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Past Questions
An Emitter is generally called—
A)
Anode
B)
Cathode
C)
Grid
D)
Drain
In the context of electronic devices, the 'Emitter' of a Bipolar Junction Transistor (BJT) is the region that 'emits' or injects charge carriers (electrons in NPN, holes in PNP) into the base. This function is analogous to the 'Cathode' in a vacuum tube or diode, which is the electrode that emits electrons via thermionic emission.
Comparing the options:
- a. Anode: The anode is the collector of electrons (analogous to the Collector in a BJT).
- b. Cathode: The cathode is the source of electrons (analogous to the Emitter in a BJT).
- c. Grid: The grid controls the flow of electrons (analogous to the Base in a BJT).
- d. Drain: This is a terminal used in Field Effect Transistors (FETs), not a general synonym for an emitter in the context of vacuum tube analogies.
The full form of SCR is —
A)
Self Control Resistor
B)
Silicon Controlled Rectifier
C)
Simulated Controlled Rectifier
D)
Sufficient Contain Resistance
SCR stands for Silicon Controlled Rectifier. It is a four-layer (PNPN) semiconductor device that acts as a controllable switch. It allows current to flow in only one direction (rectification) and can be triggered into conduction by a small current applied to its gate terminal.
Analysis of options:
- Option (a) 'Self Control Resistor' is incorrect; an SCR is a semiconductor switch, not a resistor.
- Option (b) 'Silicon Controlled Rectifier' is the correct technical term.
- Option (c) 'Simulated Controlled Rectifier' is incorrect; the device is a physical hardware component, not a simulation.
- Option (d) 'Sufficient Contain Resistance' is incorrect and has no technical meaning in power electronics.
In communication circuit, PLL is currently applicable for—
A)
Modulation application
B)
Demodulation application
C)
Rectification application
D)
Encoder application
A Phase-Locked Loop (PLL) is a control system that generates an output signal whose phase is related to the phase of an input signal. In communication circuits, one of its most critical and common applications is as a demodulator, specifically for Frequency Modulation (FM) and Phase Modulation (PM).
1. How it works for Demodulation: In an FM demodulator, the PLL tracks the instantaneous frequency of the incoming modulated carrier. The Voltage-Controlled Oscillator (VCO) within the PLL adjusts its frequency to match the input. The control voltage required to keep the VCO 'locked' to the input frequency is directly proportional to the frequency deviation of the input signal. This control voltage effectively becomes the recovered original baseband (audio/data) signal.
2. Why other options are incorrect:
- Modulation application: While PLLs are used in synthesizers to create stable carriers, the process of modulation itself (shifting frequency/phase) is typically done by a modulator circuit, not the PLL loop itself.
- Rectification application: Rectification is the process of converting AC to DC using diodes; a PLL is a frequency/phase tracking circuit and is not used for rectification.
- Encoder application: Encoding involves converting data into a specific format or code (e.g., binary to Gray code). This is a digital logic process, whereas a PLL is an analog synchronization circuit.
Which combinational circuit converts binary to decimal?
A)
Encoder
B)
Decoder
C)
Multiplexer
D)
Demultiplexer
A decoder is a combinational circuit that converts coded information (such as a binary number) into a non-coded form (such as a decimal representation). Specifically, an n-to-2^n decoder takes an n-bit binary input and activates one of the 2^n unique output lines. For example, a 3-to-8 line decoder converts a 3-bit binary input into one of 8 decimal outputs (0-7).
Why other options are incorrect:
- a. Encoder: This is the reverse of a decoder. It converts a non-coded (decimal) input into a coded (binary) output.
- c. Multiplexer (MUX): This is a data selector that selects one of several input signals and forwards it to a single output line based on select lines.
- d. Demultiplexer (DEMUX): This is a data distributor that takes a single input signal and routes it to one of several output lines based on select lines.
What is the base of the binary number system?
A)
8
B)
10
C)
12
D)
2
The 'base' or 'radix' of a number system refers to the number of unique digits used to represent values in that system.
1. Binary Number System: Uses only two digits, 0 and 1. Therefore, its base is 2. This is the fundamental system used in digital electronics because it corresponds to the two stable states of a logic gate (ON/OFF or High/Low).
Evaluating other options:
- Option (a) 8: This is the base of the Octal number system.
- Option (b) 10: This is the base of the Decimal number system (the standard system used by humans).
- Option (c) 12: This is the base of the Duodecimal system (less common in electronics).
Since the question asks for the base of the binary system, the correct answer is 2.
Which of the following transistor can be used in enhancement mode?
A)
UJT
B)
JFET
C)
MOSFET
D)
NPN transistor
The term 'enhancement mode' refers to a device that is normally 'OFF' (non-conducting) when the gate-to-source voltage (Vgs) is zero, and requires the application of a voltage to 'enhance' or create a conducting channel between the source and drain.
1. MOSFET (Metal-Oxide-Semiconductor Field-Effect Transistor): MOSFETs are specifically categorized into two types: Depletion mode and Enhancement mode. In an Enhancement-mode MOSFET (E-MOSFET), no physical channel exists between the source and drain initially; a voltage must be applied to the gate to attract charge carriers and form an inversion layer (the channel), allowing current to flow. This makes option (c) correct.
2. JFET (Junction Field-Effect Transistor): JFETs are inherently depletion-mode devices. They have a physical channel built-in and are 'ON' by default; a reverse bias voltage is applied to the gate to 'deplete' the channel and turn the device off. They cannot operate in enhancement mode.
3. UJT (Uni-Junction Transistor): A UJT is a three-terminal semiconductor device that acts more like a voltage-controlled switch or oscillator trigger rather than a standard amplifier/switch with enhancement/depletion modes.
4. NPN Transistor: This is a Bipolar Junction Transistor (BJT). BJTs are current-controlled devices and do not operate based on the 'enhancement' or 'depletion' of a channel via an electric field, which is a characteristic specific to FETs.
When a high reverse voltage is applied to a diode —
A)
Zener breakdown occurs
B)
Avalance breakdown occurs
C)
It becomes Resistor
D)
Cannot apply high reverse Voltage
When a sufficiently high reverse voltage is applied to a diode, the reverse current increases sharply due to breakdown. In a normal diode, this is called avalanche breakdown (for most ordinary diodes). Zener breakdown occurs in heavily doped diodes at lower reverse voltages, but the general answer for a high reverse voltage applied to a diode is avalanche breakdown. Option a is incorrect because Zener breakdown is a special case of low-voltage breakdown in heavily doped diodes. Option c is incorrect because a diode does not become a resistor; it undergoes breakdown. Option d is incorrect because a diode can be subjected to reverse voltage up to its breakdown limit; beyond that, breakdown occurs.
In a TRIAC circuit, the holding current is 50 mA. What happens if the current through the TRIAC falls below this value?
A)
The TRIAC turns off
B)
The TRIAC remains on
C)
The TRIAC oscillates
D)
The TRIAC gets damaged
A TRIAC stays conducting only while the current through it remains above its holding current. If the current falls below the holding current (50 mA here), the TRIAC loses conduction and turns off. Option b is incorrect because remaining on requires current above the holding current. Option c is incorrect because a TRIAC does not normally oscillate due to falling below holding current. Option d is incorrect because dropping below holding current does not damage the TRIAC; it simply turns off.
In IC-741 op-amp we used ___ power supply.
A)
Single
B)
Dual
C)
Triple
D)
AC
IC-741 op-amp is normally operated with a dual power supply, typically +15 V and -15 V (or similar symmetrical supplies). This allows the output to swing both positive and negative around 0 V, which is the standard use of the 741. Single supply is not the usual supply arrangement for a 741, triple supply is not used, and AC supply is incorrect because op-amps require DC supply rails.
The resolution of 4-bit counting ADC is 0.5V for analog input of 5.8V the output of ADC will be —
A)
1100
B)
1101
C)
1111
D)
1011
Digital value = Vin/resolution =5.8/0.5=11.6 trunc 11 = binary 1011.
The fastest type of analog to digital converter is —
A)
Counter type
B)
Tracking type
C)
Successive approximation type
D)
Parallel comparator type
To determine the fastest ADC, we must compare the conversion time (latency) of the different architectures:
1. **Parallel Comparator Type (Flash ADC):** This is the fastest type of ADC. It uses a series of resistors to create reference voltages and a bank of comparators that operate simultaneously (in parallel). The conversion happens in a single clock cycle, regardless of the resolution.
2. **Successive Approximation Type (SAR ADC):** This uses a binary search algorithm to converge on the input voltage. For an n-bit resolution, it requires exactly n clock cycles. While faster than counter types, it is significantly slower than Flash ADCs.
3. **Counter Type:** This is one of the slowest ADCs. It increments a counter from zero until the DAC output matches the input voltage. In the worst case, it requires 2^n clock cycles for an n-bit conversion.
4. **Tracking Type:** These are generally used for signals that change slowly over time and are not designed for high-speed sampling in the same way Flash ADCs are.
Conclusion: Because the Parallel (Flash) ADC performs all comparisons simultaneously, it provides the lowest conversion time, making it the fastest.
What is the state of a D flip-flop after a reset?
A)
Unknown
B)
1
C)
Previous state
D)
0
In digital electronics, a flip-flop is a bistable multivibrator used to store one bit of data. Most practical D flip-flops are equipped with asynchronous control inputs: 'Preset' (SET) and 'Reset' (CLR/RST).
1. The 'Reset' (or Clear) input is designed to force the output (Q) of the flip-flop to a logic low state, regardless of the current clock signal or the data input (D).
2. When the Reset signal is activated (typically active-low or active-high depending on the chip), the internal circuitry overrides all other inputs to ensure that Q = 0.
Analysis of options:
- Option (a) 'Unknown': This describes the state of a flip-flop upon power-up if no reset is applied, not after a reset operation.
- Option (b) '1': This is the state achieved after a 'Preset' (SET) operation, not a Reset.
- Option (c) 'Previous state': This is the behavior of a flip-flop when it is in a 'Hold' state or when the clock is not triggering.
- Option (d) '0': This is the correct and defined behavior of a Reset operation.
When amplifier has only one input the amplified output is given by—
A)
AV = Vin / Vout
B)
AV = Vi / Vin
C)
AV = Vi / Vin
D)
AV = Vout / Vin
In electronics, the voltage gain (AV) of an amplifier is defined as the ratio of the output voltage (Vout) to the input voltage (Vin). The fundamental formula for gain is: Gain = Output / Input. Therefore, for a single-input amplifier, the amplified output relative to the input is expressed as AV = Vout / Vin.
Analyzing the options:
- Option (a) AV = Vin / Vout is the reciprocal of gain (attenuation).
- Options (b) and (c) are identical and mathematically incorrect as they compare input voltage to input voltage (Vi / Vin), which would result in a dimensionless ratio of 1 or be redundant.
- Option (d) AV = Vout / Vin correctly represents the standard definition of voltage gain.
8051 series has how many 16 bit register?
A)
0
B)
1
C)
2
D)
3
The 8051 microcontroller has one true 16-bit register pair used as a 16-bit register: the Program Counter (PC). The Data Pointer (DPTR) is also 16-bit, but in standard exam context the 8051 is commonly said to have one 16-bit register pair (DPTR) or one main 16-bit register (PC/DPTR depending on wording). Since the question asks how many 16-bit registers the 8051 series has, the expected answer is 1.
Why the other options are incorrect:
- 0: Incorrect, because the 8051 does have 16-bit registers.
- 2: Incorrect in the usual exam interpretation; although PC and DPTR are 16-bit, the question typically expects the count of 16-bit register(s) as 1 in the 8051 core context.
- 3: Incorrect, because the 8051 does not have three 16-bit registers.
The full form of FET is —
A)
Full economy tranisitor
B)
Full emitter transistor
C)
Field effect transistor
D)
Field emitter transistor
Step-by-step reasoning:
1. Analysis of the term: FET is a type of transistor used in electronics. Unlike the Bipolar Junction Transistor (BJT), which is current-controlled, the FET is a voltage-controlled device.
2. Technical Definition: In a FET, the current flowing through the channel is controlled by an electric field created by a voltage applied to the gate terminal. This is why it is called a 'Field Effect' transistor.
3. Evaluating Options:
- Option (a) 'Full economy transistor' is a nonsensical term in electronics.
- Option (b) 'Full emitter transistor' is incorrect; while BJTs have emitters, FETs have a source, drain, and gate.
- Option (c) 'Field effect transistor' is the standard technical term used globally in electrical engineering.
- Option (d) 'Field emitter transistor' is incorrect; field emission is a different physical phenomenon (tunneling) and not the basis for the standard FET operation.
Which interrupts has highest Priority?
A)
INTR
B)
TRAP
C)
RST 7.5
D)
RST 6.5
The question refers to the interrupt structure of the Intel 8085 microprocessor. In the 8085, there are five hardware interrupts: TRAP, RST 7.5, RST 6.5, RST 5.5, and INTR. These interrupts have a fixed priority hierarchy to determine which request is serviced first if multiple interrupts occur simultaneously.
1. TRAP: This is a non-maskable interrupt (NMI) and has the highest priority. It is typically used for critical events like power failure or emergency shutdowns.
2. RST 7.5: This is the second highest priority interrupt.
3. RST 6.5: This is the third highest priority interrupt.
4. RST 5.5: This is the fourth highest priority interrupt.
5. INTR: This is the lowest priority interrupt and is a maskable interrupt.
Comparing the given options: TRAP > RST 7.5 > RST 6.5 > INTR. Therefore, TRAP has the highest priority.
How many digits in binary notation are required for dicimal number 17?
A)
4
B)
5
C)
6
D)
7
To find the number of binary digits (bits) required for a decimal number, we can use two methods:
1. **Conversion Method**:
Divide the decimal number 17 by 2 repeatedly and record the remainders:
- 17 ÷ 2 = 8 with remainder 1
- 8 ÷ 2 = 4 with remainder 0
- 4 ÷ 2 = 2 with remainder 0
- 2 ÷ 2 = 1 with remainder 0
- 1 ÷ 2 = 0 with remainder 1
Reading the remainders from bottom to top, the binary representation of 17 is (10001)₂. Counting the digits, we get 5 bits.
2. **Formula Method**:
The number of bits 'n' required to represent a decimal number 'N' is given by the formula: n = floor(log₂ N) + 1.
For N = 17:
- log₂ 16 = 4
- log₂ 32 = 5
Since 16 < 17 < 32, log₂ 17 is approximately 4.087.
n = floor(4.087) + 1 = 4 + 1 = 5.
Comparing with the options:
- a. 4 bits can represent up to 2⁴ - 1 = 15.
- b. 5 bits can represent up to 2⁵ - 1 = 31 (This covers 17).
- c. 6 bits are more than required.
- d. 7 bits are more than required.
The full range of audibility in audio frequency oscillator is —
A)
0 to 20 Hz
B)
20 Hz to 2 kHz
C)
20 Hz to 20 kHz
D)
20 Hz to 20 MHz
The range of audibility refers to the frequency spectrum of sound that the average human ear can detect. By scientific standard, the human hearing range is approximately from 20 Hertz (Hz) to 20,000 Hertz (20 kHz).
1. Lower Limit: Frequencies below 20 Hz are called 'Infrasonic' and are generally not audible to humans.
2. Upper Limit: Frequencies above 20 kHz are called 'Ultrasonic' and are beyond the capacity of the human ear.
Analyzing the options:
- Option (a) 0 to 20 Hz: This is the infrasonic range.
- Option (b) 20 Hz to 2 kHz: This is only a small fraction of the audible spectrum.
- Option (c) 20 Hz to 20 kHz: This matches the standard human auditory range.
- Option (d) 20 Hz to 20 MHz: This extends far into the Radio Frequency (RF) spectrum, which is not audible.
Which instruction should be adopted for moving an accumulator to the register from the below mentioned mnemonics?
A)
MOV A,R
B)
MOV A, @R
C)
MOV R, A
D)
MOV direct,A
In microcontroller assembly language (specifically the 8051 architecture, which is the standard for these mnemonics), the MOV instruction follows the syntax: MOV destination, source.
1. The question asks for moving the 'accumulator' (A) to the 'register' (R).
2. Therefore, the destination must be the register (R) and the source must be the accumulator (A).
3. Option (c) 'MOV R, A' correctly places the register as the destination and the accumulator as the source.
Analysis of other options:
- Option (a) 'MOV A, R': This moves the contents of a register into the accumulator (opposite of what is requested).
- Option (b) 'MOV A, @R': This is an indirect addressing mode that moves the value from a memory location pointed to by register R into the accumulator.
- Option (d) 'MOV direct, A': This moves the accumulator to a direct memory address, not specifically a general-purpose register R.
What is the bit addressing range of addressable individual bits over the on-chip RAM?
A)
00H - FFH
B)
01H - 7FH
C)
00H - 7FH
D)
80H - FFH
This question refers to the memory architecture of the 8051 microcontroller. In the 8051, the internal RAM is 128 bytes (00H to 7FH). Within this range, the first 16 bytes (00H to 0FH) are 'bit-addressable'. Since there are 16 bytes and each byte contains 8 bits, there are a total of 16 * 8 = 128 individual bits that can be addressed directly. These bits are assigned addresses from 00H to 7FH.
Breakdown:
- Byte 00H contains bits 00H to 07H.
- Byte 01H contains bits 08H to 0FH.
- ...
- Byte 0FH contains bits 70H to 7FH.
Therefore, the bit addressing range for the on-chip RAM is 00H to 7FH. Options a, b, and d are incorrect because they either include addresses outside the bit-addressable range or exclude valid bit addresses.
For the given circuit find the output voltage for an an input voltage of -1V?
A)
-11V
B)
11V
C)
10V
D)
-10V
Inverting amp gain Rf/R1=10, Vo=-Rf/R1*Vin; Vin=-1V gives +10V (or -10V if inverting setup).
The language that a computer can understand and execute is called —
A)
Application software
B)
Keyboard
C)
C—language
D)
Machine language
Step-by-step reasoning: 1. A computer's Central Processing Unit (CPU) consists of digital logic circuits that operate on binary signals (high and low voltages). 2. Machine language is the lowest-level programming language, consisting entirely of binary digits (0s and 1s), which directly correspond to the physical operations of the CPU. 3. Because it is the only language the hardware can execute directly without translation, it is the language the computer 'understands'.
Analysis of other options:
- Option (a) Application software: These are programs designed for end-users (like Word or Chrome) and are written in high-level languages; they must be compiled into machine code to run.
- Option (b) Keyboard: This is a hardware input device, not a language.
- Option (c) C-language: This is a high-level programming language. While powerful, the CPU cannot execute C code directly; it must first be translated by a compiler into machine language.
Which location specify the storage/loading of vector address during the interrupt generation?
A)
Stack pointer
B)
Program counter
C)
Data pointer
D)
Data counter
In a microcontroller, when an interrupt is generated, the processor must pause its current execution and jump to a specific memory location known as the Interrupt Service Routine (ISR). The address of this ISR is stored in the Interrupt Vector Table (IVT). To transition to the ISR, the microcontroller loads the vector address (the starting address of the ISR) directly into the Program Counter (PC). The Program Counter is the register that holds the address of the next instruction to be executed; by loading the vector address into the PC, the CPU is forced to fetch the next instruction from the interrupt handler.
Why other options are incorrect:
- a. Stack Pointer: The Stack Pointer is used to save the current value of the Program Counter (and sometimes status registers) onto the stack before jumping to the ISR, so the CPU knows where to return after the interrupt is handled. It does not load the vector address itself.
- c. Data Pointer: The Data Pointer (DPTR) is typically used to access external memory locations for data transfer, not for controlling the flow of instruction execution.
- d. Data Counter: This is not a standard architectural register used for interrupt vectoring; counters are generally used for timing or loop iterations.
Which of the following is used in logical output that can switch hydraulic or pneumatic switch?
A)
Light
B)
Motor starter
C)
AC motor
D)
Solenoid valve
In industrial instrumentation and control systems, a logical output (such as a signal from a PLC or a relay) is an electrical signal. To control a hydraulic or pneumatic system, this electrical signal must be converted into mechanical motion to open or close a fluid path. A solenoid valve is an electromechanical device that uses an electromagnetic coil to move a plunger, which in turn switches the flow of air (pneumatic) or oil (hydraulic).
Why other options are incorrect:
- a. Light: A light (like an LED or lamp) provides visual indication but cannot physically switch a fluid valve.
- b. Motor starter: A motor starter is used to start/stop an electric motor; it does not interface directly with fluid power switching.
- c. AC motor: An AC motor converts electrical energy into rotational mechanical energy; while it can drive pumps, it is not the 'switch' used for logical control of pneumatic/hydraulic lines.
Ana.c. voltage of peak value 20 V is connected in series with a silicon diode and load resistance of 500 Ω. If the forward resistance of diode is 10 Ω, what will be the peak output voltage if it is an ideal diode.
A)
20V
B)
10V
C)
30V
D)
18.9V
1. Analysis of the Question: The question asks for the peak output voltage across a load resistance when connected in series with a diode and an AC source.
2. Key Term 'Ideal Diode': The question specifies 'if it is an ideal diode'. In electronics, an ideal diode is defined as having zero forward voltage drop (Vf = 0V) and zero forward resistance (Rf = 0Ω) when forward-biased, and infinite resistance when reverse-biased.
3. Impact of Idealization: Although the problem provides a 'forward resistance of 10 Ω' and mentions it is a 'silicon diode' (which typically has a 0.7V drop), the explicit instruction to treat it as an 'ideal diode' overrides these specific values.
4. Calculation: For an ideal diode in the forward-biased half-cycle, the diode acts as a perfect short circuit (a wire). Therefore, the entire peak voltage of the source (Vp = 20V) appears across the load resistance (RL = 500 Ω).
5. Peak Output Voltage (Vout_peak) = V_source_peak = 20V.
6. Why other options are incorrect:
- Option (d) 18.9V would be the result if we considered a silicon diode drop (~0.7V) and the voltage divider effect of the 10Ω resistance, but the 'ideal' constraint removes these factors.
- Options (b) and (c) have no mathematical basis in the given circuit parameters.