EE 435. Lecture 16. Compensation of Feedback Amplifiers

Transcription

1 EE 435 Lecture 16 ompensation of Feedback Amplifiers

2 . Review from last lecture. Basic Two-Stae Miller ompensated Op Amp DD M 3 M 4 M 5 OUT IN M 1 M IN L I T B M 7 B3 M 6 By inspection SS A o m1 o p 1 o1 o4 05 o5 m5 o5 o6 m5 o6 p GB Will also et these results from a more complete (and time consumin) analysis This analysis was based only upon findin the poles and will miss zeros if they exist m5 L m1

3 Small Sinal Analysis of Basic Two-Stae Op Amp DD M 3 M 4 M 5 OUT IN M 1 M IN L I T B M 7 B3 M 6 SS Differential Small Sinal Equivalent M 3 M 4 M 5 OUT d M 1 M d M 6 L Norton Equivalent One-Port Two-Port

4 Small Sinal Analysis of Basic Two-Stae Op Amp Differential Small Sinal Equivalent M 3 M 4 I X X d M 1 M d o3 M M4 4 o4 I X X d o1 1 M1 1 o M d

5 Small Sinal Analysis of Basic Two-Stae Op Amp Differential Small Sinal Equivalent o3 M M4 4 o4 I X X d o1 M1 1 1 o M d I X I X X d o o4 m m44 d 0 4 m3 o1 03 m1 X I X o 04 X m 1 d m1 m m3 m4 o o3 o 04 m d

6 Small Sinal Analysis of Basic Two-Stae Op Amp Differential Small Sinal Equivalent I X M 3 M 4 I X X md d od X d M 1 M d I X X o 04 m d Since M 1 and M are matched as are M 3 and M 4 md = m1 od =

7 Small Sinal Analysis of Basic Two-Stae Op Amp Differential Small Sinal Equivalent M 5 M 6 5 M5 5 o5 I X oo = o X 6 M6 6 o6 mo = m5

8 Small Sinal Analysis of Basic Two-Stae Op Amp Differential Small Sinal Equivalent M 5 M 6 mo OO

9 Small Sinal Analysis of Basic Two-Stae Op Amp Differential Small Sinal Equivalent M 3 M 4 M 5 OUT d M 1 M d M 6 L OUT md d Od mo OO L

10 Small Sinal Analysis of Basic Two-Stae Op Amp Differential Small Sinal Equivalent OUT md d Od mo OO L Solvin we obtain: OUT This simplifies to: d OU T OUT s sl oo mo s s od mdd s OUT s d L s s md mo s mo oo od Lod oood L md s mo mo s oo od

11 Small Sinal Analysis of Basic Two-Stae Op Amp Differential Small Sinal Equivalent Summary: where A s s md L s m5 m5 s oo od md m1 m od o o4 oo o5 o6

12 Small Sinal Analysis of Two-Stae Miller- ompensated Op Amp A s s Note this is of the form: md L s m5 A 1 A s 0 m5 s s +1 z s s p p 1 This has two neative real-axis poles and one positive real-axis zero jω oo od p p 1 z 1 Re

13 How does the Gain of the Two-Stae Miller-ompensated Op Amp ompare with Internal ompensated Op Amp? P 1 P P 1 P OUT OUT IN F 1 F EFF L IN F 1 F L md m5 -s A s = md As m5 s +s + s +1 z 0 s s p1 p A 1 A s L m5 oo od L oo oo od s +s + A0 A s 1 s s p1 p jω jω Re p p 1 Re p p 1 z 1 must be developed ompensation criteria: 4β A p 0 β A0 p1

14 Review of Basic oncepts onsider a second-order factor of a denominator polynomial, P(s), P(s)=s +a 1 s+a 0 Then P(s) can be expressed in several alternative but equivalent ways s s ω s Q sξ s p s p 1 and if 0 ω 0 ω complex s α jβs α jβ 0 0 conjuate poles, These are all -paramater characterizations of the second-order factor and it is easy to map from any one characterization to any other

15 Review of Basic oncepts Im s s ω Q 0 ω 0 ω o sinθ 1 Q Re ω o = manitude of pole Q determines the anle of the pole Observe: Q=0.5 corresponds to two identical real-axis poles Q=.707 corresponds to poles makin 45 o anle with Im axis

16 Simple pole calculations for -stae op amp Since the poles of the -stae op amp must be widely separated, a simple calculation of the poles from the characteristic polynomial is possible. Assume p 1 and p are the poles and p 1 << p D(s)=s +a 1 s+a 0 but determines p 1 D(s)=(s+p 1 )(s+p )=s +s(p 1 +p )+p 1 p s + p s + p 1 p thus determines p p =-a 1 and p 1 = -a 0 /a 1

17 an now use these results to calculate poles of Basic Two-stae Miller ompensated Op Amp From small sinal analysis: A s s p m5 = L md L s m5 m5 s oo p= oo od 1 m5 A m5 md 0= oo od od md GB= m5 md p m5 md oo od md 1 oo od oo od m5 od oo m1 o o5 o4 m o6

18 From Previous Inspection DD A o m1 o o4 o5 m5 o6 M 3 M 4 M 5 OUT p 1 o GB 05 o4 m5 m1 o6 p m5 L IN M 1 M I T B M 7 SS IN B3 M 6 L Note the simple results obtained from inspection aree with the more time consumin results obtained from a small sinal analysis

19 Feedback applications of the twostae Op Amp X IN X 1 A X OUT β How does the amplifier perform with feedback? How should the amplifier be compensated?

20 Feedback applications of the twostae Op Amp Open-loop Gain A FB (s) A(s) Standard Feedback Gain 1 N(s) D(s) A(s) A(s)β(s) N FB D FB ( s ) ( s ) N(s) D(s) N(s)β(s) defn N ( s ) D ( s ) β ( s ) N ( s ) N D FB FB (s) (s) Open-loop and closed-loop zeros identical losed-loop poles different than open-loop poles Often β(s) is not dependent upon frequency

21 Open-loop Gain A(s) N(s) D(s) Alternate Feedback Gain 1 β1(s) FB(s) 1 1 A(s) β(s) Feedback applications of the twostae Op Amp Standard Feedback Gain (s) A FB A 1 A(s) 1 A(s)β(s) (often FB is not of standard form) β(s) N(s) β (s) D(s) N(s) β(s) 1 1 β(s) 1 A(s) β(s) In either case, denominators are the same and characteristic equation defined by D FB ( s ) D ( s ) β ( s ) N ( s ) Often β(s) and β 1 (s) are not dependent upon frequency and in this case N FB ( s ) N ( s )

22 Basic Two-Stae Op Amp with Feedback DD M 3 M 4 M 5 Open-loop ain -s A(s)= s +s + md mo c L mo oo od OUT IN M 1 M IN L I T B SS M 7 A FB A B3 (s) FB (s ) where s s M 6 L s Standard feedback ain with constant β md m1 od L o s mdmo sc mo βmd oood βmdmo md m0 s c mo β md β mdmo 04 mo m5 oo o5 o6

23 IN DD M 3 M 4 M 5 M 1 M IN I T B M7 B3 M 6 SS Basic Two-Stae Op Amp mdm0 s c A FB(s ) s L s mo β md β mdmo L OUT Pole Q?

24 Basic Two-Stae Op Amp mdm0 s c A FB(s ) s L s mo β md β mdmo DD M 3 M 4 M 5 It can be shown that IN M 1 M IN L OUT Q L mo mo md md B SS I T M 7 B3 M 6 where oo But what pole Q is desired?.707< Q <0.5 o5 md m1 o6 Lβ Q and β Riht Half-Plane Zero in OL Gain (from Miller ompensation) Limits Performance mo od mo o md md mo m5 04

25 End of Lecture 16