Preparing for Basic Electronics CSE 1st Sem | Most Important PYQs + New Syllabus Model Paper 2024?
This model paper includes PYQs, most important questions, and expected long answers—designed as per the new syllabus and latest exam trends.
Objective Type Questions
Q1. Which of the following is a unidirectional device?
A. Resistor
B. PN Junction Diode
C. Capacitor
D. Inductor
Answer:
B. PN Junction Diode
Explanation: A unidirectional device lets current flow in one direction only. A diode allows current from anode to cathode but blocks the reverse, unlike resistors, capacitors, or inductors, which work in both directions.
Q2. The output of a Half-Wave Rectifier is:
A. Pure DC
B. Pulsating AC
C. Pulsating DC
D. Sine wave
Answer:
C. Pulsating DC
Explanation: A half-wave rectifier passes only half of an AC signal (positive or negative). The output is DC because it’s one-directional, but it varies (pulsates), not steady like pure DC.
Q3. What is the ideal value of CMRR for an Op-Amp?
A. 0
B. 1
C. ∞
D. 100
Answer:
C. ∞
Explanation: CMRR measures how well an op-amp ignores unwanted signals. Ideally, it rejects all common signals, making CMRR infinite (∞), meaning perfect performance.
Q4. Which of the following is not a logic gate?
A. AND
B. OR
C. NOT
D. DIODE
Answer:
D. DIODE
Explanation: Logic gates (AND, OR, NOT) process digital signals. A diode is an electronic component for current control, not a logic gate.
Q5. Which type of MOSFET can operate in both depletion and enhancement mode?
A. N-channel
B. P-channel
C. Depletion-type
D. Enhancement-type
Answer:
C. Depletion-type
Explanation: Depletion-type MOSFETs have a channel that conducts without voltage and can be enhanced or depleted by gate voltage. Enhancement-type needs voltage to start conducting.
Q6. The Slew Rate of an Op-Amp refers to:
A. Output voltage per unit time
B. Input current per volt
C. Frequency gain
D. Input resistance
Answer:
A. Output voltage per unit time
Explanation: Slew rate tells how fast an op-amp’s output voltage can change, measured in volts per microsecond. It’s about speed, not current, gain, or resistance.
Q7. Which device has three terminals: Gate, Drain, Source?
A. BJT
B. SCR
C. JFET
D. MOSFET
Answer:
D. MOSFET
Explanation: A MOSFET has Gate, Drain, and Source terminals. BJT has Base, Collector, Emitter; SCR has Anode, Cathode, Gate; JFET also has Gate, Drain, Source, but MOSFET is more common.
Q8. A clipping circuit is used to:
A. Amplify signal
B. Remove noise
C. Remove a portion of input signal
D. Rectify signal
Answer:
C. Remove a portion of input signal
Explanation: A clipping circuit cuts off parts of a signal above or below a set level, removing those portions. It doesn’t amplify, filter noise, or rectify.
Q9. Which logic gate output is LOW only when all inputs are HIGH?
A. OR
B. NAND
C. AND
D. NOR
Answer:
B. NAND
Explanation: A NAND gate outputs LOW only when all inputs are HIGH. OR and AND don’t match this, and NOR is LOW when any input is HIGH.
Q10. Binary equivalent of decimal 25 is:
A. 11001
B. 10101
C. 11100
D. 10011
Answer:
A. 11001
Explanation: Divide 25 by 2 repeatedly:
- 25 ÷ 2 = 12, remainder 1
- 12 ÷ 2 = 6, remainder 0
- 6 ÷ 2 = 3, remainder 0
- 3 ÷ 2 = 1, remainder 1
- 1 ÷ 2 = 0, remainder 1
Read remainders bottom-up: 11001 = 25 (16 + 8 + 1).
OR
Very Short Answer Type Questions :
Q1. Draw the VI characteristics of a PN junction diode.
Answer: The VI characteristics graph shows the relationship between voltage (V) and current (I) in a PN junction diode.
- In forward bias, current increases rapidly after a threshold voltage (~0.7V for silicon).
- In reverse bias, a small leakage current flows until breakdown.
Q2. What is the difference between forward and reverse biasing?
Answer:
Forward bias: Positive terminal is connected to P-side, diode conducts.
Reverse bias: Positive terminal is connected to N-side, diode does not conduct (except leakage current).
Q3. Define depletion region and explain its significance.
Answer: The depletion region is the area near the PN junction where free electrons and holes recombine, creating a non-conductive zone. It acts like a barrier and controls the flow of current in the diode.
Q4. What is the use of a filter capacitor in rectifiers?
Answer: A filter capacitor smooths the output of a rectifier by reducing the ripple in the DC voltage. It stores charge and releases it during voltage dips.
Q5. Define Slew Rate of an Op-Amp.
Answer: Slew Rate is the maximum rate at which the output voltage of an Op-Amp can change, expressed in V/µs.
For example: 1 V/µs means the voltage can change by 1V in 1 microsecond.
Q6. What are universal logic gates? Give examples.
Answer: Universal logic gates are gates that can be used to build any other logic gate.
Examples: NAND and NOR gates.
Q7. Write any two differences between BJT and FET.
Answer:
BJT is current-controlled, FET is voltage-controlled.
BJT has low input impedance, FET has high input impedance.
Q8. What is a voltage regulator?
Answer: A voltage regulator is a circuit/device that maintains constant output voltage regardless of input voltage or load changes.
Q9. What is 1’s complement? Where is it used?
Answer: 1’s complement of a binary number means inverting all bits (0→1, 1→0).
It is used in binary subtraction and error detection.
Q10. Mention any two applications of LED.
Answer:
Used in indicator lights and display panels.
Used in remote controls and digital clocks.
Long Answer Type Questions:
Q1. Explain the working of a Full Wave Bridge Rectifier with filter and draw the input/output waveforms.
Answer: A Full Wave Bridge Rectifier is a circuit that converts an alternating current (AC) input into a direct current (DC) output using four diodes arranged in a bridge configuration. A filter capacitor is used to smooth the pulsating DC into a more stable DC output.
Working Principle:
- AC Input:
The input to the rectifier is a sinusoidal AC voltage that alternates between positive and negative cycles. - Bridge Rectification:
- Positive Half Cycle:
During this cycle, diodes D1 and D4 conduct, allowing current to pass through the load resistor in one direction. - Negative Half Cycle:
During this cycle, diodes D2 and D3 conduct. Current again flows through the load in the same direction as in the positive half cycle.
- Positive Half Cycle:
- Filtering Using Capacitor:
A capacitor is connected in parallel with the load resistor. It charges when the voltage rises and discharges when the voltage drops. This process smooths the output voltage, reducing the ripple, and gives a more constant DC output.
Waveforms:
- AC Input:
A sine wave oscillating above and below the zero voltage line. - Output Without Filter:
A pulsating DC waveform containing two positive pulses for each AC cycle. - Output With Filter:
A nearly steady DC voltage with small ripple (wave-like fluctuations) due to charging and discharging of the capacitor.
Basic Electronics CSE 1st Sem | Most Important PYQs + New Syllabus Model Paper 2024
Q2. Draw and explain the circuit diagram and working of a Zener diode as a voltage regulator.
Answer: A Zener diode is used as a voltage regulator to provide a constant DC output voltage, even if the input voltage or load current changes.
Circuit Diagram:
- The Zener diode is connected in reverse bias (cathode to positive terminal).
- A series resistor (Rs) is used to limit the current.
- The Zener diode maintains a constant voltage (Vz) across the load resistor (RL).
Working:
- When the input voltage (Vin) is less than the Zener voltage (Vz), the Zener diode does not conduct and acts like an open circuit.
- When Vin ≥ Vz, the Zener diode enters breakdown and starts conducting.
- The Zener voltage (Vz) remains constant across the load (RL), even if the input voltage increases.
Use:
This circuit is used to get a stable output voltage, useful for powering electronic circuits that require a fixed voltage.
Basic Electronics CSE 1st Sem | Most Important PYQs + New Syllabus Model Paper 2024
Q3. Explain the construction, operation, and characteristics of an enhancement-type MOSFET.
Answer:
1. Construction:
- An Enhancement-type MOSFET is a type of Field Effect Transistor (FET).
- It has three terminals: Gate (G), Drain (D), and Source (S).
- It uses a silicon substrate with an N-channel (for N-MOS) or P-channel (for P-MOS).
- A thin insulating layer of silicon dioxide (SiO₂) is placed between the gate terminal and the substrate.
- No channel exists between Drain and Source in the beginning — it forms only when voltage is applied.
2. Operation (N-channel E-MOSFET):
- When VGS = 0 (no gate voltage), no current flows between Drain and Source.
- When VGS > threshold voltage (Vth), an N-type channel is induced under the gate.
- This allows electrons to flow from Source to Drain, and the MOSFET turns ON.
- The more the gate voltage, the wider the channel and higher the current flow.
3. Characteristics:
- Input characteristics: Gate draws no current due to the insulating layer.
- Output characteristics:
- In the Ohmic region, current increases with VDS.
- In the Saturation region, current becomes constant (controlled by VGS).
- Enhancement MOSFET is normally OFF and turns ON only when gate voltage exceeds threshold.
Basic Electronics CSE 1st Sem | Most Important PYQs + New Syllabus Model Paper 2024
Q4. Draw and explain the inverting and non-inverting configurations of an Op-Amp with circuit and expression for gain.
Answer:
1. Inverting Amplifier:
- In this configuration, the input signal is applied to the inverting terminal (-) of the Op-Amp.
- The non-inverting terminal (+) is connected to ground.
- A feedback resistor (Rf) is connected from the output to the inverting input.
- An input resistor (Rin) is connected between the input source and the inverting terminal.
Gain Formula (Av):
$$A_v = -\frac{R_f}{R_{in}}$$
The negative sign indicates phase reversal (180° out of phase with input).
2. Non-Inverting Amplifier:
- The input signal is applied to the non-inverting terminal (+) of the Op-Amp.
- The inverting terminal (-) is connected to a voltage divider (two resistors: Rf and Rin) between output and ground.
- Feedback is used to control gain but no phase change occurs.
Gain Formula (Av):
$$A_v = 1 + \frac{R_f}{R_{in}}$$
The output is in phase with the input signal.
Basic Electronics CSE 1st Sem | Most Important PYQs + New Syllabus Model Paper 2024
Q5. Compare BJT and FET on at least five points including input impedance, switching speed, gain, and biasing.
Answer:
| BJT (Bipolar Junction Transistor) | FET (Field Effect Transistor) |
|---|---|
| A BJT has low input impedance because it draws current from the base terminal. | A FET has high input impedance since it draws almost no current at the gate terminal. |
| It is a current-controlled device where the base current controls the output. | It is a voltage-controlled device where gate voltage controls the output. |
| BJTs have slower switching speed due to charge storage in the base region. | FETs have faster switching speed because there is no charge storage. |
| It provides higher voltage and current gain, suitable for analog amplification. | It provides moderate gain and is often used in digital and low-power circuits. |
| Biasing a BJT requires more components and is slightly more complex. | Biasing a FET is easier and requires fewer components. |
| BJTs consume more power and generate more heat. | FETs consume less power and are more energy-efficient. |
Basic Electronics CSE 1st Sem | Most Important PYQs + New Syllabus Model Paper 2024
Q6. Realize the Boolean expression A + B using only NAND gates and explain with truth table.
Answer: We are given the Boolean expression:
A + B (OR gate) implement it using only NAND gates.
Step-by-step Realization:
To realize A + B using NAND gates only, we use the following logic identity: $A + B = (A’ \cdot B’)’$
This is called DeMorgan’s Theorem. It means:
OR can be implemented using inverted inputs ANDed together, then inverted again — which is exactly what a NAND gate can do.
Realization using NAND gates:
- First, invert A using NAND:
A NAND A = A′ - Then, invert B using NAND:
B NAND B = B′ - Now, AND A′ and B′, then NAND the result:
(A′ NAND B′) = (A + B)
This gives the desired OR function using only NAND gates.
Circuit Diagram (in words):
- Use three NAND gates:
- First gate: A and A → gives A′
- Second gate: B and B → gives B′
- Third gate: A′ and B′ → gives A + B
Truth Table:
| A | B | A′ | B′ | A + B |
|---|---|---|---|---|
| 0 | 0 | 1 | 1 | 0 |
| 0 | 1 | 1 | 0 | 1 |
| 1 | 0 | 0 | 1 | 1 |
| 1 | 1 | 0 | 0 | 1 |
- A + B = 1 whenever either A or B is 1.
Basic Electronics CSE 1st Sem | Most Important PYQs + New Syllabus Model Paper 2024
Q7. Convert the following:
(a) (1010.01)₂ to Decimal (b) (35)₁₀ to Binary
Answer: (a) Given: (1010.01)₂
Break into two parts:
Integer part: 1010
Fractional part: 0.01
Step 1: Convert integer part (1010): $1×2^3 + 0×2^2 + 1×2^1 + 0×2^0 = 8 + 0 + 2 + 0 = 10$
Step 2: Convert fractional part (0.01): $0×2^{-1} + 1×2^{-2} = 0 + 0.25 = 0.25$
Final Answer: $(1010.01)_2 = 10.25_{10}$
(b)Given: (35)₁₀
Step-by-step division by 2:
| Division | Quotient | Remainder |
|---|---|---|
| 35 ÷ 2 | 17 | 1 |
| 17 ÷ 2 | 8 | 1 |
| 8 ÷ 2 | 4 | 0 |
| 4 ÷ 2 | 2 | 0 |
| 2 ÷ 2 | 1 | 0 |
| 1 ÷ 2 | 0 | 1 |
Now, write the remainders in reverse order: $(35)_{10} = (100011)_2$
Basic Electronics CSE 1st Sem | Most Important PYQs + New Syllabus Model Paper 2024
Q8. Explain the working of a photodiode and a solar cell with proper diagrams and applications.
Answer:
1. Photodiode
Working:
A photodiode is a special PN junction diode that works in reverse bias. It allows current to flow only when light falls on it.
- When light (photons) strikes the junction, it creates electron-hole pairs.
- This increases the reverse current.
- The stronger the light, the more current flows.
Key Point: It converts light into electrical current.
Applications:
- Used in light sensors, IR receivers, smoke detectors, and optical communication.
2. Solar Cell
Working:
A solar cell (also called photovoltaic cell) is also a PN junction, but it works without external voltage.
- It directly converts sunlight into electricity.
- When sunlight hits the surface, electrons are released.
- These free electrons move and generate DC voltage across the terminals.
Key Point: It produces electric power from sunlight.
Applications:
- Used in solar panels, calculators, street lights, and satellites.
Basic Electronics CSE 1st Sem | Most Important PYQs + New Syllabus Model Paper 2024
Q9. Explain the operation of a Half-Wave Rectifier with input and output waveforms.
Answer: A Half-Wave Rectifier is a circuit that converts AC (alternating current) into DC (direct current) using a single PN junction diode.
Working:
- During Positive Half Cycle:
- The diode becomes forward biased.
- It conducts current, and the output appears across the load resistor.
- The waveform follows the positive half of the AC input.
- During Negative Half Cycle:
- The diode becomes reverse biased.
- It blocks current, and the output is zero.
So, the output only contains the positive half cycles of the AC input. The negative halves are removed.
Waveforms:
- Input waveform: Sine wave with both positive and negative cycles.
- Output waveform: Only positive half of the sine wave appears; negative half is missing.
Conclusion:
In this post, we provided detailed solutions for Basic Electronics CSE 1st Sem | Most Important PYQs + New Syllabus Model Paper 2024 previous year questions. Understanding these solutions will help you strengthen your core concepts and improve your exam performance. Keep practicing and exploring related topics to enhance your knowledge.
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