55:041 Electronic Circuits The University of Iowa Fall Final Exam


 Roland Thomas
 3 years ago
 Views:
Transcription
1 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 in data sheets of capacitors. What does ESR stand for? Answer: Equivalent Series Resistance c. When researching part numbers for threeterminal regulators, an engineer encounters the term LDO. What does LDO stand for? Answer: Low Drop Out d. Name one important difference between an operational amplifier and a comparator. Answer: Comparators are very similar to opamps, but comparators don t need and generally don t have internal frequency compensation. 1
2 e. What is v o in the following circuit if v REF = 1.2 V, R 1 = 680 Ω, and R 2 = 200 Ω? (3 points) Answer: The current through R 2 is 1.2/200 = 6 ma, which also flows through R 1. Thus, the output voltage is = 5.28 V f. What is the purpose of R 3 in the circuit below, and what should the value be to be effective? (2 points) Answer: This compensates for the opamp s input currents. The value should be R 1 R 2. g. The output voltage of a threeterminal voltage regulator is 5 5 ma load, and A load. What is the regulator s output resistance? (2 points) Answer: R = ΔV ΔI = = 27 mω 2
3 h. Give one phrase that describes the purpose of the capacitor, and give another phrase that describes the purpose of two series diodes in the circuit below. (2 points) Answer: Frequency compensation, and reduction of crossover distortion i. What type of negative feedback (seriesshunt, seriesseries,... ) is used in the following amplifier? Answer: Seriesshunt. j. What type of negative feedback (seriesshunt, seriesseries, ) is used in the following amplifier? Answer: Shuntshunt 3
4 k. What type of negative feedback (seriesshunt, seriesseries, ) is used in the following amplifier? Answer: Shuntshunt. l. The output voltage of a threeterminal voltage regulator is 5 5 ma load, and A load. What is the regulator s load regulation? (2 points) Answer: Load Regulation = V o(nl) V o(fl) 100 = = 0.8% V o(nl) 5 4
5 Question 2 Consider the comparator circuit below. What is the threshold voltage (i.e., the voltage where the output changes state? (5 points) R 1 = 10 kω R 2 = 20 kω V REF = 1.2 V Solution (This problem comes from a Midterm exam) The voltage at the inverting input is v = V REF v I R 1 + R 2 R 2 + v I The circuit changes state when this voltage is 0 V. That is, when Substituting values yield the threshold voltage: v I = V REF v I R 1 + R 2 R 2 v I = R 2 R 1 V REF v I = 20 (1.2 V) = 2.4 V 10 5
6 Question 3 For the amplifier below, determine the differentialmode voltage gain A d = v o3 v d. State and very briefly justify simplifying assumptions you make. (15 points) β = 200 V A = 80 V V BE(on) = 0.7 V Hints: This is a twostage amplifier that consists of a differentialinput stage and a buffer/output stage. Consider computing the gains of these stages separately. Solution To simplify calculations note that β is large, so ignore base currents. This means I C I E for all transistors. Further, using the BJT scaling property, the resistance looking into the base of Q 3 is larger than β(3.3k) = 660K. This is in parallel with the 8K collector resistance of the differential stage and it is large enough to ignore. Finally, with reasonable bias currents, BJT output resistances are ~100K, so in this circuit it is reasonable to ignore. Now I 1 = K = ma, and I E1 = I 1 2 = ma, and g m1 = 40I E1 = 38.3 ma/v. The gain of the differential input stage is A 1 = (g m1 R C ) 2 = The gain of the output stage is A 2 R C2 R E = 1.2 The total gain is then A d = A 1 A 2 = (153.2)( 1.2) = 186 6
7 Question 4 Consider the following smallsignal model of an amplifier. Determine (i.e., write down an expression) for the output resistance R o indicated in the figure. (10 points) Solution (This problem comes from a Midterm exam, 2007) 7
8 Question 5 Using the circuit below, design a current source such that I O = 0.7 ma. (10 points) Transistors are matched with And V A =, β = 80 V BE(on) = 0.7 V V + = 10 V R 2 = 10K Solution Q 1, Q 2 are matched, and in the circuit have the same baseemitter voltage, and have the same collector currents I O = 0.7 ma. They also have the same base currents I O β. Then I E3 = I O β + I O β K = 2( ) = 87.5 μa 80 10K I B3 = (17.5 μa) (1 + β) = 1.08 μa I REF = I B3 + I C1 = 1.08 μa ma = ma R 1 = = 12.27K
9 Question 6 A power MOSFET has thermal characteristics given below and dissipates 25 W. Design, by specifying the thermal resistance, a heat sink that will ensure the MOSFET does not overheat. The ambient temperature is 25 o C. Assume one can keep the thermal resistance between the MOSFET case and heat sink (θ case sink or θ CS ) below 1 o C/W. (5 points) θ juntion case = θ JC = 1.75 /W θ ase ambient = θ CA = 50 /W T j,max = 150 Solution (This problem comes from a Midterm exam, 2007) A thermal model that captures the information is shown below: T j = T A + P D θ JC + θ CS + θ SA 150 = ( θ SA ) θ A = 2.25 /W The heat sink s thermal resistance must be less than this value. Note: a number of students used θ CA in various ways (incorrectly) in their calculations. θ CA is not part of the picture, since we will replace it with a much lower θ SA 9
10 Question 7 Consider the feedback amplifier in (i) and the smallsignal model in (ii). The feedback network is modeled with a yparameter twoport as shown. A dc analysis shows I C = 1 ma β = 150 V A = I C = 1 ma (i) (ii) (a) Determine y 12 (4 points) If you can t do this, use y 12 = A/V and continue. (b) Determine the gain of the feedback amplifier, namely A = v o /i i. (10 points) (c) Determine the input impedance R if (2 points) (d) Determine R i (2 points) (e) Determine R of (2 points) 10
11 Solution (This is a simplified version of some a homework question) Part (a) y 12 i 1 = 1 v 2 v 1 120K =0 Twoport model for R f Calculation for y 12 Part (b) g m = 40I C = 0.04 A/V, and r π = β g m = 3.75 K, and r o =. We ignore y 21, and y 12 is the feedback transfer ratio β. Next turn off the feedback (set y 12 = 0). At the input, (1/y 11 R S r π ) = 2.1K, and at the output, (1/y 22 R L R C ) = 1.85K Now v o = (g m v π )(1.85K) = g m i i (2.1K) (1.85K) = i i V A = v o i i = V/A Now (1 + βa) = ( K) = 2.3 A f = A 1 + βa = = V/A Part (c) R if = 2.1K (1 + βa) = 2.1K 2.3 = 0.913K Part (d) 1 = = R i R if R s 0.913K 1 5K = 1.12K Part (e) R of = 1.85K (1 + βa) = 1.85K 2.3 = 0.904K (Over for more space) 11
12 12
13 Question 8 Below is a screen dump of a SPICE simulation of a Wien bridge oscillator that uses an incandescent lamp for gain control. The graph depicts the lamp resistance as a function of the voltage across the lamp. Determine the voltage at which the output stabilizes. (10 points) Lamp Resistance (Ohm) Lamp Voltage (V) (a) Solution (This problem comes from a Midterm exam) (b) At startup R lamp < 5 Ω, and the loop gain T = R 4 /(R lamp +39) +1 > 3.7, enough for a Wien bridge oscillator to start oscillating. As the output amplitude grows, V lamp grows, R lamp increases. However when R lamp increases, the loop gain T = R 4 /(R lamp +39) + 1 decreases. The loop stabilizes when R T lamp = R4 = 21 ( R Ω lamp + 39) = 3 From the plot R lamp = 21 Ω when the V lamp 1.25 V. The current flowing through the lamp is 1.25/21 = 60 ma. The voltage at the output is (0.06)( ) = 10.8 V. 13
14 Question 9 For the switchedcapacitor circuit below, the parameters are C 1 = 30 pf, C 2 = 5 pf, C F = 12 pf. The clock frequency is 100 khz. Determine the lowfrequency gain and the cutoff frequency. (10 points) Solution (This was a homework problem) The switched capacitors C 1 and C 2 function as resistors with values R 1 = 1 f C C 1 = kω, and R 2 = 1 f C C 2 = 2 MΩ respectively. At low frequencies C F is an open circuit and the lowfrequency gain is A V = R 2 R 1 = C 1 C 2 = 6 The cutoff frequency is determined by C F and the switched capacitor R 2 f 3dB = 1 1 = 2πR 2 C F 2π( = 6.63 khz )
15 Question 10 The loop gain function of an amplifier is given by T(f) = j f j f f j 10 5 (a) What is the frequency (to a good approximation) at which the phase is 180 o? (4 points) (b) What is the magnitude of T(f) at this frequency? (4 points) (c) Is the amplifier stable? (2 points) Solution (This was a homework problem) Part (a) One must determine f 180 such that φ = tan 1 f 10 4 f f tan 1 tan = 180o Program this value into a programmable calculator and try different values for f to find f Hz Part (b) 500 T(f) = = (8.063)(1.887)(1.28) = Part (c) The amplifier is unstable, because the magnitude of the loop gain is greater than 1 when the phase shift is 180 o. 15
16 Question 11 Consider the differential amplifier below. Determine the lowfrequency, singleended differentialmode gain A vd = v O1 v d and the upper 3dB frequency. (15 points) V A = I Q = 1 ma R C = 10K R B = 0.5K C π = 8 pf C μ = 2 pf β = 100 Hint: draw a halfamplifier and its smallsignal equivalent, and compute the Miller capacitance. Solution (This problem comes from the Review Questions) Below is a schematic of the left halfamplifier and its smallsignal model. For the smallsignal model r π = β = = g m 40I C (40)( ) = 5K The gain for this amplifier is The Miller capacitance is A vd = v O1 v d = v O1 = 1 2v 1 2 r π ( g r π + R m R c ) B = (40)( )( ) = 91 16
17 And f 3 db = C M = (1 + g m R C )C μ = 1 + (40)( )( ) (2 pf) = 402 pf 1 2π(C π + C M )[r π R B ] = 1 2π( )( = 853 khz ) 17
18 Question 12 K n = 0.5 ma/v 2, V TN = 2 V, λ = 0, C gd = 0.1 pf, C gs = 1 pf (a) Show that V GS = 3.55 V. (4 points) (b) Draw a detailed smallsignal model for the amplifier showing the numerical values for the components. (6 points) Solution (This comes from the Review Questions) 18
19 19
Final 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 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 information55:041 Electronic Circuits The University of Iowa Fall Exam 2
Exam 2 Name: Score /60 Question 1 One point unless indicated otherwise. 1. An engineer measures the (step response) rise time of an amplifier as t r = 0.35 μs. Estimate the 3 db bandwidth of the amplifier.
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 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 informationECE343 Test 2: Mar 21, :008:00, Closed Book. Name : SOLUTION
ECE343 Test 2: Mar 21, 2012 6:008:00, Closed Book Name : SOLUTION 1. (25 pts) (a) Draw a circuit diagram for a differential amplifier designed under the following constraints: Use only BJTs. (You may
More informationAssignment 3 ELEC 312/Winter 12 R.Raut, Ph.D.
Page 1 of 3 ELEC 312: ELECTRONICS II : ASSIGNMENT3 Department of Electrical and Computer Engineering Winter 2012 1. A commonemitter amplifier that can be represented by the following equivalent circuit,
More informationECE 3050A, Spring 2004 Page 1. FINAL EXAMINATION  SOLUTIONS (Average score = 78/100) R 2 = R 1 =
ECE 3050A, Spring 2004 Page Problem (20 points This problem must be attempted) The simplified schematic of a feedback amplifier is shown. Assume that all transistors are matched and g m ma/v and r ds.
More informationElectronics II. Final Examination
The University of Toledo f17fs_elct27.fm 1 Electronics II Final Examination Problems Points 1. 11 2. 14 3. 15 Total 40 Was the exam fair? yes no The University of Toledo f17fs_elct27.fm 2 Problem 1 11
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 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 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 informationUniversity of Toronto. Final Exam
University of Toronto Final Exam Date  Dec 16, 013 Duration:.5 hrs ECE331 Electronic Circuits Lecturer  D. Johns ANSWER QUESTIONS ON THESE SHEETS USING BACKS IF NECESSARY 1. Equation sheet is on last
More informationCircle the one best answer for each question. Five points per question.
ID # NAME EE255 EXAM 3 November 8, 2001 Instructor (circle one) Talavage Gray This exam consists of 16 multiple choice questions and one workout problem. Record all answers to the multiple choice questions
More informationElectronic Circuits Summary
Electronic Circuits Summary Andreas Biri, DITET 6.06.4 Constants (@300K) ε 0 = 8.854 0 F m m 0 = 9. 0 3 kg k =.38 0 3 J K = 8.67 0 5 ev/k kt q = 0.059 V, q kt = 38.6, kt = 5.9 mev V Small Signal Equivalent
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 informationECE342 Test 3: Nov 30, :008:00, Closed Book. Name : Solution
ECE342 Test 3: Nov 30, 2010 6:008:00, Closed Book Name : Solution All solutions must provide units as appropriate. Unless otherwise stated, assume T = 300 K. 1. (25 pts) Consider the amplifier shown
More informationECE343 Test 1: Feb 10, :008:00pm, Closed Book. Name : SOLUTION
ECE343 Test : Feb 0, 00 6:008:00pm, Closed Book Name : SOLUTION C Depl = C J0 + V R /V o ) m C Diff = τ F g m ω T = g m C µ + C π ω T = g m I / D C GD + C or V OV GS b = τ i τ i = R i C i ω H b Z = Z
More informationCE/CS Amplifier Response at High Frequencies
.. CE/CS Amplifier Response at High Frequencies INEL 4202  Manuel Toledo August 20, 2012 INEL 4202  Manuel Toledo CE/CS High Frequency Analysis 1/ 24 Outline.1 High Frequency Models.2 Simplified Method.3
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 informationElectronics II. Midterm #2
The University of Toledo EECS:3400 Electronics I Section sums_elct7.fm  StudentName Electronics II Midterm # Problems Points. 8. 3. 7 Total 0 Was the exam fair? yes no The University of Toledo sums_elct7.fm
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 information6.012 Electronic Devices and Circuits Spring 2005
6.012 Electronic Devices and Circuits Spring 2005 May 16, 2005 Final Exam (200 points) OPEN BOOK Problem NAME RECITATION TIME 1 2 3 4 5 Total General guidelines (please read carefully before starting):
More informationBiasing the CE Amplifier
Biasing the CE Amplifier Graphical approach: plot I C as a function of the DC baseemitter voltage (note: normally plot vs. base current, so we must return to EbersMoll): I C I S e V BE V th I S e V th
More informationChapter 9 Frequency Response. PART C: High Frequency Response
Chapter 9 Frequency Response PART C: High Frequency Response Discrete Common Source (CS) Amplifier Goal: find high cutoff frequency, f H 2 f H is dependent on internal capacitances V o Load Resistance
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 informationElectronics II. Midterm #1
The University of Toledo EECS:3400 Electronics I su3ms_elct7.fm Section Electronics II Midterm # Problems Points. 5. 6 3. 9 Total 0 Was the exam fair? yes no The University of Toledo su3ms_elct7.fm Problem
More informationCHAPTER.4: Transistor at low frequencies
CHAPTER.4: Transistor at low frequencies Introduction Amplification in the AC domain BJT transistor modeling The re Transistor Model The Hybrid equivalent Model Introduction There are three models commonly
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 information1. (50 points, BJT curves & equivalent) For the 2N3904 =(npn) and the 2N3906 =(pnp)
HW 3 1. (50 points, BJT curves & equivalent) For the 2N3904 =(npn) and the 2N3906 =(pnp) a) Obtain in Spice the transistor curves given on the course web page except do in separate plots, one for the npn
More informationDelhi Noida Bhopal Hyderabad Jaipur Lucknow Indore Pune Bhubaneswar Kolkata Patna Web: Ph:
Serial : ND_EE_NW_Analog Electronics_05088 Delhi Noida Bhopal Hyderabad Jaipur Lucknow ndore Pune Bhubaneswar Kolkata Patna Web: Email: info@madeeasy.in Ph: 04546 CLASS TEST 089 ELECTCAL ENGNEENG Subject
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 informationGeneral Purpose Transistors
General Purpose Transistors NPN and PNP Silicon These transistors are designed for general purpose amplifier applications. They are housed in the SOT 33/SC which is designed for low power surface mount
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 informationChapter 10 Feedback. PART C: Stability and Compensation
1 Chapter 10 Feedback PART C: Stability and Compensation Example: Noninverting Amplifier We are analyzing the two circuits (nmos diff pair or pmos diff pair) to realize this symbol: either of the circuits
More informationMassachusetts Institute of Technology Department of Electrical Engineering and Computer Science Electronic Circuits Fall 2000.
Massachusetts Institute of Technology Department of Electrical Engineering and Computer Science 6.002 Electronic Circuits Fall 2000 Final Exam Please write your name in the space provided below, and circle
More informationL4970A 10A SWITCHING REGULATOR
L4970A 10A SWITCHING REGULATOR 10A OUTPUT CURRENT.1 TO 40 OUTPUT OLTAGE RANGE 0 TO 90 DUTY CYCLE RANGE INTERNAL FEEDFORWARD LINE REGULA TION INTERNAL CURRENT LIMITING PRECISE.1 ± 2 ON CHIP REFERENCE
More informationUNIVERSITY OF CALIFORNIA College of Engineering Department of Electrical Engineering and Computer Sciences
UNIVERSITY OF CALIFORNIA College of Engineering Department of Electrical Engineering and Computer Sciences E. Alon Final EECS 240 Monday, May 19, 2008 SPRING 2008 You should write your results on the exam
More informationKOM2751 Analog Electronics :: Dr. Muharrem Mercimek :: YTU  Control and Automation Dept. 1 4 DC BIASING BJTS (CONT D II )
KOM2751 Analog Electronics :: Dr. Muharrem Mercimek :: YTU  Control and Automation Dept. 1 4 DC BIASING BJTS (CONT D II ) Most of the content is from the textbook: Electronic devices and circuit theory,
More informationElectronic Circuits 1. Transistor Devices. Contents BJT and FET Characteristics Operations. Prof. C.K. Tse: Transistor devices
Electronic Circuits 1 Transistor Devices Contents BJT and FET Characteristics Operations 1 What is a transistor? Threeterminal device whose voltagecurrent relationship is controlled by a third voltage
More informationStudio 9 Review Operational Amplifier Stability Compensation Miller Effect Phase Margin Unity Gain Frequency Slew Rate Limiting Reading: Text sec 5.
Studio 9 Review Operational Amplifier Stability Compensation Miller Effect Phase Margin Unity Gain Frequency Slew Rate Limiting Reading: Text sec 5.2 pp. 232242 Twostage opamp Analysis Strategy Recognize
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 information6.012 Electronic Devices and Circuits
Page 1 of 10 YOUR NAME Department of Electrical Engineering and Computer Science Massachusetts Institute of Technology 6.012 Electronic Devices and Circuits Exam No. 2 Thursday, November 5, 2009 7:30 to
More informationElectronics II. Midterm II
The University of Toledo su7ms_elct7.fm  Electronics II Midterm II Problems Points. 7. 7 3. 6 Total 0 Was the exam fair? yes no The University of Toledo su7ms_elct7.fm  Problem 7 points Equation ()
More informationConventional PaperI2011 PARTA
Conventional PaperI0 PARTA.a Give five properties of static magnetic field intensity. What are the different methods by which it can be calculated? Write a Maxwell s equation relating this in integral
More informationThe 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 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 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 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 informationLecture 7: Transistors and Amplifiers
Lecture 7: Transistors and Amplifiers Hybrid Transistor Model for small AC : The previous model for a transistor used one parameter (β, the current gain) to describe the transistor. doesn't explain many
More informationBipolar junction transistors
Bipolar junction transistors Find parameters of te BJT in CE configuration at BQ 40 µa and CBQ V. nput caracteristic B / µa 40 0 00 80 60 40 0 0 0, 0,5 0,3 0,35 0,4 BE / V Output caracteristics C / ma
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 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 informationAdvanced Current Mirrors and Opamps
Advanced Current Mirrors and Opamps David Johns and Ken Martin (johns@eecg.toronto.edu) (martin@eecg.toronto.edu) slide 1 of 26 WideSwing Current Mirrors I bias I V I in out out = I in V W L bias 
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 informationLecture 37: Frequency response. Context
EECS 05 Spring 004, Lecture 37 Lecture 37: Frequency response Prof J. S. Smith EECS 05 Spring 004, Lecture 37 Context We will figure out more of the design parameters for the amplifier we looked at in
More informationECE 2210 Final given: Spring 15 p1
ECE 2 Final given: Spring 15 Closed Book, Closed notes except preprinted yellow sheet, Calculators OK. Show all work to receive credit. Circle answers, show units, and round off reasonably 1. (15 pts)
More informationECE137B Final Exam. Wednesday 6/8/2016, 7:3010:30PM.
ECE137B Final Exam Wednesday 6/8/2016, 7:3010:30PM. There are7 problems on this exam and you have 3 hours There are pages 132 in the exam: please make sure all are there. Do not open this exam until
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 informationBiasing BJTs CHAPTER OBJECTIVES 4.1 INTRODUCTION
4 DC Biasing BJTs CHAPTER OBJECTIVES Be able to determine the dc levels for the variety of important BJT configurations. Understand how to measure the important voltage levels of a BJT transistor configuration
More informationESE319 Introduction to Microelectronics Common Emitter BJT Amplifier
Common Emitter BJT Amplifier 1 Adding a signal source to the single power supply bias amplifier R C R 1 R C V CC V CC V B R E R 2 R E Desired effect addition of bias and signal sources Starting point 
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 informationECE 202 Fall 2013 Final Exam
ECE 202 Fall 2013 Final Exam December 12, 2013 Circle your division: Division 0101: Furgason (8:30 am) Division 0201: Bermel (9:30 am) Name (Last, First) Purdue ID # There are 18 multiple choice problems
More informationBipolar Junction Transistor (BJT)  Introduction
Bipolar Junction Transistor (BJT)  Introduction It was found in 1948 at the Bell Telephone Laboratories. It is a three terminal device and has three semiconductor regions. It can be used in signal amplification
More informationElectronic Devices and Circuits Lecture 18  Single Transistor Amplifier Stages  Outline Announcements. Notes on Single Transistor Amplifiers
6.012 Electronic Devices and Circuits Lecture 18 Single Transistor Amplifier Stages Outline Announcements Handouts Lecture Outline and Summary Notes on Single Transistor Amplifiers Exam 2 Wednesday night,
More informationECE 6412, Spring Final Exam Page 1 FINAL EXAMINATION NAME SCORE /120
ECE 6412, Spring 2002 Final Exam Page 1 FINAL EXAMINATION NAME SCORE /120 Problem 1O 2O 3 4 5 6 7 8 Score INSTRUCTIONS: This exam is closed book with four sheets of notes permitted. The exam consists of
More informationEE214 Early Final Examination: Fall STANFORD UNIVERSITY Department of Electrical Engineering. SAMPLE FINAL EXAMINATION Fall Quarter, 2002
STANFORD UNIVERSITY Department of Electrical Engineering SAMPLE FINAL EXAMINATION Fall Quarter, 2002 EE214 8 December 2002 CLOSED BOOK; Two std. 8.5 x 11 sheets of notes permitted CAUTION: Useful information
More informationElectronics and Communication Exercise 1
Electronics and Communication Exercise 1 1. For matrices of same dimension M, N and scalar c, which one of these properties DOES NOT ALWAYS hold? (A) (M T ) T = M (C) (M + N) T = M T + N T (B) (cm)+ =
More informationKOM2751 Analog Electronics :: Dr. Muharrem Mercimek :: YTU  Control and Automation Dept. 1 7 DC BIASING FETS (CONT D)
KOM751 Analog Electronics :: Dr. Muharrem Mercimek :: YTU Control and Automation Dept. 1 7 DC BIASING FETS (CONT D) Most of the content is from the textbook: Electronic devices and circuit theory, Robert
More informationChapter 11 AC and DC Equivalent Circuit Modeling of the Discontinuous Conduction Mode
Chapter 11 AC and DC Equivalent Circuit Modeling of the Discontinuous Conduction Mode Introduction 11.1. DCM Averaged Switch Model 11.2. SmallSignal AC Modeling of the DCM Switch Network 11.3. HighFrequency
More informationDepartment of Electrical Engineering and Computer Sciences University of California, Berkeley. Final Exam Solutions
Electrical Engineering 42/00 Summer 202 Instructor: Tony Dear Department of Electrical Engineering and omputer Sciences University of alifornia, Berkeley Final Exam Solutions. Diodes Have apacitance?!?!
More informationECE2210 Final given: Fall 13
ECE22 Final given: Fall 3. (23 pts) a) Draw the asymptotic Bode plot (the straightline approximation) of the transfer function below. Accurately draw it on the graph provided. You must show the steps
More informationElectronics II. Final Examination
f3fs_elct7.fm  The University of Toledo EECS:3400 Electronics I Section Student Name Electronics II Final Examination Problems Points.. 3 3. 5 Total 40 Was the exam fair? yes no Analog Electronics f3fs_elct7.fm
More informationVidyalankar S.E. Sem. III [EXTC] Analog Electronics  I Prelim Question Paper Solution
. (a) S.E. Sem. [EXTC] Analog Electronics  Prelim Question Paper Solution Comparison between BJT and JFET BJT JFET ) BJT is a bipolar device, both majority JFET is an unipolar device, electron and minority
More informationEE105 Fall 2015 Microelectronic Devices and Circuits Frequency Response. Prof. Ming C. Wu 511 Sutardja Dai Hall (SDH)
EE05 Fall 205 Microelectronic Devices and Circuits Frequency Response Prof. Ming C. Wu wu@eecs.berkeley.edu 5 Sutardja Dai Hall (SDH) Amplifier Frequency Response: Lower and Upper Cutoff Frequency Midband
More informationEECS 105: FALL 06 FINAL
University of California College of Engineering Department of Electrical Engineering and Computer Sciences Jan M. Rabaey TuTh 23:30 Wednesday December 13, 12:303:30pm EECS 105: FALL 06 FINAL NAME Last
More informationGENERAL DESCRIPTION The PT5128 is a dual channel lowdropout voltage regulator designed for portable and wireless applications that require high PSRR, low quiescent current and excellent line and load
More informationAs light level increases, resistance decreases. As temperature increases, resistance decreases. Voltage across capacitor increases with time LDR
LDR As light level increases, resistance decreases thermistor As temperature increases, resistance decreases capacitor Voltage across capacitor increases with time Potential divider basics: R 1 1. Both
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 informationCapacitors Diodes Transistors. PC200 Lectures. Terry Sturtevant. Wilfrid Laurier University. June 4, 2009
Wilfrid Laurier University June 4, 2009 Capacitor an electronic device which consists of two conductive plates separated by an insulator Capacitor an electronic device which consists of two conductive
More informationMICROELECTRONIC CIRCUIT DESIGN Second Edition
MICROELECTRONIC CIRCUIT DESIGN Second Edition Richard C. Jaeger and Travis N. Blalock Answers to Selected Problems Updated 10/23/06 Chapter 1 1.3 1.52 years, 5.06 years 1.5 2.00 years, 6.65 years 1.8 113
More informationCHAPTER.6 :TRANSISTOR FREQUENCY RESPONSE
CHAPTER.6 :TRANSISTOR FREQUENCY RESPONSE To understand Decibels, log scale, general frequency considerations of an amplifier. low frequency analysis  Bode plot low frequency response BJT amplifier Miller
More informationLecture 13 MOSFET as an amplifier with an introduction to MOSFET smallsignal model and smallsignal schematics. Lena Peterson
Lecture 13 MOSFET as an amplifier with an introduction to MOSFET smallsignal model and smallsignal schematics Lena Peterson 20151013 Outline (1) Why is the CMOS inverter gain not infinite? Largesignal
More informationMod. Sim. Dyn. Sys. Amplifiers page 1
AMPLIFIERS A circuit containing only capacitors, amplifiers (transistors) and resistors may resonate. A circuit containing only capacitors and resistors may not. Why does amplification permit resonance
More informationElectronics II. Midterm II
The University of Toledo f4ms_elct7.fm  Section Electronics II Midterm II Problems Points. 7. 7 3. 6 Total 0 Was the exam fair? yes no The University of Toledo f4ms_elct7.fm  Problem 7 points Given in
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 informationESE319 Introduction to Microelectronics. BJT Biasing Cont.
BJT Biasing Cont. Biasing for DC Operating Point Stability BJT Bias Using Emitter Negative Feedback Single Supply BJT Bias Scheme Constant Current BJT Bias Scheme Rule of Thumb BJT Bias Design 1 Simple
More informationEE 321 Analog Electronics, Fall 2013 Homework #3 solution
EE 32 Analog Electronics, Fall 203 Homework #3 solution 2.47. (a) Use superposition to show that the output of the circuit in Fig. P2.47 is given by + [ Rf v N + R f v N2 +... + R ] f v Nn R N R N2 R [
More informationECE137B Final Exam. There are 5 problems on this exam and you have 3 hours There are pages 119 in the exam: please make sure all are there.
ECE37B Final Exam There are 5 problems on this exam and you have 3 hours There are pages 9 in the exam: please make sure all are there. Do not open this exam until told to do so Show all work: Credit
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 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 informationECE 220 Laboratory 4 Volt Meter, Comparators, and Timer
ECE 220 Laboratory 4 Volt Meter, Comparators, and Timer Michael W. Marcellin Please follow all rules, procedures and report requirements as described at the beginning of the document entitled ECE 220 Laboratory
More informationMod. Sim. Dyn. Sys. Amplifiers page 1
AMPLIFIERS A circuit containing only capacitors, amplifiers (transistors) and resistors may resonate. A circuit containing only capacitors and resistors may not. Why does amplification permit resonance
More informationAC Circuits Homework Set
Problem 1. In an oscillating LC circuit in which C=4.0 μf, the maximum potential difference across the capacitor during the oscillations is 1.50 V and the maximum current through the inductor is 50.0 ma.
More informationAOP606 Complementary Enhancement Mode Field Effect Transistor
AOP66 Complementary Enhancement Mode Field Effect Transistor General Description The AOP66 uses advanced trench technology MOSFETs to provide excellent and low gate charge. The complementary MOSFETs may
More informationLow Voltage(1.24V) Adjustable Precision Shunt Regulator
FEATURES Low Voltage Operation : 1.24 V Programmable Out Voltage to 18V Sink Current Capability of 1mA to 100mA Equivalent full range Temperature Coefficient of 50ppm/ Temperature Compensated for operation
More informationElectronics II. Midterm #2
The University of Toledo EECS:3400 Electronics I su4ms_elct7.fm Section Electronics II Midterm # Problems Points. 8. 7 3. 5 Total 0 Was the exam fair? yes no The University of Toledo su4ms_elct7.fm Problem
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 informationVI. Transistor amplifiers: Biasing and Small Signal Model
VI. Transistor amplifiers: iasing and Small Signal Model 6.1 Introduction Transistor amplifiers utilizing JT or FET are similar in design and analysis. Accordingly we will discuss JT amplifiers thoroughly.
More informationElectronics II. Final Examination
The University of Toledo f6fs_elct7.fm  Electronics II Final Examination Problems Points. 5. 0 3. 5 Total 40 Was the exam fair? yes no The University of Toledo f6fs_elct7.fm  Problem 5 points Given is
More information