CAPACITANCE AND INDUCTANCE

Transcription

1 APAITANE AND INDUTANE Inroduces wo passive, energy soring devices: apaciors and Inducors LEARNING GOALS APAITORS Sore energy in heir elecric field (elecrosaic energy) Model as circui elemen INDUTORS Sore energy in heir magneic field Model as circui elemen APAITOR AND INDUTOR OMBINATIONS Series/parallel combinaions of elemens R OP-AMP IRUITS Inegraion and differeniaion circuis

2 APAITORS Firs of he energy sorage devices o be discussed Typical apaciors Basic parallel-plaes capacior IRUIT REPRESENTATION NOTIE USE OF PASSIVE SIGN ONVENTION

3 A d Dielecric consan of maerial in gap PLATE SIZE FOR EQUIVALENT AIR-GAP APAITOR A 8 55F A m Normal values of capaciance are small. Microfarads is common. For inegraed circuis nano or pico farads are no unusual

4 Basic capaciance law Q f V ) ( Linear capaciors obey oulomb s law Q V is called he APAITANE of he device and has unis of charge volage One Farad(F)is he capaciance of a device ha can sore one oulomb of charge a one Vol. oulomb Farad Vol Linear capacior circui represenaion EXAMPLE V Volage across a capacior of micro Farads holding 0m of charge 3 Q *0 0* apaciance in Farads, charge in oulombs resul in volage in Vols apaciors can be dangerous!!! V

5 apaciors only sore and release ELETROSTATI energy. They do no creae The capacior is a passive elemen and follows he passive sign convenion LEARNING BY DOING Linear capacior circui represenaion dv i( ) ( ) d

6 Q V apaciance Law If he volage varies he charge varies and here is a displacemen curren One can also express he volage across in erms of he curren V ( ) Q i ( x) dx Inegral form of apaciance law Or one can express he curren hrough in erms of he volage across i dq d dv d Differenial form of apaciance law The mahemaical implicaion of he inegral form is... V ( ) V ( ); Volage across a capacior MUST be coninuous Implicaions of differenial form?? V ons i 0 D or seady sae behavior A capacior in seady sae acs as an OPEN IRUIT

7 APAITOR AS IRUIT ELEMENT LEARNING EXAMPLE i i v dvc ( ) ( ) d v ( ) i ( x) dx v v ( ) ( ) v ( 0 i ) ( x) dx ( x) dx 0 i ( x) dx 0 The fac ha he volage is defined hrough an inegral has imporan implicaions... i i v v c ( O ) R R v R Ri R Ohm s Law R 5F DETERMINE THE URRENT dv i( ) ( ) d 60mA 6 4 V i 50 [ F] 0mA 3 60 s i( ) 0 elsewhere

8 APAITOR AS ENERGY STORAGE DEVIE ) ( ) ( ) ( i v p Insananeous power ) ( ) ( d dv i c d dv v p c ) ( ) ( q dx x i v ) ( ) ( ) ( ) ( ) ( ) ( d dq q p Energy is he inegral of power ) ( ), ( dx x p w If is minus infiniy we alk abou energy sored a ime. If boh limis are infiniy hen we alk abou he oal energy sored. ) ( ) ( v d d p ) ( ) ( ), ( v v w ) ( ) ( q d d p c ) ( ) ( ), ( q q w W v i

9 5F LEARNING EXAMPLE w w Energy sored in 0-6 msec (0,6) v (6) v (0) 6 (0,6) 5*0 [ F]*(4) [ V q harge sored a 3msec q ( 3) v (3) (3) 5*0 6 [ F]*[ V ] 60 ] oal energy sored?... oal charge sored?... If charge is in oulombs and capaciance in Farads hen he energy is in.

10 4 F. FIND THE VOLTAGE v( 0) 0 v( ) v(0) i( x) dx; 0 0 v( ) v() i( x) dx; 0 v( ) 80 3 [ V 4ms ]

11 4 F. FIND THE POWER i( ) p( ) 8, 0 ms 0 v( 0) p ( ) 0, 4ms elsewhere

12 FIND THE ENERGY 3 p( ) 8, 0 ms 4ms p ( ) 0, elsewhere

13 LEARNING EXTENSION F DETERMINE THE URRENT dv i( ) ( ) d 6 i 0 F 3 40 V s 6 i 0 F 3 0 V s

14 SAMPLE PROBLEM v() v( ) 30 sin (0 ) F WHAT VARIABLES AN BE OMPUTED? E(/ 40) q ( ) v ( (/0 ) Energy sored a a given ime E( ) v ( ) harge sored a a given ime ) dv d urren hrough he capacior i () 6 *0 [ F]*30 sin 6 q *0 [ ]*sin( )[ V ] 0 6 i (/0) *0 *30 *0 cos( ) Elecric power supplied o capacior a a given ime p ( ) v ( ) i ( ) Energy sored over a given ime inerval W w(, ) v ( ) v ( ) J J A

15 SAMPLE PROBLEM If he curren is known... i v F i ( ) Volage a a given ime 0.5 e ; 0 [ ma] 0; 0 Volage a a given ime when volage a ime o< is also known v () v urren hrough capacior v ( ) i ( x) dx v (0) 0[ V ] v ( ) v ( 0) i ( x) dx 0 0.5x 0.5 ( 0) e dx 6 *0 x 6 e 0.5 e 0.63* 0 6 V 0 0 *0 0.5 q ( ) v ( q () *0. 63 v ( ) 0; 0 v ( ) i ( x) dx 0.5x v ( ) v (0) e dx ( e ); 0 p ( ) v ( ) i ( W v ( ) 0; 0 w( ) v ( J w T v ( wt *0 *(0 ) 0 J harge a a given ime ) Volage as a funcion of ime Elecric power supplied o capacior ) Energy sored in capacior a a given ime ) Toal energy sored in he capacior ) V

16 SAMPLE PROBLEM 5 0 Given curren and capaciance V ompue volage as a funcion of ime A minus infiniy everyhing is zero. Since curren is zero for <0 we have 0 5msec (0) 0 V 3*0 ( ) 4*0 In paricular (msec) V (5 5 0 ms i ( ) 0 [ A] 6 5 A 0 A 3 i ( ) 3 3*0 [ A/ s] ms 0 s 3 xdx [ V ] 3*0 3 6 [ V ];0 5*0 [ s] 0 8 ms) 3*0 3 V *(5*0 8 ( ) 0; 0 3 ) [ V ] 75 8 [ mv ] harge sored a 5ms q ( ) V ( ) 6 75 *0 q(5ms) 4*0 [ F]* 8 q( 5ms) (75/ ) [ n] Toal energy sored E V 3 V [ V ] oal means a infiniy. Hence E T 0.5* 4*0 6 5*0 8 3 (5ms) V [ J ] ( ) 75 8 [ mv ] V 75* *0 ( ) 8 3 4*0 ( 3 6 5*0 0*0 6 )[ A/ s] dx *0 [ V ]; 5*0 0*0 [ s] Before looking ino a formal way o describe he curren we will look a addiional quesions ha can be answered. Now, for a formal way o represen piecewise funcions...

17 Formal descripion of a piecewise analyical signal V c ( ) ; 3 ; ; 4 5 ; ms 0 [ ms] 0[ ms] [mv ]

18 INDUTORS NOTIE USE OF PASSIVE SIGN ONVENTION Flux lines may exend beyond inducor creaing sray inducance effecs ircui represenaion for an inducor A TIME VARYING FLUX REATES A OUNTER EMF AND AUSES A VOLTAGE TO APPEAR AT THE TERMINALS OF THE DEVIE

19 A TIME VARYING MAGNETI FLUX INDUES A VOLTAGE v L d d Inducion law FOR A LINEAR INDUTOR THE FLUX IS PROPORTIONAL TO THE URRENT Li L v L dil L d DIFFERENTIAL FORM OF INDUTION LAW THE PROPORTIONALITY ONSTANT, L, IS ALLED THE INDUTANE OF THE OMPONENT LEARNING by Doing INDUTANE IS MEASURED IN UNITS OF henry (H). DIMENSIONALLY HENRY Vol Amp sec INDUTORS STORE ELETROMAGNETI ENERGY. THEY MAY SUPPLY STORED ENERGY BAK TO THE IRUIT BUT THEY ANNOT REATE ENERGY. THEY MUST ABIDE BY THE PASSIVE SIGN ONVENTION Follow passive sign convenion

20 A direc consequence of inegral form v L dil L d i L( ) vl( x) dx L i L ( ) i i L L ( Differenial form of inducion law 0 ) vl( x) dx; L 0 Inegral form of inducion law ( ) i ( ); urren MUST be coninuous A direc consequence of differenial form il ons. vl 0 w L ( L Power and Energy sored di d L Li ( ) d d L L pl( ) vl( ) il( ) W p ( ) L ( ) il( ), ) d d Li ( x) dx w(, ) LiL( ) LiL( ) w ( ) Li L L( ) L Energy sored a ime Mus be non-negaive. Passive elemen!!! J urren in Amps, Inducance in Henrys yield energy in Joules Energy sored on he inerval an be posiive or negaive 0 D (seady sae) behavior

21 LEARNING EXAMPLE FIND THE TOTLA ENERGY STORED IN THE IRUIT In seady sae inducors ac as shor circuis and capaciors ac as open circuis W V W LI L L V V 9 A A: 3A VA 8 [ V ] 5 I 3A I I.A L L L V 9 6I V 6.V L 6 V V 0.8V A 6 3 I L VA.8A 9

22 LEARNING EXAMPLE L=0mH. FIND THE VOLTAGE 3 00 A m s A s di v( ) L ( ) d m 0 A s THE DERIVATIVE OF A STRAIGHT LINE IS ITS SLOPE di d 0( A/ s) 0 ms 0( A/ s) 4ms 0 elsewhere di ) 0( A/ s) d v( ) L 00 H ( 3 V 00mV ENERGY STORED BETWEEN AND 4 ms w(4,) w(4,) 0 0.5*0*0 Li L (4) Li L () 3 (0*0 3 THE VALUE IS NEGATIVE BEAUSE THE INDUTOR IS SUPPLYING ENERGY PREVIOUSLY STORED ) J

23 SAMPLE PROBLEM v(v) (s) L=0.H, i(0)=a. Find i(), >0 i( ) i(0) L v( x) v( x) dx ; v( x) dx L 0.H i( ) 0; 0 s v( x) 0; i( ) i(); s 4 i(a) (s) w(, ) LiL( ) LiL( ) ENERGY OMPUTATIONS Iniial energy sored in inducor w ( 0) 0.5* 0.[ H](A) 0.[ J] Toal energy sored in he inducor w( ) 0.5* 0.[ H ]*(4A) 88. J Energy sored beween 0 and sec w(,0) Li L () Li L (0) w(,0) 0.5* 0.*(4) 0.5*0.*() w(,0) 88[ J] Energy sored on he inerval an be posiive or negaive

24 LEARNING EXAMPLE FIND THE VOLTAGE AROSS AND THE ENERGY STORED (AS FUNTION OF TIME) v() FOR ENERGY STORED IN THE INDUTOR w L () NOTIE THAT ENERGY STORED AT ANY GIVEN TIME IS NON NEGATIVE -THIS IS A PASSIVE ELEMENT-

25 LEARNING EXAMPLE L 0mH DETERMINE THE VOLTAGE di v( ) L ( ) d v 00mV v 00 [ H ] 3 0 A s

26 LEARNING EXAMPLE FIND THE URRENT i() L=00mH i() i( ) i(0) v( x) dx; L v( ) 0; 0 i(0) 0 0 0

27 L=00mH FIND THE POWER NOTIE HOW POWER HANGES SIGN i() POWER p() ENERGY FIND THE ENERGY ENERGY IS NEVER NEGATIVE. THE DEVIE IS PASSIVE w()

28 LEARNING EXTENSION L=5mH FIND THE VOLTAGE 0mA 0 0 m m ( A/ s) ms v 50mV m 0 v 0V di v( ) L ( ) d 0 0 m ( A/ s) 4 3 v 50mV 3 v 50 ( H ) 0( A/ s); 0 ms 00mV

29 APAITOR SPEIFIATIONS APAITAN E IN RANGE STANDARD VALUES STANDARD 6.3V 500V STANDARD APAITOR TOLERANE 5%, 0%, 0% pf 50mF RATINGS Nominal curren 300nA ( 3) 3 V F 600nA 3 s 300nA LEARNING EXAMPLE 00nF 0% dv i( ) ( ) d GIVEN THE VOLTAGE WAVEFORM DETERMINE THE VARIATIONS IN URRENT

30 INDUTANE RANGES IN STANDARD VALUES STANDARD INDUTOR SPEIFIATIONS ma A INDUTOR nh L 00mH RATINGS URRENT WAVEFORM v 00 0 H 6 00 A S STANDARD 5%, 0% TOLERANE LEARNING EXAMPLE L 00 H 0% s di v( ) L ( ) d GIVEN THE URRENT WAVEFORM DETERMINE THE VARIATIONS IN VOLTAGE

31 L v i i v

32 IDEAL AND PRATIAL ELEMENTS i () i() i() i() v() v() v() v() IDEAL ELEMENTS dv di i( ) ( ) v( ) L ( ) d d APAITOR/INDUTOR MODELS INLUDING LEAKAGE RESISTANE v( ) dv i( ) ( ) R d leak MODEL FOR LEAKY APAITOR di v( ) Rleaki( ) L ( ) d MODEL FOR LEAKY INDUTORS

33 SERIES APAITORS s Series ombinaion of wo capaciors 6F F 3 S F NOTIE SIMILARITY WITH RESITORS IN PARALLEL

34 LEARNING EXAMPLE DETERMINE EQUIVALENT APAITANE AND THE INITIAL VOLTAGE F F 3 6 OR WE AN REDUE TWO AT A TIME V 4V V ALGEBRAI SUM OF INITIAL VOLTAGES POLARITY IS DITATED BY THE REFERENE DIRETION FOR THE VOLTAGE

35 LEARNING EXAMPLE Two uncharged capaciors are conneced as shown. Find he unknown capaciance V 8V + - 4V 30 F FIND 8V SAME URRENT. ONNETED FOR THE SAME TIME PERIOD Q ( 30 F)(8V ) 40 SAME HARGE ON BOTH APAITORS Q V Q ( F )(6V ) 7 7 8V 4 F

36 PARALLEL APAITORS i k dv ( ) k ( ) d i() LEARNING EXAMPLE P F

37 LEARNING EXTENSION 6 F 3 F F 4 F eq 4 F 3 eq F F 3 F

38 SAMPLE PROBLEM FIND EQUIVALENT APAITANE ALL APAITORS ARE 4F 8 F 8 F 4 F 8 3 eq F 8 F 8 F

39 SAMPLE PROBLEM IF ALL APAITORS HAVE THE SAME APAITANE VALUE DETERMINE THE VARIOUS EQUIVALENT APAITANES

40 Examples of inerconnecions EQ All capaciors are equal wih =8 microfarads AB

41 SERIES INDUTORS v k di ( ) Lk ( ) d LEARNING EXAMPLE di v( ) LS ( ) d Leq 7H

42 PARALLEL INDUTORS i() LEARNING EXAMPLE i( N 0 ) i j ( 0) j INDUTORS OMBINE LIKE RESISTORS APAITORS OMBINE LIKE ONDUTANES 4mH mh i( 0 ) 3A 6A A A

43 LEARNING EXTENSION b c a ALL INDUTORS ARE 4mH WHEN IN DOUBT REDRAW! b IDENTIFY ALL NODES d c mh mh ONNET OMPONENTS BETWEEN NODES L eq 6mH a 4mH 4mH d c mh mh PLAE NODES IN HOSEN LOATIONS a b L eq ( 6mH 4mH ) mh 4. 4mH d

44 LEARNING EXTENSION ALL INDUTORS ARE 6mH a a b L eq mh b 6mH 6mH c 6mH c NODES AN HAVE OMPLIATED SHAPES. KEEP IN MIND DIFFERENE BETWEEN PHYSIAL LAYOUT AND ELETRIAL ONNETIONS a 48 4 L eq (6 ) mh 66 L eq 7 mh b SELETED LAYOUT c

45 L-

46 R OPERATIONAL AMPLIFIER IRUITS INTRODUES TWO VERY IMPORTANT PRATIAL IRUITS BASED ON OPERATIONAL AMPLIFIERS THE IDEAL OP-AMP IDEAL RO 0, Ri, A R R i O 0 v O A ( v v ) A

47 R OPERATIONAL AMPLIFIER IRUITS -THE INTEGRATOR v 0 v i IDEAL OP-AMP ASSUMPTIONS v 0 ( A ) ( R i )

48 R OPERATIONAL AMPLIFIER IRUITS - THE DIFFERENTIATOR i R i KVL v 0 KL@ v i v : i i i IDEAL OP-AMP ASSUMPTIONS v 0 ( A ) ( R i ) ( ) Ri i v ( x) dx di dv R i ( ) d d i R v O 0 DIFFERENTIATE replace i in erms of v o ( i v R dvo dv R vo R ( ) d d IF R OULD BE SET TO ZERO WE WOULD HAVE AN IDEAL DIFFERENTIATOR. IN PRATIE AN IDEAL DIFFERENTIATOR AMPLIFIES ELETRI NOISE AND DOES NOT OPERATE. THE RESISTOR INTRODUES A FILTERING ATION. ITS VALUE IS KEPT AS SMALL AS POSSIBLE TO APPROXIMATE A DIFFERENTIATOR o )

49 ABOUT ELETRI NOISE ALL ELETRIAL SIGNALS ARE ORRUPTED BY EXTERNAL, UNONTROLLABLE AND OFTEN UNMEASURABLE, SIGNALS. THESE UNDESIRED SIGNALS ARE REFERRED TO AS NOISE SIMPLE MODEL FOR A NOISY 60Hz SINUSOID ORRUPTED WITH ONE MIROVOLT OF GHz INTERFERENE. y( ) sin(0 ) 0 signal 6 9 sin( 0 ) noise noise ampliude signal ampliude 0 6 THE DERIVATIVE dy 9 ( ) 0 cos(0 ) 000 cos( 0 ) d signal noise noise ampliude signal ampliude

50 LEARNING EXTENSION INPUT TO IDEAL DIFFERENTI ATOR WITH R k, F 0 m 3 50 V s IDEAL DIFFERENTI ATOR dv v o R ( ) d DIMENSIONA L ANALYSIS F V A Q V V Q s V s Q F s 3 6 R 0 0 F 0 3 s

51 LEARNING EXTENSION INPUT TO AN INTEGRATOR WITH R 5k, 0. F APAITOR IS INITIALLY DISHARGED INTEGRATOR R s v o F ( ) v V A Q V o (0) R V Q s s 0 V s Q F v ( x) dx DIMENSIONA L ANALYSIS s : v( ) 00 yo ) v( x) dx 00 V s y(0.) 0 V s : v( ) yo( ) yo(0.) v ( x) dx 0 00 ( 0.) 0. ( s V s v o ( ) R y o ( ) i

52 APPLIATION EXAMPLE ROSS-TALK IN INTEGRATED IRUITS Simplified Model REDUE ROSSTALK BY Reducing Increasing OST? EXTRA SPAE BY GROUND WIRE USING GROUND WIRE TO REDUE ROSSTALK SMALLER

53 LEARNING EXAMPLE SIMPLE IRUIT MODEL FOR DYNAMI RANDOM AESS MEMORY ELL (DRAM) REPRESENTS HARGE LEAKAGE FROM ELL APAITOR V cell. 5V NOTIE THE VALUES OF THE APAITANES v FOR ORRET OF A LOGI ONE v (0) i ( x) dx 0 STORAGE Ileak Ileak Vcell V cell 5 cell.5( V ) 500 ( F) 3 H.50 s 500 A THE ELL MUST BE REFRESHED AT A FREQUENY HIGHER THAN / H ELL AT THE BEGINNING OF A MEMORY READ OPERATION SWITHED APAITOR IRUIT THE ANALYSIS OF THE READ OPERATION GIVES FURTHER INSIGHT ON THE REQUIREMENTS

54 ELL READ OPERATION IF SWITH IS LOSED BOTH APAITORS MUST HAVE THE SAME VOLTAGE ASSUMING NO LOSS OF HARGE THEN THE HARGE BEFORE LOSING MUST BE EQUAL TO HARGE AFTER LOSING 5 5 Q before.5v F 3V 500 F Q afer V afer (5000 V afer. 65V 5 F ) Even a full charge he volage variaion is small. SENSOR amplifiers are required Afer a READ operaion he cell mus be refreshed

55 LEARNING EXAMPLE FLIP HIP MOUNTING I WITH WIREBONDS TO THE OUTSIDE GOAL: REDUE INDUTANE IN THE WIRING AND REDUE THE GROUND BOUNE EFFET A SIMPLE MODEL AN BE USED TO DESRIBE GROUND BOUNE

56 MODELING THE GROUND BOUNE EFFET V GB dig ( ) Lball ( ) d L ball 0. nh m A s IF ALL GATES IN A HIP ARE ONNETED TO A SINGLE GROUND THE URRENT AN BE QUITE HIGH AND THE BOUNE MAY BEOME UNAEPTABLE USE SEVERAL GROUND ONNETIONS (BALLS) AND ALLOATE A FRATION OF THE GATES TO EAH BALL

57 LEARNING BY DESIGN POWER OUTAGE RIDE THROUGH IRUIT APAITOR MUST MAINTAIN AT LEAST.4V FOR AT LEAST SE. v DESIGN EQUATION hp://

58 DESIGN EXAMPLE DESIGN AN OP-AMP IRUIT TO REALIZE THE EQUATION Proposed soluion v 5 v ( ) d v O 0 Needs inegraor And adder adder inegraor Design equaions Two equaions in five unknowns!! ANALYSIS OF THE SOLUTION No oo large. No oo small Seems a reasonable value R3 0 R 0k k If supply volages are 0V (or less) all currens will be less han ma, which seems reasonable