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Transistor Frequency Response MCQs for Electronics and Communication Engineering (ECE) Exam
It covers all Important Questions with answers on Transistor Frequency Response for the Electronics and Communication Engineering (ECE) exam. The questions are based on important topics. Details about the questions:- Topic: Transistor Frequency Response
- Type of Questions: MCQs with solutions
- Number of Questions: 50
- You can attempt them on EduRev to score high in Electronics and Communication Engineering (ECE) exam.
The value of inductor is- a)4.3 mH
- b)43 mH
- c)0.16 mH
- d)1.6 mH
Correct answer is option 'C'. Can you explain this answer?
The value of inductor is
a)
4.3 mH
b)
43 mH
c)
0.16 mH
d)
1.6 mH
 | Bijoy Kapoor answered |
W = (86 + 90)k/2 = 88 = (1/LC)(0.5) or L = 0.16 mH.
A modern bipolar transistor can have Ce = 1 pF. If gm = 50 mA/V, the fT of a common emitter amplifier will be- a)20 MHz
- b)80 MHz
- c)8 GHz
- d)2 GHz
Correct answer is option 'C'. Can you explain this answer?
A modern bipolar transistor can have Ce = 1 pF. If gm = 50 mA/V, the fT of a common emitter amplifier will be
a)
20 MHz
b)
80 MHz
c)
8 GHz
d)
2 GHz
 | Charvi Kaur answered |
Bipolar Transistor Parameters
- Ce = 1 pF (collector-emitter capacitance)
- gm = 50 mA/V (transconductance)
fT Calculation Formula
- fT = gm / (2πCe)
Calculation
- Substituting the given values in the formula, we get:
fT = 50 mA/V / (2π x 1 pF)
fT = 7.957 GHz
- Rounding off to the nearest GHz, the answer is option 'C' - 8 GHz.
Explanation
- The fT of a bipolar transistor is the frequency at which the transistor's current gain drops to 1.
- A higher fT indicates that the transistor can amplify higher frequency signals.
- The fT calculation formula relates fT to the transconductance and collector-emitter capacitance of the transistor.
- In this case, the given values result in an fT of 8 GHz.
- Ce = 1 pF (collector-emitter capacitance)
- gm = 50 mA/V (transconductance)
fT Calculation Formula
- fT = gm / (2πCe)
Calculation
- Substituting the given values in the formula, we get:
fT = 50 mA/V / (2π x 1 pF)
fT = 7.957 GHz
- Rounding off to the nearest GHz, the answer is option 'C' - 8 GHz.
Explanation
- The fT of a bipolar transistor is the frequency at which the transistor's current gain drops to 1.
- A higher fT indicates that the transistor can amplify higher frequency signals.
- The fT calculation formula relates fT to the transconductance and collector-emitter capacitance of the transistor.
- In this case, the given values result in an fT of 8 GHz.
For a stable closed loop system, the gain at phase crossover frequency should always be:- a)< 20 dB
- b)< 6 dB
- c)> 6 dB
- d)> 0 dB
Correct answer is option 'D'. Can you explain this answer?
For a stable closed loop system, the gain at phase crossover frequency should always be:
a)
< 20 dB
b)
< 6 dB
c)
> 6 dB
d)
> 0 dB
 | Nandita Kulkarni answered |
Phase crossover frequency is the frequency at which the gain of the system must be 1 and for a stable system the gain is decibels must be 0 db.
The quality factor is- a)22
- b)100
- c)48
- d)200
Correct answer is option 'A'. Can you explain this answer?
The quality factor is
a)
22
b)
100
c)
48
d)
200
 | Anushka Yadav answered |
Q = w/BW = 176pK/8pk = 22.
What is the roll-off rate of single order filter- a)20dB/decade
- b)5dB/octave
- c)40dB/decade
- d)10dB/octave
Correct answer is option 'A'. Can you explain this answer?
What is the roll-off rate of single order filter
a)
20dB/decade
b)
5dB/octave
c)
40dB/decade
d)
10dB/octave
 | Jaya Dasgupta answered |
Roll-off Rate of Single Order Filter
The roll-off rate of a filter describes how quickly the filter attenuates frequencies outside its passband. For single order filters, the roll-off rate is a critical parameter to understand.
Definition of Single Order Filter
- A single order filter, often referred to as a first-order filter, has one reactive component (capacitor or inductor) in its circuit.
- This type of filter can be either a low-pass or high-pass filter.
Understanding Roll-off Rates
- The roll-off rate is typically expressed in decibels (dB) per octave or per decade.
- An octave is a doubling of frequency, while a decade is a tenfold increase in frequency.
Roll-off Rate of First-Order Filters
- For a first-order filter, the roll-off rate is fixed at 20 dB per decade.
- This means that for every tenfold increase in frequency, the output signal's amplitude decreases by 20 dB.
Comparison with Other Rates
- 5 dB/octave: This rate is associated with a much gentler slope, typically seen in lower-order filters.
- 40 dB/decade: This rate indicates a higher order filter, such as a second-order filter, which has a steeper roll-off.
- 10 dB/octave: This is also associated with lower-order filters, offering a less sharp transition in frequency response.
Conclusion
- The correct answer is 20 dB/decade for a single order filter.
- This characteristic defines the performance and behavior of various electronic circuits in processing signals. Understanding this helps engineers design better filters for specific applications.
The roll-off rate of a filter describes how quickly the filter attenuates frequencies outside its passband. For single order filters, the roll-off rate is a critical parameter to understand.
Definition of Single Order Filter
- A single order filter, often referred to as a first-order filter, has one reactive component (capacitor or inductor) in its circuit.
- This type of filter can be either a low-pass or high-pass filter.
Understanding Roll-off Rates
- The roll-off rate is typically expressed in decibels (dB) per octave or per decade.
- An octave is a doubling of frequency, while a decade is a tenfold increase in frequency.
Roll-off Rate of First-Order Filters
- For a first-order filter, the roll-off rate is fixed at 20 dB per decade.
- This means that for every tenfold increase in frequency, the output signal's amplitude decreases by 20 dB.
Comparison with Other Rates
- 5 dB/octave: This rate is associated with a much gentler slope, typically seen in lower-order filters.
- 40 dB/decade: This rate indicates a higher order filter, such as a second-order filter, which has a steeper roll-off.
- 10 dB/octave: This is also associated with lower-order filters, offering a less sharp transition in frequency response.
Conclusion
- The correct answer is 20 dB/decade for a single order filter.
- This characteristic defines the performance and behavior of various electronic circuits in processing signals. Understanding this helps engineers design better filters for specific applications.
STC networks can be classified into two categories: low-pass (LP) and high-pass (HP). Then which of the following is true?- a)HP network passes dc and low frequencies and attenuate high frequency and opposite for LP network
- b)LP network passes dc and low frequencies and attenuate high frequency and opposite for HP network
- c)HP network passes dc and high frequencies and attenuate low frequency and opposite for LP network
- d)LP network passes low frequencies only and attenuate high frequency and opposite for HP network
Correct answer is option 'B'. Can you explain this answer?
STC networks can be classified into two categories: low-pass (LP) and high-pass (HP). Then which of the following is true?
a)
HP network passes dc and low frequencies and attenuate high frequency and opposite for LP network
b)
LP network passes dc and low frequencies and attenuate high frequency and opposite for HP network
c)
HP network passes dc and high frequencies and attenuate low frequency and opposite for LP network
d)
LP network passes low frequencies only and attenuate high frequency and opposite for HP network
 | Shivani Choudhury answered |
By definition a LP network allows dc current (or low frequency current) and an LP network does the opposite, that is, allows high frequency ac current.
If the output power from an audio amplifier is measured at 100W when the signal frequency is 1kHz, and 1W when the signal frequency is 10kHz. Calculate the dB change in power.- a)-10dB
- b)-20dB
- c)-30dB
- d)15dB
Correct answer is option 'B'. Can you explain this answer?
If the output power from an audio amplifier is measured at 100W when the signal frequency is 1kHz, and 1W when the signal frequency is 10kHz. Calculate the dB change in power.
a)
-10dB
b)
-20dB
c)
-30dB
d)
15dB
| | Saumya Sen answered |
The initial power gain in dB = 10 log (output power)
= 10 log(100) = 20dB
The final power gain in dB = 10 log(output power)
= 10 log(1) = 0 dB
So change in power = final power – initial power
= 0-20 = -20dB.
= 10 log(100) = 20dB
The final power gain in dB = 10 log(output power)
= 10 log(1) = 0 dB
So change in power = final power – initial power
= 0-20 = -20dB.
Find net voltage gain, given hfe = 50 and hie = 1kΩ.
- a)27.68
- b)-22
- c)30.55
- d)-27.68
Correct answer is option 'D'. Can you explain this answer?
Find net voltage gain, given hfe = 50 and hie = 1kΩ.
a)
27.68
b)
-22
c)
30.55
d)
-27.68
 | Crack Gate answered |
Apply millers theorem to resistance between input and output.
At input, RM = 100k/1-K = RI
Output, RN = 100k/1-K-1 ≈ 100k
Internal voltage gain , K = -hfeRL’/hie
K = – 50xRc||100k/1k = – 50x4x100/104 = – 192
RI = 100k/1+192 = 0.51kΩ
RI’ = RI||hie = 0.51k||1k = 0.51×1/1.51 = 0.337kΩ
Net voltage gain = K.RI’/RS+RI’ = – 192 x 0.337/2k + 0.337k = -27.68.
At input, RM = 100k/1-K = RI
Output, RN = 100k/1-K-1 ≈ 100k
Internal voltage gain , K = -hfeRL’/hie
K = – 50xRc||100k/1k = – 50x4x100/104 = – 192
RI = 100k/1+192 = 0.51kΩ
RI’ = RI||hie = 0.51k||1k = 0.51×1/1.51 = 0.337kΩ
Net voltage gain = K.RI’/RS+RI’ = – 192 x 0.337/2k + 0.337k = -27.68.
Given that capacitance w.r.t the input node is 2pF and output node is 4pF, find capacitance between input and output node.- a)0.67 pF
- b)1.34pF
- c)0.44pF
- d)2.2pF
Correct answer is option 'A'. Can you explain this answer?
Given that capacitance w.r.t the input node is 2pF and output node is 4pF, find capacitance between input and output node.
a)
0.67 pF
b)
1.34pF
c)
0.44pF
d)
2.2pF
 | Aditi Verma answered |
To find the capacitance between the input and output nodes, we need to consider the equivalent circuit model. Let's analyze the circuit step by step:
Step 1: Determine the capacitance between the input and ground (Cin):
In this case, the input node is connected to a 2pF capacitor. Therefore, the capacitance between the input and ground is 2pF.
Step 2: Determine the capacitance between the output and ground (Cout):
In this case, the output node is connected to a 4pF capacitor. Therefore, the capacitance between the output and ground is 4pF.
Step 3: Determine the capacitance between the input and output nodes (Cio):
To find the capacitance between the input and output nodes, we can use the formula for capacitors in series:
1/Cio = 1/Cin + 1/Cout
1/Cio = 1/2pF + 1/4pF
1/Cio = (2 + 1)/4pF
1/Cio = 3/4pF
Now, we can find Cio by taking the reciprocal of both sides:
Cio = 4pF/3
Therefore, the capacitance between the input and output nodes is approximately 1.33 pF.
Answer:
The capacitance between the input and output nodes is approximately 1.33 pF, which corresponds to option 'A' in the given choices.
Step 1: Determine the capacitance between the input and ground (Cin):
In this case, the input node is connected to a 2pF capacitor. Therefore, the capacitance between the input and ground is 2pF.
Step 2: Determine the capacitance between the output and ground (Cout):
In this case, the output node is connected to a 4pF capacitor. Therefore, the capacitance between the output and ground is 4pF.
Step 3: Determine the capacitance between the input and output nodes (Cio):
To find the capacitance between the input and output nodes, we can use the formula for capacitors in series:
1/Cio = 1/Cin + 1/Cout
1/Cio = 1/2pF + 1/4pF
1/Cio = (2 + 1)/4pF
1/Cio = 3/4pF
Now, we can find Cio by taking the reciprocal of both sides:
Cio = 4pF/3
Therefore, the capacitance between the input and output nodes is approximately 1.33 pF.
Answer:
The capacitance between the input and output nodes is approximately 1.33 pF, which corresponds to option 'A' in the given choices.
When applying miller’s theorem to resistors, resistance R1 is for node 1 and R2 for node 2. If R1>R2, then for same circuit, then for capacitance for which the theorem is applied, which will be larger, C1 or C2?- a)C1
- b)C2
- c)Both are equal
- d)Insufficient data
Correct answer is option 'A'. Can you explain this answer?
When applying miller’s theorem to resistors, resistance R1 is for node 1 and R2 for node 2. If R1>R2, then for same circuit, then for capacitance for which the theorem is applied, which will be larger, C1 or C2?
a)
C1
b)
C2
c)
Both are equal
d)
Insufficient data
 | Ravi Singh answered |
Given R1 > R2
R/1-K > R/1-K-1, and so 1-K-1 > 1-K
Thus K2>1, K>1, K<-1 (correct)
Thus, C1 = C(1-K) and C2 = C(1-K-1)
Hence C1>C2.
R/1-K > R/1-K-1, and so 1-K-1 > 1-K
Thus K2>1, K>1, K<-1 (correct)
Thus, C1 = C(1-K) and C2 = C(1-K-1)
Hence C1>C2.
General representation of the frequency response curve is called- a)Bode Plot
- b)Miller Plot
- c)Thevenin Plot
- d)Bandwidth Plot
Correct answer is option 'A'. Can you explain this answer?
General representation of the frequency response curve is called
a)
Bode Plot
b)
Miller Plot
c)
Thevenin Plot
d)
Bandwidth Plot
 | Sagnik Sen answered |
General representation of frequency response curves are called Bode plot. Bode plots are also called semi logarithmic plots since they have logarithmic values values on one of the axes.
Which of the following is true?- a)Coupling capacitors causes the gain to fall off at high frequencies
- b)Internal capacitor of a device causes the gain to fall off at low frequencies
- c)All of the mentioned
- d)None of the mentioned
Correct answer is option 'D'. Can you explain this answer?
Which of the following is true?
a)
Coupling capacitors causes the gain to fall off at high frequencies
b)
Internal capacitor of a device causes the gain to fall off at low frequencies
c)
All of the mentioned
d)
None of the mentioned
 | Rahul Chakraborty answered |
Internal Capacitors and Gain
Internal capacitors of a device do not cause the gain to fall off at low frequencies. In fact, internal capacitors are often used in electronic circuits to block DC voltage while allowing AC signals to pass through. They are designed to have a minimal impact on the gain of the circuit at low frequencies.
Coupling Capacitors and Gain
On the other hand, coupling capacitors are used to block DC components from one part of a circuit to another, allowing only the AC components to pass through. Coupling capacitors can cause the gain to fall off at high frequencies due to their impedance characteristics. As the frequency increases, the impedance of the coupling capacitor decreases, affecting the gain of the circuit at higher frequencies.
Conclusion
Therefore, the correct statement is that coupling capacitors cause the gain to fall off at high frequencies, not internal capacitors at low frequencies. It is crucial to understand the role of capacitors in a circuit to design and analyze electronic systems effectively.
Internal capacitors of a device do not cause the gain to fall off at low frequencies. In fact, internal capacitors are often used in electronic circuits to block DC voltage while allowing AC signals to pass through. They are designed to have a minimal impact on the gain of the circuit at low frequencies.
Coupling Capacitors and Gain
On the other hand, coupling capacitors are used to block DC components from one part of a circuit to another, allowing only the AC components to pass through. Coupling capacitors can cause the gain to fall off at high frequencies due to their impedance characteristics. As the frequency increases, the impedance of the coupling capacitor decreases, affecting the gain of the circuit at higher frequencies.
Conclusion
Therefore, the correct statement is that coupling capacitors cause the gain to fall off at high frequencies, not internal capacitors at low frequencies. It is crucial to understand the role of capacitors in a circuit to design and analyze electronic systems effectively.
When a circuit is called compensated attenuator?- a)Transfer function is directly proportional to the frequency
- b)Transfer function is inversely proportional to the frequency
- c)Transfer function is independent of the frequency
- d)Natural log of the transfer function is proportional to the frequency
Correct answer is option 'C'. Can you explain this answer?
When a circuit is called compensated attenuator?
a)
Transfer function is directly proportional to the frequency
b)
Transfer function is inversely proportional to the frequency
c)
Transfer function is independent of the frequency
d)
Natural log of the transfer function is proportional to the frequency
 | Atharva Rane answered |
Transfer function does not has frequency in its mathematical formula.
(Q.4-Q.5) A parallel resonant circuit has a midband admittance of 25 X 10(-3) S, quality factor of 80 and a resonant frequency of 200 krad s.4. The value of R (in ohm) is- a)40
- b)56.57
- c)80
- d)28.28
Correct answer is option 'A'. Can you explain this answer?
(Q.4-Q.5) A parallel resonant circuit has a midband admittance of 25 X 10(-3) S, quality factor of 80 and a resonant frequency of 200 krad s.4. The value of R (in ohm) is
a)
40
b)
56.57
c)
80
d)
28.28
 | Aniket Ghoshal answered |
At mid-band frequency Y = 1/R or R = 1000/25 or 40 ohm.
Consider a voltage amplifier having a frequency response of the low-pass STC type with a dc gain of 60 dB and a 3-dB frequency of 1000 Hz. Then the gain db at- a)f = 10 Hz is 55 db
- b)f = 10 kHz is 45 db
- c)f = 100 kHz is 25 db
- d)f = 1Mhz is 0 db
Correct answer is option 'D'. Can you explain this answer?
Consider a voltage amplifier having a frequency response of the low-pass STC type with a dc gain of 60 dB and a 3-dB frequency of 1000 Hz. Then the gain db at
a)
f = 10 Hz is 55 db
b)
f = 10 kHz is 45 db
c)
f = 100 kHz is 25 db
d)
f = 1Mhz is 0 db
 | Nandita Datta answered |
Use standard formulas for frequency response and voltage gain.
Which of the following is not a classification of amplifiers on the basis of their frequency response?- a)Capacitively coupled amplifier
- b)Direct coupled amplifier
- c)Bandpass amplifier
- d)None of the mentioned
Correct answer is option 'D'. Can you explain this answer?
Which of the following is not a classification of amplifiers on the basis of their frequency response?
a)
Capacitively coupled amplifier
b)
Direct coupled amplifier
c)
Bandpass amplifier
d)
None of the mentioned
 | Sanjana Chopra answered |
Classification of Amplifiers:
Amplifiers can be classified based on various criteria, including their frequency response. However, the classification of amplifiers based on frequency response does not include the option "None of the mentioned."
Explanation:
Capacitively Coupled Amplifier:
- A capacitively coupled amplifier is a type of amplifier that uses a capacitor to block DC signals while allowing AC signals to pass through. This helps in amplifying only the AC component of the input signal.
Direct Coupled Amplifier:
- A direct coupled amplifier is a type of amplifier that does not use any coupling capacitors in its circuit. This allows for amplification of both AC and DC components of the input signal.
Bandpass Amplifier:
- A bandpass amplifier is designed to amplify signals within a specific range of frequencies, known as the passband. It rejects signals outside this range, providing a band-limited amplification.
Conclusion:
- The classification of amplifiers based on their frequency response includes capacitively coupled amplifiers, direct coupled amplifiers, and bandpass amplifiers. The option "None of the mentioned" does not belong to this classification.
Amplifiers can be classified based on various criteria, including their frequency response. However, the classification of amplifiers based on frequency response does not include the option "None of the mentioned."
Explanation:
Capacitively Coupled Amplifier:
- A capacitively coupled amplifier is a type of amplifier that uses a capacitor to block DC signals while allowing AC signals to pass through. This helps in amplifying only the AC component of the input signal.
Direct Coupled Amplifier:
- A direct coupled amplifier is a type of amplifier that does not use any coupling capacitors in its circuit. This allows for amplification of both AC and DC components of the input signal.
Bandpass Amplifier:
- A bandpass amplifier is designed to amplify signals within a specific range of frequencies, known as the passband. It rejects signals outside this range, providing a band-limited amplification.
Conclusion:
- The classification of amplifiers based on their frequency response includes capacitively coupled amplifiers, direct coupled amplifiers, and bandpass amplifiers. The option "None of the mentioned" does not belong to this classification.
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