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
expand_more
expand_more
format_list_bulleted
Concept explainers
Question
Chapter 2, Problem D2.45P
To determine
To design: A circuit based on the functioning of the non-inverting amplifier to obtain the given gain.
The value of gain by using the given possibilities.
Expert Solution & Answer
Want to see the full answer?
Check out a sample textbook solutionStudents have asked these similar questions
Design an amplifier circuit using OPAMP with vo= (5*v1)+(v2)+(3*v3)-(2*v4). Draw vo knowing that v1 = 1 sin (1000*pi*t), v2 = 2 sin (2000*pi*t), v3 = 3sin (3000*pi*t), v4 = 4 sin (4000*pi*t)
A circuit is constructed by amplifier components as follow, R1= 4 Ω; R2= 2Ω; R3= 5 Ω; R4= 1 Ω; R5= 2 Ω,I= 0.3 A and Vs= 5 Volt. Find V1, V2,and V3.
An amplifier with 20dB gain is connected to another with 10dB gain by means of a transmission line with a loss of 4dB. If a signal with a power level of -14dBm were applied to the system, calculate the power output.
answer:12dbm pls show the correct solution
Chapter 2 Solutions
Microelectronic Circuits (The Oxford Series in Electrical and Computer Engineering) 7th edition
Ch. 2.1 - Prob. 2.1ECh. 2.1 - Prob. 2.2ECh. 2.1 - Prob. 2.3ECh. 2.2 - Prob. D2.4ECh. 2.2 - Prob. 2.5ECh. 2.2 - Prob. 2.6ECh. 2.2 - Prob. D2.7ECh. 2.2 - Prob. D2.8ECh. 2.3 - Prob. 2.9ECh. 2.3 - Prob. 2.10E
Ch. 2.3 - Prob. D2.11ECh. 2.3 - Prob. 2.12ECh. 2.3 - Prob. 2.13ECh. 2.3 - Prob. 2.14ECh. 2.4 - Prob. 2.15ECh. 2.4 - Prob. D2.16ECh. 2.4 - Prob. 2.17ECh. 2.5 - Prob. 2.18ECh. 2.5 - Prob. D2.19ECh. 2.5 - Prob. D2.20ECh. 2.6 - Prob. 2.21ECh. 2.6 - Prob. 2.22ECh. 2.6 - Prob. 2.23ECh. 2.6 - Prob. 2.24ECh. 2.6 - Prob. 2.25ECh. 2.7 - Prob. 2.26ECh. 2.7 - Prob. 2.27ECh. 2.7 - Prob. 2.28ECh. 2.8 - Prob. 2.29ECh. 2.8 - Prob. 2.30ECh. 2 - Prob. 2.1PCh. 2 - Prob. 2.2PCh. 2 - Prob. 2.3PCh. 2 - Prob. 2.4PCh. 2 - Prob. 2.5PCh. 2 - Prob. 2.6PCh. 2 - Prob. 2.7PCh. 2 - Prob. 2.8PCh. 2 - Prob. 2.9PCh. 2 - Prob. 2.10PCh. 2 - Prob. 2.11PCh. 2 - Prob. D2.12PCh. 2 - Prob. D2.13PCh. 2 - Prob. D2.14PCh. 2 - Prob. 2.15PCh. 2 - Prob. 2.16PCh. 2 - Prob. 2.17PCh. 2 - Prob. 2.18PCh. 2 - Prob. 2.19PCh. 2 - Prob. D2.20PCh. 2 - Prob. 2.21PCh. 2 - Prob. 2.22PCh. 2 - Prob. 2.23PCh. 2 - Prob. 2.24PCh. 2 - Prob. 2.25PCh. 2 - Prob. D2.26PCh. 2 - Prob. 2.27PCh. 2 - Prob. 2.28PCh. 2 - Prob. D2.29PCh. 2 - Prob. 2.30PCh. 2 - Prob. 2.31PCh. 2 - Prob. 2.32PCh. 2 - Prob. D2.33PCh. 2 - Prob. D2.34PCh. 2 - Prob. D2.35PCh. 2 - Prob. 2.36PCh. 2 - Prob. D2.37PCh. 2 - Prob. D2.38PCh. 2 - Prob. D2.39PCh. 2 - Prob. D2.40PCh. 2 - Prob. D2.41PCh. 2 - Prob. D2.42PCh. 2 - Prob. 2.43PCh. 2 - Prob. D2.44PCh. 2 - Prob. D2.45PCh. 2 - Prob. D2.46PCh. 2 - Prob. D2.47PCh. 2 - Prob. D2.48PCh. 2 - Prob. 2.49PCh. 2 - Prob. 2.50PCh. 2 - Prob. D2.51PCh. 2 - Prob. D2.52PCh. 2 - Prob. 2.53PCh. 2 - Prob. 2.54PCh. 2 - Prob. 2.55PCh. 2 - Prob. D2.56PCh. 2 - Prob. 2.57PCh. 2 - Prob. 2.58PCh. 2 - Prob. 2.59PCh. 2 - Prob. 2.60PCh. 2 - Prob. D2.61PCh. 2 - Prob. 2.62PCh. 2 - Prob. 2.63PCh. 2 - Prob. 2.64PCh. 2 - Prob. 2.65PCh. 2 - Prob. 2.66PCh. 2 - Prob. D2.67PCh. 2 - Prob. 2.68PCh. 2 - Prob. D2.69PCh. 2 - Prob. 2.70PCh. 2 - Prob. D2.71PCh. 2 - Prob. 2.72PCh. 2 - Prob. 2.73PCh. 2 - Prob. 2.74PCh. 2 - Prob. 2.75PCh. 2 - Prob. D2.76PCh. 2 - Prob. 2.77PCh. 2 - Prob. 2.78PCh. 2 - Prob. 2.79PCh. 2 - Prob. D2.80PCh. 2 - Prob. 2.81PCh. 2 - Prob. D2.82PCh. 2 - Prob. D2.83PCh. 2 - Prob. 2.84PCh. 2 - Prob. 2.85PCh. 2 - Prob. D2.86PCh. 2 - Prob. 2.87PCh. 2 - Prob. 2.88PCh. 2 - Prob. 2.89PCh. 2 - Prob. 2.90PCh. 2 - Prob. 2.91PCh. 2 - Prob. D2.92PCh. 2 - Prob. D2.93PCh. 2 - Prob. 2.94PCh. 2 - Prob. 2.95PCh. 2 - Prob. 2.96PCh. 2 - Prob. 2.97PCh. 2 - Prob. 2.98PCh. 2 - Prob. D2.99PCh. 2 - Prob. D2.100PCh. 2 - Prob. 2.101PCh. 2 - Prob. 2.102PCh. 2 - Prob. 2.103PCh. 2 - Prob. 2.104PCh. 2 - Prob. 2.105PCh. 2 - Prob. 2.106PCh. 2 - Prob. 2.107PCh. 2 - Prob. 2.108PCh. 2 - Prob. 2.109PCh. 2 - Prob. 2.110PCh. 2 - Prob. 2.111PCh. 2 - Prob. 2.112PCh. 2 - Prob. 2.113PCh. 2 - Prob. 2.114PCh. 2 - Prob. 2.115PCh. 2 - Prob. D2.116PCh. 2 - Prob. D2.117PCh. 2 - Prob. D2.118PCh. 2 - Prob. 2.119PCh. 2 - Prob. 2.120PCh. 2 - Prob. 2.121PCh. 2 - Prob. 2.122PCh. 2 - Prob. 2.123PCh. 2 - Prob. 2.124PCh. 2 - Prob. 2.125PCh. 2 - Prob. 2.126PCh. 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
- 13- Amplifiers are used to increase the value of Resistance. Select one: True Falsearrow_forwarda) A source vs = 2 V with internal source resistance Rs =10 kΩ applies a signal to a5 gain inverting amplifier with R1 =20 kΩ and R2 =100 kΩ. find the voltage ofoutput and verify that if a load is connected it will receive less than 10 V. (b) Find the value atthat we should change R2 to compensate and obtain 10 V at the output. present calculationsarrow_forwardThe op amp shown is ideal. Find vo if va=1.2 V, vb=-1.5 V, and vc=4 V.arrow_forward
- The op amp in the adder-subtracter circuit shown shown is ideal. Find vo when va=1 V, vb=2 V, vc=3 V, and vd=4 V.arrow_forwardA source vs =10V is fed to a voltage divider implemented with Ra= 120 kilo ohms and Rb= 30 kilo ohms and the voltage across Rb is fed in turn to a gain of 5 non inverting amplifier having R1 = 30 kilo ohms and R2= 120 kilo ohms. (a)Sketch the circuit and predict the amplifier Vo. (b)for a gain of 5 inverting amplifier having R1=30 kilo ohms and R2=150 kilo ohms. compre an dcomment on the differences. confirm results on simulator (multisim)arrow_forwardTwo identical amplifiers are connected in cascade, as shown .Each amplifier is described using its h parameters. The values areh11=1000Ω, h12=0.0015, h21=100, and h22=100 μS. Find the voltage gainV2/Vg.arrow_forward
- 1. In the difference amplifier shown, vb=4.0 V. What range of values for va will result in linear operation? 2. Repeat (a) with the 20 kΩ resistor decreased to 8 kΩ.arrow_forwardIn the circuit given below, if (V_1K>5V) then Vç=Vcc; else Vç=Vss. Design the opamp circuit that will fulfill the condition.arrow_forwardFor the circuit below, you may assume the op amps are ideal and V_Z = 5V. Use Ohm's Law, KCL, and/or KVL to determine values for R1,R2,R3,R4,and R5 such that V_out = 2.5V. Make surc your resistor values allow you to ignore the op amps' input resistances (1 M amps ) and output resistances (50amps).arrow_forward
- Helping tags: Electrical Engineering . . . WILL UPVOTE, just pls help me answer the question and show complete solutions. Thanks. . . . Find Rb, Ibq, Icq, Vre, Vrc , Vc , Vrb, Vb, Zb, Zin and Zo for the CE amplifier shown in fig.1 Remember Icmax = Vcc/Rc+Re Graph the DC loadline showing the Q point of the circuitarrow_forwardIs it possible to get a gain less than unity using a non-inverting amplifier configuration? If yes, sketch a circuit.''please solve it in computer font, not on paper''arrow_forwardhere im asked to find vx the way it was solved is that a node A was taken just before the op amp and nodal analysis aka kcl was applied and the node after it was called v0. but how did we name the first one A and the 2nd one v0? on what basis? also if im asked to get vx how do i think of taking the node A to get an equation so i can get Vx? what should i look for? im really bad at knowing what nodes to take and apply kcl or nodal analysis on to reach my desired solution so can you please help me with that? at the end of the answer it said vx=v0.2R / R+2R getting -2rA0 please can you address all my points so i can understand what and how to look for when solving this type of question?arrow_forward
arrow_back_ios
SEE MORE QUESTIONS
arrow_forward_ios
Recommended textbooks for you
- Delmar's Standard Textbook Of ElectricityElectrical EngineeringISBN:9781337900348Author:Stephen L. HermanPublisher:Cengage Learning
Delmar's Standard Textbook Of Electricity
Electrical Engineering
ISBN:9781337900348
Author:Stephen L. Herman
Publisher:Cengage Learning
Electrical Engineering: Ch 5: Operational Amp (2 of 28) Inverting Amplifier-Basic Operation; Author: Michel van Biezen;https://www.youtube.com/watch?v=x2xxOKOTwM4;License: Standard YouTube License, CC-BY