ESE319 Introduction to Microelectronics. Feedback Basics

Transcription

1 Feedback Basics Feedback concept Feedback in emitter follower Stability One-pole feedback and root locus Frequency dependent feedback and root locus Gain and phase margins Conditions for closed loop stability Frequency compensation

2 Operational Amplifier From Another View A vi vo vi F A vf A A ΓF vo vf A A 0 F 0 F A A F A F ideally A= v i =v f vi A vf vo F 2

3 Feedback - Block Diagram Point of View Fundamental block diagram vi Sign very important + vf - A vo βf Fundamental feedback equation We can split the ideal op amp A and resistive divider ΓF into 2 separate blocks because they do not interact with, or load, each other. 2. The op amp output is an ideal voltage source.. Sign - => Negative Feedback Sign + => Positive Feedback 3. The op amp input draws no current. 3

4 Feedback - Block Diagram Point of View Sign - => Negative Feedback vi Sign very important + vf - vo A βf Sign + => Positive Feedback vi Sign very important + A vf + vo βf ASSUME A > 0 and ΓF > 0 What if AΓF =? 4

5 Comments on the Feedback Equation vi + vf - A vo The quantity A F loop-gain and Acl = closed-loop gain. The quantity A = open-loop gain and βf vo A Acl = = v i A F vo A= v =0 vi f vf F= vo F = feedback gain. If the loop gain is much greater than one, i.e. A F (and the system is stable a topic to be discussed later!) the closed-loop gain approximates / F and is independent of A! vo Acl = vi F R2 R F= = R R 2 F R2 5

6 Feedback in the Emitter Follower C in= RS RB vo RE vi V B R B R S ib RS vi i b e ie re RE vo Small-signal ac emitter current equation: i e= vi RS V CC r e R E RE v o=r E i e = vi RS r R E e Create an artificial feedback equation, multiply num & denom by / R S and multiply denom by RE/RE: R E RS A v o= v i= vi A R E r e R E F RS RE 6

7 Emitter Follower Emitter Current The forward gain open-loop term, A: R E A= RS The feedback term, F : r e R E F= RE R E r e R E r e R E A F= = RS RE Rs R E RS A v o= v i= vi A F R E r e R E RS RE If the loop gain is large, A F RE v o v i= vi r e R E F Dependence on BJT is eliminated. 7

8 Loop Gain Sensitivities For: calculus d f x g x d f x d g x =g x f x dx dx dx Acl = A A F Open-loop gain A: dacl = d A A dacl = Acl A F 2 F A F da 2 A dacl da = 0 Acl A F A A F Feedback gain ΓF: dacl A F d F d F = Acl A F F F A F High loop gain makes system insensitive to A, but sensitive to F! 8

9 a number of other techniques developed to assist in the determination of stability of linear feedback systems by pencil-and-paper analysis, simulation and/or measured data. 9

10 vi + vf - Feedback - One-pole A(s) A vo βf vo A s Acl s, F = = vi F A s pole(s) of Acl(s) are the roots of + AΓF = 0, or F A s = = e ± j k Consider the case where: a K0 A s = s a open-loop pole (LP) a K0 ak0 s a Acl s, F = = a K 0 s a F K 0 F s a Where a & K0 are positive real quantities, and let ΓF be a positive-real variable. for k =, 3,...odd Root Locus Plot for Acl s, F s= j j very large K 0 F X -a F =0 closed-loop pole stable for all ΓF! 0

11 Feedback - One-pole A(s) cont. a K0 N s Acl s = = s a F K 0 D s a K0 A s = s a open-loop (OL) closed-loop (CL) Gain - db 20log 0 K 0 20log 0-20 db /dec - 20 db /dec K0 K FK0 0 a **In the text β = ΓF** frequency (ω) ω a F K 0 GBW OL =GBW CL=a K 0

12 One-pole Feedback - Root Locus ak0 N s A s Acl s = = = D s F A s s a F K 0 a K0 A s = s a Pole of Acl(s): D s = A s F =s a F K 0 =0 => s= a F K 0 ALTERNATIVELY a F K0 ± j k for k =, 3,...odd s a F K 0 =0 = = e s a s = r is a root of D(s=r) = 0 or A(s=r)ΓF = - iff a F K0 r a =arg [ ]=± k r a and a F K 0 for p.r. a, K0 and ΓF = r a Root locus algorithm for -pole Amplifier a, K0 and ΓF being positive-real => r = - (a + aγf K0 ) is negative-real 2

13 One-pole Feedback - Root Locus ak0 N s A s Acl s = = = D s F A s s a F K 0 s = r is a root of D(s=r) = 0 or A(s=r)ΓF = - iff a F K0 r a =arg [ ]=± k and r a a F K 0 = r a a, K0 and ΓF being positive-real => r is negative-real s= j r j r+a r a 2 r a = r a =0 ± k x j -a 2 r very large K 0 F x r -a Not allowed 3

14 Frequency-Dependent Feedback Consider the case where the open-loop gain is: s a a b c s b s c 0 a a = K =K 0 dc Gain: A s s=0=k 0 b 0 c bc A s =K The Bode plot of A(jω) : open-loop (OL) 20 log0 A(jω) - db 20log 0KK db /dec a b A j =K - 20 db /dec c ω j a j b j c a b c 4

15 Frequency-Dependent Feedback Acl s = A s F A s where a, b, c, K0 & ΓF are p.r. where r and r2 are either negative-real or complex-conjugate roots 5

16 What's the Root Locus for Acl(s)? Is Acl(s) Stable for all ΓF? s a s b s c s a Acl s = =K 0 s a s b s c F K 0 s a F K 0 s b s c K0 for p.r. a, b, c, K0 and ΓF j x -c x -b o -a a b c 6

17 What's the Root Locus for Acl(s)? Is Acl(s) Stable for all ΓF? D(s) = Since the poles of Acl(s) for ΓF > 0 will not equal any of the poles A(s), we can divide D(s) by (s + b) (s + c) to obtain s a ± jk F K0 = F K 0 f s = = e for k =, 3, odd s b s c The loop-gain terms, ΓF and K0, are positive-real numbers, so for a root s = -ri to exist, the value of f(s) must be negative-real when evaluated at s = ri, where ri is in general complex (and a conjugate). 7

18 What's the Root Locus for Acl(s)? Is Acl(s) Stable for all ΓF? e r b r r a ± 2kk r a c =± r b c r= i i i ri b ri c ri a ± jk where k = odd A s for p.r. a, b, c, s = r F AaF K r0i a a0k r i a a A s == K and Γ r c r b r i r b r c 0 F i c r i bi cl F = risubto test a potential root at s = ri, wesfirst tract the sum of the angles for the open-loop poles to the point ri from the sum of the angles for the zeros to the point (open-loop). ri. 8

19 What's the Root Locus for Acl(s)? Is Acl(s) Stable for all ΓF? D s=r i = r i b r i c F K 0 r i a =0 tot = r a r b r c =± k i i i r a =0 ; r b =0 ; r c =0 tot =0 ± k r a = ; r b =0 ; r c =0 tot = i i i2 r i3 r i2 = ; r a i4 i i3 = ; r a i2 = ; r b i4 b i3 = ; r c i4 =0 tot =0 ± k c = tot = 9

20 What's the Root Locus for Acl(s)? Is Acl(s) Stable for all ΓF? s a K0 s b s c s a Acl s = =K 0 a b c s a s b s c F K 0 s a F K 0 s b s c j x -c x -b l ri o -a Acl(s) IS stable for all ΓF! ABSOLUTELY STABLE. zeros of Acl(s) = zeros of A(s) and are independent of feedback 2. poles of Acl(s) poles of A(s) are a function of the feedback 20

21 What's the Root Locus for Acl(s)? Is Acl(s) Stable for all ΓF? K 0 bc N s Acl s = = s b s c F K 0 bc D s K 0 bc A s = s b s c b c where b, c, K0 & ΓF are p.r. complex conjugate poles of Acl(s) j F 0 X -c X -b Av(s) IS stable for all ΓF! ABSOLUTELY STABLE 2

22 What's the Root Locus for Acl(s)? Is Acl(s) Stable for all ΓF? K 0 bcd A s = s b s c s d K 0 bcd N s Acl s = = s b s c s d F K 0 bcd D s j b c d X -d X -c X -b 22

23 What's the Root Locus for Acl(s)? Is Acl(s) Stable for all ΓF? K 0 bcd A s = s b s c s d K 0 bcd N s Acl s = = s b s c s d F K 0 bcd D s j b c d Acl(s) NOT stable for all ΓF! CONDITIONALLY STABLE Root Locus Sketch X -d X -c X -b 23

24 The Root Locus Method This graphical method for finding the roots of a polynomial is known as the root locus method. It was developed before computers were available. It is still used because it gives valuable insight into the behavior of feedback systems as the loop gain is varied. Matlab (control systems toolbox) will plot root loci. In the frequency-dependent feedback (two-pole & one-zero) example, we noted that increasing feedback increases the CL bandwidth i.e. the low & high frequency break points moved in opposite directions as ΓF increased. As expected Higher ΓF reduces the closed-loop mid-band gain. 24

25 ri b F A s r i c r i a 25

26 F A s= j i F A=V f /V i F A j vf r i = j i σ ± jk ± = j k ± je 2k A j A j = F A Fj F i =i e i e for k odd F A j i = e ± jk jω for any value of ωi? F A j i for a particular ΓF = ΓFa and all ωi is determined from equations, simulation or experiment. 26

27 F A j i = e ± j 2k for any ω => onset of instability i Important Definitions arg [ Fa A j i ]= 80 o 20 log0 [ Fa A j i ]=0 db ωi ω80 = 80o phase crossover freq. ωi ω = unity (0 db) gain crossover freq. 20 log0 F A j 80 0 db o or arg [ Fa A j ] 80 [ F A80 ]db=20 log 0 [ Fa A j 80 ] 0 db GM =0 db [ F A80 ]db 0 db j =arg [ Fa A j ] 80 o PM = j 80 o 0 o 27

28 20 log0 [ Fa A j ] db ω = unity (0 db) gain crossover frequency 0 ω ω (log scale) j =arg [ Fa A j ] deg 0-90o -80o ω80 ω (log scale) ω80 = 80o phase crossover frequency -270o -360o 28

29 20 log0 [ Fa A j ] db 0 GM =0 db 20 log0 [ F A j 80 ] ω80 ω (log scale) j =arg [ Fa A j ] deg 0-90o ω80 ω (log scale) -80o -270o -360o 29

30 20 log0 [ Fa A j ] db 0 ω ω (log scale) j =arg [ Fa A j ] deg 0-90o ω ω (log scale) -80o -270o -360o PM = j 80o PM > 0o => stable amplifier 30

31 F A j 3

32 32

33 Alternative Stability Analysis. Investigating stability for a variety of feedback gains ΓF by constructing Bode plots for the loop-gain ΓFA(jω) can be tedious and time consuming. 2. A simpler approach involves constructing Bode plots for A(jω) and ΓF (or /ΓF) separately. 20 log = 20 log when 20 log F A j =20 log A j 20 log F 20 log A j =20 log 20 log F A j =0 db = F F F gain cross-over frequency 33

34 20 log0 [ Fa A j ] db 0 ω (log scale) j =arg [ Fa A j ] deg 0-90o ω (log scale) -80o -270o -360o 34

35 20 log0 [ Fa A j ] db 0 j =arg [ Fa A j ] deg 0-90o GM =0 db 20 log0 [ F A j 80 ] ω ω ω80 ω (log scale) ω80-80o -270o -360o PM = j 80o 35

36 Basic Relations 20 log A F.=20 log A 20 log F 20 log Fa A =0 f = f 20 log A =20 log Fa 36

37 37

38 0 5 A = /05 /06 /07 20log A 20 log A Fb =0 db 20log /ΓFb = 85 db GM =25 db f < f80 Basic Relations PM = GM = 0 20 log A log f φ() = -08o f80 =0 db Fb j f =arg [ A j f Fb ] 80o PM =φ() + 80o = 72o stable! 38

39 0 5 A = /0 /0 /0 20log A 20log /ΓFc = 50 db PM = GM = 0 GM GM = - 0 db f80 φ() = -20 o -40 db /dec 20 log A j f Fc =0 Basic Relations =0 db Fc j f =arg [ A j f Fb ] 80o 20 log A j f 20 log PM =φ() + 80o = -30o unstable! 39

40 Alternative Stability Analysis Observation 20log A db 00 20log /ΓFc A = /0 /0 /0 5 0 A = /0 /0 /0-20 db /dec db /dec Stable with f =arg A 02-45o -35o 0 f db /dec φ() -35o => PM 45o f (Hz) f (Hz) -90o -80o -270o 40

41 Alternative Stability Analysis - cont. 20log A db log /ΓFb /0 /0 /0 A = db /dec log /ΓFa Stable with PM 45o 0 A = /0 /0 /0-40 db /dec PM = GM = f =arg A 02 φ() = -08 o 0 f db /dec f (Hz) f (Hz) -90o -80o 72o PM -270o 4

42 Alternative Stability Analysis - cont. 5 20log A -20 db /dec stable -40 db /dec unstable unstable f80 f -60 db /dec 0 A = /0 /0 /0 20 log A F =20 log A 20 log Rule of Thumb Closed -loop amplifier will be stable if the 20log/ΓF line intersects the 20log A(j f) curve on the -20 db/dec segment. Using Rule of Thumb : arg (A() ΓF) arg (A() ΓF) - 35o => F PM 45o 42

43 Rate of Closure (RC) 20log A F db /dec RC = - 20 db/dec 0 db/dec = - 20 db/dec RC = - 40 db/dec 0 db/dec = - 40 db/dec RC = - 40 db/dec 0 db/dec = - 40 db/dec From previous slide: Rule of Thumb Closed-loop amplifier will be stable with PM 45o if 20log/ΓF line intersects 20log A(j f) curve on the -20 db/dec segment. RC =slope A j2 f db /dec slope 00 arg PM = 72o Equivalent Rule of Thumb Closed-loop amplifier will be stable with PM 45o if RC = 20 db /dec 43

44 Frequency Dependent Feedback db F db /dec RC = 20 db /dec 20 db /dec = 40 db /dec RC =slope A j2 f db /dec slope A(j2πf) 00 /Γ(j2πf) 80 RC 2= 20 db/ dec 0 db/ dec = 20 db/ dec -20 db /dec 60 /Γ RC3 = - 40 db/dec (0 db/dec) = - 40 db/dec /Γ3 RC4 = - 40 db/dec (-20 db/dec) = - 20 db/dec /Γ4(j2πf) k 0k 00k M Frequency (Hz) 0M Equivalent Rule of Thumb Closed-loop amplifier will be stable with PM 45o if RC = 20 db /dec 44

45 Quick Review A s Acl s = F A s Gain & Phase Margins A. What are the two methods for determining stability of closed loop amplifier? B. What is the Rule of thumb? 45

46 Quick Review cont. A. What are the two methods for determining stability of closed loop amplifier? 20log0 A(jω) ΓFa db A s Acl s = F A s φ(jω) = arg (A(jω)ΓFa GM ω (log scale) 80 ω (log scale) PM () A j F = = e where F = Fa j ± 2k GM =0 db 20 log 0 A j 80 Fa 0 PM =arg A j Fa 80 o 0 46

47 Quick Review cont. 20log0 A(jω) db A s Acl s = F A s 20 log (/ΓFa) 20 log (/ΓFb) 20 log (/ΓFc) φ(jω) = arg (A(jω)) (2) 20 log A j F =20 log A j 20 log F 20 log A j =20 log Fa f(hz) f(hz) where F = Fa 47

48 Quick Review cont. Rule of Thumb Closed -loop amplifier will be stable if the 20log/ΓF line intersects the 20log A(j f) curve on the -20 db/dec segment and PM 45o. 20log0 A(jω) db over 20dB/dec. segment 45ο arg (A()ΓF) -35o => PM = φ() + 80o 45o φ(jω) =arg (A(jω)) 48

49 Frequency Compensation 20log A /ΓF /ΓF /ΓF Ex: LM 74 op amp is frequency compensated to be stable with 60o PM for 20 log Acl() = 20 log /ΓF = 0 db. frequency compensation modifying the open-loop A(s) so that the closed-loop Acl(s) is stable for any desired value of Acl() by extending the -20 db/dec segment. desired implementation - minimum on-chip or external components. 49

50 Op Amp Basics or 2 Differential Stages v IN v IN2 + Differential Amplifier - Stage Compensation Circuit High Gain Stage Output Stage v OUT Bias Circuit 50

51 Compensation What if st pole is shifted lower? 20log A db 20log Acomp -20 db /dec 00 20log /ΓF = 85 db db /dec db /dec 20log /ΓF2 = 45 db db /dec db /dec VCC f (Hz) -60 db /dec 08-90o -80o 5 0 Acomp /0 3 /08 / o 5

52 Frequency Compensation Using Miller Effect VCC C comp Ii RC C comp with Comp = 0 Vo R C p2 = R2 C 2 p = Ii Vo Rin2 C in2 R V C g m V R2 C =C C g m R2 R= R B r BJT Miller Capacitance C2 R2 = RC r o Rin2 C 2 =C C in2 Vo sc comp g m R R2 = I i s[c R C 2 R2 C comp g m R R2 R R 2 ] s 2 [C C 2 C comp C C 2 ] R R2 sc comp g m R R2 sc comp g m R R2.= = where p2c pc 2 s s s s pc p2c pc p2c pc p2c 52

53 Compensation Using Miller Effect - cont. Vo = Ii sc comp g m R R2 sc comp g m R R s s s s pc p2c pc p2c pc pc p2c assumption If p2c pc pole-splitting and Vo s C comp g m R R2 = I i s [C R C 2 R2 C comp g m R R2 R R 2 ] s 2 [C C 2 C comp C C 2 ] R R2 pc = C R C 2 R 2 C comp g m R R2 R R 2 C comp g m R 2 R Compensation Miller Capacitance pc p2c C R C 2 R2 C comp g m R R2 R R2 C comp g m R R 2 gm p2c = = pc [C C 2 C comp C C 2 ] R R 2 C comp C C 2 R R 2 C C 2 If C comp C C 2 C C 2 53

54 Compensation Using Miller Effect - cont. Vo s C comp g m R R2 = I i s [C R C 2 R2 C comp g m R R2 R R 2 ] s 2 [C C 2 C comp C C 2 ] R R2 s pc = pc 2 s pc p2c C R C 2 R 2 C comp g m R R2 R R 2 C comp g m R 2 R Compensation Miller Capacitance pc p2c C R C 2 R2 C comp g m R R2 R R2 C comp g m R R 2 gm p2c = = pc [C C 2 C comp C C 2 ] R R 2 C comp C C 2 R R 2 C C 2 If C comp C C 2 C C 2 54

55 Compensation Using Miller Effect - cont. A = p f p = = 2 2 R C 0 = 5 f p f p2 f p3 p2 f p2 = = 2 2 R2 C2 Using Miller effect compensation: fp -> fpc << fp and fp2 -> fp2c >> fp2 (pole-splitting) Also fp3, determined by another stage, is unaffected by the compensation => fp3c = fp3 0 Acomp = f pc 5 f p2c pc f pc = 2 2 [C comp g m R R2 ] f p3 p2c gm f p2c = 2 2 [C C 2 ] 55

56 Given: Miller Compensation Example A = = = R C R2 C 2 f p f p where C = 20 pf, C2 = 5 pf, gm = 40 ms, R = 00/2π kω and R2 = 200/2π kω Design Objective: determine Ccomp s.t. fpc = 00 Hz and compute fp2c. Acomp = = f p2c f pc f p2c f pc =00 Hz = C comp=78.5 pf 2 [C comp g m R R 2 ] 2 [C comp 40x x05 ] gm f p2c = 6 MHz 2 C C 2 56

57 Compensation Using Miller Effect - cont. 20log A db -20 db /dec 00 20log Acomp db /dec Acomp -40 db /dec o A = PM 45 for Acl = o db /dec -pole roll-off! F 20log /ΓF = 0 db f (Hz) -60 db /dec -40 db /dec o -80o Acomp

58 Summary Feedback has many desirable features, but it can create unexpected - undesired results if the complete frequencydependent nature (phase-shift with frequency) of the feedback circuit is not taken into account. Feedback can be used to convert a well-behaved stable circuit into an oscillator. Sometimes, due to parasitics, feedback in amplifiers results in unexpected oscillations. Iff all the poles/zeros are accounted for, the root-locus visually illustrates how feedback can create the conditions for oscillation and instability. 58