ANALOG ELECTRONICS 1 DR NORLAILI MOHD NOH

Similar documents
ANALOG ELECTRONICS DR NORLAILI MOHD NOH

Exercises for Frequency Response. ECE 102, Winter 2011, F. Najmabadi

Design of Analog Integrated Circuits

Microelectronics Circuit Analysis and Design. NMOS Common-Source Circuit. NMOS Common-Source Circuit 10/15/2013. In this chapter, we will:

Is current gain generally significant in FET amplifiers? Why or why not? Substitute each capacitor with a

Faculty of Engineering

Introduction to Electronic circuits.

FYSE400 ANALOG ELECTRONICS

T-model: - + v o. v i. i o. v e. R i

CHAPTER 3 ANALYSIS OF KY BOOST CONVERTER

The two main types of FETs are the junction field effect transistor (JFET) and the metal oxide field effect transistor (MOSFET).

PHYSICS 536 Experiment 12: Applications of the Golden Rules for Negative Feedback

CIRCUIT ANALYSIS II Chapter 1 Sinusoidal Alternating Waveforms and Phasor Concept. Sinusoidal Alternating Waveforms and

Wp/Lmin. Wn/Lmin 2.5V

ANALOG ELECTRONICS I. Transistor Amplifiers DR NORLAILI MOHD NOH

Feedback Principle :-

55:041 Electronic Circuits

6. Cascode Amplifiers and Cascode Current Mirrors

, where. This is a highpass filter. The frequency response is the same as that for P.P.14.1 RC. Thus, the sketches of H and φ are shown below.

CHAPTER 11. Solutions for Exercises. (b) An inverting amplifier has negative gain. Thus L

EE 204 Lecture 25 More Examples on Power Factor and the Reactive Power

Linear Amplifiers and OpAmps

ME2142/ME2142E Feedback Control Systems. Modelling of Physical Systems The Transfer Function

Diodes Waveform shaping Circuits. Sedra & Smith (6 th Ed): Sec. 4.5 & 4.6 Sedra & Smith (5 th Ed): Sec. 3.5 & 3.6

CHAPTER 3: FEEDBACK. Dr. Wan Mahani Hafizah binti Wan Mahmud

Week 11: Differential Amplifiers

Diodes Waveform shaping Circuits

EE 221 Practice Problems for the Final Exam

Exercises for Frequency Response. ECE 102, Fall 2012, F. Najmabadi

Circuits Op-Amp. Interaction of Circuit Elements. Quick Check How does closing the switch affect V o and I o?

Waveshapping Circuits and Data Converters. Lesson #17 Comparators and Schmitt Triggers Section BME 373 Electronics II J.

Chapter 7. Systems 7.1 INTRODUCTION 7.2 MATHEMATICAL MODELING OF LIQUID LEVEL SYSTEMS. Steady State Flow. A. Bazoune

ES 330 Electronics II Homework 04 (Fall 2017 Due Wednesday, September 27, 2017)

College of Engineering Department of Electronics and Communication Engineering. Test 1 With Model Answer

Module B3. VLoad = = V S V LN

Active Load. Reading S&S (5ed): Sec. 7.2 S&S (6ed): Sec. 8.2

III. Operational Amplifiers

Week 9: Multivibrators, MOSFET Amplifiers

Section 3: Detailed Solutions of Word Problems Unit 1: Solving Word Problems by Modeling with Formulas

CHAPTER 3 Frequency Response of Basic BJT and MOSFET Amplifiers

IGEE 401 Power Electronic Systems. Solution to Midterm Examination Fall 2004

Lecture 2 Feedback Amplifier

Transfer Characteristic

Introduction of Two Port Network Negative Feedback (Uni lateral Case) Feedback Topology Analysis of feedback applications

Linearity. If kx is applied to the element, the output must be ky. kx ky. 2. additivity property. x 1 y 1, x 2 y 2

COLLEGE OF ENGINEERING PUTRAJAYA CAMPUS FINAL EXAMINATION SPECIAL SEMESTER 2013 / 2014

Copyright Paul Tobin 63

The Operational Amplifier and Application

Schedule. ECEN 301 Discussion #17 Operational Amplifiers 1. Date Day Class No. Lab Due date. Exam

6. Frequency Response

Applying Kirchoff s law on the primary circuit. V = - e1 V+ e1 = 0 V.D. e.m.f. From the secondary circuit e2 = v2. K e. Equivalent circuit :

College of Engineering Department of Electronics and Communication Engineering. Test 2

FE REVIEW OPERATIONAL AMPLIFIERS (OP-AMPS)

Chapter 3, Solution 1C.

The three major operations done on biological signals using Op-Amp:

SFDMB3638F. Specifications and Applications Information. orce LED Driver. Mass: 7 grams typ. 10/15/08 Preliminary. Package Configuration

Bipolar-Junction (BJT) transistors

WYSE Academic Challenge 2004 Sectional Physics Solution Set

Relationships Between Frequency, Capacitance, Inductance and Reactance.

VI. Transistor Amplifiers

Lab 11 LRC Circuits, Damped Forced Harmonic Motion

R th is the Thevenin equivalent at the capacitor terminals.

Transistors. Lesson #10 Chapter 4. BME 372 Electronics I J.Schesser

ECE-343 Test 2: Mar 21, :00-8:00, Closed Book. Name : SOLUTION

Driving your LED s. LED Driver. The question then is: how do we use this square wave to turn on and turn off the LED?

(b) i(t) for t 0. (c) υ 1 (t) and υ 2 (t) for t 0. Solution: υ 2 (0 ) = I 0 R 1 = = 10 V. υ 1 (0 ) = 0. (Given).

Lesson #14. Section BME 373 Electronics II J.Schesser

Common Base Configuration

Supplementary Course Notes Adding and Subtracting AC Voltages and Currents

Physics 102. Second Midterm Examination. Summer Term ( ) (Fundamental constants) (Coulomb constant)

V V. This calculation is repeated now for each current I.

IXD4902. Three-Terminal Negative Voltage Regulator FEATURES DESCRIPTION APPLICATIONS

Output Stages and Power Amplifiers

Diode. Current HmAL Voltage HVL Simplified equivalent circuit. V γ. Reverse bias. Forward bias. Designation: Symbol:

I = α I I. Bipolar Junction Transistors (BJTs) 2.15 The Emitter-Coupled Pair. By using KVL: V

Introduction to Three-phase Circuits. Balanced 3-phase systems Unbalanced 3-phase systems

( ) = ( ) + ( 0) ) ( )

55:141 Advanced Circuit Techniques Two-Port Theory

Physics Courseware Electronics

At point G V = = = = = = RB B B. IN RB f

2. Find i, v, and the power dissipated in the 6-Ω resistor in the following figure.

Bicycle Generator Dump Load Control Circuit: An Op Amp Comparator with Hysteresis

1. Transformer A transformer is used to obtain the approximate output voltage of the power supply. The output of the transformer is still AC.

Definition of Strain. Tutorial

Revision: December 13, E Main Suite D Pullman, WA (509) Voice and Fax

Revision: August 19, E Main Suite D Pullman, WA (509) Voice and Fax

Energy Storage Elements: Capacitors and Inductors

Chapter 10 Sinusoidal Steady-State Power Calculations

Supplementary Course Notes Adding and Subtracting AC Voltages and Currents

CHAPTER 13. Exercises. E13.1 The emitter current is given by the Shockley equation:

General Amplifiers. Analog Electronics Circuits Nagamani A N. Lecturer, PESIT, Bangalore 85. Cascade connection - FET & BJT

Section I5: Feedback in Operational Amplifiers

EE 330 Lecture 25. Amplifier Biasing (precursor) Two-Port Amplifier Model

ECSE Linearity Superposition Principle Superposition Example Dependent Sources. 10 kω. 30 V 5 ma. 6 kω. 2 kω

BASIC DIRECT-CURRENT MEASUREMENTS

55:141 Advanced Circuit Techniques Two-Port Theory

FE REVIEW OPERATIONAL AMPLIFIERS (OP-AMPS)( ) 8/25/2010

V = = A = ln V

Three charges, all with a charge of 10 C are situated as shown (each grid line is separated by 1 meter).

Exercises H /OOA> f Wo AJoTHS l^»-l S. m^ttrt /A/ ?C,0&L6M5 INFERENCE FOR DISTRIBUTIONS OF CATEGORICAL DATA. tts^e&n tai-ns 5 2%-cas-hews^, 27%

Transcription:

24 ANALOG LTRONIS TUTORIAL DR NORLAILI MOHD NOH

. 0 8kΩ Gen, Y β β 00 T F 26, 00 0.7 (a)deterne the dc ltages at the 3 X ternals f the JT (,, ). 0kΩ Z (b) Deterne g,r π and r? (c) Deterne the ltage gan ( IA Y / s )f ths aplfer crcut f a sgnal surce wth a ltage s anda2kω nternal resstr s a) dc ltage at, and. cnnected t X and an 8 kω lad resstr I α I β F I 0.99A s cnnected t Y. Z s cnnected t β F grund. I 0.99A 9.9μA (d) If the r s neglected, calculate the β 00 percentage f errr n yur calculatn f the ltage gan. I A I 099A I 0 I R 0 0.99A(8k) 2.08 I 0-0k 9.9μA(0k) 0.099 0.7 0.099 0.70.799 A 2

b) Gen: β β 00 T A F 26 00 alculated: I α I βf I 0.99A β F I 0.99 A g 0.038 T 26 β 00 g 0.038 A 00 I 099 0.99 rπ 2624.679Ω r 000.0Ω 3

2kΩ Y s 0kΩ g rπ rο 8kΩ 8kΩ X y 8kΩ 0 Y 0kΩ Z c) Y a g r //8k //8k s s (r π //0k)s g (r//4k) 2kr π //0k s 2079.002 0.038 3847.6337 002 4079.002 74.77 I 0.99 A g 0.038 T 26 β 00 g 0.038 A 00 I 0.99 rπ 2624.679Ω r 000.0Ω Ω IA 4

d) If r s neglected, A 0.038(0.5097)4 k 77.6783 Hence, the percentage f errr 77.678374.77 74.77 00% 3.96% 396% 2kΩ Y s 0kΩ g rπ 8kΩ rο 8kΩ y 5

2. Deterne the sgnal ltage at the. The sgnal surce ges a ltage f 0 rs and has a resstance f 300Ω. 22kΩ 2 3 ut 22kΩ dc equalent crcut : 2 I n 6.8kΩ 560Ω 2 R L 6.8kΩ I 560Ω ac equalent crcut : 300Ω Gen β F 50; β 60; 0.7 and 26 T n 6.8kΩ 22kΩ g be rπ be rο R L 6

2 22kΩ 3 ut s 22kΩ 6.8kΩ 2 560Ω 2 R L Methd I R R IN() I I Assue ( β ) I R <<, IN() I I R R where I ( β )I R 2 Assue β F >> R,R βf R 6.8kΩ R R IN() I 560Ω R Snce β F 50, RIN() IN() 50(560) 84kΩ F 7

A cn rule-f-thub s that f tw resstrs are n parallel and ne s at least ten tes the ther, the ttal resstance s taken t be apprxately equal t the saller alue. 6.8k//R IN() (2) 6 6.8k//R 22k IN() (The exact alue s 6.8k//84k 6.2907k Ω) IN 6.8k () 84kΩ, (2) 68k 6.8k 22k Snce R 2.8333 2.8333 0.77 2.333 2.333 560 Ω βf α βf I 3.7843A T 26A β 60 g 0.456 I 3.8095 I I I 3.7843A g 0.456 rπ 098.90Ω A 22kΩ R 2 68kΩ 6.8kΩ R 2 I 560Ω R R IN() 8

6.8k (2) 6.8k 22k 2.8333 2.333 I 3.8095 I 3.7843A g 0.456A r 098.90Ω π 6.8k//22k // rπ be 6.8k//22k // rπ 300 R 6.8k 22k.0989k T.025 0 3 n Hence, R T 907.085Ω Ω 0.7545 be 7.545 be n rs 300 kω n 6.8kΩ 22kΩ rπ g be be rο R L 9

Methd 2 TH TH 6.8k (2) 6.8k 22k 2.8333 R 22k//6.8k 594.4444 Ω 2.8333 I (594.4444) I (560) 0 I (594.4444( )560)0.7 2.8333 βf 2.8333 0.7 89754.4444 I 2.3768 0 5 A 2.0098 I 3.5889 A R I αi 3.565A g I 0.37A rπ 2.8333 (594.4444 2.3768 0 5) 2.7098 T β.670 kω g 0.7606 7.6 rs be n s 22kΩ 6.8kΩ 560Ω R TH 2 3 2 ut 2 R L I TH 560Ω 0

nclusn Usng R (nce t s knwn that R ) R R IN() 0 R 2 wll result n a be that s ery clse t the ne btaned fr Methd 2 n

3. Select a nu alue fr the etter-bypass capactr f the aplfer ust perate er a frequency range fr 2kHz t 0kHz Slutn The alue f the bypass capactr ust be large enugh s that ts reactance er the frequency saple f the aplfer s ery sall (deally 0Ω) cpared t R. A gd rule f thub s that X f the bypass capactr shuld be at least 0 tes saller than R at the nu frequency fr whch the aplfer ust perate. 2 Fr ths crcut, 0X R X R 560 56 Ω 0 0 X 2 π f 2πfX 2 π(2k)56.42μ F 22kΩ 6.8kΩ Ths s the nu alue fr the bypass capactr fr ths crcut. A larger alue can be used, althugh cst and physcal sze usually pse ltatns. 3 560Ω 2 R 2

At the axu frequency f 0kHz, 2 π (0k).42 μ X.2Ω At the axu frequency f 0kHz, X.2Ω and ths alue s R /0 where R 560 Ω If the axu frequency s used nstead t deterne, then 0.2842μF 2 π (0k)56 At the nu frequency f 2kHz, X 280 2 π (2k)0.2842μ Ω 0 X 2800 and ths s > R. Hence, n rder t calculate the nu alue f the bypass capactr, the X ust be 0 tes saller than R at the nu frequency. 3

4. alculate the ltage gan wth and wthut the bypass capactr. 2 a) Wth capactr 22kΩ 3 n 300Ω 68kΩ 6.8kΩ 560Ω 2 ac equalent crcut : 300Ω n 22kΩ 6.8kΩ g be rπ be rο 4

300Ω n 22kΩ 6.8kΩ g be rπ be rο Open crcut ltage gan : a ( ) gbe r //k n n Assung r >>k, We knw be a g be( k) n 22k//6.8k//rπ 22k//6.8k//r 300 π n r 098.90 π Ω be 0.7545n Fr Questn 2 (ethd ), and hence g( k ) 0.754n a n Fr Questn 2 (ethd ), g 0.456A a 09.4038 5

If there s n 2 2 22kΩ 3 n 300Ω 6.8kΩ 560Ω ac equalent crcut 300Ω n 22kΩ 6.8kΩ g be rππ be 560Ω rο 6

b) Wthut bypass capactr : 300 Ω b g be rπ be n 68kΩ 6.8kΩ 22kΩ 560Ω R' e rο If the effect f r s neglected, the pen crcut ltage gan : a n g be ( k) be e ( g ) be b be 560 be be ( g ) 560 r π ( g )560 r be π ( 0.456)560 098.90 83.0456 be be ( ) 0.456 k 83.0456.7532 7

300Ω n 22kΩ b g be rπ be 68kΩ 6.8kΩ R' 560Ω e ( 22k//6.8k//R' ) n 22k//6.8k//R'300 83.0456be( rπ R' ) 9.2589kΩ b be 494.7n 0.9425 524.7 n.7532 0.9425.6524 n n Hence, f there s n 2, the pen crcut ltage gan s ery uch reduced. 8

5. nnect a 5kΩ lad t the aplfer shwn belw. Fnd the erall gan. ac equalent crcut 2 300Ω n a R be T 300Ω 22kΩ g be n R be Τ 3 6.8kΩ 560Ω g k//5k n n RT R 300 n T be( ) 6.8k//22k//r π 5kΩ 2 5kΩ be 0.754n a 0.456(833.3333)(0.754) 3333)(0 9.699 Fr Questn 4, when there s n lad, a 09.4038 Thus, pen-crcut ltage gan s the axu gan that can be acheed by an aplfer. Fr Questn 2, g 0.456A, rπ 098.90, RT 907.085 Ω Ω 9

nclusn nnectng a lad t the utput wll reduce the gan. When there s n lad (.e. R L ), the gan f the aplfer s at ts axu. At R L,the gan s sad t be the pen crcut t ltage gan. 20

n 6. Fnd R, a, a, and a p. Gen β F β 75 and assue that the capactances reactances are neglgble at the frequency f peratn. T 26 and 0.7. Neglect r. rs 0kΩ 0kΩ 0 2 kω 0 kω n 5kΩ R ac equalent crcut R b b g rπ be be β b 0k//k 909.0909Ω R RIN() R β 75kΩ F (k) 0R IN() 2 0k (0) 20k 5 0.7 4.3 I 4.3 k 4.3A I 75 I (4.3) 4.2756A R α b 064.78 76 (76)909.09090909 75 r π (26) 064.78Ω 6.0642kΩ 4.2756 R n br π ( b β b)909.0909 5k//6.0642k R b b 4.8494kΩ b R r ( )909.0909 b π β 2

0 n rs 0kΩ 0kΩ kω 2 0 kω a n ( b β b )909.0909 br π ( b β b)909.0909 ( β)909.0909 r π( β)909.0909 59999.9984 6064.765 0.9934 R ac equalent crcut n R b 5kΩ β g rπ g be rπ be β b 0k//k 909.0909Ω 75 (26) I 22

a n b g rπ be be β b 5kΩ 0k//k 909.0909Ω R ( b β) br π ( β)b909.0909 b 5k ( b β) br π ( β) b(909.0909) b 5k 76 064.78 76(909.0909) 5k 76 33.228 5.2992 23

a p n b rπ g be be β b 5kΩ 0k//k R 909.0909Ω0909Ω ( b β) br π ( β) b(909.0909) b 5k 76 064.7878 76(909.0909) 0909) 5k 76 33.228 5.2992 a a a 5.2992 0.9934 5.2642 Fr a cnfguratn, p a a r π r a 064.78Ω ( b β ) n R 76 r π ( β )909.0909 4.8494k 5.299 24

Usng Methd 2 t deterne I : 0 R Usng KL at the nput lp : 0kΩ p R 2 0kΩ 0k 20k (0) 5 0k//0k 5kΩ 0.7 0 a a a 5.2992 0.9934 5.2642 5 I (5k) 0.7 I (k) 0 I (5k) I ( β)(k) 4.3 I 4.3 8k 0.0238A I βi 4.65A rπ 092.437 Rb 092.437 76(909.0909) 6.0924kΩ R 5k//6.0924k 4.8495kΩ 76(909.0909) 092.4 37 76(909.0909) 76(4.8495k) 092.437 76(909.0909) a 0.9932 a 5.2983 25

7. Gen : β F 250, T 26 and 0.7. Fnd the nput resstance, shrt crcut current gan and pwer gan fr the aplfer shwn belw. ac equalent crcut 00μF 56kΩ 2 μf 2kΩ 0 2.2kΩ 3 μf 0kΩ rπ 56 k Ω //2 k Ω be n g be 2.2kΩ 0kΩ n The 56 kω//2 kω s shrted t grund and can be neglected n the sall sgnal del. The analyss beces dffcult because the current surce s between the utput and the nput nde. Hence, use the T-del. 26

T-del e k k n R IN() re be Re e β R α F 250kΩ 250k >0 2k 2 k 0 2 k56 k.7647.7647 0.7.0647 I.0647A I.0604A ( ) r ge e 2.2kΩ R L 0kΩ e g g I T g r e r π I.0604A g 0 0408 g 0.0408 A 250 e 25 0.0408 a Ntce that the 56 kω//2 kω are shrted t grund and can be neglected n the sall sgnal del. 0 n g e(2.2k//0k) e g 2.2k//0k2k//0k ( ) 73.5738 a g 89.76 r 24.422Ω R k//24.422422 23.8304 Ω R r e ( 2.2k) 56kΩ 2 μ F 00μF n 2kΩ 2.2kΩ 3 μf 0kΩ 27

e n k re ge e 2.2kΩ R L 0kΩ a 0 0408 s ge 0.0408 a 0 s 0.042 e g k r e be KL at nde : e k e e k r e k r e e 0.974 56kΩ 2 μf 00 μ F n Shrt-crcut current gan f a aplfer. s a current buffer. 2kΩ Open-crcut ltage gan, Hence, fr a aplfer, pwer gan ltage gan. a a a 89.76(0.974) 87.929 p s a g (2.2k) 89.76 0 2.2kΩ 3 μf 28 0kΩ

μnx W 8. Gen k 0.24 0 3A 2, GSQ 6.4, 2 L I DQ 2.75 and r O 50kΩ. Deterne g, R wth and wthut r O (cpare the result), and a wth and wthut r O (cpare the result). 2 2kΩ 0MΩ μf Slutn W gμnx ( GS t ) 2k ( GSt ) L μnx W 2 2 I D ( GS t ) k ( GS t ) 2 L 2 μf g ID k 2k g 3 2.75 4 0.24 0.6248S 29

2 2kΩ G 0MΩ D 0MΩ μf μf gs ggs 50kΩ 2kΩ R gs KL at nde D : ggs 50k//2k (0M) gs g gs (0M) 50k//2k gs R S 0M g 50k//2k 50k//2k 0M 50k//2k 520 2.448 g 50k//2k R 2.425M Ω 30

G 0MΩ D gs ggs 2kΩ R S Wthut r : 0M 2k 500 g 2.248 2k R 2.3536MΩ Obseratn : Snce r >0R D, R wth r wthut r wll be qute clse. 3

a gs G 0MΩ D g 50k//2k (0M) gs gs 50k//2k 0M ( g (0M) ) 50k//2k.6248(0M) 6247 0M 520 923.0769 g 0M ggs 50kΩ 2kΩ S a 3238 3.238 Wthut r, [ g (0M) ] 6247 0M 500 2k 3.2488 2k The tw results are clse. Hence, the effect f r s nt ery sgnfcant t the calculatn f a. 32

9. Deterne the utput ltage f 0.8 and r 40kΩ. Gen k0.4x0-3 A/ 2 and t 3. The dc ltage at surce s.2. Assue that bdy effect s neglgble. 30 40MΩ 3.3kΩ 3 0MΩ.2kΩ 2 ac equalent crcut G D g gs gs 0MΩ 40MΩ S 40kΩ 3.3kΩ 33

G D ggs gs 0MΩ 40MΩ 40kΩ 3.3kΩ dc-analyss a g gs(40k//3.3k) gs ag g? ( 3048.4988) W ( ) L μnx W 3 0.4 0 2 L 3( ) g μ S n x GS t k g 0.8 0 3 GS 30 30 40MΩ 0MΩ.2kΩ 3.3kΩ 3 2 40MΩ 0MΩ 3.3kΩ.2kΩ? GS G 0M ( ) 0M40M 30 6 S.2 GS 6.2 4.8 3 g 0.8 0 (.8).44S a.44 ( 3048.49884988 ) 4.3898 a 4.3898(0.8) 3.58 34

0 (a)????? D del R sg sg D O GS G S sg R n R D DD SS 2 R G 0 ID 0.5A D DDIDRD 0 0.5(5k)2.5 5 D I R L Gen: DD SS 0 I 0.5A RD 5kΩ.5 t μnxw A A L 75 μ W ( ) 2 L μ W ( O ) 2 L 2 O 2 (0.5)2 I n x 2 D GS t n x 2 ID 0.5 2 R D DD SS S O I GS t GS.5 2.5 GS G S 2.5 2.5 35

(b) g? r O? R sg sg DD R D 2 I R n SS I μnx W 2 D ( GS t ) 2 L W g μnx ( GSt ) L R L g ( ) I 2 D GS t 2ID 2(0.5) A g r GSt A 50k I Ω D 36

R? R? a? (c) O R sg sg R D DD 2 R L R sg gs G ggs sg S r sg R R D L sg I R g R b bs n SS S sg r D T-del R sg sg R g D sg gsg R // D R L G 37

R sg S sg r D sg gsg R // D R L g sg R Gen : G R sg KL at nde S, sg g sg r KL at nde D: RL 5kΩ R g sg 50Ω r 50k Ω g A R 5kΩ D Snce χ s nt gen, assue the bdy effect s neglgble. R? a sg r R D//RL g r r R D//RL r R D//RL g r RR r r R D//RL g r g r ( ) 38

r r R D//RL R g ( r) g r r D L R r R //R ( gr) r R D//R L g r g r 50k R 50k 7500 ( 5) 50k 7500 50k 50k 2 R D L 3 0.02.0067 0 R 045.245Ω R//R 7.5kΩ r 50kΩ g A 39

R sg S r D sg g sg gsg R D R L R' t O t 0 sg KL at nde S : g R sg t sg sg sg r sg sg g Rsg r r t g r Rsg r t G R sg R ' sg g S R r sg G D t gsg R ' t 40

S r D t R sg sg g sg gsg t KL at nde D : t sg r t sg g r r t g t r r g r Rsg r g t sg t t G R ' t t g t r r g Rsg r t 50k 50k 50 50k 6 t 6.347 0 t t R ' 57.5528k5528k Ω O O t R 5k//57.5528k 3.696kΩ 4

a sg R sg sg R R sg g sg D// L r sg S r r D g sg gsg R // D R L R G R D //RL g // r r g R D //R L r R D//R L r 7500 50k 7500 50k R R sg D L sg sg 755 7.55 sg.05 R sg R Rsg 045.245 sg 095.245 0.9543 a sg 7.905 sg sg sg sg sg sg sg a 7.905 0.9543 a 6.869 42