FE REVIEW OPERATIONAL AMPLIFIERS (OP-AMPS)
|
|
- Horace Welch
- 6 years ago
- Views:
Transcription
1 FE EIEW OPEATIONAL AMPLIFIES (OPAMPS) 1 The Opamp An opamp has two nputs and one output. Note the opamp below. The termnal labeled wth the () sgn s the nvertng nput and the nput labeled wth the () sgn s the nonnvertng termnal. The standard way to show the devce s wth the nvertng termnal on top but ths s not always the case. BE CAEFUL!! deal The Ideal assumptons When we assume Ideal characterstcs or an opamp we are able to make several assumptons whch make analyss MUCH easer. The rst o these s that the nput resstance o each o the two termnals s Innte. n deal 1
2 Ideal assumptons contnued) The next major assumpton s that the output mpedance s equal to zero. deal out 0 Ths assumpton allows us to make the urther assumpton that ut s ndependent o the load. Ths means that t can not be loaded down. Ideal assumptons contnued) The next assumpton s that the openloop gan (A) s nnte (n real le s n the mllons so ths s a sae assumpton). Open Loop Gan deal A 5 Ideal assumptons contnued) The assumpton that A s nnte allows us to make a urther assumpton that the voltage on each termnal s equal to each other. Ths s concept s called a rtual Short. Ths assumpton turns out to be the KEY to opamp crcut analyss. ut A( ) Ad ut ut d so, lm A 0 deal A A t ollows that 6
3 Ideal assumptons contnued) 7 A rtual short s not lke a real short (or t wouldn t be vrtual). A vrtual short only aects voltage, not current. The Invertng Opamp = 0 volts due to the rtual 8 I Short to the ground I I = 0A Ideal Wth the addton o two resstors and a ground on the nonnvertng termnal we get an nvertng opamp. Note that snce the nonnvertng termnal s grounded the voltage on the nvertng termnal s also grounded due to the rtual short. The Invertng Opamp contnued) 9 Note that as long as there sn t a voltage source on the nonnvertng termnal there could be a dozen resstors on t and t stll would not aect the voltage on that termnal. Snce the current s assumed to be zero on the nvertng termnal, KCL proves that current I s equal to the eedback current I. 0 I I I I I I I 0
4 The Invertng Opamp contnued) Another aspect o the vrtual short s that the nput resstance o the nvertng opamp s equal to. 0 v I 10 = 0 volts I due to the rtual Short to the ground I I = 0A Ideal The Invertng Opamp contnued) The closedloop gan o the deal nvertng opamp s: I I I = 0A Ideal 11 A CL The FE exam uses a (A) or closedloop gan. The Invertng Opamp contnued) Fnd, I, I, and n or the crcut shown: k I o v 10k v 5v 10 k v v 0v I I 1mA k k n 1 10k Ideal I
5 The Nonnvertng opamp I you sht the nput voltage down to the nonnvertng termnal and ground you now have a Nonnvertng opamp. = I 1 10k 5k I 0A 1k deal The Nonnvertng opamp contnued) A The gan o the Nonnvertng opamp s: 1 cl 1 5k I Note that snce there s 0A on the nonnvertng termnal, there wll be 0 volts across the resstor so the voltage on the nvertng termnal wll be. = I 0A 1k 10k deal The Nonnvertng opamp contnued) The nput resstance o an deal nonnvertng opamp wll be nnte. 15 = I 10k n 5k I 0A 1k deal 5
6 The ltage Follower A ltage ollower s a specal case o the Nonnvertng opamp wth a voltage gan o A CL n 1 The Ideal Opamp Integrator I Ic C deal 17 1 d C 0 t t t o 0 v (t) t, s wth: =10kΩ C=00 μ v o(t) v t, s v The Ideal Opamp Derentator 18 d C t C I dt Ic deal wth 50k C.1n v (t) v o(t) t, s t, s 6
7 Example What s the output voltage,, o the deal opamp crcut shown? The crcut s congured as An Invertng Opamp. 1k 10k deal o 10k 1k 0 Example What s the current, I, (nvertng termnal current) n I the crcut shown? 8 a) 0.88A v b) 0.5A 8 I c) 0A v deal d) 0.5A Snce the nput resstance o the nvertng and nonnvertng nputs s deally nnte, the current nto those nputs s deally 0A. So the answer s: c) 0A Example What s the output voltage o the crcut shown? v v 8 v 1 7v o v v 8 I 8 I deal The crcut s an opamp summng crcut or a Lnear Combnaton Crcut. 7
8 Example For the Derence Ampler crcut shown, determne the output voltage. a) 8.v 0v b) v c) 15.5v d) 18.1v We haven t talked about a derence amp but t can be easly solved wth KCL, the rtual short, and Ohm s law v 5v 5 I 0 I deal Frst, the 5v n the equaton s the voltage on the nonnvertng leg voltage dvded between the 5 and the ohm resstors to gve us. Contnued on (Example contnued) Next, due to the vrtual short, the nvertng termnal also has 5 volts on t. Now we can determne I, whch s the derence between the voltages on each sde =5v dvded by 15 ohms. I 15 0v 5 5v 5v I 0 I I deal 15 0v 5v A (Example contnued) Now we note that snce the current nto the nvertng termnal s zero, I = 1.667A by KCL. 0v 5v 0 I I I A 0A I I 1.667A 8
9 (Example contnued) 5 0v 5v Fnally we wrte an equaton or n terms o I. I 0 5v 1.667A 0 5v 1.667A 0 5v.v 8.v Example For the deal operatonal ampler below, what should the value o be n order to obtan a gan o 5? 6 Example contnued on. (Example contnued) 7 1k k I k deal F IF Frst, note that the opamp s upsde down rom the way we normally look at t. Ths s the way that the exam seems to show t a Lot, but t s not ndustry standard. So, n the next slde we wll lp t over. Example contnued on 9
10 (Example contnued) 8 1k k I k deal F IF Now that t s lpped over, note that the voltage on the nonnvertng termnal s a voltage dvded. k 1k k Example contnued on (Example contnued) The voltage dvder has been replaced by the result o the dvder. Next, note that due to the vrtual short the voltage on the nvertng termnal, s also /. 9 due to the vrtual short Example contnued on (Example contnued) We can now nd an equaton or I. The voltage on the butt o the current arrow s / and the voltage at the tp o the current arrow s 0 volts. Dvde that derence voltage by k ohms and you get I. 0 I k k k 0 due to the vrtual short Example contnued on 10
11 (Example contnued) 1 F By KCL, 0 I I I 0 I I 0 I I k I I 0A deal Now, remember that the INFINITE mpedance o the Invertng and nonnvertng termnals means that the current out o those termnals s assumed to be zero. Usng ths and KCL (above) we nd that the eed back current, I s equal to I. Problem contnued on. (Example contnued) F I I k I I 0A deal Now we can dene ths current n terms o. The butt o the arrow s whle voltage at the tp o the arrow s (/). Problem contnued on. (Example contnued) The problem statement states that the gan s 5. So, that means that =5 I I 5 I I So, the answer to the queston s that t an 5 value o 19.5k ohms wll result n k a gan o k 19. 5k k 11
12 Example Evaluate the ollowng ampler to determne the value o requred to obtan a voltage gan o M deal Contnued on (Example contnued) 5 Note that the nvertng termnal voltage s 0 volts due to vrtual short whch exsts between t and the nonnvertng termnal. Contnued on (Example contnued) We can now nd an equaton or the current I. Note that the voltage at the butt o the arrow s and the voltage at the tp o the arrow s 0v. We use ths voltage derence and Ohms law to nd an equaton or I. 6 0v I 1M 1M Contnued on 1
13 (Example contnued) Now, usng the act that the current nto both the nvertng and nonnvertng termnals s assumed to be 0A due to the assumed nnte nput resstance, we can use KCL to show that I equals I. 0 I I I 0 I I 0A I I 7 Contnued on We can use the same technque as beore to dene I n terms o x. and then equate agan to I. (Example contnued) 8 0 1M 0v x I I x I so x I I = 0 due to rtual short x I 100 I I deal Contnued on (Example contnued) 9 Next dene the current I n terms o and x. I x 100 I I I x 1M I deal Contnued on 1
14 (Example contnued) 0 And now use KCL to dene the currents n terms o voltage. I x 100 I I 0 I I I I I I 100 By KCL x o x x 1M I deal Contnued on (Example contnued) 1 From the problem statement: and I = I 1 so we can expand that to x 500 k x 1M 1M x x x M 1M 1M 100 Contnued on (Example contnued) 10 1M 1M 1M M 1M 1M 1 M M M 119.5M.895k k 1 5k 1 5k So, the value o requred to have a voltage gan o 10 s k ohms. 1
FE REVIEW OPERATIONAL AMPLIFIERS (OP-AMPS)( ) 8/25/2010
FE REVEW OPERATONAL AMPLFERS (OP-AMPS)( ) 1 The Op-amp 2 An op-amp has two nputs and one output. Note the op-amp below. The termnal labeled l wth the (-) sgn s the nvertng nput and the nput labeled wth
More informationChapter 6. Operational Amplifier. inputs can be defined as the average of the sum of the two signals.
6 Operatonal mpler Chapter 6 Operatonal mpler CC Symbol: nput nput Output EE () Non-nvertng termnal, () nvertng termnal nput mpedance : Few mega (ery hgh), Output mpedance : Less than (ery low) Derental
More informationMAE140 - Linear Circuits - Winter 16 Midterm, February 5
Instructons ME140 - Lnear Crcuts - Wnter 16 Mdterm, February 5 () Ths exam s open book. You may use whatever wrtten materals you choose, ncludng your class notes and textbook. You may use a hand calculator
More informationMAE140 - Linear Circuits - Fall 13 Midterm, October 31
Instructons ME140 - Lnear Crcuts - Fall 13 Mdterm, October 31 () Ths exam s open book. You may use whatever wrtten materals you choose, ncludng your class notes and textbook. You may use a hand calculator
More informationLecture 5: Operational Amplifiers and Op Amp Circuits
Lecture 5: peratonal mplers and p mp Crcuts Gu-Yeon We Dson o Engneerng and ppled Scences Harard Unersty guyeon@eecs.harard.edu We erew eadng S&S: Chapter Supplemental eadng Background rmed wth our crcut
More informationSelected Student Solutions for Chapter 2
/3/003 Assessment Prolems Selected Student Solutons for Chapter. Frst note that we know the current through all elements n the crcut except the 6 kw resstor (the current n the three elements to the left
More informationDepartment of Electrical and Computer Engineering FEEDBACK AMPLIFIERS
Department o Electrcal and Computer Engneerng UNIT I EII FEEDBCK MPLIFIES porton the output sgnal s ed back to the nput o the ampler s called Feedback mpler. Feedback Concept: block dagram o an ampler
More informationMAE140 - Linear Circuits - Winter 16 Final, March 16, 2016
ME140 - Lnear rcuts - Wnter 16 Fnal, March 16, 2016 Instructons () The exam s open book. You may use your class notes and textbook. You may use a hand calculator wth no communcaton capabltes. () You have
More informationFEEDBACK AMPLIFIERS. v i or v s v 0
FEEDBCK MPLIFIERS Feedback n mplers FEEDBCK IS THE PROCESS OF FEEDING FRCTION OF OUTPUT ENERGY (VOLTGE OR CURRENT) BCK TO THE INPUT CIRCUIT. THE CIRCUIT EMPLOYED FOR THIS PURPOSE IS CLLED FEEDBCK NETWORK.
More informationProf. Paolo Colantonio a.a
Pro. Paolo olantono a.a. 3 4 Let s consder a two ports network o Two ports Network o L For passve network (.e. wthout nternal sources or actve devces), a general representaton can be made by a sutable
More informationLinearity. If kx is applied to the element, the output must be ky. kx ky. 2. additivity property. x 1 y 1, x 2 y 2
Lnearty An element s sad to be lnear f t satsfes homogenety (scalng) property and addte (superposton) property. 1. homogenety property Let x be the nput and y be the output of an element. x y If kx s appled
More informationWhy working at higher frequencies?
Advanced course on ELECTRICAL CHARACTERISATION OF NANOSCALE SAMPLES & BIOCHEMICAL INTERFACES: methods and electronc nstrumentaton. MEASURING SMALL CURRENTS When speed comes nto play Why workng at hgher
More informationCOLLEGE OF ENGINEERING PUTRAJAYA CAMPUS FINAL EXAMINATION SPECIAL SEMESTER 2013 / 2014
OLLEGE OF ENGNEENG PUTAJAYA AMPUS FNAL EXAMNATON SPEAL SEMESTE 03 / 04 POGAMME SUBJET ODE SUBJET : Bachelor of Electrcal & Electroncs Engneerng (Honours) Bachelor of Electrcal Power Engneerng (Honours)
More informationLesson 16: Basic Control Modes
0/8/05 Lesson 6: Basc Control Modes ET 438a Automatc Control Systems Technology lesson6et438a.tx Learnng Objectves Ater ths resentaton you wll be able to: Descrbe the common control modes used n analog
More informationG = G 1 + G 2 + G 3 G 2 +G 3 G1 G2 G3. Network (a) Network (b) Network (c) Network (d)
Massachusetts Insttute of Technology Department of Electrcal Engneerng and Computer Scence 6.002 í Electronc Crcuts Homework 2 Soluton Handout F98023 Exercse 21: Determne the conductance of each network
More informationDiode. Current HmAL Voltage HVL Simplified equivalent circuit. V γ. Reverse bias. Forward bias. Designation: Symbol:
Dode Materal: Desgnaton: Symbol: Poste Current flow: ptype ntype Anode Cathode Smplfed equalent crcut Ideal dode Current HmAL 0 8 6 4 2 Smplfed model 0.5.5 2 V γ eal dode Voltage HVL V γ closed open V
More informationE40M Device Models, Resistors, Voltage and Current Sources, Diodes, Solar Cells. M. Horowitz, J. Plummer, R. Howe 1
E40M Devce Models, Resstors, Voltage and Current Sources, Dodes, Solar Cells M. Horowtz, J. Plummer, R. Howe 1 Understandng the Solar Charger Lab Project #1 We need to understand how: 1. Current, voltage
More informationECSE Linearity Superposition Principle Superposition Example Dependent Sources. 10 kω. 30 V 5 ma. 6 kω. 2 kω
S-00 Lnearty Superposton Prncple Superposton xample Dependent Sources Lecture 4. sawyes@rp.edu www.rp.edu/~sawyes 0 kω 6 kω 8 V 0 V 5 ma 4 Nodes Voltage Sources Ref Unknown Node Voltage, kω If hae multple
More informationEE C245 ME C218 Introduction to MEMS Design
EE C45 ME C8 Introducton to MEM Desgn Fall 7 Prof. Clark T.C. Nguyen Dept. of Electrcal Engneerng & Computer cences Unersty of Calforna at Berkeley Berkeley, C 947 Dscusson: eew of Op mps EE C45: Introducton
More informationINDUCTANCE. RC Cicuits vs LR Circuits
INDUTANE R cuts vs LR rcuts R rcut hargng (battery s connected): (1/ )q + (R)dq/ dt LR rcut = (R) + (L)d/ dt q = e -t/ R ) = / R(1 - e -(R/ L)t ) q ncreases from 0 to = dq/ dt decreases from / R to 0 Dschargng
More informationEE 330 Lecture 24. Small Signal Analysis Small Signal Analysis of BJT Amplifier
EE 0 Lecture 4 Small Sgnal Analss Small Sgnal Analss o BJT Ampler Eam Frda March 9 Eam Frda Aprl Revew Sesson or Eam : 6:00 p.m. on Thursda March 8 n Room Sweene 6 Revew rom Last Lecture Comparson o Gans
More informationCHAPTER 13. Exercises. E13.1 The emitter current is given by the Shockley equation:
HPT 3 xercses 3. The emtter current s gen by the Shockley equaton: S exp VT For operaton wth, we hae exp >> S >>, and we can wrte VT S exp VT Solng for, we hae 3. 0 6ln 78.4 mv 0 0.784 5 4.86 V VT ln 4
More information55:141 Advanced Circuit Techniques Two-Port Theory
55:4 Adanced Crcut Technques Two-Port Theory Materal: Lecture Notes A. Kruger 55:4: Adanced Crcut Technques The Unersty of Iowa, 03 Two-Port Theory, Slde What Are Two-Ports? Basc dea: replace a complex
More information55:141 Advanced Circuit Techniques Two-Port Theory
55:4 Adanced Crcut Technques Two-Port Theory Materal: Lecture Notes A. Kruger 55:4: Adanced Crcut Technques The Unersty of Iowa, 205 Two-Port Theory, Slde Two-Port Networks Note, the BJT s all are hghly
More informationUNIVERSITY OF UTAH ELECTRICAL & COMPUTER ENGINEERING DEPARTMENT. 10k. 3mH. 10k. Only one current in the branch:
UNIERSITY OF UTH ELECTRICL & COMPUTER ENGINEERING DEPRTMENT ECE 70 HOMEWORK #6 Soluton Summer 009. fter beng closed a long tme, the swtch opens at t = 0. Fnd (t) for t > 0. t = 0 0kΩ 0kΩ 3mH Step : (Redraw
More informationIntroduction to circuit analysis. Classification of Materials
Introducton to crcut analyss OUTLINE Electrcal quanttes Charge Current Voltage Power The deal basc crcut element Sgn conventons Current versus voltage (I-V) graph Readng: 1.2, 1.3,1.6 Lecture 2, Slde 1
More informationAdvanced Circuits Topics - Part 1 by Dr. Colton (Fall 2017)
Advanced rcuts Topcs - Part by Dr. olton (Fall 07) Part : Some thngs you should already know from Physcs 0 and 45 These are all thngs that you should have learned n Physcs 0 and/or 45. Ths secton s organzed
More informationDesigning Information Devices and Systems II Spring 2018 J. Roychowdhury and M. Maharbiz Discussion 3A
EECS 16B Desgnng Informaton Devces and Systems II Sprng 018 J. Roychowdhury and M. Maharbz Dscusson 3A 1 Phasors We consder snusodal voltages and currents of a specfc form: where, Voltage vt) = V 0 cosωt
More informationGeneral Tips on How to Do Well in Physics Exams. 1. Establish a good habit in keeping track of your steps. For example, when you use the equation
General Tps on How to Do Well n Physcs Exams 1. Establsh a good habt n keepng track o your steps. For example when you use the equaton 1 1 1 + = d d to solve or d o you should rst rewrte t as 1 1 1 = d
More information(8) Gain Stage and Simple Output Stage
EEEB23 Electoncs Analyss & Desgn (8) Gan Stage and Smple Output Stage Leanng Outcome Able to: Analyze an example of a gan stage and output stage of a multstage amplfe. efeence: Neamen, Chapte 11 8.0) ntoducton
More informationI 2 V V. = 0 write 1 loop equation for each loop with a voltage not in the current set of equations. or I using Ohm s Law V 1 5.
Krchoff s Laws Drect: KL, KL, Ohm s Law G G Ohm s Law: 6 (always get equaton/esor) Ω 5 Ω 6Ω 4 KL: : 5 : 5 eq. are dependent (n general, get n ndep. for nodes) KL: 4 wrte loop equaton for each loop wth
More informationMAE140 - Linear Circuits - Fall 10 Midterm, October 28
M140 - Lnear rcuts - Fall 10 Mdterm, October 28 nstructons () Ths exam s open book. You may use whatever wrtten materals you choose, ncludng your class notes and textbook. You may use a hand calculator
More informationELECTRONICS. EE 42/100 Lecture 4: Resistive Networks and Nodal Analysis. Rev B 1/25/2012 (9:49PM) Prof. Ali M. Niknejad
A. M. Nknejad Unversty of Calforna, Berkeley EE 100 / 42 Lecture 4 p. 1/14 EE 42/100 Lecture 4: Resstve Networks and Nodal Analyss ELECTRONICS Rev B 1/25/2012 (9:49PM) Prof. Al M. Nknejad Unversty of Calforna,
More informationPhysics 4B. A positive value is obtained, so the current is counterclockwise around the circuit.
Physcs 4B Solutons to Chapter 7 HW Chapter 7: Questons:, 8, 0 Problems:,,, 45, 48,,, 7, 9 Queston 7- (a) no (b) yes (c) all te Queston 7-8 0 μc Queston 7-0, c;, a;, d; 4, b Problem 7- (a) Let be the current
More informationENGR-4300 Electronic Instrumentation Quiz 4 Fall 2010 Name Section. Question Value Grade I 20 II 20 III 20 IV 20 V 20. Total (100 points)
ENGR-43 Electronc Instrumentaton Quz 4 Fall 21 Name Secton Queston Value Grade I 2 II 2 III 2 IV 2 V 2 Total (1 ponts) On all questons: SHOW LL WORK. EGIN WITH FORMULS, THEN SUSTITUTE VLUES ND UNITS. No
More information6.01: Introduction to EECS I Lecture 7 March 15, 2011
6.0: Introducton to EECS I Lecture 7 March 5, 20 6.0: Introducton to EECS I Crcuts The Crcut Abstracton Crcuts represent systems as connectons of elements through whch currents (through arables) flow and
More informationEECE 301 Signals & Systems Prof. Mark Fowler
-T Sytem: Ung Bode Plot EEE 30 Sgnal & Sytem Pro. Mark Fowler Note Set #37 /3 Bode Plot Idea an Help Vualze What rcut Do Lowpa Flter Break Pont = / H ( ) j /3 Hghpa Flter c = / L Bandpa Flter n nn ( a)
More informationII. PASSIVE FILTERS. H(j ω) Pass. Stop
II. PASSIE FILTES Frequency-selectve or flter crcuts pass to the output only those nput sgnals that are n a desred range of frequences (called pass band). The ampltude of sgnals outsde ths range of frequences
More information36.1 Why is it important to be able to find roots to systems of equations? Up to this point, we have discussed how to find the solution to
ChE Lecture Notes - D. Keer, 5/9/98 Lecture 6,7,8 - Rootndng n systems o equatons (A) Theory (B) Problems (C) MATLAB Applcatons Tet: Supplementary notes rom Instructor 6. Why s t mportant to be able to
More informationDC Circuits. Crossing the emf in this direction +ΔV
DC Crcuts Delverng a steady flow of electrc charge to a crcut requres an emf devce such as a battery, solar cell or electrc generator for example. mf stands for electromotve force, but an emf devce transforms
More informationSections begin this week. Cancelled Sections: Th Labs begin this week. Attend your only second lab slot this week.
Announcements Sectons begn ths week Cancelled Sectons: Th 122. Labs begn ths week. Attend your only second lab slot ths week. Cancelled labs: ThF 25. Please check your Lab secton. Homework #1 onlne Due
More informationEnergy Storage Elements: Capacitors and Inductors
CHAPTER 6 Energy Storage Elements: Capactors and Inductors To ths pont n our study of electronc crcuts, tme has not been mportant. The analyss and desgns we hae performed so far hae been statc, and all
More informationI. INTRODUCTION. 1.1 Circuit Theory Fundamentals
I. INTRODUCTION 1.1 Crcut Theory Fundamentals Crcut theory s an approxmaton to Maxwell s electromagnetc equatons n order to smplfy analyss of complcated crcuts. A crcut s made of seeral elements (boxes
More informationELG 2135 ELECTRONICS I SECOND CHAPTER: OPERATIONAL AMPLIFIERS
ELG 35 ELECTONICS I SECOND CHAPTE: OPEATIONAL AMPLIFIES Sesson Wnter 003 Dr. M. YAGOUB Second Chapter: Operatonal amplfers II - _ After reewng the basc aspects of amplfers, we wll ntroduce a crcut representng
More informationFundamental loop-current method using virtual voltage sources technique for special cases
Fundamental loop-current method usng vrtual voltage sources technque for specal cases George E. Chatzaraks, 1 Marna D. Tortorel 1 and Anastasos D. Tzolas 1 Electrcal and Electroncs Engneerng Departments,
More informationElectrochemistry Thermodynamics
CHEM 51 Analytcal Electrochemstry Chapter Oct 5, 016 Electrochemstry Thermodynamcs Bo Zhang Department of Chemstry Unversty of Washngton Seattle, WA 98195 Former SEAC presdent Andy Ewng sellng T-shrts
More informationKey component in Operational Amplifiers
Key component n Operatonal Amplfers Objectve of Lecture Descrbe how dependent voltage and current sources functon. Chapter.6 Electrcal Engneerng: Prncples and Applcatons Chapter.6 Fundamentals of Electrc
More informationFoundations of Arithmetic
Foundatons of Arthmetc Notaton We shall denote the sum and product of numbers n the usual notaton as a 2 + a 2 + a 3 + + a = a, a 1 a 2 a 3 a = a The notaton a b means a dvdes b,.e. ac = b where c s an
More information55:041 Electronic Circuits
55:04 Electronc Crcuts Feedback & Stablty Sectons of Chapter 2. Kruger Feedback & Stablty Confguraton of Feedback mplfer Negate feedback β s the feedback transfer functon S o S S o o S S o f S S S S fb
More informationComplex Numbers. x = B B 2 4AC 2A. or x = x = 2 ± 4 4 (1) (5) 2 (1)
Complex Numbers If you have not yet encountered complex numbers, you wll soon do so n the process of solvng quadratc equatons. The general quadratc equaton Ax + Bx + C 0 has solutons x B + B 4AC A For
More informationANALOG ELECTRONICS I. Transistor Amplifiers DR NORLAILI MOHD NOH
241 ANALO LTRONI I Lectures 2&3 ngle Transstor Amplfers R NORLAILI MOH NOH 3.3 Basc ngle-transstor Amplfer tages 3 dfferent confguratons : 1. ommon-emtter ommon-source Ib B R I d I c o R o gnal appled
More informationElectrical Circuits 2.1 INTRODUCTION CHAPTER
CHAPTE Electrcal Crcuts. INTODUCTION In ths chapter, we brefly revew the three types of basc passve electrcal elements: resstor, nductor and capactor. esstance Elements: Ohm s Law: The voltage drop across
More informationPhysics Courseware Electronics
Physcs ourseware Electroncs ommon emtter amplfer Problem 1.- In the followg ommon Emtter mplfer calculate: a) The Q pot, whch s the D base current (I ), the D collector current (I ) and the voltage collector
More informationComplex Numbers, Signals, and Circuits
Complex Numbers, Sgnals, and Crcuts 3 August, 009 Complex Numbers: a Revew Suppose we have a complex number z = x jy. To convert to polar form, we need to know the magntude of z and the phase of z. z =
More informationPhysics 114 Exam 2 Fall 2014 Solutions. Name:
Physcs 114 Exam Fall 014 Name: For gradng purposes (do not wrte here): Queston 1. 1... 3. 3. Problem Answer each of the followng questons. Ponts for each queston are ndcated n red. Unless otherwse ndcated,
More informationmatter consists, measured in coulombs (C) 1 C of charge requires electrons Law of conservation of charge: charge cannot be created or
Basc Concepts Oerew SI Prefxes Defntons: Current, Voltage, Power, & Energy Passe sgn conenton Crcut elements Ideal s Portland State Unersty ECE 221 Basc Concepts Ver. 1.24 1 Crcut Analyss: Introducton
More information6.01: Introduction to EECS 1 Week 6 October 15, 2009
6.0: ntroducton to EECS Week 6 October 5, 2009 6.0: ntroducton to EECS Crcuts The Crcut Abstracton Crcuts represent systems as connectons of component through whch currents (through arables) flow and across
More informationWeek 11: Differential Amplifiers
ELE 0A Electronc rcuts Week : Dfferental Amplfers Lecture - Large sgnal analyss Topcs to coer A analyss Half-crcut analyss eadng Assgnment: hap 5.-5.8 of Jaeger and Blalock or hap 7. - 7.3, of Sedra and
More information(b) i(t) for t 0. (c) υ 1 (t) and υ 2 (t) for t 0. Solution: υ 2 (0 ) = I 0 R 1 = = 10 V. υ 1 (0 ) = 0. (Given).
Problem 5.37 Pror to t =, capactor C 1 n the crcut of Fg. P5.37 was uncharged. For I = 5 ma, R 1 = 2 kω, = 5 kω, C 1 = 3 µf, and C 2 = 6 µf, determne: (a) The equvalent crcut nvolvng the capactors for
More informationCollege of Engineering Department of Electronics and Communication Engineering. Test 1 With Model Answer
Name: Student D Number: Secton Number: 01/0/03/04 A/B Lecturer: Dr Jamaludn/ Dr Jehana Ermy/ Dr Azn Wat Table Number: College of Engneerng Department of Electroncs and Communcaton Engneerng Test 1 Wth
More informationPHYS 1441 Section 002 Lecture #16
PHYS 1441 Secton 00 Lecture #16 Monday, Mar. 4, 008 Potental Energy Conservatve and Non-conservatve Forces Conservaton o Mechancal Energy Power Today s homework s homework #8, due 9pm, Monday, Mar. 31!!
More information8.6 The Complex Number System
8.6 The Complex Number System Earler n the chapter, we mentoned that we cannot have a negatve under a square root, snce the square of any postve or negatve number s always postve. In ths secton we want
More informationEndogenous timing in a mixed oligopoly consisting of a single public firm and foreign competitors. Abstract
Endogenous tmng n a mxed olgopoly consstng o a sngle publc rm and oregn compettors Yuanzhu Lu Chna Economcs and Management Academy, Central Unversty o Fnance and Economcs Abstract We nvestgate endogenous
More informationChapter 6 Electrical Systems and Electromechanical Systems
ME 43 Systems Dynamcs & Control Chapter 6: Electrcal Systems and Electromechancal Systems Chapter 6 Electrcal Systems and Electromechancal Systems 6. INTODUCTION A. Bazoune The majorty of engneerng systems
More informationBoise State University Department of Electrical and Computer Engineering ECE 212L Circuit Analysis and Design Lab
Bose State Unersty Department of Electrcal and omputer Engneerng EE 1L rcut Analyss and Desgn Lab Experment #8: The Integratng and Dfferentatng Op-Amp rcuts 1 Objectes The objectes of ths laboratory experment
More information55:041 Electronic Circuits
55:04 Electronc Crcuts Feedback & Stablty Sectons of Chapter 2. Kruger Feedback & Stablty Confguraton of Feedback mplfer S o S ε S o ( S β S ) o Negate feedback S S o + β β s the feedback transfer functon
More informationDepartment of Statistics University of Toronto STA305H1S / 1004 HS Design and Analysis of Experiments Term Test - Winter Solution
Department of Statstcs Unversty of Toronto STA35HS / HS Desgn and Analyss of Experments Term Test - Wnter - Soluton February, Last Name: Frst Name: Student Number: Instructons: Tme: hours. Ads: a non-programmable
More informationPHYS 1443 Section 004 Lecture #12 Thursday, Oct. 2, 2014
PHYS 1443 Secton 004 Lecture #1 Thursday, Oct., 014 Work-Knetc Energy Theorem Work under rcton Potental Energy and the Conservatve Force Gravtatonal Potental Energy Elastc Potental Energy Conservaton o
More information2.3 Nilpotent endomorphisms
s a block dagonal matrx, wth A Mat dm U (C) In fact, we can assume that B = B 1 B k, wth B an ordered bass of U, and that A = [f U ] B, where f U : U U s the restrcton of f to U 40 23 Nlpotent endomorphsms
More informationModule 9. Lecture 6. Duality in Assignment Problems
Module 9 1 Lecture 6 Dualty n Assgnment Problems In ths lecture we attempt to answer few other mportant questons posed n earler lecture for (AP) and see how some of them can be explaned through the concept
More information( ) = ( ) + ( 0) ) ( )
EETOMAGNETI OMPATIBIITY HANDBOOK 1 hapter 9: Transent Behavor n the Tme Doman 9.1 Desgn a crcut usng reasonable values for the components that s capable of provdng a tme delay of 100 ms to a dgtal sgnal.
More informationMassachusetts Institute of Technology Department of Electrical Engineering and Computer Science Circuits and Electronics Spring 2001
Massachusetts Insttute of Technology Department of Electrcal Engneerng and Computer Scence Read Chapters 11 through 12. 6.002 Crcuts and Electroncs Sprng 2001 Homework #5 Handout S01031 Issued: 3/8/2001
More informationVI. Transistor Amplifiers
VI. Transstor Amplfers 6. Introducton In ths secton we wll use the transstor small-sgnal model to analyze and desgn transstor amplfers. There are two ssues that we need to dscuss frst: ) What are the mportant
More informationUnit 1. Current and Voltage U 1 VOLTAGE AND CURRENT. Circuit Basics KVL, KCL, Ohm's Law LED Outputs Buttons/Switch Inputs. Current / Voltage Analogy
..2 nt Crcut Bascs KVL, KCL, Ohm's Law LED Outputs Buttons/Swtch Inputs VOLTAGE AND CRRENT..4 Current and Voltage Current / Voltage Analogy Charge s measured n unts of Coulombs Current Amount of charge
More informationQuantum Mechanics I - Session 4
Quantum Mechancs I - Sesson 4 Aprl 3, 05 Contents Operators Change of Bass 4 3 Egenvectors and Egenvalues 5 3. Denton....................................... 5 3. Rotaton n D....................................
More informationPhysics 2A Chapter 3 HW Solutions
Phscs A Chapter 3 HW Solutons Chapter 3 Conceptual Queston: 4, 6, 8, Problems: 5,, 8, 7, 3, 44, 46, 69, 70, 73 Q3.4. Reason: (a) C = A+ B onl A and B are n the same drecton. Sze does not matter. (b) C
More informationPHYS 1441 Section 002 Lecture #15
PHYS 1441 Secton 00 Lecture #15 Monday, March 18, 013 Work wth rcton Potental Energy Gravtatonal Potental Energy Elastc Potental Energy Mechancal Energy Conservaton Announcements Mdterm comprehensve exam
More informationLectures - Week 4 Matrix norms, Conditioning, Vector Spaces, Linear Independence, Spanning sets and Basis, Null space and Range of a Matrix
Lectures - Week 4 Matrx norms, Condtonng, Vector Spaces, Lnear Independence, Spannng sets and Bass, Null space and Range of a Matrx Matrx Norms Now we turn to assocatng a number to each matrx. We could
More informationLecture 2 Solution of Nonlinear Equations ( Root Finding Problems )
Lecture Soluton o Nonlnear Equatons Root Fndng Problems Dentons Classcaton o Methods Analytcal Solutons Graphcal Methods Numercal Methods Bracketng Methods Open Methods Convergence Notatons Root Fndng
More information3.2 Terminal Characteristics of Junction Diodes (pp )
/9/008 secton3_termnal_characterstcs_of_juncton_odes.doc /6 3. Termnal Characterstcs of Juncton odes (pp.47-53) A Juncton ode I.E., A real dode! Smlar to an deal dode, ts crcut symbol s: HO: The Juncton
More informationCircuits II EE221. Instructor: Kevin D. Donohue. Instantaneous, Average, RMS, and Apparent Power, and, Maximum Power Transfer, and Power Factors
Crcuts II EE1 Unt 3 Instructor: Ken D. Donohue Instantaneous, Aerage, RMS, and Apparent Power, and, Maxmum Power pp ransfer, and Power Factors Power Defntons/Unts: Work s n unts of newton-meters or joules
More informationKIRCHHOFF CURRENT LAW
KRCHHOFF CURRENT LAW ONE OF THE FUNDAMENTAL CONSERATON PRNCPLES N ELECTRCAL ENGNEERNG CHARGE CANNOT BE CREATED NOR DESTROYED NODES, BRANCHES, LOOPS A NODE CONNECTS SEERAL COMPONENTS. BUT T DOES NOT HOLD
More informationBoise State University Department of Electrical and Computer Engineering ECE 212L Circuit Analysis and Design Lab
Bose State Unersty Department of Electrcal and omputer Engneerng EE 1L rcut Analyss and Desgn Lab Experment #8: The Integratng and Dfferentatng Op-Amp rcuts 1 Objectes The objectes of ths laboratory experment
More informationV V. This calculation is repeated now for each current I.
Page1 Page2 The power supply oltage V = +5 olts and the load resstor R = 1 k. For the range of collector bas currents, I = 0.5 ma, 1 ma, 2.5 ma, 4 ma and 4.5 ma, determne the correspondng collector-to-emtter
More informationP A = (P P + P )A = P (I P T (P P ))A = P (A P T (P P )A) Hence if we let E = P T (P P A), We have that
Backward Error Analyss for House holder Reectors We want to show that multplcaton by householder reectors s backward stable. In partcular we wsh to show fl(p A) = P (A) = P (A + E where P = I 2vv T s the
More informationWork is the change in energy of a system (neglecting heat transfer). To examine what could
Work Work s the change n energy o a system (neglectng heat transer). To eamne what could cause work, let s look at the dmensons o energy: L ML E M L F L so T T dmensonally energy s equal to a orce tmes
More informationThe optimal delay of the second test is therefore approximately 210 hours earlier than =2.
THE IEC 61508 FORMULAS 223 The optmal delay of the second test s therefore approxmately 210 hours earler than =2. 8.4 The IEC 61508 Formulas IEC 61508-6 provdes approxmaton formulas for the PF for smple
More informationME2142/ME2142E Feedback Control Systems. Modelling of Physical Systems The Transfer Function
Mdellng Physcal Systems The Transer Functn Derental Equatns U Plant Y In the plant shwn, the nput u aects the respnse the utput y. In general, the dynamcs ths respnse can be descrbed by a derental equatn
More informationTHE CHINESE REMAINDER THEOREM. We should thank the Chinese for their wonderful remainder theorem. Glenn Stevens
THE CHINESE REMAINDER THEOREM KEITH CONRAD We should thank the Chnese for ther wonderful remander theorem. Glenn Stevens 1. Introducton The Chnese remander theorem says we can unquely solve any par of
More informationUNIT I BASIC CIRCUIT CONCEPTS
UNIT I BASIC CIRCUIT CONCEPTS Crcut elements Krchhoff s Law V-I Relatonshp of R,L and C Independent and Dependent sources Smple Resstve crcuts Networks reducton Voltage dvson current source transformaton.
More informationNovember 5, 2002 SE 180: Earthquake Engineering SE 180. Final Project
SE 8 Fnal Project Story Shear Frame u m Gven: u m L L m L L EI ω ω Solve for m Story Bendng Beam u u m L m L Gven: m L L EI ω ω Solve for m 3 3 Story Shear Frame u 3 m 3 Gven: L 3 m m L L L 3 EI ω ω ω
More information8.022 (E&M) Lecture 8
8.0 (E&M) Lecture 8 Topcs: Electromotve force Crcuts and Krchhoff s rules 1 Average: 59, MS: 16 Quz 1: thoughts Last year average: 64 test slghtly harder than average Problem 1 had some subtletes math
More informationTransfer Functions. Convenient representation of a linear, dynamic model. A transfer function (TF) relates one input and one output: ( ) system
Transfer Functons Convenent representaton of a lnear, dynamc model. A transfer functon (TF) relates one nput and one output: x t X s y t system Y s The followng termnology s used: x y nput output forcng
More informationElectrical Engineering Department Network Lab.
Electrcal Engneerng Department Network Lab. Objecte: - Experment on -port Network: Negate Impedance Conerter To fnd the frequency response of a smple Negate Impedance Conerter Theory: Negate Impedance
More informationPhysics 114 Exam 2 Spring Name:
Physcs 114 Exam Sprng 013 Name: For gradng purposes (do not wrte here): Queston 1. 1... 3. 3. Problem Answer each of the followng questons. Ponts for each queston are ndcated n red wth the amount beng
More informationIntroduction to Vapor/Liquid Equilibrium, part 2. Raoult s Law:
CE304, Sprng 2004 Lecture 4 Introducton to Vapor/Lqud Equlbrum, part 2 Raoult s Law: The smplest model that allows us do VLE calculatons s obtaned when we assume that the vapor phase s an deal gas, and
More informationChapter Newton s Method
Chapter 9. Newton s Method After readng ths chapter, you should be able to:. Understand how Newton s method s dfferent from the Golden Secton Search method. Understand how Newton s method works 3. Solve
More informationIntroduction to Electronic circuits.
Intrductn t Electrnc crcuts. Passve and Actve crcut elements. Capactrs, esstrs and Inductrs n AC crcuts. Vltage and current dvders. Vltage and current surces. Amplfers, and ther transfer characterstc.
More informationSolution of Linear System of Equations and Matrix Inversion Gauss Seidel Iteration Method
Soluton of Lnear System of Equatons and Matr Inverson Gauss Sedel Iteraton Method It s another well-known teratve method for solvng a system of lnear equatons of the form a + a22 + + ann = b a2 + a222
More informationONE-DIMENSIONAL COLLISIONS
Purpose Theory ONE-DIMENSIONAL COLLISIONS a. To very the law o conservaton o lnear momentum n one-dmensonal collsons. b. To study conservaton o energy and lnear momentum n both elastc and nelastc onedmensonal
More information