Explanation Given Information: Voltage gain of the op-amp inverting integrator at 10 kHz = − 100 V/V Gain after reduction is = − 1 V/V Calculation: An inverting based op-amp integrating circuit is shown below. For this circuit, the input and output voltages are Vin and Vout respectively and the ratio of output to input voltage in the complex frequency domain is called transfer function, the magnitude of which represents voltage gain of the circuit. The basic formula for the transfer function and inverting amplifier shown below. Vout(s) Vin(s) = − R 1 R2 Where Vout(s) and Vin(s) represents the Laplace transform of output and input voltages. In the case where the feedback resistor is replaced by a capacitor, the integrator circuit obtained is shown below, whose transfer function gets modified as Vout(s) Vin(s) = − X c R1 Where, X c represents capacitive impedance given by X c = 1 s C 1 = 1 j 2 π f C1 (Applying physical frequencies s = j ω = j 2 π f and considering magnitude) Here, f is the frequency of operation in Hz. Hence, Vout(j ω ) Vin(j ω ) = − 1 j ω R1C1 The magnitude of this function gives the voltage gain of the integrator, which is | Vout Vin | = 1 2 π f R1C1 (Using ω = 2 π f ) Given the voltage gain corresponding to 10 kHz , the frequency at which voltage gain becomes − 1 V/V is to be found. In terms of the frequency response, this implies the frequency at which the gains fall to 0 dB . This can be done by forming two equations for the given conditions and finding the unknown factor which is the frequency here. Secondly, the integrator time constant is to be found which is given by the general formula as τ = R1 × C1= 1 ω = 1 2 π f Here ω − integrator frequency in radians , corresponds to the frequency at which the gain attains − 1 V/V or 0 dB , which can be found from the previous step. At 10 kHz, | Vout Vin | = 100 V/V So, | Vout Vin | = 1 2 π f R1C1 = 1 2 π × 10 × 10 3 × R1C1 1 2 π × 10 × 10 3 × R1C1 = 100

Microelectronic Circuits (The Oxford Series in Electrical and Computer Engineering) 7th edition

7th Edition

ISBN: 9780199339136

Author: Adel S. Sedra, Kenneth C. Smith

Publisher: Oxford University Press

See similar textbooks

Concept explainers

Operational Amplifier

An operational amplifier also called op amp is an integrated circuit or analog circuit that takes input as a differential voltage and produces output as a single-ended voltage. It is an IC that can amplify weak electric signals.An op amp has one output p…

Videos

Question

Chapter 2, Problem 2.81P

To determine

The frequency at which the gain of the op-amp based inverting integrator reduces to the given value and the time constant of the integrator.

Expert Solution & Answer

Want to see the full answer?

Check out a sample textbook solution

See solution

Chapter 2, Problem D2.80P

Chapter 2, Problem D2.82P

Students have asked these similar questions

The upper critical frequency of an op-amp’s open-loop response is 200 Hz. If the midrange gain is 175,000, answer the following: What is the ideal gain at 200 Hz? What is the actual gain? What is the op-amp’s open-loop bandwidth?

An op-amp has an open-loop gain of 104 and a dominant open-loop pole at 4Hz. If an inverting amplifier with a gain of -100 is made from this op-amp, what is the resulting amplifier bandwidth? What is the gain-bandwidth product?

Suppose the op amp shown has an fT of 3 MHz. What is the phase margin of the ampliﬁer?

Chapter 2 Solutions

Microelectronic Circuits (The Oxford Series in Electrical and Computer Engineering) 7th edition

Ch. 2.1 - Prob. 2.1E Ch. 2.1 - Prob. 2.2E Ch. 2.1 - Prob. 2.3E Ch. 2.2 - Prob. D2.4E Ch. 2.2 - Prob. 2.5E Ch. 2.2 - Prob. 2.6E Ch. 2.2 - Prob. D2.7E Ch. 2.2 - Prob. D2.8E Ch. 2.3 - Prob. 2.9E Ch. 2.3 - Prob. 2.10E

Ch. 2.3 - Prob. D2.11E Ch. 2.3 - Prob. 2.12E Ch. 2.3 - Prob. 2.13E Ch. 2.3 - Prob. 2.14E Ch. 2.4 - Prob. 2.15E Ch. 2.4 - Prob. D2.16E Ch. 2.4 - Prob. 2.17E Ch. 2.5 - Prob. 2.18E Ch. 2.5 - Prob. D2.19E Ch. 2.5 - Prob. D2.20E Ch. 2.6 - Prob. 2.21E Ch. 2.6 - Prob. 2.22E Ch. 2.6 - Prob. 2.23E Ch. 2.6 - Prob. 2.24E Ch. 2.6 - Prob. 2.25E Ch. 2.7 - Prob. 2.26E Ch. 2.7 - Prob. 2.27E Ch. 2.7 - Prob. 2.28E Ch. 2.8 - Prob. 2.29E Ch. 2.8 - Prob. 2.30E Ch. 2 - Prob. 2.1P Ch. 2 - Prob. 2.2P Ch. 2 - Prob. 2.3P Ch. 2 - Prob. 2.4P Ch. 2 - Prob. 2.5P Ch. 2 - Prob. 2.6P Ch. 2 - Prob. 2.7P Ch. 2 - Prob. 2.8P Ch. 2 - Prob. 2.9P Ch. 2 - Prob. 2.10P Ch. 2 - Prob. 2.11P Ch. 2 - Prob. D2.12P Ch. 2 - Prob. D2.13P Ch. 2 - Prob. D2.14P Ch. 2 - Prob. 2.15P Ch. 2 - Prob. 2.16P Ch. 2 - Prob. 2.17P Ch. 2 - Prob. 2.18P Ch. 2 - Prob. 2.19P Ch. 2 - Prob. D2.20P Ch. 2 - Prob. 2.21P Ch. 2 - Prob. 2.22P Ch. 2 - Prob. 2.23P Ch. 2 - Prob. 2.24P Ch. 2 - Prob. 2.25P Ch. 2 - Prob. D2.26P Ch. 2 - Prob. 2.27P Ch. 2 - Prob. 2.28P Ch. 2 - Prob. D2.29P Ch. 2 - Prob. 2.30P Ch. 2 - Prob. 2.31P Ch. 2 - Prob. 2.32P Ch. 2 - Prob. D2.33P Ch. 2 - Prob. D2.34P Ch. 2 - Prob. D2.35P Ch. 2 - Prob. 2.36P Ch. 2 - Prob. D2.37P Ch. 2 - Prob. D2.38P Ch. 2 - Prob. D2.39P Ch. 2 - Prob. D2.40P Ch. 2 - Prob. D2.41P Ch. 2 - Prob. D2.42P Ch. 2 - Prob. 2.43P Ch. 2 - Prob. D2.44P Ch. 2 - Prob. D2.45P Ch. 2 - Prob. D2.46P Ch. 2 - Prob. D2.47P Ch. 2 - Prob. D2.48P Ch. 2 - Prob. 2.49P Ch. 2 - Prob. 2.50P Ch. 2 - Prob. D2.51P Ch. 2 - Prob. D2.52P Ch. 2 - Prob. 2.53P Ch. 2 - Prob. 2.54P Ch. 2 - Prob. 2.55P Ch. 2 - Prob. D2.56P Ch. 2 - Prob. 2.57P Ch. 2 - Prob. 2.58P Ch. 2 - Prob. 2.59P Ch. 2 - Prob. 2.60P Ch. 2 - Prob. D2.61P Ch. 2 - Prob. 2.62P Ch. 2 - Prob. 2.63P Ch. 2 - Prob. 2.64P Ch. 2 - Prob. 2.65P Ch. 2 - Prob. 2.66P Ch. 2 - Prob. D2.67P Ch. 2 - Prob. 2.68P Ch. 2 - Prob. D2.69P Ch. 2 - Prob. 2.70P Ch. 2 - Prob. D2.71P Ch. 2 - Prob. 2.72P Ch. 2 - Prob. 2.73P Ch. 2 - Prob. 2.74P Ch. 2 - Prob. 2.75P Ch. 2 - Prob. D2.76P Ch. 2 - Prob. 2.77P Ch. 2 - Prob. 2.78P Ch. 2 - Prob. 2.79P Ch. 2 - Prob. D2.80P Ch. 2 - Prob. 2.81P Ch. 2 - Prob. D2.82P Ch. 2 - Prob. D2.83P Ch. 2 - Prob. 2.84P Ch. 2 - Prob. 2.85P Ch. 2 - Prob. D2.86P Ch. 2 - Prob. 2.87P Ch. 2 - Prob. 2.88P Ch. 2 - Prob. 2.89P Ch. 2 - Prob. 2.90P Ch. 2 - Prob. 2.91P Ch. 2 - Prob. D2.92P Ch. 2 - Prob. D2.93P Ch. 2 - Prob. 2.94P Ch. 2 - Prob. 2.95P Ch. 2 - Prob. 2.96P Ch. 2 - Prob. 2.97P Ch. 2 - Prob. 2.98P Ch. 2 - Prob. D2.99P Ch. 2 - Prob. D2.100P Ch. 2 - Prob. 2.101P Ch. 2 - Prob. 2.102P Ch. 2 - Prob. 2.103P Ch. 2 - Prob. 2.104P Ch. 2 - Prob. 2.105P Ch. 2 - Prob. 2.106P Ch. 2 - Prob. 2.107P Ch. 2 - Prob. 2.108P Ch. 2 - Prob. 2.109P Ch. 2 - Prob. 2.110P Ch. 2 - Prob. 2.111P Ch. 2 - Prob. 2.112P Ch. 2 - Prob. 2.113P Ch. 2 - Prob. 2.114P Ch. 2 - Prob. 2.115P Ch. 2 - Prob. D2.116P Ch. 2 - Prob. D2.117P Ch. 2 - Prob. D2.118P Ch. 2 - Prob. 2.119P Ch. 2 - Prob. 2.120P Ch. 2 - Prob. 2.121P Ch. 2 - Prob. 2.122P Ch. 2 - Prob. 2.123P Ch. 2 - Prob. 2.124P Ch. 2 - Prob. 2.125P Ch. 2 - Prob. 2.126P Ch. 2 - Prob. D2.127P

Knowledge Booster

Learn more about

Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, electrical-engineering and related others by exploring similar questions and additional content below.

The frequency at which the gain of the op-amp based inverting integrator reduces to the given value and the time constant of the integrator.

Solution Summary: The author explains the basic formula for the transfer function and inverting amplifier.

Concept explainers

Videos

The frequency at which the gain of the op-amp based inverting integrator reduces to the given value and the time constant of the integrator.

Want to see the full answer?

Chapter 2 Solutions