ESE319 Introduction to Microelectronics. Feedback Basics

Transcription

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

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3 Operational Amplifier From Another View 3

4 Feedback Block Diagram Point of View Fundamental block diagram v i + v f - v o = A β F A 1 A v i v o Fundamental feedback equation We can split the ideal op amp and resistive divider into 2 separate blocks because they do not interact with, or load, each other. 1. The op amp output is an ideal voltage source. 2. The op amp input draws no current. 4

5 v i Comments on the Feedback Equation + v f - G= v o v i = A= v o v i A β F = v f v o A 1 A v f =0 v o The quantity A ' loop-gain and G = closed-loop gain. The quantity A = open-loop gain and = feedback gain. If the loop gain is much greater than one, i.e. A 1 (and the system is stable a topic to be discussed later!) the closed-loop gain approximates 1/ &and is independent of A! G= v o 1 v i 5

6 Feedback in the Emitter Follower ib ie T ib Emitter current equation: 1 i e = v R i s T 1 r e R E Note: T =, i.e. the BJT beta. Create an artificial feedback equation, multiply by : T 1 R s T 1 / R s i e = 1 v i = A v T 1 1 A i r R e R E F s 6

7 Emitter Follower Emitter Current The forward gain term, A: ib ie T ib A= T 1 R s The feedback term, : i e = T 1 R s 1 T 1 R s r e R E v i = A 1 A v i =r e R E If the loop gain is large, i e 1 v i = 1 v F r e R i E Dependence on T is eliminated. A 1 7

8 Loop Gain Sensitivities Forward gain: dg da = 1 1 A 2 dg G = 1 1 A 1 A 2 A da = 1 1 A da A 0 A For: Feedback gain: dg d = A2 1 A 2 dg G = A2 1 A 1 A 2 A d A 1 A G= A 1 A d d A High loop gain makes system insensitive to A, but sensitive to! = 8

9 v i + v f - Feedback - One-pole A(s) A v o G= v o v i = A 1 A β F Consider the case where: A s = a K 0 s a G s, = Where open-loop pole a K 0 s a 1 a K 0 1 s a = a K 0 s a 1 K 0 and K 0 are positive real quantities. Root Locus Plot for G s, s= j j X -a =0 closed-loop pole stable for all Γ F! 9

10 Feedback - One-pole A(s) cont. A s = a K 0 s a open-loop (OL) Gain - db a K G s = 0 s a 1 K 0 = N s D s closed-loop (CL) 20 log 10 K 0 K 20log K F K 0 0 a - 20 db /dec - 20 db /dec a 1 K 0 frequency ω (ω) GBW OL =GBW CL =a K 0 10

11 A s = a K 0 s a Pole of G(s): ESE319 Introduction to Microelectronics One-pole Feedback - Root Locus a K 0 G s = s a 1 K 0 = N s D s D s =s a 1 K 0 =s a a K 0 =0 s a a K 0 =0 a K 0 s a a K 0 s a =1 2 = s2 a and s a = ± 2k 1 s1 a =0 ± 2 k j ± 2k = 1=1 e j s= j s i +a x -a s i si a j 2 1 s 2 K 0 0 -a x s 1 Not allowed 11

12 Frequency-Dependent Feedback Consider the case where the open-loop gain is: s a A s =K s b s c a b c For convenient Bode plotting we can rewrite A(s) as: dc gain s a 1 s a 1 A s =K a bc s b 1 s =K 0 c 1 s b 1 s c 1 where K 0 =K a bc 12

13 Frequency-Dependent Feedback A sketch of the Bode plot would look something like: 20 log 10 K 0 20 log 10 A(s) - db + 20 db /dec a b c - 20 db /dec ω open-loop (OL) s a 1 A s =K 0 s b 1 s c 1 a b c Sketching the frequency response is easy, if the polynomial is factored into its roots! 13

14 Frequency-Dependent Feedback If the frequency-dependent quantity, A(s), is embedded in a feedback loop, the poles are not so easily factored: A s =K 0 G s = s a K 0 s b s c G s = 1 K 0 s a s b s c s a s b s c A s 1 A s open-loop (OL) closed-loop (CL) =K 0 s a s b s c K 0 s a frequency dependent feedback 1. zeros of G(s) = zeros of A(s) and are independent of feedback 2. poles of G(s) poles of A(s) are a function of the feedback 14

15 Frequency-Dependent Feedback Root s a K 0 s b s c G s = s a 1 K 0 s b s c Rationalizing this expression leads to: G s =K 0 s a s b s c K 0 s a = N s D s Locus The numerator is factored, but the denominator is not. We have a new quadratic D(s) for G(s). It could be factored using the quadratic formula, but we will do it another way! Factoring a polynomial is equivalent to finding its roots, i.e. the values of s that make the polynomial equal zero. D s = s b s c K 0 s a =0 15

16 Frequency-Dependent Feedback Root Locus s b s c K 0 s a =0 Since the poles of G(s) will not equal any of the poles of A(s), we can divide by (s + b) (s + c) and obtain: s a 1 K 0 s b s c =0 K 0 s a s b s c = 1 The loop-gain terms, and K 0, are positive real numbers, so for a root, say s = -r 1 to exist, the value of the frequency-dependent terms must be real and negative when evaluated at s = -r 1! 16

17 Frequency-Dependent Feedback Root Locus K 0 s a j ± 2k = 1=1 e where k = 0, 1,... s b s c Working with the complex numbers in polar form: K 0 s a s b s c =1 s a s b s c = 1 K 0 and si a s i b s i c = ± 2k we add the angles for zeros to the point s i in the complex plane and subtract the angles for poles. 17

18 Frequency-Dependent Feedback Root Locus A sketch of the operations looks like: s i +c si c. s i s i +b s i b s= j s i +a si a x x o -c -b -a j Root Locus Protocol s i a s i b s i c = 1 K 0 si a si b si c = ± 2k Move point s i until the phase angle relation is satisfied. The only values of s i that satisfy the phase angle relationship lie on the negative - axis. 18

19 Frequency-Dependent Feedback - Root Locus Only D(s) root locations where the angles with respect to open-loop pole/zero locations of A(s) are odd multiples of ±180 o are candidates. D s = s b s c K 0 s a =0 tot = s a s b s c = ± 2k j G(s) stable for all Γ F! 4 s 4 s 3 0 s 1 x 0 -c 3 x -b o 2 s -a 1 1. All roots of D(s) must lie along the green loci. 2. Bandwidth of G(s) > bandwidth of A(s) 3. If s 2 = -a there is a zero/pole cancellation 2 19

20 Frequency-Dependent Feedback Root Locus Now, to find the gain K 0 required for a given root or pole of G(s), measure the distances from each critical point and form the ratio: s a s b s c = 1 Where s = - s r is on the root-locus K 0 r = s b s c s a K 0 s r c s= s r s r b s r a s r c x -c x l -b o s r -a The only allowed poles {e.g. s r } of G(s) are on the green loci. 20

21 Note: no zero at s = -a K 0 bc A s = s b s c G s = ESE319 Introduction to Microelectronics Root Locus Two-pole A(s) K 0 bc s b s c K 0 bc = N s D s b c complex conjugate poles of G(s) 0 20 log 10 K 0 K 20 log 10 A(s) - db j X -c X -b - 20 db /dec - 40 db /dec ω G(s) stable for all Γ F! 21

22 Root Locus Three-pole A(s) K 0 bcd A s = s b s c s d G s = K 0 bcd s b s c s d K 0 bcd = N s D s j G(s) NOT stable for all Γ F! X -d X -c X -b 22

23 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 example, we noted that increasing feedback increases the closed-loop bandwidth i.e. the low and high frequency break points moved in opposite directions as Γ F increases. Higher Γ F, as a trade-off, reduces the closed-loop mid-band gain. 23

24 G j = V j o V i j = A j 1 A j loop-gain oscillation or instability GM mutually consistent stability conditions 24

25 G j G j PM 50 o 25

26 G j 70 o PM 55 o 26

27 Alternative Stability Analysis 1. Investigating stability for a variety of feedback gains Γ F by constructing Bode plots for the loop-gain A(jω)Γ F can be tedious and time consuming. 2. A simpler approach involves constructing Bode plots for A(jω) and 1/Γ F separately. 20log A j =20 log A j 20log 1 20 log = 20 log 1 when 20 log A j = 1 =20log 1 A j 1 =1 RECALL: A j A j 1 G j = 1 1 A j closed-loop gain 27

28 Alternative Stability Analysis - cont. A jf 20log 1/Γ F = 85 db 20log 1/Γ F = 60 db 20log 1/Γ F = 50 db db o -180 o -270 o 20 log A zero PM & GM -20 db /dec f o 72 o PM stable 25 db GM -40 db /dec unstable arg A jf f 1 PM < 0 o 10 6 f 1 f A jf = 1 jf / jf / jf / db /dec log A jf f (Hz) f (Hz).=20 log A jf 20 log 1 stable iff 20log 1/Γ F > 60 db or 20log Γ F < -60 db or Γ F <

29 Alternative Stability Analysis - cont. A jf Rule of Thumb Closed -loop amplifier will be stable if the 20log1/Γ F line intersects the 20log A(j f) curve on the -20 db/dec segment. Using Rule of Thumb : arg A jf arg A jf o => PM 45 o 29

30 Frequency Compensation Ex: LM 741 op amp is frequency compensated to be stable with 60 o PM for G(jf) = 1/Γ F = 1. frequency compensation modifying the open-loop A(s) so that the closed-loop G(s) is stable for any desired value of G(jf). desired implementation - minimum on-chip or external components. 30

31 Compensation What if 1st pole is shifted lower? A jf db db /dec A comp jf 20log 1/Γ F1 = 85 db db /dec db /dec 20log 1/Γ F2 = 45 db db /dec -60 db /dec 20 V CC C comp db /dec 0-90 o f (Hz) -180 o 10 5 A comp jf 1 jf / jf / jf / o 31

32 Frequency Compensation Using Miller Effect I i C comp V o I i = V CC R C R in2 C 2 V o I i C comp V o R 1 V C 1 g R 2 C 2 m V C 1 =C C 1 g m R 2 R 1 =R B r BJT Miller Capacitance s C comp g m R 1 R 2 R 2 =R C r o R in2 with C omp = 0 p1 = 1 R 1 C 1 p2 = 1 R 2 C 2 1 s[c 1 R 1 C 2 R 2 C comp g m R 1 R 2 R 1 R 2 ] s 2 [C 1 C 2 C comp C 1 C 2 ] R 1 R 2.= sc comp g m R 1 R 2 1 s 1 s = p1c p2c 1 s sc comp g m R 1 R p1c p2c s 2 p1c p2c where p2c p1c 32

33 Compensation Using Miller Effect - V o sc comp g m R 1 R 2 sc comp g m R 1 R 2 = I i 1 1 s 1 s s s 2 1 p1c p2c p1c p2c p1c p1c p2c and cont. assumption If p2c p1c pole-splitting V o s C comp g m R 1 R 2 = I i 1 s[c 1 R 1 C 2 R 2 C comp g m R 1 R 2 R 1 R 2 ] s 2 [C 1 C 2 C comp C 1 C 2 ] R 1 R 2 1 p1c = C 1 R 1 C 2 R 2 C comp g m R 1 R 2 R 1 R 2 1 C comp g m R 1 R 2 Compensation Miller Capacitance p2c = p1c p2c = C R C R C g R R R R comp m C comp g m p1c [C 1 C 2 C comp C 1 C 2 ] R 1 R 2 C 1 C 2 C comp C 1 C 2 g m C 1 C 2 If C comp C 2 33

34 Compensation Using Miller Effect - cont. f p1c = p1c [C comp g m R 1 R 2 ] f p2c = p2c 2 g m 2 [C 1 C 2 ] 10 5 A jf = 1 jf / jf / jf /10 7 = jf 1 jf 1 jf f p1 f p2 f p3 Using Miller effect compensation: f p1 -> f p1c << f p1 and f p2 -> f p2c >> f p2 (pole-splitting) Also f p3 is unaffected by the compensation => f p3c = f p A comp jf = 1 jf / jf / f p2c 1 jf /10 7 = jf 1 jf 1 jf f p1c f p2c f p3c 34

35 Miller Compensation Example 10 5 Given: A jf = 1 jf / jf / jf /10 7 = jf 1 jf 1 jf f p1 f p2 f p3 where C 1 = 100 pf, C 2 = 5 pf, g m = 40 ms, R 1 = 100/2π kω and R 2 = 100/π kω Design Objective: determine C comp s.t. f p1c = 10 2 Hz 10 5 A comp jf = 1 jf / jf / f p2c 1 jf /10 7 = jf 1 jf 1 jf f p1c f p2c f p3c 1 f p1c =100 Hz 2 [C comp g m R 1 R 2 ] C comp =78.5 pf g m f p2c = 60 MHz 2 C 1 C 2 35

36 Compensation Using Miller Effect - cont. A jf db db /dec A comp jf db /dec -40 db /dec pole -60 db /dec roll-off! 20log 1/Γ 0 F2 = 0 db f (Hz) db /dec -40 db /dec o PM 45 o for G jf = o 10 5 A comp jf 1 jf / jf / jf / o 36

37 Summary Feedback has many desirable features, but it can create unexpected - undesired results if the full frequency-dependent nature (phase-shift with frequency) of the feedback circuit is not taken into account. We next will use feedback to convert a well-behaved stable circuit into an oscillator. Sometimes, because of parasitics, feedback in amplifiers turns them into unexpected oscillators. The root-locus method was used to help us understand how feedback can create the conditions for oscillation and instability. 37