Final Exam. 55:041 Electronic Circuits. The University of Iowa. Fall 2013.
|
|
- Chloe Payne
- 5 years ago
- Views:
Transcription
1 Final Exam Name: Max: 130 Points Question 1 In the circuit shown, the op-amp 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 t = 0. What is the output voltage after 5 seconds? (5 points) Solution For t 0, the voltage across the capacitor is v C = (±I b Δt) C which is ((± ) (5)) ( ) = ±5 mv for t = 5 s. The gain of the amplifier is 1 + R F R 1 = 101, so that the output voltage is ± 501 mv. Question 2 Consider the voltage regulator. The input voltage is V C. What should V REF be for an output voltage of 5 V? (4 points) Solution The amplifier and feedback loop maintains a voltage V REF across R 1 so the current through both resistors is I = V REF R 1. The output voltage is then V O = I(R 1 + R 2 ) = V REF R 1 (R 1 + R 2 ) = 5.0 Substituting R 1 and R 2 and solving for V REF yields V REF = 2 V 1
2 Question 3 Determine the h 11 and h 21 parameters for the circuit. Be sure to supply the units and proper sign for each parameter. (8 points) Solution Setting v 2 = 0 h 11 = v 1 i 1 v 2 =0 h 21 = i 2 i 1 v 2 =0 With v 2 = 0, the input voltage is equal to the voltage across the 20K resistor, and Use KCL to find i 2 : v 1 = (20K)i 1 h 11 = (20K)i 1 i 1 = 20K i i 1 + i 2 = 0 i 2 = 51i 1 and h 21 = 51i 1 i 1 = 51 A/A Question 4 For the circuit, assume I DQ = 28 ma, and you may ignore r o. Further, assume the capacitors are shorts at the operating frequency. Determine the voltage gain A v = v i v o. (5 points) K n = 93 ma V 2 V TN = 2 V λ = 0 C S = 100 μf C C = 1 μf The small-signal model is shown where R G = 100K 56K and g m = 2 K n I D = = A/V. From the model the voltage gain is A V = g m ( K) =
3 Question 5 Consider the 555-based astable oscillator, which has component values so that the duty cycle and frequency are 60% and 5 khz respectively. Assume you have 2N7000 FETs, 2N2222 BJTs, and a collection of standard resistors and capacitors available. Provide a circuit along with the component values that will use the 555 timer s output and drive an IR LED so that the average current through the IR LED is 30 ma. (8 points) Hint: You encountered the circuit in both the IR Tx and Rx labs The peak- and average currents, and the duty cycle D in % are related as follows I ave = D 100 I peak I peak = 100 D I ave = 100 (30 ma) = 50 ma 60 Assume that for the BJT, V BE(ON) = 0.7 V, then R limit = 0.7 (50 ma) = 14 Ω. Use the closest standard value of 15 Ω. 3
4 Question 6 The diagram below shows part of the IR receiver of the final lab. Assume that all the capacitors appear/function as shorts at 5 khz. The exception is C c1, which has a capacitance of 1 nf. The BJT has a quiescent collector current of I C = 1 ma. (a) What is the voltage gain of the BJT amplifier? That is, the voltage gain from the transistor s base to its collector? Ignore loading effects at the collector. (2 points) A v = g m R C = 40I C R C = 40( )( ) = 224 (b) Calculate a value for C E so it appears as a short. (3 points) (c) What is the 3-dB bandwidth of the circuit shown? (4 points) C c1 sees a resistance R T of (R ph + R 1 R 2 ) or 10K + 135K = 145K. The time constant of the circuit is τ = R T C C1 = ( )( ) = 145 μs. The 3-dB bandwidth is B = 1 (2πτ) = 1.1 khz. 4
5 (d) Assume 5-kHz signal photocurrent is 10 μa rms. Estimate the 5-kHz signal rms voltage at B. (2 points) v B ( )(10K) = 100 mv rms (e) Assume an incandescent lamp generates a 100 μa rms, 120-Hz noise signal in the photodetector. Estimate the noise voltage at B. (Hint: use the bandwidth result from (c) above) (4 points) The 120-Hz noise signal is log 10 ( ) = 0.92 decades below the 1-kHz, 3-dB cutoff of the high-pass network formed by C c1, R 1, R 2, and R ph. Thus, the 120-kHz signal is attenuated by 3 + (0.92)(20) = db. This is equivalent to a factor 11.77, so that the 120-Hz signal at B will be v B ( )(10K) mv (f) In order to extend the range, a student places another photodiode in series with the one already in place. Explain very briefly why this does not work. (1 point) The photodiode functions as a current source, so placing two in series is nonsensical, and probably limits the current to the smaller of the two current sources. 5
6 Question 7 For the BJT in the amplifier shown, β = 100. Ignore the BJT s parasitic capacitances. Further, C C2, and R S = 100K. (a) Find I CQ. (3 points) If you can t do this use I CQ = 1 ma for the rest of the problem. For a dc analysis, we open the capacitors. KVL gives I B R B (1 + β)i B R E 10 = 0 Solving yields I B = 8.38 μa and consequently I B (1 + β)i B I C = βi B = ma. (b) Estimate the lower 3-dB bandwidth if C c1 = 10 nf. (5 points) g m = 40I C, r π = β g m = 2.5K. Using BJT scaling, C C1 sees a resistance R i to its right where R i = (r π + (1 + β)(10k 10K)) 100K = 84K The total resistance the capacitor sees is R S + R i = 184K, and the time constant is τ = (184K)( ) = 1.84 ms. The 3-dB bandwidth is thus B = 1 (2πτ) = 86.5 Hz. (c) Estimate the overall midband voltage gain A v = v o v i. (2 points) This is an emitter follower with voltage gain from base to emitter of slightly less than 1. Further, R i, the resistance to the right of C 1 is 84K and this forms a voltage divider with R s so that the overall gain is R A v (1) i 84K = R i + R s 100K + 84K =
7 Question 8 The open-loop gain and input resistance of the opamp below is 10 6 and 1 MΩ respectively. The op-amp s output resistance is 100 Ω. Further, R 1 = 99K, R 2 = 1K. What is the closed-loop gain and input resistance? (5 points) Solution This is series-shunt (voltage-voltage) feedback, with β = R 2 (R 1 + R 2 ) = Further, 1 + βa OL = 1 + (0.01)(10 6 ) Thus A f = A OL 1 + βa OL = = 100 R if = R i (1 + βa OL ) = (10 6 )(10 4 ) = 10 4 MΩ R of = βa OL = 10 mω Question 9 For the circuit below β = 300 and I C = 1 ma R 1 = 120 kω R 2 = 75 kω V BE(ON ) = 0.62 V Determine the output resistance R o using BJT scaling. You may ignore r o. (4 points) g m = 40 I C = 40 ms, and r π = β g m = 7.5K r π R o = R E ( ) = 4K (7.5K) = 24.7Ω 25 Ω 1 + β 301 7
8 Question 10 Show that the output resistance R o in the circuit is R o = 1 g m R S You may ignore r o (6 points) R o To find R O, follow the standard procedure: turn off all independent sources (v i in this case), add a test source V x and determine I x. Then, R O = V x I x. The small-signal model for the amplifier is shown in (a) below. The gate current is zero, so that v g is zero, and the small-signal model becomes as is shown in (b). Note that v gs = V x and that r o is in parallel with R O. (a) (b) KCL at the source gives Rearranging gives g m v gs I X + V x R s = 0 g m ( V x ) I x + V x R s = 0 R o = V X I X = 1 g m R S 8
9 Question 11 Consider the amplifier below. λ = 0, C gd = 0.3 pf, C gs = 2 pf V GS = 3.55 V, and g m = 1.55 ma/v (a) Draw a complete (showing voltage polarities etc.) small-signal model for the amplifier that includes the FET capacitances. (3 points) (b) Calculate the Miller capacitance. (4 points) C M = C gd (1 + g m (R D R L )) C M = 0.3( (4 20)) = 1.85 pf (c) Calculate the 3-dB frequency of the small-signal gain (3 points) R G = R 1 R 2 = 97.1 kω τ = (R G R i )(C gs + C M ) = ( )( ) = 34.9 ns f 3dB = 1 = 4.56 MHz 2πτ (d) Calculate the overall midband gain A v = v o v i (2 points) R 1 R 2 A v = g m (R D R L ) = R 1 R 2 + R i 9
10 Question 12 Consider a BJT with a rated power of 20 W, and a maximum allowable junction temperature T j,max = 175. The transistor is mounted on a heat sink with parameters θ case sink = 1 /W, and θ sink amb = 5 /W. Determine how much power the BJT can safely dissipate. Assume an ambient temperature of T A = 25. (10 points) Hint: first determine θ dev case. Solution The thermal resistance from the device/junction to the case is not given explicitly, so we need to determine it before proceeding. The BJT is rated at 20 W at T j,max = 175, and an ambient temperature of T A = 25 is assumed. A thermal model and the calculation of θ dev case is then T j = T A + P D (θ dev case ) 175 = (θ dev case ) θ dev case = 7.5 W Now we can determine the maximum allowable power dissipation when the BJT is mounted on a heat skink with the given parameters. A thermal model for the problem is shown below. T j = T A + P D (θ dev case + θ case sink + θ sink amb ) P D,max = = T j,max T A (θ dev case + θ case sink + θ sink amb ) = 11.1 W 10
11 Question 13 The maximum current, voltage, and power ratings for the MOSFET are 4 A, 60 V, and 40 W, respectively. (a) Sketch and label the SOA for the MOSFET using linear voltage and current scales. (3 points) (b) For the amplifier, determine R D and sketch the load line that produces maximum power in the transistor for V DD = 55 V. Be sure to clearly indicate all important points on the load line. (4 points) Solution The plot below shows the SOA and a load line that is anchored at V DD = 55 V and touches the maximum power hyperbola but still stays in the SOA. It touches the SOA at V DD 2 = 27.5 V. The power dissipation is 40 W so that I D at this point is = A. From this it follows that I D(max) = A as indicated. The slope of the load line is and this equals 1 R D so that R D = = 18.9 Ω. I D (A) 5 4 SOA Boundary A Load line V DS (V) 11
12 Question 14 The open loop gain A(f) of an amplifier is shown below. What is the phase margin and bandwidth if closed loop gain = 12 db? What will the rise time for a step input be? (10 points) Using the graphical subtraction method, draw 1 β, which is practically the same as the closedloop gain. Read the phase as 95, so the phase margin is about 85. The closed-loop bandwidth is about 100 MHz, so that the rise time is t r 0.35 ( ) = 3.5 ns. 12
13 Question 15 The figure is a plot of the open-loop gain function for the LT1007 voltage amplifier. An engineer will use the amplifier as a non-inverting amplifier with a mid-frequency voltage gain of 10. (a) What is the GBP of the LT1007? (2 points) (b) Use the plot and estimate the bandwidth of the feedback amplifier. (2 points) (c) Write an expression for the gain A(f) for the feedback amplifier. (2 points) (d) (d) By how much (μs) does the amplifier delay a 250-kHz sine wave? (2 points) Solution (a) The open loop gain is 120 db ( ) at f = 10 Hz, so the GBP is 10 MHz. Alternatively, the BW is about 10 MHz when the open loop gain is 0, so the GBP is 10 MHz. (b) A voltage gain of 10 is equivalent to a gain of 20 log 10 (10) = 20 db. A horizontal line at 20 db intercepts the LT1007 gain curve at 950 khz. Alternatively, from the GBP, with a gain of 10, the bandwidth is 1 MHz. (c) The closed-loop response is 10 A(f) = f 1 + j (d) The amplifier s phase is θ = tan 1 (f ) and at 250 khz this is 14. Further, the period of a 250-kHz sine wave is 4 μs and the delay is therefore: Δt = 14 4 μs = μs
14 Question 16 The open-loop gain of an amplifier is modeled by A(f) = 10 5 (1 + j f f 102) (1 + j 10 4) (1 + j f ) An engineer uses the amplifier and negative feedback so that the gain of the feedback amplifier is 220. (a) What is the feedback factor β? (4 significant figures). (1 points) β 1 = (b) Provide an expression for the loop gain T of the amplifier. (1 point) T = βa(f) = (1 + j f f 102) (1 + j 10 4) (1 + j f ) (c) Find the crossover frequency f x. That is, the frequency where the magnitude of the loop gain is 1. If you cannot do this, take f x = 10 khz for the rest of the problem. (8 points) T(f) = ( f 10 2) ( f ) f 1 + ( ) Using trail-and-error we find that f x = 20 khz results in T(f) (d) Determine, by calculating the phase, if the amplifier is stable. (5 points) The phase at f x = 20 khz is φ = tan 1 f x 10 2 tan 1 f x 10 4 tan 1 f x 10 5 Substituting f = 20 khz shows that φ = The phase margin is φ = = The amplifier is stable. 14
55:041 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 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 Final Exam
Final Exam Name: Score Max: 135 Question 1 (1 point unless otherwise noted) a. What is the maximum theoretical efficiency for a class-b amplifier? Answer: 78% b. The abbreviation/term ESR is often encountered
More informationHomework Assignment 09
Homework Assignment 09 Question 1 (Short Takes) Two points each unless otherwise indicated. 1. What is the 3-dB bandwidth of the amplifier shown below if r π = 2.5K, r o = 100K, g m = 40 ms, and C L =
More informationECE-343 Test 2: Mar 21, :00-8:00, Closed Book. Name : SOLUTION
ECE-343 Test 2: Mar 21, 2012 6:00-8:00, 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 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 informationHomework Assignment 11
Homework Assignment Question State and then explain in 2 3 sentences, the advantage of switched capacitor filters compared to continuous-time active filters. (3 points) Continuous time filters use resistors
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 informationAssignment 3 ELEC 312/Winter 12 R.Raut, Ph.D.
Page 1 of 3 ELEC 312: ELECTRONICS II : ASSIGNMENT-3 Department of Electrical and Computer Engineering Winter 2012 1. A common-emitter amplifier that can be represented by the following equivalent circuit,
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 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 informationECE-342 Test 3: Nov 30, :00-8:00, Closed Book. Name : Solution
ECE-342 Test 3: Nov 30, 2010 6:00-8:00, 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 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 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 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 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 informationElectronic Circuits Summary
Electronic Circuits Summary Andreas Biri, D-ITET 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 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 informationCircle the one best answer for each question. Five points per question.
ID # NAME EE-255 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. Prof. Dr. Qiuting Huang Integrated Systems Laboratory
Electronic Circuits Prof. Dr. Qiuting Huang 6. Transimpedance Amplifiers, Voltage Regulators, Logarithmic Amplifiers, Anti-Logarithmic Amplifiers Transimpedance Amplifiers Sensing an input current ii in
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: E-mail: info@madeeasy.in Ph: 0-4546 CLASS TEST 08-9 ELECTCAL ENGNEENG Subject
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 informationECE-343 Test 1: Feb 10, :00-8:00pm, Closed Book. Name : SOLUTION
ECE-343 Test : Feb 0, 00 6:00-8:00pm, 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 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. Op-amp Circuits 2. Differential Amplifiers 3. Comparator Circuits
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 informationChapter7. FET Biasing
Chapter7. J configurations Fixed biasing Self biasing & Common Gate Voltage divider MOS configurations Depletion-type Enhancement-type JFET: Fixed Biasing Example 7.1: As shown in the figure, it is the
More informationOperational Amplifiers
Operational Amplifiers A Linear IC circuit Operational Amplifier (op-amp) An op-amp is a high-gain amplifier that has high input impedance and low output impedance. An ideal op-amp has infinite gain and
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 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 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 informationBiasing the CE Amplifier
Biasing the CE Amplifier Graphical approach: plot I C as a function of the DC base-emitter voltage (note: normally plot vs. base current, so we must return to Ebers-Moll): I C I S e V BE V th I S e V th
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 informationECE137B Final Exam. There are 5 problems on this exam and you have 3 hours There are pages 1-19 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 informationID # NAME. EE-255 EXAM 3 April 7, Instructor (circle one) Ogborn Lundstrom
ID # NAME EE-255 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.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 informationL4970A 10A SWITCHING REGULATOR
L4970A 10A SWITCHING REGULATOR 10A OUTPUT CURRENT.1 TO 40 OUTPUT OLTAGE RANGE 0 TO 90 DUTY CYCLE RANGE INTERNAL FEED-FORWARD LINE REGULA- TION INTERNAL CURRENT LIMITING PRECISE.1 ± 2 ON CHIP REFERENCE
More informationDC Biasing. Dr. U. Sezen & Dr. D. Gökçen (Hacettepe Uni.) ELE230 Electronics I 15-Mar / 59
Contents Three States of Operation BJT DC Analysis Fixed-Bias Circuit Emitter-Stabilized Bias Circuit Voltage Divider Bias Circuit DC Bias with Voltage Feedback Various Dierent Bias Circuits pnp Transistors
More informationESE319 Introduction to Microelectronics. Feedback Basics
Feedback Basics Stability Feedback concept Feedback in emitter follower One-pole feedback and root locus Frequency dependent feedback and root locus Gain and phase margins Conditions for closed loop stability
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 informationChapter 13 Small-Signal Modeling and Linear Amplification
Chapter 13 Small-Signal Modeling and Linear Amplification Microelectronic Circuit Design Richard C. Jaeger Travis N. Blalock 1/4/12 Chap 13-1 Chapter Goals Understanding of concepts related to: Transistors
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 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 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 cut-off frequency, f H 2 f H is dependent on internal capacitances V o Load Resistance
More informationDESIGN MICROELECTRONICS ELCT 703 (W17) LECTURE 3: OP-AMP CMOS CIRCUIT. Dr. Eman Azab Assistant Professor Office: C
MICROELECTRONICS ELCT 703 (W17) LECTURE 3: OP-AMP CMOS CIRCUIT DESIGN Dr. Eman Azab Assistant Professor Office: C3.315 E-mail: eman.azab@guc.edu.eg 1 TWO STAGE CMOS OP-AMP It consists of two stages: First
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 Wide-Swing Current Mirrors I bias I V I in out out = I in V W L bias ------------
More informationV in (min) and V in (min) = (V OH -V OL ) dv out (0) dt = A p 1 V in = = 10 6 = 1V/µs
ECE 642, Spring 2003 - Final Exam Page FINAL EXAMINATION (ALLEN) - SOLUTION (Average Score = 9/20) Problem - (20 points - This problem is required) An open-loop comparator has a gain of 0 4, a dominant
More informationECE 6412, Spring Final Exam Page 1
ECE 64, Spring 005 Final Exam Page FINAL EXAMINATION SOLUTIONS (Average score = 89/00) Problem (0 points This problem is required) A comparator consists of an amplifier cascaded with a latch as shown below.
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 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 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 informationChapter 2. - DC Biasing - BJTs
Chapter 2. - DC Biasing - BJTs Objectives To Understand : Concept of Operating point and stability Analyzing Various biasing circuits and their comparison with respect to stability BJT A Review Invented
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 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 informationRIB. ELECTRICAL ENGINEERING Analog Electronics. 8 Electrical Engineering RIB-R T7. Detailed Explanations. Rank Improvement Batch ANSWERS.
8 Electrical Engineering RIB-R T7 Session 08-9 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 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 informationESE319 Introduction to Microelectronics Bode Plot Review High Frequency BJT Model
Bode Plot Review High Frequency BJT Model 1 Logarithmic Frequency Response Plots (Bode Plots) Generic form of frequency response rational polynomial, where we substitute jω for s: H s=k sm a m 1 s m 1
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 informationRefinements to Incremental Transistor Model
Refinements to Incremental Transistor Model This section presents modifications to the incremental models that account for non-ideal transistor behavior Incremental output port resistance Incremental changes
More informationLecture 140 Simple Op Amps (2/11/02) Page 140-1
Lecture 40 Simple Op Amps (2//02) Page 40 LECTURE 40 SIMPLE OP AMPS (READING: TextGHLM 425434, 453454, AH 249253) INTRODUCTION The objective of this presentation is:.) Illustrate the analysis of BJT and
More informationIFB270 Advanced Electronic Circuits
IFB270 Advanced Electronic Circuits Chapter 0: Ampliier requency response Pro. Manar Mohaisen Department o EEC Engineering Review o the Precedent Lecture Reviewed o the JFET and MOSFET Explained and analyzed
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 linear-log scales for the phase plots. On the magnitude
More informationAON4605 Complementary Enhancement Mode Field Effect Transistor
AON5 Complementary Enhancement Mode Field Effect Transistor General Description The AON5 uses advanced trench technology to provide excellent R DS(ON) and low gate charge. The complementary MOSFETs form
More informationELECTRONIC SYSTEMS. Basic operational amplifier circuits. Electronic Systems - C3 13/05/ DDC Storey 1
Electronic Systems C3 3/05/2009 Politecnico di Torino ICT school Lesson C3 ELECTONIC SYSTEMS C OPEATIONAL AMPLIFIES C.3 Op Amp circuits» Application examples» Analysis of amplifier circuits» Single and
More informationCCS050M12CM2 1.2kV, 25mΩ All-Silicon Carbide Six-Pack (Three Phase) Module C2M MOSFET and Z-Rec TM Diode
CCS5M2CM2.2kV, 25mΩ All-Silicon Carbide Six-Pack (Three Phase) Module C2M MOSFET and Z-Rec TM Diode Features Ultra Low Loss Zero Reverse Recovery Current Zero Turn-off Tail Current High-Frequency Operation
More informationPT5108. High-PSRR 500mA LDO GENERAL DESCRIPTION FEATURES APPLICATIONS TYPICAL APPLICATIONS. Ripple Rejection vs Frequency. Ripple Rejection (db)
GENERAL DESCRIPTION The PT5108 is a low-dropout voltage regulator designed for portable applications that require both low noise performance and board space. Its PSRR at 1kHz is better than 70dB. The PT5108
More informationECE137B Final Exam. Wednesday 6/8/2016, 7:30-10:30PM.
ECE137B Final Exam Wednesday 6/8/2016, 7:30-10:30PM. There are7 problems on this exam and you have 3 hours There are pages 1-32 in the exam: please make sure all are there. Do not open this exam until
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 informationChapter 4 Field-Effect Transistors
Chapter 4 Field-Effect Transistors Microelectronic Circuit Design Richard C. Jaeger Travis N. Blalock 5/5/11 Chap 4-1 Chapter Goals Describe operation of MOSFETs. Define FET characteristics in operation
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 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 informationECE 546 Lecture 11 MOS Amplifiers
ECE 546 Lecture MOS Amplifiers Spring 208 Jose E. Schutt-Aine Electrical & Computer Engineering University of Illinois jesa@illinois.edu ECE 546 Jose Schutt Aine Amplifiers Definitions Used to increase
More informationSwitched-Capacitor Circuits David Johns and Ken Martin University of Toronto
Switched-Capacitor 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 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. 232-242 Two-stage op-amp Analysis Strategy Recognize
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. 863-866. Reader, Chapter 8 Noninverting Amp http://www.electronics-tutorials.ws/opamp/opamp_3.html Inverting Amp http://www.electronics-tutorials.ws/opamp/opamp_2.html
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. Small-Signal AC Modeling of the DCM Switch Network 11.3. High-Frequency
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 informationand V DS V GS V T (the saturation region) I DS = k 2 (V GS V T )2 (1+ V DS )
ECE 4420 Spring 2005 Page 1 FINAL EXAMINATION NAME SCORE /100 Problem 1O 2 3 4 5 6 7 Sum Points INSTRUCTIONS: This exam is closed book. You are permitted four sheets of notes (three of which are your sheets
More informationc Copyright 2009. W. Marshall Leach, Jr., Professor, Georgia Institute of Technology, School of Electrical and Computer Engineering. Feedback Amplifiers CollectionofSolvedProblems A collection of solved
More informationAO4620 Complementary Enhancement Mode Field Effect Transistor
AO46 Complementary Enhancement Mode Field Effect Transistor General Description The AO46 uses advanced trench technology MOSFETs to provide excellent and low gate charge. The complementary MOSFETs may
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 informationAO4607, AO4607L(Lead-Free) Complementary Enhancement Mode Field Effect Transistor
Rev : Feb 3 Rev : Jan 4 AO467, AO467L(Lead-Free) Complementary Enhancement Mode Field Effect Transistor General Description The AO467 uses advanced trench technology MOSFETs to provide excellen R DS(ON)
More information500V N-Channel MOSFET
830 / 830 500V N-Channel MOSFET General Description This Power MOSFET is produced using SL semi s advanced planar stripe DMOS technology. This advanced technology has been especially tailored to minimize
More informationSwitching circuits: basics and switching speed
ECE137B notes; copyright 2018 Switching circuits: basics and switching speed Mark Rodwell, University of California, Santa Barbara Amplifiers vs. switching circuits Some transistor circuit might have V
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 informationErrata of CMOS Analog Circuit Design 2 nd Edition By Phillip E. Allen and Douglas R. Holberg
Errata 2 nd Ed. (5/22/2) Page Errata of CMOS Analog Circuit Design 2 nd Edition By Phillip E. Allen and Douglas R. Holberg Page Errata 82 Line 4 after figure 3.2-3, CISW CJSW 88 Line between Eqs. (3.3-2)
More informationECEN 607 (ESS) Op-Amps Stability and Frequency Compensation Techniques. Analog & Mixed-Signal Center Texas A&M University
ECEN 67 (ESS) Op-Amps Stability and Frequency Compensation Techniques Analog & Mixed-Signal Center Texas A&M University Stability of Linear Systems Harold S. Black, 97 Negative feedback concept Negative
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 informationMaxim Integrated Products 1
19-4187; Rev 4; 7/1 μ μ PART AMPS PER PACKAGE PIN- PACKAGE + * TOP MARK MAX965AZK+ 1 5 SOT3 ADSI MAX965AZK/V+ 1 5 SOT3 ADSK MAX965AUA+ 1 8 μmax-ep* AABI MAX965ATA+ 1 8 TDFN-EP* BKX MAX9651AUA+ 8 μmax-ep*
More informationAO7401 P-Channel Enhancement Mode Field Effect Transistor
Nov P-Channel Enhancement Mode Field Effect Transistor General Description The uses advanced trench technology to provide excellent R DS(ON), low gate charge, and operation with gate voltages as low as.5v,
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 informationTO-247-3L Inner Circuit Product Summary I C) R DS(on)
Silicon Carbide Power MOSFET N-CHANNEL ENHANCEMENT MODE TO-247-3L Inner Circuit Product Summary V DS I D(@25 C) R DS(on) 1200V 20A 120mΩ Features u Low On-Resistance u Low Capacitance u Avalanche Ruggedness
More informationExact Analysis of a Common-Source MOSFET Amplifier
Exact Analysis of a Common-Source MOSFET Amplifier Consider the common-source MOSFET amplifier driven from signal source v s with Thévenin equivalent resistance R S and a load consisting of a parallel
More information3. Basic building blocks. Analog Design for CMOS VLSI Systems Franco Maloberti
Inverter with active load It is the simplest gain stage. The dc gain is given by the slope of the transfer characteristics. Small signal analysis C = C gs + C gs,ov C 2 = C gd + C gd,ov + C 3 = C db +
More informationn-channel Solar Inverter Induction Heating G C E Gate Collector Emitter
INSULATED GATE BIPOLAR TRANSISTOR WITH ULTRAFAST SOFT RECOVERY DIODE Features C Low V CE (ON) trench IGBT technology Low switching losses Square RBSOA 1% of the parts tested for I LM Positive V CE (ON)
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 informationECE 255, Frequency Response
ECE 255, Frequency Response 19 April 2018 1 Introduction In this lecture, we address the frequency response of amplifiers. This was touched upon briefly in our previous lecture in Section 7.5 of the textbook.
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 informationCHAPTER 7 - CD COMPANION
Chapter 7 - CD companion 1 CHAPTER 7 - CD COMPANION CD-7.2 Biasing of Single-Stage Amplifiers This companion section to the text contains detailed treatments of biasing circuits for both bipolar and field-effect
More informationFrequency Dependent Aspects of Op-amps
Frequency Dependent Aspects of Op-amps Frequency dependent feedback circuits The arguments that lead to expressions describing the circuit gain of inverting and non-inverting amplifier circuits with resistive
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