Homework Assignment 09
|
|
- Britton Griffin
- 5 years ago
- Views:
Transcription
1 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 = 1 nf? (a) khz (b) 10 khz (c) 1.59 khz (d) 10.4 khz Answer: The capacitor sees an equivalent resistance r o = 100K. (If one turns off V I, g m v π = 0, and the current source is effectively removed from the circuit.) The timeconstant is τ = RC = 100 μs. The bandwidth is 1 (2πτ) = 1.59 khz, so the answer is (c). 2. Explain why one cannot use BJT scaling to determine R O in the circuit below. Answer: The transistor is a MOSFET and not a BJT so one cannot use BJT scaling. (Actually, with the appropriate interpretation of the BJT scaling equations, one could use them, but we did not cover this.) 3. Consider the following drive circuit for an IR remote control. The drive signal is a 0 5 V square wave and V CC = 9 V. The MOSFET is replaced with another MOSFET with a V TN that is 20% higher. The new peak current through the IR diode will be a) Increased by 20% b) Decreased by 20% c) Unchanged d) Decreased much more than 20%, since I D = K n (V T V GS ) 2 Answer: (c) 1
2 4. Consider the following drive circuit for an IR remote control. The drive signal is a 0 5 V square wave and V CC = 9 V. The battery voltage drops from 9 V to 5 V, i.e., by a factor 1.8. The average IR diode current will be a) Unchanged b) Increased by about 18% c) Decreased by 18% d) Decreased significantly more than 18%, since I D = K n (V T V GS ) 2 Answer: (a) 5. Consider the following drive circuit for an IR remote control. The drive signal is a 0 5 V square wave and V CC = 9 V. The IR diode is replaced with another IR diode that has a turn-on voltage that is 20% lower. The new peak current through the IR diode will be a) Unchanged b) Increased by 20% c) Decreased by 20% d) Decreased much than 20%, since I D = I S e V D/V T Answer: (a) 6. What is the output voltage V OUT at the end of the 2.82 ms pulse? (3 points) (a) 0 V (b) 0.75 V (c) 5.55 V (d) V Answer: On the rising edge the capacitor is uncharged and 15V appears across R 1. The t τ voltage across the capacitor is V C = 15 1 e where τ = RC = 940 μs is the time t τ constant. The voltage across R 1 is 15e 2.82 ms 940 μs. At t = 2.82 ms, this is 5e = 15e 3 = V, so the answer is (b). 2
3 Question 2 Consider the circuit shown. Determine the current through R L when R L is 1K, 3K, and 5K. Assume V BE(ON) = 0.64 V, β = 200, and V CC = 9 V. (5 points) Since β is large, we can neglect the base current. The voltage at the base is then V B = (9)(1 11) = V. Then V E = = V. The emitter current is then ma. Not that this value is independent of R L as long as the BJT does not saturate. Taking V CE(SAT) = 0.2 V, saturation occurs when the drop across R L is around = 8.62 V. This will occur if when R L = 8.62K. All the R L values in the problem statement are less than this. Thus, the current for all the cases is 1 ma. Question 3 The threshold voltage for each transistor below is 0.4 V. Determine the region of operation (Ohmic, saturation, etc.) of the transistor in each circuit. (6 points) These are n-mosfets, so that the transition between Ohmic and saturation region is at v DS (sat) = v GS V TN. Circuit (a): v DS (sat) = = 1.8 V. Note that in the circuit, v DS = 2.2 V which is more than v DS (sat) MOSFET is in the saturation region. Circuit (b): v DS (sat) = = 0.6 V. Note that in the circuit, v DS = 0.4 V wich is less than v DS (sat) MOSFET is in the Ohmic region. Circuit (c): v GS = 0 V MOSFET is off. 3
4 Question 4 In the amplifier shown, R 1 = R 2 = 470K, R Sig = 10K, R S = 10K, and I DQ = 0.4 ma. For the transistor K n = 50 ma V 2, V TN = 2 V, and λ = Show that an expression for the output resistance is (8 points): 1 1 R o = g m + 1 = R s r o R s r o Determine the numerical value for the output resistance. (8 points) g m A small-signal model for the amplifier is shown. A test source V x was added to determine the output resistance R o. Assuming v i is turned off, then R o = V x I x. KCL at the source (using the convention that currents into the node is positive) gives g m v gs + I x V x V x = 0 R s r o g m v gs + I x V x = 0 R s r o From the circuit, with v i turned off R sig is in parallel with R 1 R 2. However, since the gate current is zero, v g = 0. That is, the gate is at signal ground. Thus, v gs = V x, so that V x g m V x + I x = 0 R s r o R o = V x 1 1 = I x g m + 1 = R s r o R s r o Next, determine numerical values for r o, g m, and R o : r o = 1 1 = λi IDQ ( )( ) = 9.6M g m = 2 K n I DQ = 2 ( )( ) = A V R o = R s r o 1 = (10K) (9.6M) (112) 112 Ω g m g m 4
5 Checking results with SPICE (Not Required) The following circuit was used to check the calculation with SPICE. For the MOSFET the SPICE parameters were set as follows: KP = 0.1, W = L = 100μ and λ = This translates to K n = A V 2. The amplitude for v i was set to zero, and the amplitude for V X was 10 mv. Further, C C = C X = 1F which are essentially shorts at V X s frequency of 1 khz. A transient analysis was performed and the ratio RMS(V(V X )) RMS(I(V X )) gives the output resistance. The value measured was Ω 5
6 Question 5 In the amplifier below, determine R O as indicated. Use the results from the previous question. (5 points) R i = 2.5K R S = 10K R 1 = R 2 = 470K R Sig = 10K I D = 0.4 ma K N = 50 ma V 2 r o The MOSFET is a follower and its output resistance is from the previous question R o = R S 1 g m. Now g m = 2 K n I DQ = A V, so that 1 g m 112 Ω, and consequently R o 112 Ω. Question 6 Shown is the symbol a popular SPICE computer simulation program uses for an n-channel MOSFET, along with labels indicating the drain, source and gate terminals. There is also a 4 th terminal, indicated with an arrow. The physical device has three, not four terminals. A perplexed student asks her professor what is this terminal, and what should she do with in when she builds her circuit in SPICE. Provide a short (3 4 sentence) answer to the student. (3 points) 6
7 Question 7 The transistor in the circuit shows has parameters V TN = 0.8 V and K n = 0.5 ma V 2. Write an expression and sketch the load line for (a) V DD = 4 V, R D = 1K and (b) V DD = 5 V, R D = 3K. Additionally, calculate the Q-point for each case and indicate these on the plot. Finally, for each case determine if the transistor is operating in the saturation or non-saturation region. (12 points) Part (a) Load line is I D = V DS R D + V DD R D = V DS 1K + 4 1K. Assume saturation region operation, then I D = K n (V GS V TN ) 2. From the circuit V GS = V DD = 4 V, so that I D = (0.5)(4 0.8) 2 = 5.12 ma. This current will result in V DS = 4 (5.12)(1) = 1.12 V. This is not a valid so solution assumption of saturation region operation wrong and MOSFET is in Ohmic region. Thus V DD = I D R D + V DS = K N R D [2(V GS V T )V DS V 2 DS ] + V DS Substituting values and simplifying results in 0.5V 2 DS 4.2V DS + 4 = 0 Valid solution is V DS = V, I D = (V DD V DS ) R D = (4 1.1) 1K = 2.92 ma. Part (b) Similar to Part (a), load line is I D = V DS 3K + 5 3K. Assume saturation region operation and find V DS = V, which is not valid. Thus, MOSFET is in Ohmic region. Following the same procedure as for Part (a), we find 1.5V 2 DS 13.6V DS + 5 = 0 The valid solution is V DS = 0.38 V, from which it follows that I D = 1.54 ma I D (ma) Q-point Part (b) Q-point, Part (a) V DS (V) 7
8 Question 8 Find v E, v C1, and v C2 in the circuit shown. Also, find the current though the top 1K resistor. Assume that v BE(on) = 0.7 V, and that β is large. (6 points) The base of Q 1 is at 0.5 V, while the base of Q 2 is at 0 V, so that Q 2 s base-emitter voltage is much larger than that of Q 1, and Q 2 is turned on hard, while Q 1 is essentially off. From this it follows that v C1 = 5 V, and v E = 0.7 V. Further, the current through the top 1K resistor is = (5 0.7) 1K = 4.3 ma v C2 = 5 + (1K)(4.3 ma) = 0.7 V. Question 9 The transistor in the amplifier shown has β = 350 and V BE(ON) = 0.65 V. (a) Make reasonable assumptions and show that I CQ 1 ma (3 points) (b) Show that R i 13.7K (5 points) Part (a) Since β is large, ignore I BQ so that V B = (9)(27 ( ) ) = 1.9V. Since V BE(ON) = 0.65 V, then V RE = = 1.25 V. Consequently, I CQ I E = K = ma 1 ma. Part (b) r π = β g m = I CQ = 8.75K. Using BJT scaling, R i = 65K 18K r π + (1 + β)(1.3k) = 13.68K 8
9 Question 10 Shown is the functional diagram of dual current source IC, REF200. In addition to two 100-μA sources, the IC incorporates a current mirror. Below are a collection of circuits that use the IC. For each of the circuits, determine I out. (12 points) I out I out I out (a) (b) (c) I out I out I out (d) (e) (f) (a) 50 μa, (b) 400 μa, (c) 50 μa, (d) (N + 1)100 μa (e) (N + 1) 100 μa (f) 300 μa 9
10 Question 11 Consider the amplifier below, which amplifies the signal from a sensor with an internal resistance of 1K. Ignore BJT s output resistance, and assume C 1 = C 2 = C 3. β = 100 I C = ma (a) Determine g m, r π (4 points) (b) Using BJT scaling, determine R i see figure (4 points) (c) Using the ratio of the collector and emitter resistors, estimate the overall voltage gain A v = v o v s (4 points) (d) Calculate the voltage gain A v = v o v s, but do not use the approximation that involves the ratio of the collector and emitter resistors, but rather incorporate the β of the transistor (4 points) Part (a) Part (b) g m = 40 I C = 9.8 ms, r π = β g m = 10.2 K R i = R 1 R 2 [r π + (1 + β)r E ] = 300K 160K [10.2K K] = 78.3K Part (c) The effective collector resistance is R C resistance and R i form a voltage divider. Thus = 22K 100K = 18K and the sensor s internal A v = v o R i R C = 78.3 v s R S + R i R E K 3K =
11 Question 12 For the amplifier below, R L = 500 Ω. Determine R o R ib, and estimate A v. (8 points) Hint, use BJT scaling. R S = 10K V + = 3 V V = 3 V I Q = 2 ma β = 300 V A = 100 V C C Since β is large, I C I E = I Q = 2 ma. Then g m = 40I C = 80 ma V and r π = β g m = 3.75K. Using BJT scaling: and R o = R S + r π 1 + β This is an Emitter Follower, so A v 1. = 10K K 301 = 45.7 Ω R i = r π + (β + 1)R L = 3.75K + (301)(500) = 154.3K 11
12 Question 13 The figure is a plot of the open-loop gain function for the LF357 voltage amplifier. An engineer will use the amplifier as an inverting amplifier with a mid-frequency voltage gain at 100. Use the plot and estimate the bandwidth of the feedback amplifier. (3 points) Write expressions for the transfer function A(f) for the open loop amplifier as well as the closed loop, inverting amplifier. (6 points). A gain of 100 is equivalent to a gain of 20 log 10 (100) = 40 db. A horizontal line at 40 db intercepts the LF357 gain curve at 100 khz Thus, the bandwidth ~ 100 khz and the GBP is (100)( ) = The pole for the open-loop amplifier is about 90 Hz, and the low-frequency gain is 105 db, so the open-loop gain is A(f) Open Loop = 1 + j f = j f 90 The closed-loop gain is 100 A(f) Closed Loop = f 1 + j
13 Question 14 Use BJT impedance scaling and determine the input and output impedances of the flowing circuits. Assume β = 300. (12 points) (a) (b) Circuit (a) g m = 40I C = 40 ms and 1 g m = 25 Ω, r π = β g m = 7.5 K. Using BJT scaling R i = (R 1 R 2 ) (r π + (1 + β)r L ) = (33K) (7.5K + (301)R L ) R o = r π (1 + β) = 7.5K Ω (6 points) Circuit (b) We need to find I C first. Since β is large, we will ignore I b and V B = 2.75 V using voltage division. Then V E 2 V and I E = 1 ma. Thus, g m = 40I C = 40 ms and 1 g m = 25 Ω, r π = β g m = 7.5 K. Using BJT scaling: R i = (R 1 R 2 ) r π + (1 + β)(r L 2K) = (15.27K) 7.5K + (301)(R L 2K) r π R o = (1 + β) R E = 25 2K 25 Ω (6 points) 13
14 Question 15 For the circuits below, assume β = 100 and use BJT impedance scaling to find the missing circuit parameters. (2 points for each parameter) I C = 1.7 ma A v (R C R L ) R E = r π = β g m = 100 (40I C ) = 1.47K R ib = r π + (1 + β) = 304.5K I C = 15 ma r π = β g m = 100 (40I C ) = 166 Ω R ib = r π + (1 + β)(r E R L ) = 15.32K R o = R S R 1 R 2 + r π R 1 + β E = 9.7 Ω I C 2 ma A v 1 r π = β g m = 100 (40I C ) = 1.25K R o = R S + r π (1 + β) = Ω 14
Homework 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 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 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
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 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 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 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 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 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 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 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 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 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 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 informationSOME USEFUL NETWORK THEOREMS
APPENDIX D SOME USEFUL NETWORK THEOREMS Introduction In this appendix we review three network theorems that are useful in simplifying the analysis of electronic circuits: Thévenin s theorem Norton s theorem
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 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 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 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 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 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 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 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 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 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 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 information6.012 Electronic Devices and Circuits
Page 1 of 12 YOUR NAME Department of Electrical Engineering and Computer Science Massachusetts Institute of Technology 6.012 Electronic Devices and Circuits FINAL EXAMINATION Open book. Notes: 1. Unless
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 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 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 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 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 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 informationChapter 6: Field-Effect Transistors
Chapter 6: Field-Effect Transistors slamic University of Gaza Dr. Talal Skaik FETs vs. BJTs Similarities: Amplifiers Switching devices mpedance matching circuits Differences: FETs are voltage controlled
More informationLecture 11: J-FET and MOSFET
ENE 311 Lecture 11: J-FET and MOSFET FETs vs. BJTs Similarities: Amplifiers Switching devices Impedance matching circuits Differences: FETs are voltage controlled devices. BJTs are current controlled devices.
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 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 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 informationTransistor Characteristics and A simple BJT Current Mirror
Transistor Characteristics and A simple BJT Current Mirror Current-oltage (I-) Characteristics Device Under Test DUT i v T T 1 R X R X T for test Independent variable on horizontal axis Could force current
More informationGEORGIA INSTITUTE OF TECHNOLOGY School of Electrical and Computer Engineering
NAME: GEORGIA INSTITUTE OF TECHNOLOGY School of Electrical and Computer Engineering ECE 4430 First Exam Closed Book and Notes Fall 2002 September 27, 2002 General Instructions: 1. Write on one side of
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 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 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 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 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 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 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 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 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 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 information6.301 Solid-State Circuits Recitation 14: Op-Amps and Assorted Other Topics Prof. Joel L. Dawson
First, let s take a moment to further explore device matching for current mirrors: I R I 0 Q 1 Q 2 and ask what happens when Q 1 and Q 2 operate at different temperatures. It turns out that grinding through
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 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? Three-terminal device whose voltage-current relationship is controlled by a third voltage
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 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 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 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 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 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 informationLecture 04: Single Transistor Ampliers
Lecture 04: Single Transistor Ampliers Analog IC Design Dr. Ryan Robucci Department of Computer Science and Electrical Engineering, UMBC Spring 2015 Dr. Ryan Robucci Lecture IV 1 / 37 Single-Transistor
More information(Refer Slide Time: 1:49)
Analog Electronic Circuits Professor S. C. Dutta Roy Department of Electrical Engineering Indian Institute of Technology Delhi Lecture no 14 Module no 01 Midband analysis of FET Amplifiers (Refer Slide
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 informationFinal Examination EE 130 December 16, 1997 Time allotted: 180 minutes
Final Examination EE 130 December 16, 1997 Time allotted: 180 minutes Problem 1: Semiconductor Fundamentals [30 points] A uniformly doped silicon sample of length 100µm and cross-sectional area 100µm 2
More informationLecture 12: MOSFET Devices
Lecture 12: MOSFET Devices Gu-Yeon Wei Division of Engineering and Applied Sciences Harvard University guyeon@eecs.harvard.edu Wei 1 Overview Reading S&S: Chapter 5.1~5.4 Supplemental Reading Background
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 informationLecture 15: MOS Transistor models: Body effects, SPICE models. Context. In the last lecture, we discussed the modes of operation of a MOS FET:
Lecture 15: MOS Transistor models: Body effects, SPICE models Context In the last lecture, we discussed the modes of operation of a MOS FET: oltage controlled resistor model I- curve (Square-Law Model)
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 informationReview of Band Energy Diagrams MIS & MOS Capacitor MOS TRANSISTORS MOSFET Capacitances MOSFET Static Model
Content- MOS Devices and Switching Circuits Review of Band Energy Diagrams MIS & MOS Capacitor MOS TRANSISTORS MOSFET Capacitances MOSFET Static Model A Cantoni 2009-2013 Digital Switching 1 Content- MOS
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 informationCM400DY-24A. APPLICATION AC drive inverters & Servo controls, etc CM400DY-24A. IC...400A VCES V Insulated Type 2-elements in a pack
CM00DY-A CM00DY-A IC...00A CES... 0 Insulated Type -elements in a pack APPLICATION AC drive inverters & Servo controls, etc OUTLINE DRAWING & CIRCUIT DIAGRAM Dimensions in mm 9±0. G 80 6±0. CE E C G E
More informationEE105 Fall 2014 Microelectronic Devices and Circuits. NMOS Transistor Capacitances: Saturation Region
EE105 Fall 014 Microelectronic Devices and Circuits Prof. Ming C. Wu wu@eecs.berkeley.edu 511 Sutardja Dai Hall (SDH) 1 NMOS Transistor Capacitances: Saturation Region Drain no longer connected to channel
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 informationT C MEASURED POINT G1 E1 E2 G2 W - (4 PLACES) G2 E2 E1 G1
CMDU-3KA Powerex, Inc., Hillis Street, Youngwood, Pennsylvania 15697-1 (7) 95-77 Dual IGBTMOD KA-Series Module Amperes/17 Volts B F A G T C MEASURED POINT M C L T - ( TYP.) N R Z CE1 E C1 C E AA S - (3
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 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 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 informationMMIX4B22N300 V CES. = 3000V = 22A V CE(sat) 2.7V I C90
Advance Technical Information High Voltage, High Gain BIMOSFET TM Monolithic Bipolar MOS Transistor (Electrically Isolated Tab) C G EC3 Symbol Test Conditions Maximum Ratings G3 C2 G2 E2C V CES = 25 C
More informationTutorial #4: Bias Point Analysis in Multisim
SCHOOL OF ENGINEERING AND APPLIED SCIENCE DEPARTMENT OF ELECTRICAL AND COMPUTER ENGINEERING ECE 2115: ENGINEERING ELECTRONICS LABORATORY Tutorial #4: Bias Point Analysis in Multisim INTRODUCTION When BJTs
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 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 informationMetal-Oxide-Semiconductor Field Effect Transistor (MOSFET)
Metal-Oxide-Semiconductor ield Effect Transistor (MOSET) Source Gate Drain p p n- substrate - SUB MOSET is a symmetrical device in the most general case (for example, in an integrating circuit) In a separate
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 informationEE 330 Lecture 22. Small Signal Modelling Operating Points for Amplifier Applications Amplification with Transistor Circuits
EE 330 Lecture 22 Small Signal Modelling Operating Points for Amplifier Applications Amplification with Transistor Circuits Exam 2 Friday March 9 Exam 3 Friday April 13 Review Session for Exam 2: 6:00
More informationSection 1: Common Emitter CE Amplifier Design
ECE 3274 BJT amplifier design CE, CE with Ref, and CC. Richard Cooper Section 1: CE amp Re completely bypassed (open Loop) Section 2: CE amp Re partially bypassed (gain controlled). Section 3: CC amp (open
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 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 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 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 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 informationThe Miller Approximation
The Miller Approximation The exact analysis is not particularly helpful for gaining insight into the frequency response... consider the effect of C µ on the input only I t C µ V t g m V t R'out = r o r
More informationEECS 105: FALL 06 FINAL
University of California College of Engineering Department of Electrical Engineering and Computer Sciences Jan M. Rabaey TuTh 2-3:30 Wednesday December 13, 12:30-3:30pm EECS 105: FALL 06 FINAL NAME Last
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 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 informationEECS 141: FALL 05 MIDTERM 1
University of California College of Engineering Department of Electrical Engineering and Computer Sciences D. Markovic TuTh 11-1:3 Thursday, October 6, 6:3-8:pm EECS 141: FALL 5 MIDTERM 1 NAME Last SOLUTION
More informationMMIX4B12N300 V CES = 3000V. = 11A V CE(sat) 3.2V. High Voltage, High Gain BIMOSFET TM Monolithic Bipolar MOS Transistor
High Voltage, High Gain BIMOSFET TM Monolithic Bipolar MOS Transistor Preliminary Technical Information V CES = 3V 11 = 11A V CE(sat) 3.2V C1 C2 (Electrically Isolated Tab) G1 E1C3 G2 E2C G3 G E3E C1 C2
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 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 information7. DESIGN OF AC-COUPLED BJT AMPLIFIERS FOR MAXIMUM UNDISTORTED VOLTAGE SWING
à 7. DESIGN OF AC-COUPLED BJT AMPLIFIERS FOR MAXIMUM UNDISTORTED VOLTAGE SWING Figure. AC coupled common emitter amplifier circuit ü The DC Load Line V CC = I CQ + V CEQ + R E I EQ I EQ = I CQ + I BQ I
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 informationECE 205: Intro Elec & Electr Circuits
ECE 205: Intro Elec & Electr Circuits Final Exam Study Guide Version 1.00 Created by Charles Feng http://www.fenguin.net ECE 205: Intro Elec & Electr Circuits Final Exam Study Guide 1 Contents 1 Introductory
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 information