ECEN326: Electronic Circuits Fall 2017

Transcription

1 ECEN36: Electronic Circuits Fall 07 Lecture 7: Feedback Sa Palero Analo & Mixed-Sal Center Texas A&M University

2 Announceents Hoework 7 due /9 Exa 3 / 8:00-0:00 Closed book w/ one standard note sheet 8.5 x front & back Br your calculator Ephasis will be on feedback and put staes Saple Exa 3 is posted on the website ead azavi Chapter

3 ECEN 474/704 Spr 08 will be teach ECEN 474/704 (Analo) LS Circuit es Spr 08 Lecture T :0AM-:5PM Lab W 0:0AM-:0PM, :50PM-4:40PM portant class for jobs seiconductor dustry Essential class for an analo C des position Topics Transistor odels, lay techniques, analo C des ethodoloies, noise analysis, differential aplifiers, current irrors, put staes, analo systes 3

4 Aenda Feedback Overview Feedback Properties Aplifier Types Sense and eturn Techniques Feedback Polarity Feedback Topoloies Effect of Nonideal /O pedances Stability Two-Stae Miller OTA 4

5 Neative Feedback Syste Feedback Factor A neative feedback syste consists of four coponents: ) feedforward syste, ) sense echanis, 3) feedback network, and 4) coparison echanis. CH Feedback 5 K Y F

6 Close-loop Transfer Function Y F A YK Y KA A A Closed-Loop Ga Y CH Feedback 6 A KA

7 Feedback Exaple K Y A A A is the feedforward network, and provide the sens and feedback capabilities, and coparison is provided by differential put of A. CH Feedback 7

8 Coparison Error E E E F A K EA K E A K As A K creases, the error between the put and fed back sal decreases. Or the fed back sal approaches a ood replica of the put. CH Feedback 8

9 CH Feedback 9 Coparison Error What happens to the closed-loop and error transfer function as A? K Y KA A Y 0 KA E

10 Loop Ga N test K A 0 KA N test Loop Ga N test KA When the put is rounded, and the loop is broken at an arbitrary location, the loop a is easured to be -KA. CH Feedback 0

11 Exaple: Alternative Loop Ga Measureent KA N test esult should be the sae wherever we break the loop as lon as we analyze the loop the proper sal direction CH Feedback

12 ncorrect Calculation of Loop Ga Sal naturally flows fro the put to the put of a feedforward/feedback syste. f we apply the put the other way around, the put sal we et is not a result of the loop a, but due to poor isolation. CH Feedback

13 Aenda Feedback Overview Feedback Properties Aplifier Types Sense and eturn Techniques Feedback Polarity Feedback Topoloies Effect of Nonideal /O pedances Stability Two-Stae Miller OTA 3

14 Ga esensitization Y A KA A K Y K A lare loop a is needed to create a precise a, one that does not depend on A, which can vary by ±0% with process and teperature variations. Can we ake a feedback factor K with low variations? CH Feedback 4

15 atio of esistors When two resistors are coposed of the sae unit resistor, their ratio is very accurate. Sce when they vary, they will vary toether and ata a constant ratio. Consider the previous circuit K w/ variations ( ) (ideally not chaned) CH Feedback 5

16 Merits of Neative Feedback Bandwidth enhanceent Modification of /O ipedances educed sensitivity to load ipedance Learization CH Feedback 6

17 Bandwidth Enhanceent A Open Loop s A0 s 0 Neative Feedback Y s Closed Loop A0 Closed - Loop"C" Ga KA Closed - Loop Bandwidth Constant Ga - Bandwith Product (GBW) A 0 KA s 0 KA 0 0 KA A 0 0 Althouh neative feedback lowers the a by (+KA 0 ), it also extends the bandwidth by the sae aount. CH Feedback 7

18 Bandwidth Extension Exaple Open-Loop Aplifier Ga A 0 0 0dB As the loop a creases, we can see the decrease of the overall a and the extension of the bandwidth. CH Feedback 8

19 Exaple: Open Loop Paraeters Assue: λ 0 A 0 CH Feedback 9

20 CH Feedback 0 Exaple: Closed Loop oltae Ga KA A v v 0 0 K F v v i we have neative feedback due to Note, Closed-loop a decreases by +KA 0 factor

21 CH Feedback Exaple: Closed Loop /O pedance nput esistance x x x x F x F x v i i v v v i i v i v Assu that nput resistance creases by +KA 0 factor Sae factor as the closedloop a decreases

22 CH Feedback Exaple: Closed Loop /O pedance Output esistance F x v v v v i Assu that Output resistance decreases by +KA 0 factor Sae factor as the closed-loop a decreases

23 CH Feedback 3 Exaple: Load esensitization 3 / 3 W/O Feedback Lare ifference With Feedback Sall ifference

24 Learization Sificant distortion with lare put sal Ga at A KA A KA A << A Ga at x CH Feedback 4 x K K KA KA f KA and KA rea lare, overall a is ~ /K

25 Aenda Feedback Overview Feedback Properties Aplifier Types Sense and eturn Techniques Feedback Polarity Feedback Topoloies Effect of Nonideal /O pedances Stability Two-Stae Miller OTA 5

26 Four Types of Aplifiers oltae Aplifier A 0 Transipedance Aplifier 0 0 Transconductance Aplifier Current Aplifier G A CH Feedback 6

27 deal Models of the Four Aplifier Types oltae Aplifier A 0 Transipedance Aplifier 0 Transconductance Aplifier Current Aplifier G A CH Feedback 7

28 ealistic Models of the Four Aplifier Types CH Feedback 8

29 Exaples of the Four Aplifier Types Assu 0 oltae Aplifier Transipedance Aplifier A0 Transconductance Aplifier 0 Current Aplifier/Buffer G A CH Feedback 9

30 Aenda Feedback Overview Feedback Properties Aplifier Types Sense and eturn Techniques Feedback Polarity Feedback Topoloies Effect of Nonideal /O pedances Stability Two-Stae Miller OTA 30

31 Sens a oltae n order to sense a voltae across two terals, a volteter with ideally fite ipedance is used. CH Feedback 3

32 Sens and eturn a oltae Feedback Network Siilarly, for a feedback network to correctly sense the put voltae, its put ipedance needs to be lare. and also provide a ean to return the voltae. CH Feedback 3

33 Sens a Current A current is easured by sert a current eter with ideally zero ipedance series with the conduction path. The current eter is coposed of a sall resistance r parallel with a volteter. CH Feedback 33

34 Sens and eturn a Current Feedback Network S 0 Siilarly for a feedback network to correctly sense the current, its put ipedance has to be sall. S has to be sall so that its voltae drop will not chane. CH Feedback 34

35 Addition (Subtraction) of Two oltae Sources Feedback Network v e v v F n order to add or subtract two voltae sources, we place the series. So the feedback network is placed series with the put source. CH Feedback 35

36 Practical Circuits to Subtract Two oltae Sources i i i v v F i vf v i Althouh not directly series, and F are be subtracted sce the resultant currents, differential and sle-ended, are proportional to the difference of and F. CH Feedback 36

37 Addition (Subtraction) of Two Current Sources Feedback Network i e i i F n order to add two current sources, we place the parallel. So the feedback network is placed parallel with the put sal. CH Feedback 37

38 Practical Circuits to Subtract Two Current Sources i e i i F Sce M and F are parallel with the put current source, their respective currents are be subtracted. Note, F has to be lare enouh to approxiate a current source. CH Feedback 38

39 CH Feedback 39 Exaple: Sense a oltae and eturn a oltae and sense and return the put voltae to feedforward network consist of M -M 4. M and M also act as a voltae subtractor. F v v is lare f r r r r K A A v v K r r A o o o o o o

40 Exaple: Feedback Factor K i v F F This circuit senses a voltae and returns a current CH Feedback 40

41 nput pedance of an deal Feedback Network To sense a voltae, the put ipedance of an ideal feedback network ust be fite. To sense a current, the put ipedance of an ideal feedback network ust be zero. CH Feedback 4

42 Output pedance of an deal Feedback Network To return a voltae, the put ipedance of an ideal feedback network ust be zero. To return a current, the put ipedance of an ideal feedback network ust be fite. CH Feedback 4

43 Aenda Feedback Overview Feedback Properties Aplifier Types Sense and eturn Techniques Feedback Polarity Feedback Topoloies Effect of Nonideal /O pedances Stability Two-Stae Miller OTA 43

44 eter the Polarity of Feedback ) Assue the put oes either up or down. ) Follow the sal throuh the loop. 3) etere whether the returned quantity enhances or opposes the orial chane. CH Feedback 44

45 Polarity of Feedback Exaple,, x, Neative Feedback CH Feedback 45

46 Polarity of Feedback Exaple, A,, x A Neative Feedback CH Feedback 46

47 Polarity of Feedback Exaple Positive Feedback,,, f we are try to build a lear aplifier, positive feedback is bad Circuit can latch up or oscillate CH Feedback 47

48 Aenda Feedback Overview Feedback Properties Aplifier Types Sense and eturn Techniques Feedback Polarity Feedback Topoloies Effect of Nonideal /O pedances Stability Two-Stae Miller OTA 48

49 oltae-oltae Feedback A voltae aplifier requires sens of the put voltae to produce a feedback voltae Output voltae is sensed parallel and feedback voltae applied series with the put A 0 F A 0 A0 KA 0 K K K F CH Feedback 49

50 CH Feedback 50 Exaple: oltae-oltae Feedback ) is lare ( f r r r r K A A v v K r r A r r o o o o o o o o

51 nput pedance of a - Feedback A better put voltae sensor, as the put ipedance creases by +A 0 K F CH Feedback 5 A 0 K A 0 K

52 CH Feedback 5 Exaple: - Feedback nput pedance closed open K A, 0, Assu Break Feedback

53 Output pedance of a - Feedback i 0 F K Assu Feedback has lare put ipedance A A 0 0 K K A better put voltae source, as has been reduced by (+A 0 K) - CH Feedback 53

54 CH Feedback 54 Exaple: - Feedback Output pedance 4 4, , Assu O O O O closed O O O O O O open r r r r K r r r r A r r

55 oltae-current Feedback A transipedance aplifier requires sens of the put voltae to produce a feedback current Output voltae is sensed parallel and feedback current applied parallel with the put K F 0 0 K 0 0 K F A CH Feedback 55

56 CH Feedback 56 Exaple: oltae-current Feedback F F F F F F F F F K 0 is lare Assue F

57 nput pedance of a -C Feedback 0 KCL at put node K 0 F K 0 K 0 A better put current sensor, as has been reduced by (+K 0 ) - CH Feedback 57

58 CH Feedback 58 Exaple: -C Feedback nput pedance F closed,. 0, F open K

59 Output pedance of a -C Feedback F K A A F A 0 K Nelect the sall feedback current K K A better put voltae source, as has been reduced by (+K 0 ) - CH Feedback 59

60 CH Feedback 60 Exaple: -C Feedback Output pedance F closed, F open K 0,

61 Current-oltae Feedback A transconductance aplifier requires sens of the put current to produce a feedback voltae Output current is sensed series and feedback voltae applied series with the put G G K F G KG K F CH Feedback 6

62 CH Feedback 6 Exaple: Current-oltae Feedback M M O O O O closed r r r r Laser M O O K r r G 5 3 3

63 nput pedance of a C- Feedback F GK G K KG ( ) A better put voltae sensor, as creases by +KG CH Feedback 63

64 Output pedance of a C- Feedback K G K For correct easureent of a feedback current syste s, ust be serted series KCL at the put node G K ( KG ) A better put current source, as creases by +KG CH Feedback 64

65 CH Feedback 65 Laser Exaple: Current-oltae Feedback ) ( ) ( M closed M closed M closed M open K G, expression assues that M is sall

66 Current-Current Feedback A current aplifier requires sens of the put current to produce a feedback current Output current is sensed series and feedback current applied parallel with the put A A K F A KA A F CH Feedback K A 66

67 nput pedance of C-C Feedback F KA KA A better put current sensor, as decreases by (+KA ) - CH Feedback 67

68 Output pedance of C-C Feedback KA A KCL at put node KA ( KA ) A better put current source, as creases by (+KA ) CH Feedback 68

69 Exaple: Test of Neative Feedback Laser,,, CH Feedback 69 P Neative Feedback F

70 CH Feedback 70 Exaple: C-C Neative Feedback )] / ( [ ) / (. ) / ( F O closed F M closed F M closed r A M Laser F M M F M M F open K A Assu, expression assues that M is sall

71 Aenda Feedback Overview Feedback Properties Aplifier Types Sense and eturn Techniques Feedback Polarity Feedback Topoloies Effect of Nonideal /O pedances Stability Two-Stae Miller OTA 7

72 Feedback Network Load n the previous exaples, we ade a lot of siplify assuptions that nelect the load the feedback network has on the aplifiers /O ports However, the fite feedback network ipedance ay alter the overall circuit s perforance n order to clude the feedback network load effects on the /O ipedances, the follow ethodoloy can be eployed 7

73 Feedback Analysis Methodoloy with /O Load. dentify the forward aplifier. dentify the feedback network 3. Break the feedback network correctly 4. Calculate the open-loop paraeters 5. etere the feedback factor correctly 6. Calculate the closed-loop paraeters 73

74 How to Break a Loop The correct way of break a loop is such that the loop does not know it has been broken. Therefore, we need to present the feedback network to both the put and the put of the feedforward aplifier. CH Feedback 74

75 ules for Break the Loop of Aplifier Types Sense duplicate put is loaded with the ideal put ipedance of the forward aplifier eturn duplicate put is set with the ideal put ipedance of the forward aplifier CH Feedback 75

76 ntuitive Understand of these ules oltae-oltae Feedback Sce ideally, the put of the feedback network sees zero ipedance (Z of an ideal voltae source), the return replicate needs to be rounded. Siilarly, the put of the feedback network sees an fite ipedance (Z of an ideal voltae sensor), the sense replicate needs to be open. Siilar ideas apply to the other types. CH Feedback 76

77 ules for Calculat Feedback Factor oltae feedback: feedback network put is opened Current feedback: feedback network put is shorted CH Feedback 77

78 ntuitive Understand of these ules oltae-oltae Feedback Sce the feedback senses voltae, the put of the feedback is a voltae source. Moreover, sce the return quantity is also voltae, the put of the feedback is left open (a short eans the put is always zero). Siilar ideas apply to the other types. CH Feedback 78

79 Break the Loop Exaple A v, open, open, open / CH Feedback 79

80 CH Feedback 80 Feedback Factor Exaple ) /( ) ( ) /( ) /(,,,,,,,,, open v open closed open v open closed open v open v closed v KA KA KA A A K

81 Break the Loop Exaple A v, open, open, open N r ON r ON r CH Feedback 8 OP r OP

82 Feedback Factor Exaple K A v, closed, closed, closed /( A v, open /( /( CH Feedback 8 ), open KA v, open KA ) v, open )

83 Break the Loop Exaple, open open, open F F F CH Feedback 83 F. F

84 Feedback Factor Exaple K / closed, closed, closed F, open, open /( K /( K /( K CH Feedback 84 open open open ) open ) )

85 Break the Loop Exaple open, open, open r O 3 ro 3 r O M r L O5 M r O For a current put with feedback, the put resistance ust be obtaed with the test stiulus series CH Feedback 85

86 CH Feedback 86 ] ) / ( [ ] ) / ( /[ ) / ( ) / (,,, open open closed closed open open closed M K K K Feedback Factor Exaple

87 Break the Loop Exaple open, open, open / L (/ M ) / M CH Feedback 87

88 CH Feedback 88 Feedback Factor Exaple ] ) / ( [ ] ) / ( [ ] ) / ( /[ ) / ( ) / (,,,, open open closed open open closed open open closed M K K K K

89 CH Feedback 89 Break the Loop Exaple M F O open M F open F M L O O M F M F open r r r A ) (. ) (,,, = 0 > 0 Eq. Circuit

90 Feedback Factor Exaple K A, closed, closed, closed M /( A, open, open, open /( /( ( CH Feedback 90 F M ) KA KA, open KA ), open ), open )

91 Break the Loop Exaple = 0, open open, open F F F / F M CH Feedback 9 [ ( F M )]

92 CH Feedback 9 Feedback Factor Exaple ] ) / ( /[ ] ) / ( /[ ] ) / ( /[ ) / ( ) / ( /,,,, open open closed open open closed open open closed F K K K K

93 Aenda Feedback Overview Feedback Properties Aplifier Types Sense and eturn Techniques Feedback Polarity Feedback Topoloies Effect of Nonideal /O pedances Stability Two-Stae Miller OTA 93

94 Bode Plot Alorith - Phase. Calculate low frequency value of Phase(H(j)) a. An neative s troduces -80 phase shift b. A C pole troduces -90 phase shift c. A C zero troduces +90 phase shift. Where are the poles and zeros? a. For neative poles: dec. before the pole freq., the phase will decrease with a slope of -45/dec. until dec. after the pole freq., for a total phase shift of -90 b. For neative zeros: dec. before the zero freq., the phase will crease with a slope of +45/dec. until dec. after the zero freq., for a total phase shift of +90 c. Note, if you have positive poles or zeros, the phase chane polarity is verted 3. Note, the above alorith is only valid for real poles and zeros. 94

95 Exaple: Phase esponse Assu z 00 p and p 00 z (db) Assu eneral neative (left-half plane) poles and zeros, the phase of H(jω) starts to drop at /0 of the pole, hits -45 o at the pole, and approaches -90 o at 0 ties the pole. CH Feedback 95

96 Exaple: Three-Pole Syste (db) For a three-pole syste, a fite frequency produces a phase of -80 o, which eans an put sal that operates at this frequency will have its put verted. CH Feedback 96

97 nstability of a Neative Feedback Loop f Y ( s) H ( s) KH ( s) KH ( j ), then s j H ( j) KH ( j) ( j ) H ( j) 0 Substitute jω for s. f for a certa ω, KH(jω ) reaches -, the closed loop a becoes fite. This iplies for a very sall put sal (or herent syste noise) at ω, the put can be very lare. Thus the syste becoes unstable. CH Feedback 97 Y

98 Barkhausen s Criteria for Oscillation KH ( KH ( j ) j ) 80 We want our lear aplifiers to be stable (not oscillate) Thus, we don t want this criteria to be satisfied CH Feedback 98

99 Tie Evolution of nstability CH Feedback 99

100 Oscillation Exaple (db) This syste oscillates, sce there s a fite frequency at which the phase is -80 o and the a is reater than unity. n fact, this syste exceeds the iu oscillation requireent. CH Feedback 00

101 Condition for Oscillation (db) (db) When Un-Stable KH 80, KH When KH Stable 80, KH Althouh, this syste will probably still display a bad transient response CH Feedback 0

102 Condition for Stability (db) For Stability : G P ω P, ( phase crossover ), is the frequency at which KH=-80 o ( above) ω G, ( a crossover ), is the frequency where KH = ( above) CH Feedback 0

103 For Stability with the worst - case feedback factor the open - loop anitude response.the KH KH s where 3 p C s p. Fd the phase crossover frequency, H P 3tan p anitude at P p 3. This iplies that the low - frequency a j P p P P Stability Exaple ust be less than unity 3 H p when there is a 80 P K, we need to fd phase shift CH Feedback 03 3

104 Stability Exaple Now the feedback factor has been reduced to K. The 0.5 KH s where 3 p s C p. The phase crossover frequency is the sae H P 3 3tan P p 3. This iplies that the low - frequency a j P 3 p p P KH anitude at P ust be less than unity 3 04

105 Un-Stable vs. Marally Stable vs. Stable G P G P, but close G P sufficiently Un-Stable Marally Stable Stable While the iddle syste is Marally Stable, it has a poor transient step response, that it displays lare r which takes a lon tie to die CH Feedback 05

106 Phase Mar The "phase ar" quantifies the separation of the loop - a phase shift fro the unstable 80 value. Plot assues that G p Phase Mar KH 80 KH G G 80 The worst - case phase ar is when K and is often an aplifier des spec., Worst - Case Phase Mar H G 80 Worst - Case PM H G CH Feedback 06

107 Frequency Copensation p p3 Orial syste is not stable G P Mov ' p causes a lower to a G lower frequency and allows stability Phase ar can be iproved by ov ω G closer to ori while ata ω P unchaned. CH Feedback 07

108 CH Feedback 08 Frequency Copensation Exaple C cop is added to lower the doant pole so that ω G occurs at a lower frequency than before, which eans phase ar creases. O O p B O O O O B p A A p C r r C r r r r C 6 5, 4 3 3,, with C cop

109 Frequency Copensation Procedure Assued that G p ) We identify a PM, then -80 o +PM ives us the new ω G, or ω PM. ) On the anitude plot at ω PM, we extrapolate up with a slope of +0dB/dec until we hit the low frequency a then we look down and the frequency we see is our new doant pole, ω P. A slope of 0dB/dec is used, as we assue that we want a PM 45º CH Feedback 09

110 Exaple: 45 o Phase Mar Copensation p H p 0 PM p CH Feedback 0

111 Miller Copensation C )] C eq [ ( r r 5 O5 O6 c To save chip area, Miller ultiplication of a saller capacitance creates an equivalent effect. CH Feedback

112 Aenda Feedback Overview Feedback Properties Aplifier Types Sense and eturn Techniques Feedback Polarity Feedback Topoloies Effect of Nonideal /O pedances Stability Two-Stae Miller OTA

113 Two-Stae Miller OTA C Ga C Ga o o v o o C o o o o o o o o v v C A G G A A A A

114 Two-Stae Miller OTA Frequency esponse Stae is a differential aplifier with an active load Stae is a coon-source aplifier with a lare iller capacitor Us a Theven equivalent for Stae, we can use the coon-source equations fro Lecture 5 4

115 Two-Stae Miller OTA Frequency esponse The aplifier should be desed to yield one doant pole, so we use the doant pole approxiation equations Lecture 5, Slide 44 p p 8 CC CC CL C C C where Nelect Transistor 8 r O r O4 C C C and Capacitances C L C r O7 L r O8 8 C 8 L C C 5

116 Two-Stae Miller OTA Unity Ga Frequency The Two -Stae Miller OTA has the follow transfer function G where A C o i p s p H s AC s s p and C 8 G 8 C C p p C For a well - desed doant - pole aplifer, the unity - a frequency is C 8 8 L C C 6

117 Two-Stae Miller OTA Phase Mar The worst - case phase ar f PM H G G tan tan p p Us tan x 90 tan x p PM tan G p 8CC PM tan tan G CL G 90 tan p G Note: We are nelect a rihthalf plane (positive) zero that is troduced with C C, which can potentially derade the aplifier stability We will talk ore ab this 474 7

118 Two-Stae Miller OTA Exaple Aa, we are nelect a riht-half plane (positive) zero, which can potentially derade the aplifier stability z 8 C C This is OK, as lon as C C << C L More ab this 474 f p G A 8 0, ro 7 ro 8 0k,. What should C be for f C. What is 8 C f Assue that for the first stae p A 0.5 and ro ro 4 0k and for the second stae A 0.5 C PM tan, CC A 0 8 C pf f C p f f L p G f 500MHz p C, and the phase ar? 0k 0 5k59 ff tan G and A.59GHz.59GHz 500MHz C pf 500MHz? C L C 59 ff 7.5 MHz 8

119 Next Tie Output Staes and Power Aplifiers azavi Chapter 4 9