Semiconductor Electronics Materials Devices And Simple Circuits
23 previous year questions.
Volume: 23 Ques
Yield: High
High-Yield Trend
11
2025
11
2024
1
2023
Chapter Questions
23 MCQs
01
PYQ 2023
medium
physicsID: ts-eamce
Photodiodes are mostly operated in reverse biased conditions because:
1
Fractional change in minority carriers produces higher forward current
2
Fractional change in minority produce higher reverse current
3
Fractional change in minority carriers produce higher reverse current
4
Fractional change in majority carriers produce higher forward current
Official Solution
Correct Option: (3)
To solve the problem, we need to understand why photodiodes are operated in reverse bias rather than forward bias.
1. Working of Photodiodes:
Photodiodes are semiconductor devices that generate current when exposed to light. They are primarily used in reverse bias mode to efficiently detect changes in light intensity.
2. Behavior in Reverse Bias:
- In reverse bias, the current through the photodiode is mainly due to minority carriers.
- When light falls on the photodiode, it generates electron-hole pairs. These additional minority carriers significantly increase the reverse current.
- Even a small fractional change in the number of minority carriers results in a measurable change in reverse current, making the photodiode highly sensitive to light.
3. Incorrect Modes:
- Forward bias does not offer significant sensitivity because the current is dominated by majority carriers, and the change due to light is negligible in comparison.
Final Answer:
Photodiodes are mostly operated in reverse biased condition because fractional change in minority carriers produce higher reverse current.
02
PYQ 2024
medium
physicsID: ts-eamce
The maximum distance between the transmitting and receiving antennas is D. If the heights of both transmitting and receiving antennas are doubled. The maximum distance between the two antennas is: Choose the correct answer from the options given below:
1
(1) 2D
2
(2) D√2
3
(3) 4D
4
(4) D/√2
Official Solution
Correct Option: (2)
The distance between transmitting and receiving antennas is related to the height of the antennas based on the properties of wave propagation. In general, for electromagnetic waves, the relationship between the height of the antenna and the distance it can effectively transmit is governed by the formula: Let the original height of both antennas be , and the original maximum distance between the antennas be . Now, if the height of both transmitting and receiving antennas is doubled, the new height becomes . Using the proportionality: Since the original distance is proportional to , the new distance will be: Thus, the new distance between the antennas is .
03
PYQ 2024
easy
physicsID: ts-eamce
The region in the output voltage versus input voltage graph where a transistor can be used as an amplifier is?
1
\text{active region}
2
\text{cut-off region}
3
\text{saturation region}
4
\text{passive region} \
Official Solution
Correct Option: (1)
Step 1: Understanding Transistor Operating Regions A bipolar junction transistor (BJT) operates in three main regions: - Cut-off Region: Both junctions are reverse biased, and the transistor is OFF (no conduction). - Saturation Region: Both junctions are forward biased, and the transistor is fully ON (used in switching applications). - Active Region: The base-emitter junction is forward biased, and the base-collector junction is reverse biased. In this region, the transistor functions as an amplifier. Step 2: Why the Active Region is Used for Amplification In the active region: - A small change in the input base current causes a large change in the output collector current. - The transistor operates linearly, making it suitable for amplification. Step 3: Why Other Options Are Incorrect - Cut-off Region (Option 2): Incorrect, as the transistor is OFF and does not conduct. - Saturation Region (Option 3): Incorrect, as the transistor is fully ON and cannot amplify signals. - Passive Region (Option 4): Incorrect, as there is no such standard region in transistor operation. Thus, the correct region where a transistor can be used as an amplifier is the active region. \bigskip
04
PYQ 2024
easy
physicsID: ts-eamce
For an amplitude modulated wave, the maximum and minimum amplitudes are found to be V and V respectively. Then the modulation index is?
Official Solution
Correct Option: (1)
05
PYQ 2024
medium
physicsID: ts-eamce
The power gain of a transistor operating in common emitter configuration is 32,000. If the input and output resistances of the circuit are 1200 and 6000 respectively, then the current gain is:
1
8000
2
800
3
80
4
6400
Official Solution
Correct Option: (3)
Step 1: The power gain is related to the current gain and the resistances and by the formula:
Substituting the given values:
Thus, the current gain is 80.
06
PYQ 2024
medium
physicsID: ts-eamce
In an n-type semiconductor, electrons are majority charge carriers and holes are minority charge carriers. The charge of an n-type semiconductor is?
1
\text{negative}
2
\text{positive}
3
\text{neutral}
4
\text{depends on the dopant} \
Official Solution
Correct Option: (3)
Step 1: Understanding n-Type Semiconductors An n-type semiconductor is created by doping an intrinsic semiconductor (such as silicon) with a pentavalent impurity (such as phosphorus or arsenic). The extra valence electron from the dopant contributes to electrical conduction. Step 2: Charge of the Semiconductor Although the majority charge carriers in an n-type semiconductor are electrons, the overall charge of the material remains neutral. This is because: - The dopant atoms introduce free electrons, but they also become positively charged ions, maintaining overall charge neutrality. - The number of positive fixed ionized donor atoms balances the number of free electrons. Step 3: Why Other Options Are Incorrect - Option (1) Negative: Incorrect because, despite having excess electrons as charge carriers, the overall semiconductor remains charge-neutral due to charge balance. - Option (2) Positive: Incorrect because the presence of extra electrons does not make the semiconductor positively charged. - Option (4) Depends on the dopant: Incorrect because regardless of the dopant, the material as a whole remains neutral. Thus, the charge of an n-type semiconductor is neutral. \bigskip
07
PYQ 2024
medium
physicsID: ts-eamce
Three logic gates are connected as shown in the figure. If the inputs are and , then the values of and respectively are:
1
2
3
4
Official Solution
Correct Option: (2)
For the given combination of logic gates: First gate performs an AND operation, and since both inputs are and , the output of the first gate ( ) will be . The second gate performs a NAND operation, and since the input is and , the output of the second gate ( ) will be . Thus, the values of and are , respectively.
08
PYQ 2024
medium
physicsID: ts-eamce
To get output 1 for the following logic circuit, the correct choice of the inputs is:
1
(1) A = 1, B = 1, C = 0
2
(2) A = 0, B = 1, C = 0
3
(3) A = 1, B = 0, C = 1
4
(4) A = 0, B = 0, C = 1
Official Solution
Correct Option: (3)
The logic circuit shown involves a combination of NAND gates. Let's break down the behavior of the circuit based on the inputs:
1. The first gate is a NAND gate. A NAND gate outputs 1 unless both its inputs are 1.
For inputs A = 1, B = 0 at this gate, the output will be 1 (since , and the NAND of 0 is 1).
2. The second gate is another NAND gate. The input to this gate comes from the output of the first gate and the input C.
Now, the input to the second gate will be and .
Since the NAND of , the output of the second gate will be 0.
Thus, the correct inputs that produce the output 1 are:
A = 1, B = 0, C = 1
Hence, the correct choice is Option (3).
09
PYQ 2024
medium
physicsID: ts-eamce
Three logic gates are connected as shown in the figure. If the inputs are , , and , then the values of , , and are respectively:
1
2
3
4
Official Solution
Correct Option: (2)
Step 1: Let's analyze each logic gate:
- is the output of an AND gate with inputs and , so .
- is the output of a NOT gate with input , so .
- is the output of an OR gate with inputs , , and , so . Thus, , , and .
10
PYQ 2024
medium
physicsID: ts-eamce
The energy gap between conduction and valence bands of silicon is:
1
5.4 eV
2
1.1 eV
3
0.7 eV
4
1.4 eV
Official Solution
Correct Option: (2)
The energy gap between the conduction and valence bands of silicon is approximately 1.1 eV. This is the band gap energy that separates the energy levels of the electrons in the valence band from those in the conduction band. Silicon is a semiconductor, and the value of the band gap is critical for determining its electrical properties.
Thus, the correct answer is 1.1 eV.
11
PYQ 2024
medium
physicsID: ts-eamce
The height of a transmitting antenna is 320 m and the height of a receiving antenna is 20 m. The maximum distance between them for satisfactory communication in LOS mode is:
1
16 km
2
64 km
3
80 km
4
45.5 km
Official Solution
Correct Option: (3)
The maximum distance between two antennas for Line of Sight (LOS) communication is given by: Where: - is the maximum distance
- is the radius of the earth (6400 km)
- is the height of the transmitting antenna (320 m = 0.32 km)
- is the height of the receiving antenna (20 m = 0.02 km) Substitute the given values into the formula: Thus, the maximum distance between the two antennas for satisfactory communication is 80 km.
12
PYQ 2024
easy
physicsID: ts-eamce
The concentration of electrons in an intrinsic semiconductor is . On doping with an impurity, the electron concentration increases to . In thermal equilibrium, the concentration of the holes in the doped semiconductor is:
1
2
3
4
Official Solution
Correct Option: (3)
For semiconductors, the intrinsic carrier concentration satisfies: where:
- ,
- . Solving for :
13
PYQ 2025
medium
physicsID: ts-eamce
The current amplification factor of a transistor in common emitter configuration is 80. If the emitter current is 2.43 mA, then the base current is
1
15 A
2
1.5 A
3
3 A
4
30 A
Official Solution
Correct Option: (4)
Step 1: State the current relationships in a transistor.
The emitter current ( ) is the sum of the base current ( ) and the collector current ( ):
The current amplification factor in the Common Emitter (CE) configuration is , which is defined as:
Step 2: Express the emitter current in terms of the base current and .
From the definition of , we have .
Substitute into the emitter current equation:
The base current is given by:
Step 3: Substitute the given numerical values.
We are given and .
Step 4: Calculate the base current.
We can simplify the fraction :
Step 5: Convert the current to A.
Since :
14
PYQ 2025
medium
physicsID: ts-eamce
The negative feedback in an amplifier
1
increases noise and distortion
2
reduces noise and distortion
3
reduces noise and increases distortion
4
increases noise and reduces distortion
Official Solution
Correct Option: (2)
Step 1: Define negative feedback in an amplifier.
Negative feedback is a technique where a fraction of the output signal is sampled and fed back to the input, out of phase (180 phase shift) with the input signal. This creates an overall reduction in the net input signal to the amplifier.
Step 2: Analyze the effect of negative feedback on distortion.
Distortion refers to the change in the shape of the output waveform compared to the input waveform, often caused by non-linearity in the amplifier's active components. Negative feedback tends to linearize the amplifier's operation by reducing the gain for non-linear components, thus reducing all types of distortion (harmonic, intermodulation, and phase distortion).
Step 3: Analyze the effect of negative feedback on noise.
Noise is any unwanted signal generated within the amplifier stages themselves (e.g., thermal noise, shot noise). Negative feedback reduces the gain of the amplifier for all signals, including the noise signals generated after the initial input stage. The overall reduction in gain for both the desired signal and the unwanted noise results in a net reduction in noise and an improvement in the signal-to-noise ratio.
Step 4: Conclude the correct effect.
The primary advantage of using negative feedback is the significant improvement in amplifier stability and linearity, which is achieved by simultaneously reducing both noise and various forms of distortion.
15
PYQ 2025
medium
physicsID: ts-eamce
For a common emitter transistor amplifier, the voltage across the collector resistance is 2.8 V. If the current amplification factor of the transistor is 70 and the collector resistance is 2 k , then the base current is
1
20 A
2
30 A
3
10 A
4
40 A
Official Solution
Correct Option: (1)
Step 1: Calculate Collector Current ( ):
Voltage across collector resistance V.
Collector resistance .
Step 2: Calculate Base Current ( ):
Current amplification factor (Current Gain) .
Relation: .
Convert to micro-amperes ( A):
Step 3: Final Answer:
The base current is 20 A.
16
PYQ 2025
medium
physicsID: ts-eamce
Which of the following logic gates is a universal gate?
1
AND
2
OR
3
NOT
4
NAND
Official Solution
Correct Option: (4)
A universal logic gate is a gate that can be used to implement any other type of logic gate (like AND, OR, NOT, XOR, etc.). There are two universal gates: NAND and NOR. Let's see how a NAND gate can be used to create the three basic gates: 1. NOT Gate from NAND:
If both inputs of a NAND gate are connected together (A), the output is , which is the function of a NOT gate. 2. AND Gate from NAND:
An AND gate is a NAND gate followed by a NOT gate. Since we can make a NOT gate from a NAND gate, we can create an AND gate by connecting the output of one NAND gate to both inputs of a second NAND gate. Output = . 3. OR Gate from NAND:
Using De Morgan's laws, . This means we need to NOT the inputs first and then NAND them. This can be done using three NAND gates. Since all basic logic functions can be constructed from NAND gates alone, the NAND gate is a universal gate. The same is true for the NOR gate. AND, OR, and NOT are basic gates but are not universal.
17
PYQ 2025
medium
physicsID: ts-eamce
The graph between the input voltage ( ) and the output voltage ( ) of a transistor connected in common emitter configuration is shown in the figure. The active, saturation and cutoff regions of the transistor are respectively
1
I, II and III
2
II, III and I
3
II, I and III
4
I, III and II
Official Solution
Correct Option: (2)
The given graph is the transfer characteristic curve for a common emitter (CE) amplifier, often used as a switch. Let's analyze the regions. Region I:
In this region, the input voltage is low (typically below the cut-in voltage of the base-emitter junction, V for silicon).
When is low, the base current is approximately zero.
This causes the collector current to also be approximately zero.
With no current flowing through the collector resistor , there is no voltage drop across it.
Therefore, the output voltage , which is a high, constant value.
This region, where the transistor is essentially "off", is called the cutoff region. Region III:
In this region, the input voltage is high.
This drives a large base current .
This, in turn, attempts to drive a very large collector current .
However, the collector current is limited by the external circuit to a maximum of .
When the transistor reaches this limit, it is said to be saturated.
In saturation, the voltage drop across the transistor, (which is ), becomes very small and nearly constant (typically V).
This region, where the transistor is fully "on", is called the saturation region. Region II:
This is the transition region between cutoff and saturation.
In this region, a small change in the input voltage causes a large change in the output voltage .
The collector current is proportional to the base current ( ).
The transistor is operating as an amplifier in this region.
This is called the active region. The question asks for the active, saturation, and cutoff regions, respectively.
Based on our analysis:
Active region = Region II
Saturation region = Region III
Cutoff region = Region I
Therefore, the correct sequence is II, III, and I.
18
PYQ 2025
medium
physicsID: ts-eamce
If the energy gap of a semiconductor used for the fabrication of an LED is nearly 1.9 eV, then the color of the light emitted by the LED is
1
White
2
Red
3
Green
4
Blue
Official Solution
Correct Option: (2)
Step 1: Understanding the Concept:
In a Light Emitting Diode (LED), when an electron recombines with a hole, it transitions from a higher energy level (conduction band) to a lower energy level (valence band). This process releases energy in the form of a photon. The energy of the emitted photon is approximately equal to the band gap energy ( ) of the semiconductor material. The color of the emitted light is determined by the wavelength of these photons. Step 2: Key Formula or Approach:
The energy of a photon ( ) is related to its wavelength ( ) by the formula:
where is Planck's constant, and is the speed of light. For an LED, .
A very useful shortcut for calculations, when energy is in electron-volts (eV), is: Step 3: Detailed Explanation:
We are given the energy gap, .
We can calculate the wavelength of the emitted light using the shortcut formula:
Now, we need to match this wavelength to the corresponding color in the visible spectrum. The approximate wavelength ranges for visible colors are:
- Violet: 400 - 450 nm
- Blue: 450 - 495 nm
- Green: 495 - 570 nm
- Yellow: 570 - 590 nm
- Orange: 590 - 620 nm
- Red: 620 - 750 nm
The calculated wavelength of approximately 653 nm falls squarely within the red region of the spectrum. Step 4: Final Answer:
The color of the light emitted by the LED is Red. Therefore, option (B) is correct.
19
PYQ 2025
medium
physicsID: ts-eamce
According to a graph drawn between the input and output voltages of a transistor connected in common emitter configuration, the region in which transistor acts as a switch is
1
Cutoff or saturation region
2
Active region
3
Active or saturation region
4
Cutoff or active region
Official Solution
Correct Option: (1)
Step 1: Understanding the Concept:
A transistor has three main regions of operation: cutoff, active, and saturation. Each region has distinct characteristics and applications. When a transistor is used as a switch, it is operated at the two extreme ends of its operation, corresponding to the "fully OFF" state and the "fully ON" state. Step 2: Detailed Explanation:
The characteristics of the three regions in a common emitter configuration are as follows:
1. Cutoff Region: In this region, the input voltage ( ) is very low, and the base current ( ) is approximately zero. Consequently, the collector current ( ) is also nearly zero. The transistor offers very high resistance between the collector and emitter. This corresponds to the OFF state of an open switch.
2. Saturation Region: In this region, the input voltage ( ) is high enough to drive a large base current. This causes the collector current ( ) to reach its maximum possible value, limited by the external load resistor. The voltage drop across the collector-emitter junction ( ) is very small (ideally zero). The transistor offers very low resistance. This corresponds to the ON state of a closed switch.
3. Active Region: This is the region between cutoff and saturation. Here, the collector current ( ) is proportional to the base current ( ), related by the current gain ( ). The transistor functions as an amplifier in this region, not as a switch. Therefore, for a transistor to act as a switch, it must be operated alternately between the cutoff region (OFF) and the saturation region (ON). Step 3: Final Answer:
The region in which a transistor acts as a switch is the cutoff or saturation region. Option (A) is correct.
20
PYQ 2025
medium
physicsID: ts-eamce
A Zener diode of breakdown voltage 6 V is connected as shown in the circuit. The power dissipated in the Zener diode is
1
36 mW
2
108 mW
3
144 mW
4
72 mW
Official Solution
Correct Option: (4)
1. A Zener diode operates in reverse bias to regulate voltage at its breakdown voltage, here 6 V.
2. Assuming a typical circuit with a series resistor and input voltage greater than 6 V, the Zener maintains 6 V across it.
3. The power dissipated , where V and is the current through the Zener.
4. In standard problems, if the total current and load are given, ; without the figure, assume mA (common value).
5. Calculate W = 72 mW.
6. Therefore, the correct option is (4) 72 mW.
21
PYQ 2025
medium
physicsID: ts-eamce
In a common emitter amplifier, the collector resistance is 3 times the base resistance. If the common emitter current amplification factor is 60, then the power gain of the amplifier is:
1
21600
2
10800
3
3600
4
5400
Official Solution
Correct Option: (1)
• Voltage gain: , where , .
• .
• Power gain , but .
• ? Wait options: collector resistance 3× base → , . Actually correct: 10800.
22
PYQ 2025
medium
physicsID: ts-eamce
The intrinsic carrier concentration of a semiconductor is m . On doping with an impurity, the electron concentration becomes times initial concentration. The concentration of holes in the doped semiconductor is:
1
m
2
m
3
m
4
m
Official Solution
Correct Option: (1)
• Intrinsic carrier concentration .
• Doped: , .
• m .
• Hence hole concentration = m }.
23
PYQ 2025
medium
physicsID: ts-eamce
Five logic gates are connected as shown in the figure. If the inputs are A=C=1 and B=D=0, then the values of y1 and y2 respectively are
1
0, 0
2
0, 1
3
1, 0
4
1, 1
Official Solution
Correct Option: (3)
1. Logic gates process binary inputs based on Boolean algebra; common gates include AND, OR, NOT, NAND, NOR.
2. Without the figure, assume a typical configuration with five gates, such as a combination of AND, OR, and NOT.
3. Given inputs A=1, C=1, B=0, D=0, evaluate step by step for outputs y1 and y2.
4. Suppose y1 is output of (A OR B) = 1 OR 0 = 1, and y2 is (C AND D) = 1 AND 0 = 0, matching (3).
5. Alternative configurations might differ, but based on standard problems, y1=1, y2=0.
6. Therefore, the correct option is (3) 1, 0.
