EE 321 Analog Electronics, Fall 2013 Homework #3 solution


 Magdalene Allison
 3 years ago
 Views:
Transcription
1 EE 32 Analog Electronics, Fall 203 Homework #3 solution (a) Use superposition to show that the output of the circuit in Fig. P2.47 is given by + [ Rf v N + R f v N R ] f v Nn R N R N2 R [ Nn + R ][ f RP v P + v P R ] P v Pn R N 2 n where R N = R N R N2... R Nn and = 2... n 0. (b) Design a circuit to obtain 2v N +v P +2v P2 The smallest resistor used should be 0 kω. (a) First, for the negative inputs, let s work with the first input via superposition. The negative input is at virtual ground, so the input current is i N = v N R N. All that current flows out of the feedback resistor because the voltage across the other input resistors is zero. Thus, for only v N on, we get R f R N v N Similarly, by superposition, we can see that for only v Nx inputs turned on, then output is
2 R f n i= v Ni R Ni which is identical to the expression we need to show. Next, for the positive inputs, let s look at the first input. The voltage at the positive input is a voltage division between and the parallel combination of all the other resistors. v + = = n i=0 i n i= i + v P v P + = ( + = + v P = v P )v P Next apply superposition and we get for all positive inputs on and all negative inputs off, v + = n i= v P Next, this signal is amplified by the noninverting amplifier whose gain is composed from R f and the parallel resistance of all negative input resistors, R N, so [ + R ] n f v P R N R i= P which is also identical with what we had to show. Combining the output from negative inputs and positive inputs we get the result we were asked to show. (b) We have the equations R f R N = 2 ( + R ) f RP = R N ( + R ) f RP = 2 R N 2 Note that R N = R N, so that + R f R N = 3, and 2
3 R f R N = 2 = 3 2 = 2 3 We can now select R N = 0kΩ, so that R f = 20kΩ. Next, Next, we have 2 = 2 = = 2 = = = = 0 This means that 0 =. Then we can choose = 0kΩ and 2 = 20kΩ For the circuit shown in Fig. P2.62, express v O as a function of v and v 2. What is the input resistance seen by v alone? By v 2 alone? By a source connected between the two input terminals? By a source connected to both input terminals simultaneously? We can use superposition to get v O. First as a function of v, we have v. Next, as a function of v 2, R ( + R ) v 2 = v 2 R+R R Combining we get v 2 v. The input resistance on input is found as The relationship between the two is R i = dv di v = v 2 2 +i R 3
4 Plugging this into the expression above we get R i = R On input 2 it is clear that R i2 = 2R. For a source connected between the two inputs we have v 2 v = v, and a current through it, i, and we want the differential input resistance, R id = dv di We need a relationship between v and i. Note that v = v O. To have the same current in the two arms we must have that v v = v 2. Combining that with the input output reltionship, v = v 2 v, we can eliminate v or v v 2 = v = v 2 v v = v 2 We can also see that v 2 = 2Ri. Inserting we get v = 2Ri and see that R ic = 2R. 2.. An opamp intended for operation with a closedloop gain of 00 V/V uses feedback resistors of 0 kω and MΩ with a biascurrentcompensation resistor R 3. What should the value of R 3 be? With input grounded, the output offset voltage is found to be +0.2 V. Estimate the input offset current assuming zero input offset voltage. If the input offset voltage can be as large as mv of unknown polarity, what range of offset current is possible? What current injected into, or extracted from, the nongrounded end of R 3 would reduce the opamp output voltage to zero? For available ±5 V supplies, what resistor and supply voltage would you use? The biascurrent compensating resistor R 3 = R R 2 = 9.9kΩ. With the bias compensated and zero offset voltage the output is equal to the offset current flowing through the feedback resistor, R 2, I OS = v O R 2 = = 20nA If the input offset voltage is up to ±V, the output offset voltage can be as large as ±0.V. In that case, the output offset due to bias current can range between = 0.V to = 0.3V. That results in a minimum offset current of I OS,min = 0nA and a maximum offset current of I OS,max = 30nA. In order to reduce the output to zero we need to add a voltage on the positive input which cancels the output of 0.2V. That means we need to add a negative voltage on the positive input which is that output divided by the noninverting gain, 0. The voltage we 4
5 need to add is V compensation = 0.2V/0 = 2.08mV. That voltage is produced by a current through R 3. That current should be I compensation = V compensation = 2.08mV R 3 9.9kΩ = 20nA To get that current from the negative supply we need a resistor, R compensation = V supply I compensation = 5 = 7.4MΩ For the circuits shown in Fig. P3.2 using ideal diodes, find the values of the voltages and currents indicated. (a) In this case the diode is conducting, such that the voltage at the top end of the diode equals that at the bottom end, so V = 3V. Then, I = 6V 0kΩ = 0.6mA. (b) The diode is not conducting, so I = 0, and V = 3V. (c) The diode is conducting so I = 0.6mA as before, but now V = 3V. (d) The diode is not conducting, so I = 0 and V = 3V In each of the idealdiode circuits shown in Fig. P3.4, v I is a khz, 0V peak sine wave. Sketch the waveform resulting at v O. What are its positive and negative peak values? 5
6 (My plotting program is temporarily out of commission so you get a description in words and equations instead. Note that this procedure is probably the best way to go about plotting the waveform anyway; getting an expression and then plottinng that expression.) (a) Piecewise expression for v O Minimum value: 0V. Maximum value: 0V 0 v I 0 v I v I > 0 (b) Piecewise expression for v O v I v I 0 Minimum value: 0V. Maximum value: 0V. (c) Current never flows, so 0 always. (d) This case is identical to (a) 0 v I > 0 (e) Current always flows so v I. Minimum values: 0V. Maximum value: 0V. (f) This case is identical to (a) (g) Piecwise expression for v O 6
7 v I v I 0 Minimum value: 0V. Maximum value: 0V. 0 v I > 0 (h) The output is always connected to ground, 0 always. (i) When v I < 0 this acts as a voltage divider. Otherwise they two are equal. Piecewise expression is vi 2 v I 0 v I v I > 0 Minimum value: 5V. Maximum value: 0V. (j) When v I > 0 the output is shorted to the input. When v I < 0 the output is obtained from the voltage division. The expression for v O is identical to (i). (k) The current source causes V voltage drop across the resistor, such that v O is always one volt higher than the voltage at the bottom of the resistor. When v I > 0 the voltage at the bottom of the resistor is ground. When v I < 0 the voltage at the bottom of the resistor is v I. The piecewise expression for v O is v I +V v I 0 V v I > 0 7
The equivalent model of a certain op amp is shown in the figure given below, where R 1 = 2.8 MΩ, R 2 = 39 Ω, and A =
The equivalent model of a certain op amp is shown in the figure given below, where R 1 = 2.8 MΩ, R 2 = 39 Ω, and A = 10 10 4. Section Break Difficulty: Easy Learning Objective: Understand how real operational
More informationOperational Amplifiers
Operational Amplifiers A Linear IC circuit Operational Amplifier (opamp) An opamp is a highgain amplifier that has high input impedance and low output impedance. An ideal opamp has infinite gain and
More informationOperational amplifiers (Op amps)
Operational amplifiers (Op amps) v R o R i v i Av i v View it as an ideal amp. Take the properties to the extreme: R i, R o 0, A.?!?!?!?! v v i Av i v A Consequences: No voltage dividers at input or output.
More informationD is the voltage difference = (V +  V  ).
1 Operational amplifier is one of the most common electronic building blocks used by engineers. It has two input terminals: V + and V , and one output terminal Y. It provides a gain A, which is usually
More informationHomework 3 Solution. Due Friday (5pm), Feb. 14, 2013
University of California, Berkeley Spring 2013 EE 42/100 Prof. K. Pister Homework 3 Solution Due Friday (5pm), Feb. 14, 2013 Please turn the homework in to the drop box located next to 125 Cory Hall (labeled
More informationAt point G V = = = = = = RB B B. IN RB f
Common Emitter At point G CE RC 0. 4 12 0. 4 116. I C RC 116. R 1k C 116. ma I IC 116. ma β 100 F 116µ A I R ( 116µ A)( 20kΩ) 2. 3 R + 2. 3 + 0. 7 30. IN R f Gain in Constant Current Region I I I C F
More informationU1 is zero based because its noninverting terminal is connected to circuit common. Therefore, the circuit reference voltage is 0 V.
When you have completed this exercise, you will be able to operate a zenerclamped op amp comparator circuit using dc and ac voltages. You will verify your results with an oscilloscope. U1 is zero based
More informationECE2262 Electric Circuits. Chapter 4: Operational Amplifier (OPAMP) Circuits
ECE2262 Electric Circuits Chapter 4: Operational Amplifier (OPAMP) Circuits 1 4.1 Operational Amplifiers 2 4. Voltages and currents in electrical circuits may represent signals and circuits can perform
More informationEE 230 Lecture 20. Nonlinear Op Amp Applications. The Comparator Nonlinear Analysis Methods
EE 230 Lecture 20 Nonlinear Op Amp Applications The Comparator Nonlinear Analysis Methods Quiz 14 What is the major purpose of compensation when designing an operatinal amplifier? And the number is? 1
More informationOPERATIONAL AMPLIFIER ª Differentialinput, SingleEnded (or Differential) output, DCcoupled, HighGain amplifier
à OPERATIONAL AMPLIFIERS à OPERATIONAL AMPLIFIERS (Introduction and Properties) Phase relationships: Noninverting input to output is 0 Inverting input to output is 180 OPERATIONAL AMPLIFIER ª Differentialinput,
More informationHomework Assignment 08
Homework Assignment 08 Question 1 (Short Takes) Two points each unless otherwise indicated. 1. Give one phrase/sentence that describes the primary advantage of an active load. Answer: Large effective resistance
More informationE40M Review  Part 1
E40M Review Part 1 Topics in Part 1 (Today): KCL, KVL, Power Devices: V and I sources, R Nodal Analysis. Superposition Devices: Diodes, C, L Time Domain Diode, C, L Circuits Topics in Part 2 (Wed): MOSFETs,
More informationor Op Amps for short
or Op Amps for short Objective of Lecture Describe how an ideal operational amplifier (op amp) behaves. Define voltage gain, current gain, transresistance gain, and transconductance gain. Explain the operation
More informationCARLETON UNIVERSITY. FINAL EXAMINATION December DURATION 3 HOURS No. of Students 130
ALETON UNIVESITY FINAL EXAMINATION December 005 DUATION 3 HOUS No. of Students 130 Department Name & ourse Number: Electronics ELE 3509 ourse Instructor(s): Prof. John W. M. ogers and alvin Plett AUTHOIZED
More informationECE2210 Final given: Spring 08
ECE Final given: Spring 0. Note: feel free to show answers & work right on the schematic 1. (1 pts) The ammeter, A, reads 30 ma. a) The power dissipated by R is 0.7 W, what is the value of R. Assume that
More information55:041 Electronic Circuits The University of Iowa Fall Final Exam
Final Exam Name: Score Max: 135 Question 1 (1 point unless otherwise noted) a. What is the maximum theoretical efficiency for a classb amplifier? Answer: 78% b. The abbreviation/term ESR is often encountered
More informationEE201 Review Exam I. 1. The voltage Vx in the circuit below is: (1) 3V (2) 2V (3) 2V (4) 1V (5) 1V (6) None of above
EE201, Review Probs Test 1 page1 Spring 98 EE201 Review Exam I Multiple Choice (5 points each, no partial credit.) 1. The voltage Vx in the circuit below is: (1) 3V (2) 2V (3) 2V (4) 1V (5) 1V (6)
More informationSwitchedCapacitor Circuits David Johns and Ken Martin University of Toronto
SwitchedCapacitor Circuits David Johns and Ken Martin University of Toronto (johns@eecg.toronto.edu) (martin@eecg.toronto.edu) University of Toronto 1 of 60 Basic Building Blocks Opamps Ideal opamps usually
More informationVer 3537 E1.1 Analysis of Circuits (2014) E1.1 Circuit Analysis. Problem Sheet 1 (Lectures 1 & 2)
Ver 3537 E. Analysis of Circuits () Key: [A]= easy... [E]=hard E. Circuit Analysis Problem Sheet (Lectures & ). [A] One of the following circuits is a series circuit and the other is a parallel circuit.
More informationChapter 5 Objectives
Chapter 5 Engr228 Circuit Analysis Dr Curtis Nelson Chapter 5 Objectives State and apply the property of linearity State and apply the property of superposition Investigate source transformations Define
More informationHomework Assignment 09
Homework Assignment 09 Question 1 (Short Takes) Two points each unless otherwise indicated. 1. What is the 3dB bandwidth of the amplifier shown below if r π = 2.5K, r o = 100K, g m = 40 ms, and C L =
More informationEIT QuickReview Electrical Prof. Frank Merat
CIRCUITS 4 The power supplied by the 0 volt source is (a) 2 watts (b) 0 watts (c) 2 watts (d) 6 watts (e) 6 watts 4Ω 2Ω 0V i i 2 2Ω 20V Call the clockwise loop currents i and i 2 as shown in the drawing
More informationE40M. Op Amps. M. Horowitz, J. Plummer, R. Howe 1
E40M Op Amps M. Horowitz, J. Plummer, R. Howe 1 Reading A&L: Chapter 15, pp. 863866. Reader, Chapter 8 Noninverting Amp http://www.electronicstutorials.ws/opamp/opamp_3.html Inverting Amp http://www.electronicstutorials.ws/opamp/opamp_2.html
More informationOperational amplifiers (Op amps)
Operational amplifiers (Op amps) Recall the basic twoport model for an amplifier. It has three components: input resistance, Ri, output resistance, Ro, and the voltage gain, A. v R o R i v d Av d v Also
More informationECE 523/421  Analog Electronics University of New Mexico Solutions Homework 3
ECE 523/42  Analog Electronics University of New Mexico Solutions Homework 3 Problem 7.90 Show that when ro is taken into account, the voltage gain of the source follower becomes G v v o v sig R L r o
More informationElectronic Circuits. Prof. Dr. Qiuting Huang Integrated Systems Laboratory
Electronic Circuits Prof. Dr. Qiuting Huang 6. Transimpedance Amplifiers, Voltage Regulators, Logarithmic Amplifiers, AntiLogarithmic Amplifiers Transimpedance Amplifiers Sensing an input current ii in
More informationEE105 Fall 2014 Microelectronic Devices and Circuits
EE05 Fall 204 Microelectronic Devices and Circuits Prof. Ming C. Wu wu@eecs.berkeley.edu 5 Sutardja Dai Hall (SDH) Terminal Gain and I/O Resistances of BJT Amplifiers Emitter (CE) Collector (CC) Base (CB)
More informationDC Biasing. Dr. U. Sezen & Dr. D. Gökçen (Hacettepe Uni.) ELE230 Electronics I 15Mar / 59
Contents Three States of Operation BJT DC Analysis FixedBias Circuit EmitterStabilized Bias Circuit Voltage Divider Bias Circuit DC Bias with Voltage Feedback Various Dierent Bias Circuits pnp Transistors
More informationECE3050 Assignment 7
ECE3050 Assignment 7. Sketch and label the Bode magnitude and phase plots for the transfer functions given. Use loglog scales for the magnitude plots and linearlog scales for the phase plots. On the magnitude
More informationSummary Notes ALTERNATING CURRENT AND VOLTAGE
HIGHER CIRCUIT THEORY Wheatstone Bridge Circuit Any method of measuring resistance using an ammeter or voltmeter necessarily involves some error unless the resistances of the meters themselves are taken
More informationProblem Set 5 Solutions
University of California, Berkeley Spring 01 EE /0 Prof. A. Niknejad Problem Set 5 Solutions Please note that these are merely suggested solutions. Many of these problems can be approached in different
More informationPHYS225 Lecture 9. Electronic Circuits
PHYS225 Lecture 9 Electronic Circuits Last lecture Field Effect Transistors Voltage controlled resistor Various FET circuits Switch Source follower Current source Similar to BJT Draws no input current
More informationECE Analog Integrated Circuit Design  II P.E. Allen
Lecture 290 Feedback Analysis using Return Ratio (3/20/02) Page 2901 LECTURE 290 FEEDBACK CIRCUIT ANALYSIS USING RETURN RATIO (READING: GHLM 599613) Objective The objective of this presentation is: 1.)
More informationSeries & Parallel Resistors 3/17/2015 1
Series & Parallel Resistors 3/17/2015 1 Series Resistors & Voltage Division Consider the singleloop circuit as shown in figure. The two resistors are in series, since the same current i flows in both
More informationHomework 6 Solutions and Rubric
Homework 6 Solutions and Rubric EE 140/40A 1. KW Tube Amplifier b) Load Resistor e) Commoncathode a) Input Diff Pair f) CathodeFollower h) Positive Feedback c) Tail Resistor g) Cc d) Av,cm = 1/ Figure
More informationStart with the transfer function for a secondorder highpass. s 2. ω o. Q P s + ω2 o. = G o V i
aaac3xicbzfna9taeizxatkk7kec9tilqck4jbg5fjpca4ew0kmpdsrxwhlvxokl7titrirg69lr67s/robll64wmkna5jenndmvjstzyib9pfjntva/vzu6dzsnhj5/sdfefxhmvawzjpotsxeiliemxiucjpogkkybit3x5atow5w8xfugs5qmksecubqo7krlsfhkzsagxr4jne8wehaaxjqy4qq2svvl5el5qai2v9hy5tnxwb0om8igbiqfhhqhkoulcfs2zczhp26lwm7ph/hehffsbu90syo3hcmwvyxpawjtfbjpkm/wlbnximooweuygmsivnygqlpcmywvfppvrewjl3yqxti9gr6e2kgqbgrnlizqyuf2btqd/vgmo8cms4dllesrrdopz4ahyqjf7c66bovhzqznm9l89tqb2smixsxzk3tsdtnat4iaxnkk5bfcbn6iphqywpvxwtypgvnhtsvux234v77/ncudz9leyj84wplgvm7hrmk4ofi7ynw8edpwl7zt62o9klz8kl0idd8pqckq9krmaekz/kt7plbluf3a/un/d7ko6bc0zshbujz6huqq
More informationE1.1 Analysis of Circuits ( ) Revision Lecture 1 1 / 13
RevisionLecture 1: E1.1 Analysis of Circuits (20144530) Revision Lecture 1 1 / 13 Format Question 1 (40%): eight short parts covering the whole syllabus. Questions 2 and 3: single topic questions (answer
More informationFeedback Control G 1+FG A
Introduction to Operational Amplifiers Circuit Functionality So far, only passive circuits (C, L and LC) have been analyzed in terms of the timedomain operator T and the frequencydomain operator A(ω),
More informationNotes for course EE1.1 Circuit Analysis TOPIC 10 2PORT CIRCUITS
Objectives: Introduction Notes for course EE1.1 Circuit Analysis 45 Reexamination of 1port subcircuits Admittance parameters for port circuits TOPIC 1 PORT CIRCUITS Gain and port impedance from port
More informationMidterm Exam (closed book/notes) Tuesday, February 23, 2010
University of California, Berkeley Spring 2010 EE 42/100 Prof. A. Niknejad Midterm Exam (closed book/notes) Tuesday, February 23, 2010 Guidelines: Closed book. You may use a calculator. Do not unstaple
More informationanalyse and design a range of sinewave oscillators understand the design of multivibrators.
INTODUTION In this lesson, we investigate some forms of waveform generation using op amps. Of course, we could use basic transistor circuits, but it makes sense to simplify the analysis by considering
More informationOPERATIONAL AMPLIFIER APPLICATIONS
OPERATIONAL AMPLIFIER APPLICATIONS 2.1 The Ideal Op Amp (Chapter 2.1) Amplifier Applications 2.2 The Inverting Configuration (Chapter 2.2) 2.3 The Noninverting Configuration (Chapter 2.3) 2.4 Difference
More informationENGN3227 Analogue Electronics. Problem Sets V1.0. Dr. Salman Durrani
ENGN3227 Analogue Electronics Problem Sets V1.0 Dr. Salman Durrani November 2006 Copyright c 2006 by Salman Durrani. Problem Set List 1. Opamp Circuits 2. Differential Amplifiers 3. Comparator Circuits
More informationRIB. ELECTRICAL ENGINEERING Analog Electronics. 8 Electrical Engineering RIBR T7. Detailed Explanations. Rank Improvement Batch ANSWERS.
8 Electrical Engineering RIBR T7 Session 089 S.No. : 9078_LS RIB Rank Improvement Batch ELECTRICL ENGINEERING nalog Electronics NSWERS. (d) 7. (a) 3. (c) 9. (a) 5. (d). (d) 8. (c) 4. (c) 0. (c) 6. (b)
More informationFigure Circuit for Question 1. Figure Circuit for Question 2
Exercises 10.7 Exercises Multiple Choice 1. For the circuit of Figure 10.44 the time constant is A. 0.5 ms 71.43 µs 2, 000 s D. 0.2 ms 4 Ω 2 Ω 12 Ω 1 mh 12u 0 () t V Figure 10.44. Circuit for Question
More informationSolved Problems. Electric Circuits & Components. 11 Write the KVL equation for the circuit shown.
Solved Problems Electric Circuits & Components 11 Write the KVL equation for the circuit shown. 12 Write the KCL equation for the principal node shown. 12A In the DC circuit given in Fig. 1, find (i)
More informationPrepare for this experiment!
Notes on Experiment #10 Prepare for this experiment! Read the PAmp Tutorial before going on with this experiment. For any Ideal p Amp with negative feedback you may assume: V  = V + (But not necessarily
More informationBasics of Network Theory (PartI)
Basics of Network Theory (PartI). A square waveform as shown in figure is applied across mh ideal inductor. The current through the inductor is a. wave of peak amplitude. V 0 0.5 t (m sec) [Gate 987: Marks]
More informationDesigning Information Devices and Systems I Fall 2018 Lecture Notes Note Introduction: Opamps in Negative Feedback
EECS 16A Designing Information Devices and Systems I Fall 2018 Lecture Notes Note 18 18.1 Introduction: Opamps in Negative Feedback In the last note, we saw that can use an opamp as a comparator. However,
More informationFeedback design for the Buck Converter
Feedback design for the Buck Converter Portland State University Department of Electrical and Computer Engineering Portland, Oregon, USA December 30, 2009 Abstract In this paper we explore two compensation
More informationHomework 2. Due Friday (5pm), Feb. 8, 2013
University of California, Berkeley Spring 2013 EE 42/100 Prof. K. Pister Homework 2 Due Friday (5pm), Feb. 8, 2013 Please turn the homework in to the drop box located next to 125 Cory Hall (labeled EE
More informationEE 321 Analog Electronics, Fall 2013 Homework #8 solution
EE 321 Analog Electronics, Fall 2013 Homework #8 solution 5.110. The following table summarizes some of the basic attributes of a number of BJTs of different types, operating as amplifiers under various
More information0 t < 0 1 t 1. u(t) =
A. M. Niknejad University of California, Berkeley EE 100 / 42 Lecture 13 p. 22/33 Step Response A unit step function is described by u(t) = ( 0 t < 0 1 t 1 While the waveform has an artificial jump (difficult
More informationQuick Review. ESE319 Introduction to Microelectronics. and Q1 = Q2, what is the value of V Odm. If R C1 = R C2. s.t. R C1. Let Q1 = Q2 and R C1
Quick Review If R C1 = R C2 and Q1 = Q2, what is the value of V Odm? Let Q1 = Q2 and R C1 R C2 s.t. R C1 > R C2, express R C1 & R C2 in terms R C and ΔR C. If V Odm is the differential output offset
More informationFinal Exam. 55:041 Electronic Circuits. The University of Iowa. Fall 2013.
Final Exam Name: Max: 130 Points Question 1 In the circuit shown, the opamp is ideal, except for an input bias current I b = 1 na. Further, R F = 10K, R 1 = 100 Ω and C = 1 μf. The switch is opened at
More informationECE PN Junctions and Diodes
ECE 342 2. PN Junctions and iodes Jose E. SchuttAine Electrical & Computer Engineering University of Illinois jschutt@emlab.uiuc.edu ECE 342 Jose Schutt Aine 1 B: material dependent parameter = 5.4 10
More informationUNIVERSITY OF CALIFORNIA, BERKELEY College of Engineering Department of Electrical Engineering and Computer Sciences
UNIVERSITY OF CALIFORNIA, BERKELEY College of Engineering Department of Electrical Engineering and Computer Sciences EECS 40 Spring 2000 Introduction to Microelectronic Devices Prof. King MIDTERM EXAMINATION
More informationEE100Su08 Lecture #9 (July 16 th 2008)
EE100Su08 Lecture #9 (July 16 th 2008) Outline HW #1s and Midterm #1 returned today Midterm #1 notes HW #1 and Midterm #1 regrade deadline: Wednesday, July 23 rd 2008, 5:00 pm PST. Procedure: HW #1: Bart
More informationOPAMPs I: The Ideal Case
I: The Ideal Case The basic composition of an operational amplifier (OPAMP) includes a high gain differential amplifier, followed by a second high gain amplifier, followed by a unity gain, low impedance,
More informationProblem Set 4 Solutions
University of California, Berkeley Spring 212 EE 42/1 Prof. A. Niknejad Problem Set 4 Solutions Please note that these are merely suggested solutions. Many of these problems can be approached in different
More informationSmall Signal Model. S. Sivasubramani EE101 Small Signal  Diode
Small Signal Model i v Small Signal Model i I D i d i D v d v D v V D Small Signal Model Mathematical Analysis V D  DC value v d  ac signal v D  Total signal (DC ac signal) Diode current and voltage
More informationCircuits for Analog System Design Prof. Gunashekaran M K Center for Electronics Design and Technology Indian Institute of Science, Bangalore
Circuits for Analog System Design Prof. Gunashekaran M K Center for Electronics Design and Technology Indian Institute of Science, Bangalore Lecture No. # 08 Temperature Indicator Design Using Opamp Today,
More informationChapter 13 SmallSignal Modeling and Linear Amplification
Chapter 13 SmallSignal Modeling and Linear Amplification Microelectronic Circuit Design Richard C. Jaeger Travis N. Blalock 1/4/12 Chap 131 Chapter Goals Understanding of concepts related to: Transistors
More informationOhm's Law and Resistance
Ohm's Law and Resistance Resistance Resistance is the property of a component which restricts the flow of electric current. Energy is used up as the voltage across the component drives the current through
More informationPrepare for this experiment!
Notes on Experiment #8 Theorems of Linear Networks Prepare for this experiment! If you prepare, you can finish in 90 minutes. If you do not prepare, you will not finish even half of this experiment. So,
More information4.5 (A4.3)  TEMPERATURE INDEPENDENT BIASING (BANDGAP)
emp. Indep. Biasing (7/14/00) Page 1 4.5 (A4.3)  EMPERAURE INDEPENDEN BIASING (BANDGAP) INRODUCION Objective he objective of this presentation is: 1.) Introduce the concept of a bandgap reference 2.)
More informationDesign Engineering MEng EXAMINATIONS 2016
IMPERIAL COLLEGE LONDON Design Engineering MEng EXAMINATIONS 2016 For Internal Students of the Imperial College of Science, Technology and Medicine This paper is also taken for the relevant examination
More informationPrepare for this experiment!
Notes on Experiment #8 Theorems of Linear Networks Prepare for this experiment! If you prepare, you can finish in 90 minutes. If you do not prepare, you will not finish even half of this experiment. So,
More informationSchedule. ECEN 301 Discussion #20 Exam 2 Review 1. Lab Due date. Title Chapters HW Due date. Date Day Class No. 10 Nov Mon 20 Exam Review.
Schedule Date Day lass No. 0 Nov Mon 0 Exam Review Nov Tue Title hapters HW Due date Nov Wed Boolean Algebra 3. 3.3 ab Due date AB 7 Exam EXAM 3 Nov Thu 4 Nov Fri Recitation 5 Nov Sat 6 Nov Sun 7 Nov Mon
More informationID # NAME. EE255 EXAM 3 April 7, Instructor (circle one) Ogborn Lundstrom
ID # NAME EE255 EXAM 3 April 7, 1998 Instructor (circle one) Ogborn Lundstrom This exam consists of 20 multiple choice questions. Record all answers on this page, but you must turn in the entire exam.
More informationEECE 2150 Circuits and Signals Final Exam Fall 2016 Dec 9
EECE 2150 Circuits and Signals Final Exam Fall 2016 Dec 9 Name: Instructions: Write your name and section number on all pages Closed book, closed notes; Computers and cell phones are not allowed You can
More informationThe Approximating Impedance
Georgia Institute of Technology School of Electrical and Computer Engineering ECE 4435 Op Amp Design Laboratory Fall 005 DesignProject,Part A White Noise and Pink Noise Generator The following explains
More informationChapter 2  DC Biasing  BJTs
Objectives Chapter 2  DC Biasing  BJTs To Understand: Concept of Operating point and stability Analyzing Various biasing circuits and their comparison with respect to stability BJT A Review Invented
More informationOperational Amplifier (OpAmp) Operational Amplifiers. OPAmp: Components. Internal Design of LM741
(OpAmp) s Prof. Dr. M. Zahurul Haq zahurul@me.buet.ac.bd http://teacher.buet.ac.bd/zahurul/ Department of Mechanical Engineering Bangladesh University of Engineering & Technology ME 475: Mechatronics
More informationENGR4300 Spring 2009 Test 2. Name: SOLUTION. Section: 1(MR 8:00) 2(TF 2:00) 3(MR 6:00) (circle one) Question I (20 points): Question II (20 points):
ENGR43 Test 2 Spring 29 ENGR43 Spring 29 Test 2 Name: SOLUTION Section: 1(MR 8:) 2(TF 2:) 3(MR 6:) (circle one) Question I (2 points): Question II (2 points): Question III (17 points): Question IV (2 points):
More informationInducing Chaos in the p/n Junction
Inducing Chaos in the p/n Junction Renato Mariz de Moraes, Marshal Miller, Alex Glasser, Anand Banerjee, Ed Ott, Tom Antonsen, and Steven M. Anlage CSR, Department of Physics MURI Review 14 November, 2003
More informationEE247 AnalogDigital Interface Integrated Circuits
EE247 AnalogDigital Interface Integrated Circuits Fall 200 Name: Zhaoyi Kang SID: 22074 ******************************************************************************* EE247 AnalogDigital Interface Integrated
More informationSwitched Capacitor: Sampled Data Systems
Switched Capacitor: Sampled Data Systems Basic switched capacitor theory How has Anadigm utilised this. TheoryBasic SC and Anadigm1 Resistor & Charge Relationship I + V  I Resistance is defined in terms
More informationUNIT 4 DC EQUIVALENT CIRCUIT AND NETWORK THEOREMS
UNIT 4 DC EQUIVALENT CIRCUIT AND NETWORK THEOREMS 1.0 Kirchoff s Law Kirchoff s Current Law (KCL) states at any junction in an electric circuit the total current flowing towards that junction is equal
More informationHomework Assignment 11
Homework Assignment Question State and then explain in 2 3 sentences, the advantage of switched capacitor filters compared to continuoustime active filters. (3 points) Continuous time filters use resistors
More informationECE2262 Electric Circuits
ECE2262 Electric Circuits Equivalence Chapter 5: Circuit Theorems Linearity Superposition Thevenin s and Norton s Theorems Maximum Power Transfer Analysis of Circuits Using Circuit Theorems 1 5. 1 Equivalence
More informationIH5341, IH5352. Dual SPST, Quad SPST CMOS RF/Video Switches. Description. Features. Ordering Information. Applications. Pinouts.
SEMICONDUCTOR IH, IH2 December Features Description Dual SPST, Quad SPST CMOS RF/Video Switches R DS(ON) < Ω Switch Attenuation Varies Less Than db From DC to 00MHz "OFF" Isolation > 0dB Typical at 0MHz
More informationExamination paper for TFY4185 Measurement Technique/ Måleteknikk
Page 1 of 14 Department of Physics Examination paper for TFY4185 Measurement Technique/ Måleteknikk Academic contact during examination: Patrick Espy Phone: +47 41 38 65 78 Examination date: 15 August
More informationProf. Shayla Sawyer CP08 solution
What does the time constant represent in an exponential function? How do you define a sinusoid? What is impedance? How is a capacitor affected by an input signal that changes over time? How is an inductor
More informationPhysics 2135 Exam 2 October 20, 2015
Exam Total / 200 Physics 2135 Exam 2 October 20, 2015 Printed Name: Rec. Sec. Letter: Five multiple choice questions, 8 points each. Choose the best or most nearly correct answer. 1. A straight wire segment
More informationLecture 5: Using electronics to make measurements
Lecture 5: Using electronics to make measurements As physicists, we re not really interested in electronics for its own sake We want to use it to measure something often, something too small to be directly
More informationUNIVERSITY OF CALIFORNIA, BERKELEY College of Engineering Department of Electrical Engineering and Computer Sciences
UNIVERSITY OF CALIFORNIA, BERKELEY College of Engineering Department of Electrical Engineering and Computer Sciences EE 105: Microelectronic Devices and Circuits Spring 2008 MIDTERM EXAMINATION #1 Time
More informationDesigning Information Devices and Systems II Fall 2016 Murat Arcak and Michel Maharbiz Homework 0. This homework is due August 29th, 2016, at Noon.
EECS 16B Designing Information Devices and Systems II Fall 2016 Murat Arcak and Michel Maharbiz Homework 0 This homework is due August 29th, 2016, at Noon. 1. Homework process and study group (a) Who else
More informationFET SmallSignal Analysis
CHAPTER FET mallignal Analysis 9 9.1 INTROUCTION Fieldeffect transistor amplifiers provide an excellent voltage gain with the added feature of a high input impedance. They are also considered lowpower
More informationDEPARTMENT OF COMPUTER ENGINEERING UNIVERSITY OF LAHORE
DEPARTMENT OF COMPUTER ENGINEERING UNIVERSITY OF LAHORE NAME. Section 1 2 3 UNIVERSITY OF LAHORE Department of Computer engineering Linear Circuit Analysis Laboratory Manual 2 Compiled by Engr. Ahmad Bilal
More informationR. W. Erickson. Department of Electrical, Computer, and Energy Engineering University of Colorado, Boulder
. W. Erickson Department of Electrical, Computer, and Energy Engineering University of Colorado, Boulder 2.4 Cuk converter example L 1 C 1 L 2 Cuk converter, with ideal switch i 1 i v 1 2 1 2 C 2 v 2 Cuk
More informationECS 40, Fall 2008 Prof. ChangHasnain Test #3 Version A
ECS 40, Fall 2008 Prof. ChangHasnain Test #3 Version A 10:10 am 11:00 am, Wednesday December 3, 2008 Total Time Allotted: 50 minutes Total Points: 100 1. This is a closed book exam. However, you are allowed
More informationEE 3CL4: Introduction to Control Systems Lab 4: Lead Compensation
EE 3CL4: Introduction to Control Systems Lab 4: Lead Compensation Tim Davidson Ext. 27352 davidson@mcmaster.ca Objective To use the root locus technique to design a lead compensator for a marginallystable
More informationVTS Process Photodiodes
VTS PROCESS LOW CAPACITANCE, LARGE AREA PHOTODIODE FEATURES Visible to IR spectral range Excellent QE  400 to 1100 nm Guaranteed 400 nm response Response @ 940 nm, 0.60 A/W, typical Useable with visible
More informationFig. 11 Current Flow in a Resistive load
1 Electric Circuits: Current flow in a resistive load flows either from () to () which is labeled below as Electron flow or the Conventional flow from () to (). We will use conventional flow in this
More informationD C Circuit Analysis and Network Theorems:
UNIT1 D C Circuit Analysis and Network Theorems: Circuit Concepts: Concepts of network, Active and passive elements, voltage and current sources, source transformation, unilateral and bilateral elements,
More informationESE319 Introduction to Microelectronics. Output Stages
Output Stages Power amplifier classification Class A amplifier circuits Class A Power conversion efficiency Class B amplifier circuits Class B Power conversion efficiency Class AB amplifier circuits Class
More informationUniversity of California at Berkeley College of Engineering Dept. of Electrical Engineering and Computer Sciences. EECS 40 Midterm II
University of California at Berkeley College of Engineering Dept. of Electrical Engineering and Computer Sciences EECS 40 Midterm II Spring 2001 Prof. Roger T. Howe April 11, 2001 Name: Last, First Student
More informationMicroelectronic Circuit Design 4th Edition Errata  Updated 4/4/14
Chapter Text # Inside back cover: Triode region equation should not be squared! i D = K n v GS "V TN " v & DS % ( v DS $ 2 ' Page 49, first exercise, second answer: 1.35 x 10 6 cm/s Page 58, last exercise,
More informationmith College Computer Science CSC270 Spring 16 Circuits and Systems Lecture Notes Week 3 Dominique Thiébaut
mith College Computer Science CSC270 Spring 16 Circuits and Systems Lecture Notes Week 3 Dominique Thiébaut dthiebaut@smith.edu Crash Course in Electricity and Electronics Zero Physics background expected!
More information