Chapter 10 Stability and Frequency...

37
Chapter 10 Stability and Frequency Compensation Compensation

Transcript of Chapter 10 Stability and Frequency...

Page 1: Chapter 10 Stability and Frequency Compensationocw.snu.ac.kr/sites/default/files/NOTE/3663.pdfMicrosoft PowerPoint - 10장_Stability and Frequency Compensation.ppt [호환 모드]

Chapter 10Stability and Frequency

CompensationCompensation

Page 2: Chapter 10 Stability and Frequency Compensationocw.snu.ac.kr/sites/default/files/NOTE/3663.pdfMicrosoft PowerPoint - 10장_Stability and Frequency Compensation.ppt [호환 모드]

Basic Stabilityy

YX

(s) =H (s)

1+ βH (s)X 1+ βH (s)May oscillate at ω if |βH ( jω) |=1 and ∠βH ( jω) = −180 (Barkhausen criteria)

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Stable and Unstable Systemsy

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Stability and Complex Polesy p

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Basic One Pole Bode Plot

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Multipole Systemsp y

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Loot Locus for a 2-Pole Systemy

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Phase Marging

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Phase Margin (cont.)g ( )

-180Phase Margin, θm = 45°

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Phase Margin (cont.)g ( )

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Small Signal Analysis Limitationsg y

Small signal analysis yields 65 ° of θm, but large i l t i t i diff t L i lsignal transient response is different. Large signal

simulation includes effects not seen in small signal analysisanalysis.

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Frequency Compensationq y p

Compensation is the manipulation of gain and/or pole positions to improve phase margin.

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Compensation (cont.)p ( )

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Single Ended Single Stage Ampg g g p

D i t P l t O t tDominant Pole at Output

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Pre- and Post-Compensationp

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Compensation (cont.)p ( )

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Compensation Examplep p

Given T0 = 5000V /V, fp1 = 2MHz, fp2 = 25MHz, fp3 = 50MHzDesire φm = 70°, find f0db and new f p1∴φm , f f0db f p1

70° =180°− tan−1( f0dbf p1

) − tan−1( f0dbf p2

) − tan−1( f0dbf p 2

)p p p

70° = 90 − tan−1( f0dbf p 2

) − tan−1( f0dbf p2

), assume 90° from f p1

f0db ≈ 6MHzf0db = T0 f p1 → f p1 = f0db /T0 =1.2KHzf0db 0 f p1 f p1 f0db 0

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Fully Differential Telescopic Op-AAmp

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Cascode Current Source I dImpedance

Z ||1

(1+ gm 5ro5)ro7

1+ sro7CN

1sCLZout ||

sCL

= o7 N L

(1+ gm 5ro5)ro7

1+ sro7CN

+ 1sCLo7 N L

= (1+ gm 5ro5)ro7

1+ s (1+ gm 5ro5)ro7CL + ro7CN[ ]( gm 5 o5) o7 L o7 N[ ]

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Compensation of Two-Stage Op Ampsp g p p

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Compensation (cont.)p ( )

Miller Effect C = CE + (1+ A 2)CCMiller Effect Ceq CE + (1+ Av2)CC

f pE =1

2πR [C + (1+ A )C ]2πRout[CE + (1+ Av2)CC ]

Pro: Need a smaller capacitor

Con: Exact effect of bridging capacitor involve a zero If open-loop gain is highinvolve a zero. If open-loop gain is high, zero may prevent -20db/dec curve from

going all the way to 0db

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Compensation (cont.)p ( )

Recall from Chap 6:

f1

Recall, from Chap. 6:(assume CC includes CGD9)

f p,in = 2π RS CE + (1+ gmRL )CC[ ]+ RL (CC + CL)( )

f p,in ≈1

2πRS CE + (1+ gmRL )CC[ ]

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Compensation (cont.)p ( )

fRS (1+ gm 9RL )CC + RSCE + RL(CC + CL)

f p,out =S m 9 L C S E L C L

2πRS RL(CECC + CECL + CC CL)

f p,out ≈RSgm 9RLCC + RLCC

2πRS RL(CECC + CCCL )=

gm 9

2π (CE + CL )2πRS RL(CECC + CCCL ) 2π (CE + CL )

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Compensation (cont.)p ( )

Recall, transfer function includes(1− s /ωz ) numerator term and

f (RHP) gm9fZ(RHP) = gm9

2πCC

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Phase and Magnitude of RHP ZZero

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RHP Zero Removal

fZ =gm 9

2 C (1/ R )fZ 2πCC (1/gm 9 − RZ )

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RHP Zero Removal (cont.)( )fZ = 1fZ 2πCC (1/ gm 9 − RZ )

Could set R z =1/gm9, or cancel other non - dominant pole

1C (1/g − R )

= −gm9

C + CCC (1/gm 9 RZ ) CL + CE

RZ =CL + CE + CC

g C≈ CL + CC

g Cgm 9CC gm 9CC

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Miller Compensation (cont.)p ( )Temp. and Process

Tracking

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Miller Compensation (cont.)p ( )

gm 9 ∝ ID 9 ∝ ID11 ∝1/ RS

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Load Capacitance Effectsp

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Slewing in Two-Stage Op Ampsg g p p

Basic Two-Stage Op Ampg p p

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Slewing in Two-Stage Op Ampg g p p

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Other Compensation Techniques p q

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Other Compensation Techniques ( t )(cont.)

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Other Compensation Techniques ( t )(cont.)

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Slewing with Common-Gate C tiCompensation

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Other Compensation Techniques ( t )(cont.)