Explanation Given: The circuit is given as: The virtual short circuit at the op-amp input causes the current through the R 1 resistors to be v i d 2 R 1 . Marking the current and voltage in above circuit: Applying Kirchhoff’s voltage law to the v i d v 1 v 2 : − v i d + i 1 R 1 + i 1 R 1 + 0 = 0 − v i d + 2 i 1 = 0 i 1 = v i d 2 R 1 Applying Kirchhoff’s voltage law to the v 2 : i 1 + v 2 − v y R 2 + 0 = 0 i R 2 + v 2 − v y R 2 = 0 i R 2 + v 2 − v y = 0 ( v i d 2 R 1 ) R 2 + v 2 − v y = 0 v 2 = v y − ( v i d 2 R 1 ) R 2 Applying the nodal analysis at inverting node: − i 1 + v 1 − v x R 2 + 0 = 0 − i 1 R 2 + v 1 − v x R 2 = 0 − i 1 R 2 + v 1 − v x = 0 − ( v i d 2 R 1 ) R 2 + v 1 − v x = 0 v 1 = v x + ( v i d 2 R 1 ) R 2 By the use of virtual ground concept, the voltage at the inverting node is equal to the non-inverting terminal. v 1 = v 2 v x + ( v i d 2 R 1 ) R 2 = v y − ( v i d 2 R 1 ) R 2 v x − v y = − ( v i d 2 R 1 ) R 2 − ( v i d 2 R 1 ) R 2 v x − v y = − 2 ( v i d 2 R 1 ) R 2 v x − v y = − v i d ( R 2 R 1 ) Applying the nodal analysis at node v y : v y − v x R G + v y − v 2 R 2 + v 2 R 2 = 0 v i d ( R 2 R 1 ) R G + v y R 2 − v 2 R 2 + v y R 2 = 0 v i d ( R 2 R 1 R G ) + 2 v y R 2 − 1 R 2 ( v y − ( v i d 2 R 1 ) R 2 ) = 0 v i d ( R 2 R 1 R G + 1 2 R 1 ) + v y R 2 = 0 v y R 2 = − v i d ( R 2 R 1 R G + 1 2 R 1 ) v y = − v i d R 2 ( R 2 R 1 R G + 1 2 R 1 ) Applying the nodal analysis at node, v x : v x − v 1 R 2 + v x − v y R G + v x − v o R 2 = 0 v x R 2 − v 1 R 2 + v x − v y R G + v x R 2 − v o R 2 = 0 2 v x R 2 − v 1 R 2 − v i d ( R 2 R 1 ) R G − v o R 2 = 0 2 v x R 2 − 1 R 2 ( v x + ( v i d

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

Differential amplifier

There are two amplifying stages with common degeneration in a differential amplifier.

Videos

Question

Chapter 2, Problem 2.70P

To determine

To prove: The differential voltage gain is given by vovid=−2R2R1[1+R2RG] .

Expert Solution & Answer

Want to see the full answer?

Check out a sample textbook solution

See solution

Chapter 2, Problem D2.69P

Chapter 2, Problem D2.71P

Students have asked these similar questions

1) Op-amp limiting factors - Current Saturation U2 Vin R1 OUT OPAMP R2 RLoad The saturation output current for the above op-amp is +/- 15mA. For the following input signals, determine if the output appears as would be expected in an ideal circuit. Include sketches of the output voltage. a) R1 = 100, R2 = 900, an open circuit load, and Vin is 2Vpp triangle wave with zero offset voltage (Vmax 1V, Vmin = -1V) and a period of 2ms. b) R1 = 100, R2 = 900, a 1k load, and Vin is 2Vpp triangle wave with a 1V offset voltage (Vmax=2V, Vmin = 0V) and a period of 2ms. c) R1 = 100, R2 = 900, an open circuit load, and Vin is 4Vpp triangle wave with zero offset voltage (Vmax=2V, Vmin = -2V) and a period of 2ms.

2)For the following circuit, assuming the Op-Amp is ideal, fill in the table for the missing values when vi = 2V, 12 10 k [mA] [mA] [mA] [mA] ...... I kn lo I2 ...... IL Io o vo vi ........

For the op-amp shown in the following figure, if R1 = 100 2 and Rf = 1 k2, the overall voltage gain is equal to ---------. V₁ O a. -10 O b.-1 O c. 11 O d. 9 Rf V₂

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.

Similar questions

SEE MORE QUESTIONS

Recommended textbooks for you

Introductory Circuit Analysis (13th Edition)
Electrical Engineering
ISBN:9780133923605
Author:Robert L. Boylestad
Publisher:PEARSON
Delmar's Standard Textbook Of Electricity
Electrical Engineering
ISBN:9781337900348
Author:Stephen L. Herman
Publisher:Cengage Learning
Programmable Logic Controllers
Electrical Engineering
ISBN:9780073373843
Author:Frank D. Petruzella
Publisher:McGraw-Hill Education
Fundamentals of Electric Circuits
Electrical Engineering
ISBN:9780078028229
Author:Charles K Alexander, Matthew Sadiku
Publisher:McGraw-Hill Education
Electric Circuits. (11th Edition)
Electrical Engineering
ISBN:9780134746968
Author:James W. Nilsson, Susan Riedel
Publisher:PEARSON
Engineering Electromagnetics
Electrical Engineering
ISBN:9780078028151
Author:Hayt, William H. (william Hart), Jr, BUCK, John A.
Publisher:Mcgraw-hill Education,