal runaway, not frequency response.
- c. Parasitic inductive elements: While inductances exist in lead wires and packaging, they typically cause instability or oscillations at extremely high (microwave) frequencies, but the primary cause of the gain roll-off in standard BJT frequency response is capacitance.
- d. The Early effect: This refers to base-width modulation caused by the collector-base voltage, which affects the output resistance and DC gain, but it is not the cause of high-frequency gain drop.
A microprocessor has 24 address lines and 32 data lines. If it uses 10 bits of Op-Code, the size of its memory Buffer register is
A)
22 bits
B)
24 bits
C)
32 bits
D)
14 bits
To determine the size of the Memory Buffer Register (MBR), also known as the Memory Data Register (MDR), we must look at the architecture of the microprocessor.
1. **Address Lines**: The 24 address lines determine the addressable memory space (2^24 locations), but they do not determine the size of the data stored at each location.
2. **Data Lines**: The data lines (bus width) determine the amount of data that can be transferred between the microprocessor and memory in a single memory cycle. In this case, there are 32 data lines.
3. **Op-Code**: The 10-bit Op-Code determines the number of unique instructions the CPU can recognize, but it does not dictate the hardware width of the data buffer.
4. **Memory Buffer Register (MBR)**: The MBR acts as a temporary holding area for data being read from or written to memory. For the microprocessor to efficiently utilize its 32-bit data bus, the MBR must be wide enough to hold one full word of data from that bus. Therefore, the size of the MBR must match the number of data lines.
Since there are 32 data lines, the MBR size is 32 bits.
If gate current is increased, the anode-cathode voltage at which SCR closes is
A)
increased
B)
decreased
C)
maximum
D)
least
An SCR (Silicon Controlled Rectifier) is a four-layer PNPN device. To turn on (trigger) the SCR, it must reach its 'breakover voltage' (VBO). There are two primary ways to trigger an SCR: 1) By increasing the anode-to-cathode voltage until it reaches the forward breakover voltage, or 2) By applying a positive gate current (Ig) while the device is forward-biased.
When a gate current is injected into the gate terminal, it increases the charge carrier concentration in the P-layer (the gate-cathode junction). This reduces the depletion region width and lowers the barrier for current to flow from the anode to the cathode. Consequently, the amount of anode-to-cathode voltage required to trigger the device into conduction (the 'closing' or triggering voltage) decreases.
Mathematically, the relationship is inverse: as the gate current (Ig) increases, the required anode-to-cathode voltage (Va) for triggering decreases. If the gate current is sufficiently high, the SCR can be triggered even at a very low anode voltage. Therefore, increasing the gate current decreases the anode-cathode voltage at which the SCR closes (turns on).
Turn-on and turn-off times of transistor depend on
A)
Static characteristic
B)
Junction capacitances
C)
Current gain
D)
None of these
The turn-on and turn-off times of a transistor (switching speed) are primarily determined by the time it takes to charge and discharge the internal parasitic capacitances of the device.
1. **Junction Capacitances**: A Bipolar Junction Transistor (BJT) consists of two PN junctions (Emitter-Base and Collector-Base). These junctions act as capacitors. When the transistor switches states (e.g., from cutoff to saturation), these capacitances must be charged or discharged. The time required for this process directly determines the switching speed (turn-on and turn-off times).
2. **Diffusion Capacitance**: In addition to depletion capacitance, there is diffusion capacitance caused by the storage of minority carriers in the base region. This 'charge storage' effect is a major contributor to the turn-off delay (storage time).
Why other options are incorrect:
- **Static characteristics**: These describe the DC behavior of the transistor (like $I_C$ vs $V_{CE}$) and do not provide information about the time-dependent (dynamic) response of the device.
- **Current gain**: While current gain ($\beta$ or $\alpha$) affects the amplification capability and the amount of current flowing, it does not fundamentally dictate the time constants associated with the charging/discharging of the junctions.
The effective channel length of a MOSFET in saturation decreases with
A)
gate voltage
B)
drain voltage
C)
source voltage
D)
body voltage
This question refers to the phenomenon known as Channel Length Modulation (CLM) in MOSFETs. When a MOSFET operates in the saturation region, the pinch-off point occurs where the channel ends. As the drain-to-source voltage (Vds) increases beyond the saturation voltage (Vdsat), the pinch-off point moves slightly toward the source. This effectively reduces the length of the inverted channel (the distance over which the carriers flow before entering the depletion region near the drain). The effective channel length (Leff) is expressed as Leff = L - ΔL, where ΔL is the length of the depletion region at the drain end. Since ΔL increases as the drain voltage increases, the effective channel length decreases. This effect leads to a non-zero slope in the I-d vs V-ds curve in the saturation region, meaning the output resistance is finite rather than infinite. Gate voltage primarily affects the channel depth/width (inversion layer), and source/body voltages affect the threshold voltage (body effect), but the specific reduction of the effective length in saturation is driven by the drain voltage.
Time division multiplex
A)
can be used with PCM only
B)
combines five groups into a supergroup
C)
stacks 24 channels in adjacent frequency slots
D)
interleaves pulses belonging to different transmissions
Time Division Multiplexing (TDM) is a technique where multiple signals are combined into one single transmission medium by assigning each signal a specific, non-overlapping time slot.
1. **Why Option (d) is correct**: In TDM, the time axis is divided into slots. The multiplexer takes a sample (pulse) from the first channel, then the second, then the third, and so on. This process of alternating samples from different sources in a sequence is known as 'interleaving'.
2. **Why Option (a) is incorrect**: While TDM is extensively used with Pulse Code Modulation (PCM) in digital telephony, it is not used *only* with PCM. TDM can be applied to analog signals (Pulse Amplitude Modulation - PAM) as well.
3. **Why Option (b) is incorrect**: This is a specific detail of the North American T-carrier system (where 24 channels form a group, and groups form supergroups), but it is not a general definition or characteristic of Time Division Multiplexing itself.
4. **Why Option (c) is incorrect**: Stacking channels in 'adjacent frequency slots' is the definition of Frequency Division Multiplexing (FDM), not TDM. TDM shares the same frequency band but divides the time.
Quantizing noise can be reduced by increasing
A)
bandwidth
B)
sampling rate
C)
number of standard quantum levels
D)
all the these
Quantization noise (or quantization error) occurs during the process of converting a continuous-time analog signal into a discrete-time digital signal. It is the difference between the actual analog value and the nearest quantized level.
1. **Quantization Step Size (Δ):** The range of the signal is divided into 'L' discrete levels. The step size is given by Δ = (Vmax - Vmin) / L.
2. **Quantization Noise Power:** For a uniformly quantized signal, the quantization noise power (mean square error) is given by the formula: Pq = (Δ²)/12.
3. **Relationship with Levels:** Since Δ is inversely proportional to the number of quantization levels (L), increasing the number of standard quantum levels (L) decreases the step size (Δ). As Δ decreases, the quantization noise power (Pq) decreases significantly.
Why other options are incorrect:
- **Bandwidth (a):** Increasing bandwidth may affect the signal's frequency range or transmission speed, but it does not reduce the error introduced by the quantization process itself.
- **Sampling Rate (b):** Increasing the sampling rate helps avoid 'aliasing' (according to the Nyquist theorem), but it does not change the vertical precision (quantization levels) of each sample. Sampling is a time-domain discretization, while quantization is an amplitude-domain discretization.
If firing angle in a SCR rectifier is increased, output is
A)
increased
B)
maximum
C)
decreased
D)
unaffected
In an SCR (Silicon Controlled Rectifier) rectifier, the firing angle (denoted by α) is the angle measured from the zero crossing of the input AC voltage to the point where the SCR is triggered into conduction.
1. **Mechanism**: An SCR remains in the 'OFF' state (blocking state) until a trigger pulse is applied to its gate. Once triggered, it conducts current until the current falls below the holding current level (usually at the end of the half-cycle).
2. **Effect of Firing Angle**:
- If the firing angle α is 0°, the SCR conducts for the entire half-cycle, acting like a standard diode rectifier (maximum output).
- As the firing angle α increases, the SCR is triggered later in the cycle. This means the period during which the SCR is 'OFF' increases, and the period during which it is 'ON' (the conduction angle, which is 180° - α) decreases.
3. **Output Voltage**: The average output voltage (Vdc) of a single-phase half-wave SCR rectifier is given by the formula: Vdc = (Vm / π) * (1 + cos α). As α increases from 0° to 180°, the value of cos α decreases, thereby reducing the average output voltage.
Therefore, increasing the firing angle reduces the area under the voltage curve delivered to the load, resulting in a decreased output voltage.
The regenerative breaking is applicable for a single-phase full-wave bridge converter at the SCR firing angle:
A)
equal to 90°
B)
greater than 90°
C)
less than 90°
D)
greater than 180°
In a single-phase full-wave bridge converter (fully controlled rectifier), the average output voltage (Vdc) is given by the formula: Vdc = (2Vm / π) * cos(α), where α is the firing angle.
1. When 0° < α < 90°, cos(α) is positive, meaning Vdc is positive. Power flows from the AC source to the DC load (Rectification mode).
2. When α = 90°, cos(α) = 0, and the average output voltage is zero.
3. When 90° < α < 180°, cos(α) becomes negative, meaning Vdc is negative. In this region, if the load has a source (like a DC motor acting as a generator during braking), power can flow from the DC side back to the AC source. This process is known as regenerative braking or inversion.
Therefore, for regenerative braking to occur, the firing angle α must be greater than 90° to reverse the polarity of the average voltage and allow power to flow back into the supply.
In common base amplifier which of the following is wrong?
A)
Input impedance low
B)
Voltage gain about 150
C)
Used for audio frequency application
D)
Current gain is less than one
In a common-base (CB) amplifier, the current gain is less than 1 (approximately \u03b1 < 1), the input impedance is low, and it is mainly used for high-frequency applications rather than audio-frequency amplification. Therefore, the statement "Voltage gain about 150" is wrong because CB amplifiers can have high voltage gain, but it is not a fixed characteristic value like 150; the exact gain depends on circuit parameters and load. Hence, option b is the incorrect statement. Option d is correct because CB current gain is less than one. Option a is correct because CB input impedance is low. Option c is incorrect as written if taken strictly, since CB amplifiers are generally preferred for RF/high-frequency use, not audio frequency.
Which of the following flag condition is used for BCD arithmetic operations in microprocessor?
A)
Sign flag
B)
Auxiliary carry flag
C)
Parity flag
D)
Zero flag
In microprocessors (such as the 8085 or 8086), BCD (Binary Coded Decimal) arithmetic requires a way to detect when a carry occurs from the lower nibble (the first 4 bits) to the upper nibble (the next 4 bits).
1. **Auxiliary Carry Flag (AC):** This flag is specifically designed for BCD operations. It is set to 1 if there is a carry from bit 3 to bit 4 during an addition operation. This flag is used by the processor (or the programmer via instructions like DAA - Decimal Adjust Accumulator) to determine if a correction is needed to keep the result in valid BCD format (where each 4-bit group represents a digit from 0-9).
2. **Sign Flag (SF):** This flag indicates whether the result of an operation is positive or negative (based on the Most Significant Bit). It is not used for BCD correction.
3. **Parity Flag (PF):** This flag indicates whether the number of set bits (1s) in the result is even or odd. It is used for error checking in data transmission, not for BCD arithmetic.
4. **Zero Flag (ZF):** This flag is set if the result of an operation is zero. While useful for loops and comparisons, it does not provide the specific nibble-level carry information required for BCD adjustments.
In a zener regulator, the change in load current produces, change in
A)
Zener voltage
B)
Zener current
C)
Both (a) & (b)
D)
None of these
In a Zener voltage regulator circuit, the Zener diode is connected in parallel with the load resistor. The total current flowing from the source through the series resistor (Rs) is given by I_total = (V_in - V_z) / Rs. According to Kirchhoff's Current Law (KCL), this total current splits into two paths: the Zener current (Iz) and the load current (Il). Therefore, I_total = Iz + Il. Since the input voltage (Vin) and the series resistance (Rs) are typically constant, the total current (I_total) remains constant. Consequently, any change in the load current (ΔIl) must be compensated for by an equal and opposite change in the Zener current (ΔIz) to keep the sum constant (ΔIz = -ΔIl). The primary purpose of the Zener regulator is to keep the Zener voltage (Vz) constant despite changes in load current or input voltage, provided the Zener diode remains in its breakdown region. Thus, the change in load current produces a change in the Zener current, not the Zener voltage.
The program counter in a 8085 microprocessor is a 16-bit register, because
A)
Small systems
B)
Large systems
C)
Both large and small systems
D)
Very large systems
The 8085 has a 16-bit program counter because it must hold the address of the next instruction in memory, and the 8085 can address up to 64 KB of memory. A 16-bit register is required to represent all possible memory addresses in such a system. Hence, the correct option is 'Large systems'.
Why others are incorrect:
- a) Small systems: small systems do not require a 16-bit PC specifically; the 16-bit size is chosen to support a larger address space.
- c) Both large and small systems: this is too general and does not explain the need for 16-bit width.
- d) Very large systems: 8085 is an 8-bit microprocessor with a 16-bit address space, suitable for relatively large but not very large systems.
An ‘Assembler’ for a microprocessor is used for
A)
Assembly of processors in a production line
B)
Creation of new programs using different modules
C)
Translation of a program from assembly language to machine language
D)
Translation of a higher level language into English text
In microprocessor architecture, an 'Assembler' is a specialized utility software used to translate source code written in Assembly Language (which uses mnemonics like MOV, ADD, JMP) into Machine Language (binary code consisting of 0s and 1s). Microprocessors can only execute instructions in binary format (machine code); therefore, the assembler acts as the bridge between human-readable symbolic code and the hardware-executable format.
Analysis of other options:
- Option (a) is incorrect because 'assembly' in a production line refers to physical manufacturing, not software translation.
- Option (b) describes the role of a 'Linker', which combines different object modules into a single executable program.
- Option (d) is incorrect because translating higher-level languages (like C++ or Java) into machine code is done by a 'Compiler' or 'Interpreter', and they do not translate code into 'English text'.
Output voltage of a single-phase bridge inverter, fed from a fixed DC source, is varied by
A)
varying the switching frequency
B)
pulse-width modulation
C)
pulse amplitude modulation
D)
All of these
In a single-phase bridge inverter fed from a fixed DC source, the output voltage magnitude and frequency need to be controlled. Since the DC source is fixed, the average output voltage cannot be changed by varying the input. Pulse-Width Modulation (PWM) is the standard technique used to vary the output voltage. By adjusting the duty cycle (the ratio of 'on' time to the total switching period), the effective RMS value of the output voltage can be controlled while maintaining a constant DC input.
Why other options are incorrect:
- (a) Varying the switching frequency primarily affects the harmonic content and the size of the output filter, but it does not directly control the fundamental output voltage magnitude in a standard inverter.
- (c) Pulse Amplitude Modulation (PAM) involves varying the amplitude of the pulses. Since the inverter is fed by a 'fixed DC source', the amplitude of the pulses is constrained by that fixed voltage and cannot be varied.
- (d) Since only PWM is the correct method for voltage control in this context, 'All of these' is incorrect.