← MES Jr. Grade
Past Questions
A 1 kHz audio tone is used to amplitude modulate a 1 MHz carrier. If the modulation index is 0.5 and carrier power is 1 kW, calculate the sideband power and total transmitted power.
Define: Pulse Amplitude modulation, Pulse Width modulation and Pulse Code modulation with suitable graph.
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'.
Explain with figure the working of a snubber Circuit used in SCR subjected to Rs and Cs.
A flipflop is popularly known as:
A)
Operating circuit
B)
Memory circuit
C)
Adder
D)
Amplifier
A flip-flop is a bistable multivibrator, meaning it has two stable states (0 or 1). Because it can maintain its output state indefinitely as long as power is supplied, it is capable of storing one bit of binary information. This fundamental ability to 'remember' a state makes it the basic building block of digital memory, such as registers and RAM.
Why other options are incorrect:
- a. Operating circuit: This is a generic term and does not describe the specific function of a flip-flop.
- c. Adder: An adder is a combinational logic circuit used to perform arithmetic addition; it does not store state.
- d. Amplifier: An amplifier is an analog circuit used to increase the power or amplitude of a signal, which is entirely different from the digital switching/storage function of a flip-flop.
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.
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.
The figure below shows a logic circuit and its incomplete truth table below. Complete the truth table.
Explain with waveform, the working principle of a Single-phase Half wave thyristor Circuit with RL load.
If for a Si npn transistor, the VBE = 0.7 V and VCB = 0.2 V, then the transistor is operating in the:
A)
Cut-off region
B)
Active region
C)
Saturation region
D)
Inverse active mode
To determine the operating region of an npn transistor, we must examine the biasing of the two junctions: the Base-Emitter (BE) junction and the Base-Collector (BC) junction.
1. **Base-Emitter (BE) Junction**: The given $V_{BE} = 0.7\text{ V}$. For a Silicon (Si) transistor, the threshold voltage for forward bias is approximately $0.7\text{ V}$. Since $V_{BE} > 0$, the BE junction is **Forward Biased**.
2. **Base-Collector (BC) Junction**: We are given $V_{CB} = 0.2\text{ V}$. In an npn transistor, the Collector is n-type and the Base is p-type. For the BC junction to be reverse biased, the Collector must be more positive than the Base ($V_{CB} > 0$). Since $V_{CB} = 0.2\text{ V}$ is positive, the BC junction is **Reverse Biased**.
**Operating Region Summary for npn:**
- **Cut-off**: Both junctions reverse biased ($V_{BE} < 0.7\text{ V}$, $V_{CB} > 0$).
- **Active**: BE junction forward biased, BC junction reverse biased ($V_{BE} \approx 0.7\text{ V}$, $V_{CB} > 0$).
- **Saturation**: Both junctions forward biased ($V_{BE} \approx 0.7\text{ V}$, $V_{CB} < 0$ or $V_{BC} > 0$).
- **Inverse Active**: BE junction reverse biased, BC junction forward biased.
Since the BE junction is forward biased and the BC junction is reverse biased, the transistor is operating in the **Active region**.
Which device is most suitable for AC phase control in lighting?
A)
BJT
B)
MOSFET
C)
TRIAC
D)
GTO
Step-by-step reasoning: 1. AC Phase Control requires a device that can conduct current in both directions (bidirectional) to control both the positive and negative half-cycles of the AC mains. 2. A TRIAC (Triode for Alternating Current) is specifically designed as a bidirectional thyristor, making it ideal for AC power control applications like light dimmers and fan speed regulators. 3. BJT (Bipolar Junction Transistor) and MOSFET (Metal-Oxide-Semiconductor Field-Effect Transistor) are primarily DC switches or used in high-frequency switching power supplies; they are unidirectional and would require complex bridge configurations to handle AC phase control. 4. GTO (Gate Turn-Off thyristor) is a high-power unidirectional device used in heavy industrial drives, not suitable for simple domestic lighting control. Therefore, the TRIAC is the most suitable and cost-effective device for this application.
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.
Explain the concept of differential amplifier. Derive expressions for differential gain and common-mode gain.
In 8085, the Program Counter is a:
A)
8-bit register
B)
16-bit register
C)
32-bit register
D)
Special flag register
In the Intel 8085 microprocessor, the Program Counter (PC) is a dedicated 16-bit register. Its primary function is to keep track of the memory address of the next instruction to be executed. Since the 8085 has a 16-bit address bus, it can address up to 64 KB (2^16 bytes) of memory; therefore, the Program Counter must be 16 bits wide to hold a full memory address.
Analysis of options:
- Option (a) is incorrect because an 8-bit register (like the Accumulator or B, C, D, E, H, L registers) cannot hold a full memory address.
- Option (c) is incorrect because the 8085 is an 8-bit microprocessor with a 16-bit address bus; 32-bit registers are found in much more advanced architectures.
- Option (d) is incorrect because the Program Counter is a pointer to memory, not a status/flag register (the 8085 has a separate 8-bit Flag Register for status indicators).
The bandwidth of an AM signal is:
A)
Equal to the carrier frequency
B)
Twice the highest modulating frequency
C)
Sum of carrier and modulating frequency
D)
Half the modulating frequency
In Amplitude Modulation (AM), the modulated signal consists of a carrier frequency (fc) and two sidebands: the Upper Sideband (USB) and the Lower Sideband (LSB). If the modulating signal (the message signal) has a maximum frequency of fm, the resulting AM spectrum contains frequencies at fc, (fc + fm), and (fc - fm). The bandwidth (BW) is defined as the difference between the highest and lowest frequency components of the signal. Therefore, BW = (fc + fm) - (fc - fm) = 2fm. This means the bandwidth of a standard AM signal is exactly twice the highest frequency present in the modulating signal.
Analysis of other options:
- Option (a) is incorrect because the carrier frequency is the center frequency, not the width of the spectrum.
- Option (c) is incorrect as the sum of frequencies does not represent the spectral width.
- Option (d) is incorrect as it underestimates the required bandwidth by a factor of four.
Which flip-flop is commonly used as a frequency divider?
A)
SR
B)
JK
C)
D
D)
T
A frequency divider is a circuit that takes an input signal and produces an output signal with a lower frequency. The T (Toggle) flip-flop is specifically designed for this purpose. When the T input is held high (logic 1), the flip-flop toggles its output state on every active clock edge. This means the output (Q) completes one full cycle (low to high and high to low) for every two clock pulses. Consequently, the output frequency is exactly half of the input clock frequency (f_out = f_in / 2). While a JK flip-flop can act as a T flip-flop if both J and K inputs are tied to logic 1, the T flip-flop is the fundamental architectural answer for frequency division. SR and D flip-flops do not inherently toggle their state based on a single clock pulse without external feedback circuitry.
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'.
A good inverter should have:
A)
Non sinusoidal output
B)
Low Gain
C)
High operating cost
D)
Minimum switching loss
An inverter is a power electronic circuit that converts DC power to AC power. To evaluate the quality of an inverter, we look for efficiency and power quality.
1. **Switching Loss (Option d):** Inverters use semiconductor switches (like MOSFETs or IGBTs). Every time a switch toggles between ON and OFF states, energy is lost as heat due to the overlap of voltage and current. Minimizing these switching losses is critical to increasing the overall efficiency of the inverter and reducing the need for bulky cooling systems. Therefore, 'Minimum switching loss' is a primary goal for a good inverter.
2. **Output Waveform (Option a):** A good inverter should strive for a **sinusoidal** output (Pure Sine Wave) to ensure compatibility with most electrical loads and to minimize Total Harmonic Distortion (THD). A non-sinusoidal output is generally undesirable.
3. **Gain (Option b):** Gain refers to the ability of the inverter to maintain the desired output voltage relative to the input. Low gain is not a desirable characteristic; rather, stable and controllable voltage regulation is required.
4. **Operating Cost (Option c):** High operating costs are always undesirable in engineering design. A good inverter should be cost-effective and energy-efficient.
Maximum directly addressable memory in 8085 is:
A)
32 KB
B)
64 KB
C)
128 KB
D)
1 MB
To determine the maximum directly addressable memory of a microprocessor, we look at the size of its address bus.
1. The Intel 8085 is an 8-bit microprocessor, but it has a 16-bit address bus (A0 to A15).
2. The number of unique memory locations that can be addressed is calculated as 2 raised to the power of the number of address lines: 2^16.
3. 2^16 = 65,536 bytes.
4. Since 1 KB (Kilobyte) is defined as 1,024 bytes, we divide 65,536 by 1,024: 65,536 / 1,024 = 64 KB.
Therefore, the 8085 can directly address up to 64 KB of memory.
Analysis of other options:
- 32 KB: This is half of the 8085's capacity.
- 128 KB: This would require a 17-bit address bus (2^17).
- 1 MB: This is the addressable memory of the 8086 microprocessor, which has a 20-bit address bus (2^20).
The diode with a faster operation is:
A)
Vacuum diode
B)
Zener Diode
C)
Schottky diode
D)
Normal p-n diode
The speed of a diode is primarily determined by its reverse recovery time (trr), which is the time it takes for the diode to switch from a conducting state to a non-conducting state.
1. **Normal p-n diode**: These are bipolar devices. When forward-biased, minority carriers are injected into the opposite regions. When switched to reverse bias, these stored charges must be removed or recombine before the diode can block current, leading to a significant reverse recovery time.
2. **Schottky diode**: This is a majority-carrier device formed by a metal-semiconductor junction. Because it does not rely on minority carrier injection, there is virtually no stored charge to be removed during switching. This results in an extremely fast switching speed (nanoseconds), making it ideal for high-frequency applications.
3. **Zener Diode**: These are specialized p-n junction diodes designed to operate in the breakdown region. Their switching speed is similar to or slower than normal p-n diodes depending on the application.
4. **Vacuum diode**: While they operate via thermionic emission and can be fast, they are bulky, inefficient, and generally slower or less practical than modern semiconductor Schottky diodes in high-speed switching circuits.
Therefore, the Schottky diode is the fastest among the given options.
The universal gates are:
A)
AND, OR, NOT
B)
NAND and NOR
C)
X-OR and X-NOR
D)
All of the above
In digital electronics, a 'universal gate' is a gate that can be used to implement any other Boolean function without the need for any other gate type.
1. **NAND Gate**: By combining NAND gates, one can create NOT, AND, OR, and XOR functions. For example, a NAND gate with tied inputs acts as a NOT gate.
2. **NOR Gate**: Similarly, by combining NOR gates, one can implement all basic logic functions (NOT, OR, AND, etc.).
Evaluating the options:
- Option (a): AND, OR, and NOT are 'Basic Gates'. While they can implement any function together, no single one of them can implement all others on its own.
- Option (b): NAND and NOR are the only two gates that possess the property of universality.
- Option (c): XOR and XNOR are 'Special Purpose Gates' used primarily for parity checking and arithmetic operations; they are not universal.
- Option (d): Incorrect because only option (b) satisfies the definition.
In a thyristor at which layer gate terminal is connected and why? Explain dv/dt triggering of the thyristor.
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)
The ripple frequency of single-phase full-wave rectifier:
A)
F
B)
2f
C)
3f
D)
4f
1. Definition of Ripple Frequency: The ripple frequency is the frequency of the AC component remaining in the output of a rectifier after the rectification process.
2. Single-Phase Half-Wave Rectifier: In a half-wave rectifier, only one half-cycle of the input sine wave is passed to the load. Therefore, the output repeats every one full cycle of the input. The ripple frequency is equal to the supply frequency (f).
3. Single-Phase Full-Wave Rectifier: In a full-wave rectifier (whether center-tapped or bridge), both the positive and negative half-cycles of the input AC signal are converted into positive DC pulses. This means for every one cycle of the input frequency (f), there are two pulses (peaks) in the output voltage.
4. Calculation: Since the output waveform repeats twice within one period of the input supply, the frequency of the output ripple is twice the input frequency.
Ripple Frequency = 2 × Supply Frequency = 2f.
5. Analysis of Options:
- Option (a) 'f' is the ripple frequency for a half-wave rectifier.
- Option (b) '2f' is the correct ripple frequency for a full-wave rectifier.
- Options (c) and (d) are incorrect as they do not correspond to standard single-phase rectification patterns.
In renewable energy integration, microprocessors help in:
A)
Grid synchronization and control
B)
Manual operation
C)
Power loss increase
D)
None
In the context of renewable energy integration (such as solar and wind), microprocessors are essential for managing the interface between the energy source and the electrical grid.
1. Why 'a' is correct: Renewable energy sources typically produce DC power (solar) or variable AC power (wind). Microprocessors are used in smart inverters and controllers to perform 'Grid Synchronization,' which ensures that the voltage, frequency, and phase angle of the generated power exactly match those of the utility grid before connection. They also handle real-time control loops for Maximum Power Point Tracking (MPPT) and stability management.
2. Why 'b' is incorrect: The goal of integrating microprocessors is to automate complex processes. Manual operation is inefficient and dangerous for high-speed synchronization tasks.
3. Why 'c' is incorrect: Microprocessors are implemented to optimize efficiency and reduce power losses through precise switching and control; they are not used to increase power loss.
4. Why 'd' is incorrect: Since option 'a' is a primary function of microprocessors in this field, 'None' is incorrect.