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(Q.9-Q.10) In designing with op amps one has to check the limitations on the voltage and frequency ranges of operation of the closed-loop amplifier, imposed by the op-amp finite bandwidth (ft), slew rate (SR), and output saturation (Vo max). Consider the use of an op amp with ftt = 2 MHz, SR = 1 V/µs, and V0 max = 10 V in the design of a non-inverting amplifier with a nominal gain of 10. Assume a sine-wave input with peak amplitude Vi.
Q. If Vi = 0.5 V, what is the maximum frequency before the output distorts?
  • a)
    31.8 kHz
  • b)
    318 kHz
  • c)
    3.18 kHz
  • d)
     3.18 MHz
Correct answer is option 'A'. Can you explain this answer?
Verified Answer
(Q.9-Q.10) In designing with op amps one has to check the limitations ...
 Vi = 0.5v, V0 = 0.5 X 10 = 5V
2πf V0 = SR or f = 31.8 kHz.
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(Q.9-Q.10) In designing with op amps one has to check the limitations ...
Μs, and Vo max = ±10 V.

Q.9) What is the maximum closed-loop gain that can be achieved without causing the op amp to saturate?

To determine the maximum closed-loop gain without causing saturation, we need to consider the output voltage limitation (Vo max). The maximum output voltage swing is ±10V.

The formula for closed-loop gain (A_cl) is A_cl = -(R_f/R_in), where R_f is the feedback resistor and R_in is the input resistor.

Let's assume R_in = 1kΩ (1 kiloohm). By rearranging the formula, we can find the maximum value of R_f:

R_f = -A_cl * R_in

To avoid saturation, we need to make sure that the maximum output voltage (Vo max) is not exceeded. Therefore:

Vo max = A_cl * Vin

Since Vin is typically small (in the range of millivolts), we can assume it to be negligible.

Therefore:

Vo max = A_cl * 0

This implies that the maximum closed-loop gain (A_cl) is limited by Vo max.

Using the given Vo max = ±10V, we can calculate the maximum closed-loop gain:

±10V = A_cl * 0
±10V = 0

Since 0 times any value is 0, there is no maximum closed-loop gain that can be achieved without causing the op amp to saturate. The maximum closed-loop gain is limited by the output voltage saturation.

Q.10) What is the maximum frequency at which the op amp can operate in a closed-loop configuration without significant distortion, assuming the gain is set to 10?

To determine the maximum frequency without significant distortion, we need to consider the op amp's finite bandwidth (ft) and the gain setting of 10.

The formula for the frequency at which the op amp's gain starts to roll off is given by:

f_gain_roll-off = ft / A_cl

Given ft = 2 MHz and A_cl = 10, we can calculate the maximum frequency without significant distortion:

f_gain_roll-off = 2 MHz / 10
f_gain_roll-off = 200 kHz

Therefore, the maximum frequency at which the op amp can operate in a closed-loop configuration without significant distortion, assuming the gain is set to 10, is 200 kHz.
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(Q.9-Q.10) In designing with op amps one has to check the limitations on the voltage and frequency ranges of operation of the closed-loop amplifier, imposed by the op-amp finite bandwidth (ft), slew rate (SR), and output saturation (Vo max). Consider the use of an op amp with ftt= 2 MHz, SR = 1 V/µs, and V0 max= 10 V in the design of a non-inverting amplifier with a nominal gain of 10. Assume a sine-wave input with peak amplitude Vi.Q. If Vi = 0.5 V, what is the maximum frequency before the output distorts?a)31.8 kHzb)318 kHzc)3.18 kHzd)3.18 MHzCorrect answer is option 'A'. Can you explain this answer?
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(Q.9-Q.10) In designing with op amps one has to check the limitations on the voltage and frequency ranges of operation of the closed-loop amplifier, imposed by the op-amp finite bandwidth (ft), slew rate (SR), and output saturation (Vo max). Consider the use of an op amp with ftt= 2 MHz, SR = 1 V/µs, and V0 max= 10 V in the design of a non-inverting amplifier with a nominal gain of 10. Assume a sine-wave input with peak amplitude Vi.Q. If Vi = 0.5 V, what is the maximum frequency before the output distorts?a)31.8 kHzb)318 kHzc)3.18 kHzd)3.18 MHzCorrect answer is option 'A'. Can you explain this answer? for Electrical Engineering (EE) 2024 is part of Electrical Engineering (EE) preparation. The Question and answers have been prepared according to the Electrical Engineering (EE) exam syllabus. Information about (Q.9-Q.10) In designing with op amps one has to check the limitations on the voltage and frequency ranges of operation of the closed-loop amplifier, imposed by the op-amp finite bandwidth (ft), slew rate (SR), and output saturation (Vo max). Consider the use of an op amp with ftt= 2 MHz, SR = 1 V/µs, and V0 max= 10 V in the design of a non-inverting amplifier with a nominal gain of 10. Assume a sine-wave input with peak amplitude Vi.Q. If Vi = 0.5 V, what is the maximum frequency before the output distorts?a)31.8 kHzb)318 kHzc)3.18 kHzd)3.18 MHzCorrect answer is option 'A'. Can you explain this answer? covers all topics & solutions for Electrical Engineering (EE) 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for (Q.9-Q.10) In designing with op amps one has to check the limitations on the voltage and frequency ranges of operation of the closed-loop amplifier, imposed by the op-amp finite bandwidth (ft), slew rate (SR), and output saturation (Vo max). Consider the use of an op amp with ftt= 2 MHz, SR = 1 V/µs, and V0 max= 10 V in the design of a non-inverting amplifier with a nominal gain of 10. Assume a sine-wave input with peak amplitude Vi.Q. If Vi = 0.5 V, what is the maximum frequency before the output distorts?a)31.8 kHzb)318 kHzc)3.18 kHzd)3.18 MHzCorrect answer is option 'A'. Can you explain this answer?.
Solutions for (Q.9-Q.10) In designing with op amps one has to check the limitations on the voltage and frequency ranges of operation of the closed-loop amplifier, imposed by the op-amp finite bandwidth (ft), slew rate (SR), and output saturation (Vo max). Consider the use of an op amp with ftt= 2 MHz, SR = 1 V/µs, and V0 max= 10 V in the design of a non-inverting amplifier with a nominal gain of 10. Assume a sine-wave input with peak amplitude Vi.Q. If Vi = 0.5 V, what is the maximum frequency before the output distorts?a)31.8 kHzb)318 kHzc)3.18 kHzd)3.18 MHzCorrect answer is option 'A'. Can you explain this answer? in English & in Hindi are available as part of our courses for Electrical Engineering (EE). Download more important topics, notes, lectures and mock test series for Electrical Engineering (EE) Exam by signing up for free.
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