(a) Explanation Given: | T ( f ) | = 1 f ′ P 1 = 10 Hz f P 2 = 5 MHz f P 3 = 10 MHz T ( 0 ) = 100 dB Calculation: The expression for the transfer function is given by, T ( f ) = 10 5 ( 1 + j ( f f ′ p 1 ) ) ( 1 + j ( f f p 2 ) ) ( 1 + j ( f f p 3 ) ) Substitute 10 for f ′ P 1 , 5 × 10 6 for f P 2 and 10 × 10 6 for f P 3 in the above equation. T ( f ) = 10 5 ( 1 + j ( f 10 ) ) ( 1 + j ( f 5 × 10 6 ) ) ( 1 + j ( f 10 × 10 6 ) ) The expression to determine the value of the frequency at | T ( f ) | = 1 in the above equation. | T ( f ) | = ( 1 + j ( f 10 ) ) ( 1 + j ( f 5 × 10 6 ) ) ( 1 + j ( f 10 × 10 6 ) ) Substitute 1 for | T ( f ) | in the above equation (b) To determine The new dominant pole frequency and the value of the phase margin.

Microelectronics: Circuit Analysis and Design

4th Edition

ISBN: 9780073380643

Author: Donald A. Neamen

Publisher: McGraw-Hill Companies, The

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Chapter 12, Problem 12.87P

(a)

To determine

The frequency and the value of the phase margin.

(b)

To determine

The new dominant pole frequency and the value of the phase margin.

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Chapter 12, Problem 12.86P

Chapter 12, Problem 12.88P

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Chapter 12 Solutions

Microelectronics: Circuit Analysis and Design

Ch. 12 - (a) The open-loop gain of an amplifier is A=5104...Ch. 12 - (a) Consider a general feedback system with...Ch. 12 - (a) A feedback amplifier has an open-loop...Ch. 12 - (a) Consider the circuit shown in Figure...Ch. 12 - (a) The closed-loop gain of a feedback amplifier...Ch. 12 - The gain factors in a feedback system are A=5105...Ch. 12 - Prob. 12.3TYU Ch. 12 - An ideal series-shunt feedback amplifier is shown...Ch. 12 - Consider the ideal shunt-series feedback amplifier...Ch. 12 - An ideal series-series feedback amplifier is shown...

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The frequency and the value of the phase margin.

Solution Summary: The expression for the transfer function is given by, T(f)|=1 in the above equation.

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The frequency and the value of the phase margin.

The new dominant pole frequency and the value of the phase margin.

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Chapter 12 Solutions