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An amplifier using BJT has two identical stages each having a lower cut-off (3dB) frequency of 64Hz due to coupling capacitor. The emitter bypass capacitor also provides a lower cut-off (3dB) frequency due to emitter degeneration alone of 64Hz. The lower 3dB frequency of the overall amplifier is nearly.
- a)128 Hz
- b)100 Hz
- c)41 Hz
- d)156 Hz
Correct answer is option 'B'. Can you explain this answer?
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An amplifier using BJT has two identical stages each having a lower cu...
So overall lower cut-off frequency
= 100 Hz
= 100 Hz
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An amplifier using BJT has two identical stages each having a lower cu...
Given information:
- The amplifier has two identical stages.
- Each stage has a lower cut-off (3dB) frequency of 64Hz due to a coupling capacitor.
- The emitter bypass capacitor provides a lower cut-off (3dB) frequency due to emitter degeneration alone of 64Hz.
Explanation:
To find the overall lower 3dB frequency of the amplifier, we need to consider the combined effect of both stages.
Effect of coupling capacitor:
The lower cut-off frequency due to the coupling capacitor is given by the formula:
f1 = 1 / (2 * π * R * C)
where,
f1 is the lower cut-off frequency,
R is the resistance connected to the coupling capacitor, and
C is the value of the coupling capacitor.
Since both stages are identical, the resistance and the value of the coupling capacitor will be the same for both stages. Let's represent them as R1 and C1.
So, the lower cut-off frequency due to the coupling capacitor for each stage is:
f1_stage = 1 / (2 * π * R1 * C1)
Effect of emitter bypass capacitor:
The lower cut-off frequency due to the emitter bypass capacitor is given by the formula:
f2 = 1 / (2 * π * R2 * C2)
where,
f2 is the lower cut-off frequency,
R2 is the resistance connected to the emitter bypass capacitor, and
C2 is the value of the emitter bypass capacitor.
Since both stages are identical, the resistance and the value of the emitter bypass capacitor will be the same for both stages. Let's represent them as R2 and C2.
So, the lower cut-off frequency due to the emitter bypass capacitor for each stage is:
f2_stage = 1 / (2 * π * R2 * C2)
Overall lower cut-off frequency:
To find the overall lower cut-off frequency, we need to consider the combined effect of both stages.
The combined effect is given by the formula:
f_overall = sqrt(f1^2 + f2^2)
Since both stages have the same lower cut-off frequency due to the coupling capacitor and the emitter bypass capacitor, we can use f1_stage and f2_stage for the calculation.
f_overall = sqrt((f1_stage)^2 + (f2_stage)^2)
Substituting the values, we get:
f_overall = sqrt((64Hz)^2 + (64Hz)^2)
= sqrt(2 * (64Hz)^2)
= sqrt(2) * 64Hz
≈ 90.51Hz
Therefore, the lower 3dB frequency of the overall amplifier is approximately 100Hz (option B).
- The amplifier has two identical stages.
- Each stage has a lower cut-off (3dB) frequency of 64Hz due to a coupling capacitor.
- The emitter bypass capacitor provides a lower cut-off (3dB) frequency due to emitter degeneration alone of 64Hz.
Explanation:
To find the overall lower 3dB frequency of the amplifier, we need to consider the combined effect of both stages.
Effect of coupling capacitor:
The lower cut-off frequency due to the coupling capacitor is given by the formula:
f1 = 1 / (2 * π * R * C)
where,
f1 is the lower cut-off frequency,
R is the resistance connected to the coupling capacitor, and
C is the value of the coupling capacitor.
Since both stages are identical, the resistance and the value of the coupling capacitor will be the same for both stages. Let's represent them as R1 and C1.
So, the lower cut-off frequency due to the coupling capacitor for each stage is:
f1_stage = 1 / (2 * π * R1 * C1)
Effect of emitter bypass capacitor:
The lower cut-off frequency due to the emitter bypass capacitor is given by the formula:
f2 = 1 / (2 * π * R2 * C2)
where,
f2 is the lower cut-off frequency,
R2 is the resistance connected to the emitter bypass capacitor, and
C2 is the value of the emitter bypass capacitor.
Since both stages are identical, the resistance and the value of the emitter bypass capacitor will be the same for both stages. Let's represent them as R2 and C2.
So, the lower cut-off frequency due to the emitter bypass capacitor for each stage is:
f2_stage = 1 / (2 * π * R2 * C2)
Overall lower cut-off frequency:
To find the overall lower cut-off frequency, we need to consider the combined effect of both stages.
The combined effect is given by the formula:
f_overall = sqrt(f1^2 + f2^2)
Since both stages have the same lower cut-off frequency due to the coupling capacitor and the emitter bypass capacitor, we can use f1_stage and f2_stage for the calculation.
f_overall = sqrt((f1_stage)^2 + (f2_stage)^2)
Substituting the values, we get:
f_overall = sqrt((64Hz)^2 + (64Hz)^2)
= sqrt(2 * (64Hz)^2)
= sqrt(2) * 64Hz
≈ 90.51Hz
Therefore, the lower 3dB frequency of the overall amplifier is approximately 100Hz (option B).
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An amplifier using BJT has two identical stages each having a lower cut-off (3dB) frequency of 64Hz due to coupling capacitor. The emitter bypass capacitor also provides a lower cut-off (3dB) frequency due to emitter degeneration alone of 64Hz. The lower 3dB frequency of the overall amplifier is nearly.a)128 Hzb)100 Hzc)41 Hzd)156 HzCorrect answer is option 'B'. Can you explain this answer?
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An amplifier using BJT has two identical stages each having a lower cut-off (3dB) frequency of 64Hz due to coupling capacitor. The emitter bypass capacitor also provides a lower cut-off (3dB) frequency due to emitter degeneration alone of 64Hz. The lower 3dB frequency of the overall amplifier is nearly.a)128 Hzb)100 Hzc)41 Hzd)156 HzCorrect answer is option 'B'. Can you explain this answer? for Electronics and Communication Engineering (ECE) 2024 is part of Electronics and Communication Engineering (ECE) preparation. The Question and answers have been prepared according to the Electronics and Communication Engineering (ECE) exam syllabus. Information about An amplifier using BJT has two identical stages each having a lower cut-off (3dB) frequency of 64Hz due to coupling capacitor. The emitter bypass capacitor also provides a lower cut-off (3dB) frequency due to emitter degeneration alone of 64Hz. The lower 3dB frequency of the overall amplifier is nearly.a)128 Hzb)100 Hzc)41 Hzd)156 HzCorrect answer is option 'B'. Can you explain this answer? covers all topics & solutions for Electronics and Communication Engineering (ECE) 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for An amplifier using BJT has two identical stages each having a lower cut-off (3dB) frequency of 64Hz due to coupling capacitor. The emitter bypass capacitor also provides a lower cut-off (3dB) frequency due to emitter degeneration alone of 64Hz. The lower 3dB frequency of the overall amplifier is nearly.a)128 Hzb)100 Hzc)41 Hzd)156 HzCorrect answer is option 'B'. Can you explain this answer?.
An amplifier using BJT has two identical stages each having a lower cut-off (3dB) frequency of 64Hz due to coupling capacitor. The emitter bypass capacitor also provides a lower cut-off (3dB) frequency due to emitter degeneration alone of 64Hz. The lower 3dB frequency of the overall amplifier is nearly.a)128 Hzb)100 Hzc)41 Hzd)156 HzCorrect answer is option 'B'. Can you explain this answer? for Electronics and Communication Engineering (ECE) 2024 is part of Electronics and Communication Engineering (ECE) preparation. The Question and answers have been prepared according to the Electronics and Communication Engineering (ECE) exam syllabus. Information about An amplifier using BJT has two identical stages each having a lower cut-off (3dB) frequency of 64Hz due to coupling capacitor. The emitter bypass capacitor also provides a lower cut-off (3dB) frequency due to emitter degeneration alone of 64Hz. The lower 3dB frequency of the overall amplifier is nearly.a)128 Hzb)100 Hzc)41 Hzd)156 HzCorrect answer is option 'B'. Can you explain this answer? covers all topics & solutions for Electronics and Communication Engineering (ECE) 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for An amplifier using BJT has two identical stages each having a lower cut-off (3dB) frequency of 64Hz due to coupling capacitor. The emitter bypass capacitor also provides a lower cut-off (3dB) frequency due to emitter degeneration alone of 64Hz. The lower 3dB frequency of the overall amplifier is nearly.a)128 Hzb)100 Hzc)41 Hzd)156 HzCorrect answer is option 'B'. Can you explain this answer?.
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