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Consider series-shunt amplifier in which the open loop gain is Av = 105 and the closed loop gain is AVf = 50. Assume the input and output resistance of the basic amplifier are Ri = 10 kΩ and Ro = 20 kΩ. Determine the input resistance of series input connecting and the output resistance of a shunt output connection for an ideal feedback voltage amplifier.
- a)Rin = 20 MΩ and Rout = 20 Ω
- b)Rin = 40 MΩ and Rout = 10 Ω
- c)Rin = 20 MΩ and Rout = 10 Ω
- d)Rin = 30 MΩ and Rout = 200 Ω
Correct answer is option 'C'. Can you explain this answer?
Verified Answer
Consider series-shunt amplifier in which the open loop gain is Av= 105...
The ideal closed-loop voltage transfer function
The input resistance
The output resistance is
The input resistance
The output resistance isMost Upvoted Answer
Consider series-shunt amplifier in which the open loop gain is Av= 105...
Ω and Ro= 100 Ω respectively. Determine the values of the feedback resistor Rf and the shunt resistor Rs.
We can use the formula for closed loop gain in a series-shunt amplifier:
AVf = Av / (1 + Av*(Rs/Ri) + (Rs/Ro) + (Rs/Rf))
Substituting the given values:
50 = 105 / (1 + 105*(Rs/10kΩ) + (Rs/100Ω) + (Rs/Rf))
Simplifying:
50 = 105 / (1 + 10.5*(Rs/10kΩ) + (Rs/1000Ω) + (Rs/Rf))
Multiplying both sides by the denominator:
1 + 10.5*(Rs/10kΩ) + (Rs/1000Ω) + (Rs/Rf) = 2.1
Subtracting 1 from both sides:
10.5*(Rs/10kΩ) + (Rs/1000Ω) + (Rs/Rf) = 1.1
We have one equation with two variables, so we need another equation. We know that the feedback network should provide a feedback factor of β = AVf / Av = 50 / 105 = 0.476. We can use the formula for the feedback factor in a series-shunt amplifier:
β = (Rs / (Rs + Ri)) * (Rf / (Rf + Ro))
Substituting the given values:
0.476 = (Rs / (Rs + 10kΩ)) * (Rf / (Rf + 100Ω))
Simplifying:
0.476 = (Rs / (Rs + 10kΩ)) * (Rf / (Rf + 100Ω))
Multiplying both sides by the denominators:
0.476 * (Rs + 10kΩ) * (Rf + 100Ω) = Rs * Rf
Expanding:
0.476 * Rs * Rf + 0.0476 * Rs * 100Ω + 0.476 * 10kΩ * Rf + 0.0476 * 10kΩ * 100Ω = Rs * Rf
Simplifying:
0.4292 * Rs * Rf + 476Ω * Rs + 4.76MΩ * Rf + 476kΩ = Rs * Rf
Subtracting 476kΩ from both sides:
0.4292 * Rs * Rf + 476Ω * Rs + 4.76MΩ * Rf = Rs * Rf - 476kΩ
We now have two equations with two variables, which we can solve simultaneously. We can rearrange the first equation to get Rs/Rf in terms of the other variables:
Rs/Rf = (105/50 - 1) / (1 + 10.5*(Rs/10kΩ) + (Rs/1000Ω))
Rs/Rf = 0.1 - 0.00952*(Rs/10kΩ) - 0.000952*(Rs/1000Ω)
Substituting this into the second equation:
0.4292 * Rs * (
We can use the formula for closed loop gain in a series-shunt amplifier:
AVf = Av / (1 + Av*(Rs/Ri) + (Rs/Ro) + (Rs/Rf))
Substituting the given values:
50 = 105 / (1 + 105*(Rs/10kΩ) + (Rs/100Ω) + (Rs/Rf))
Simplifying:
50 = 105 / (1 + 10.5*(Rs/10kΩ) + (Rs/1000Ω) + (Rs/Rf))
Multiplying both sides by the denominator:
1 + 10.5*(Rs/10kΩ) + (Rs/1000Ω) + (Rs/Rf) = 2.1
Subtracting 1 from both sides:
10.5*(Rs/10kΩ) + (Rs/1000Ω) + (Rs/Rf) = 1.1
We have one equation with two variables, so we need another equation. We know that the feedback network should provide a feedback factor of β = AVf / Av = 50 / 105 = 0.476. We can use the formula for the feedback factor in a series-shunt amplifier:
β = (Rs / (Rs + Ri)) * (Rf / (Rf + Ro))
Substituting the given values:
0.476 = (Rs / (Rs + 10kΩ)) * (Rf / (Rf + 100Ω))
Simplifying:
0.476 = (Rs / (Rs + 10kΩ)) * (Rf / (Rf + 100Ω))
Multiplying both sides by the denominators:
0.476 * (Rs + 10kΩ) * (Rf + 100Ω) = Rs * Rf
Expanding:
0.476 * Rs * Rf + 0.0476 * Rs * 100Ω + 0.476 * 10kΩ * Rf + 0.0476 * 10kΩ * 100Ω = Rs * Rf
Simplifying:
0.4292 * Rs * Rf + 476Ω * Rs + 4.76MΩ * Rf + 476kΩ = Rs * Rf
Subtracting 476kΩ from both sides:
0.4292 * Rs * Rf + 476Ω * Rs + 4.76MΩ * Rf = Rs * Rf - 476kΩ
We now have two equations with two variables, which we can solve simultaneously. We can rearrange the first equation to get Rs/Rf in terms of the other variables:
Rs/Rf = (105/50 - 1) / (1 + 10.5*(Rs/10kΩ) + (Rs/1000Ω))
Rs/Rf = 0.1 - 0.00952*(Rs/10kΩ) - 0.000952*(Rs/1000Ω)
Substituting this into the second equation:
0.4292 * Rs * (
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Question Description
Consider series-shunt amplifier in which the open loop gain is Av= 105and the closed loop gain is AVf= 50. Assume the input and output resistance of the basic amplifier are Ri= 10 kΩ and Ro= 20 kΩ. Determine the input resistance of series input connecting and the output resistance of a shunt output connection for an ideal feedback voltage amplifier.a)Rin= 20 MΩ and Rout= 20 Ωb)Rin= 40 MΩ and Rout= 10 Ωc)Rin= 20 MΩ and Rout= 10 Ωd)Rin= 30 MΩ and Rout= 200 ΩCorrect answer is option 'C'. Can you explain this answer? for Electronics and Communication Engineering (ECE) 2026 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 Consider series-shunt amplifier in which the open loop gain is Av= 105and the closed loop gain is AVf= 50. Assume the input and output resistance of the basic amplifier are Ri= 10 kΩ and Ro= 20 kΩ. Determine the input resistance of series input connecting and the output resistance of a shunt output connection for an ideal feedback voltage amplifier.a)Rin= 20 MΩ and Rout= 20 Ωb)Rin= 40 MΩ and Rout= 10 Ωc)Rin= 20 MΩ and Rout= 10 Ωd)Rin= 30 MΩ and Rout= 200 ΩCorrect answer is option 'C'. Can you explain this answer? covers all topics & solutions for Electronics and Communication Engineering (ECE) 2026 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Consider series-shunt amplifier in which the open loop gain is Av= 105and the closed loop gain is AVf= 50. Assume the input and output resistance of the basic amplifier are Ri= 10 kΩ and Ro= 20 kΩ. Determine the input resistance of series input connecting and the output resistance of a shunt output connection for an ideal feedback voltage amplifier.a)Rin= 20 MΩ and Rout= 20 Ωb)Rin= 40 MΩ and Rout= 10 Ωc)Rin= 20 MΩ and Rout= 10 Ωd)Rin= 30 MΩ and Rout= 200 ΩCorrect answer is option 'C'. Can you explain this answer?.
Consider series-shunt amplifier in which the open loop gain is Av= 105and the closed loop gain is AVf= 50. Assume the input and output resistance of the basic amplifier are Ri= 10 kΩ and Ro= 20 kΩ. Determine the input resistance of series input connecting and the output resistance of a shunt output connection for an ideal feedback voltage amplifier.a)Rin= 20 MΩ and Rout= 20 Ωb)Rin= 40 MΩ and Rout= 10 Ωc)Rin= 20 MΩ and Rout= 10 Ωd)Rin= 30 MΩ and Rout= 200 ΩCorrect answer is option 'C'. Can you explain this answer? for Electronics and Communication Engineering (ECE) 2026 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 Consider series-shunt amplifier in which the open loop gain is Av= 105and the closed loop gain is AVf= 50. Assume the input and output resistance of the basic amplifier are Ri= 10 kΩ and Ro= 20 kΩ. Determine the input resistance of series input connecting and the output resistance of a shunt output connection for an ideal feedback voltage amplifier.a)Rin= 20 MΩ and Rout= 20 Ωb)Rin= 40 MΩ and Rout= 10 Ωc)Rin= 20 MΩ and Rout= 10 Ωd)Rin= 30 MΩ and Rout= 200 ΩCorrect answer is option 'C'. Can you explain this answer? covers all topics & solutions for Electronics and Communication Engineering (ECE) 2026 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Consider series-shunt amplifier in which the open loop gain is Av= 105and the closed loop gain is AVf= 50. Assume the input and output resistance of the basic amplifier are Ri= 10 kΩ and Ro= 20 kΩ. Determine the input resistance of series input connecting and the output resistance of a shunt output connection for an ideal feedback voltage amplifier.a)Rin= 20 MΩ and Rout= 20 Ωb)Rin= 40 MΩ and Rout= 10 Ωc)Rin= 20 MΩ and Rout= 10 Ωd)Rin= 30 MΩ and Rout= 200 ΩCorrect answer is option 'C'. Can you explain this answer?.
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