ECEN 607 (ESS) Op-Amps Stability and Frequency Compensation Techniques. Analog & Mixed-Signal Center Texas A&M University

Transcription

1 ECEN 67 (ESS) Op-Amps Stability and Frequency Compensation Techniques Analog & Mixed-Signal Center Texas A&M University

2 Stability of Linear Systems Harold S. Black, 97 Negative feedback concept Negative feedback provides: Gain stabilization Reduction of nonlinearity Impedance transformation But also brings: Potential stability problems Causes accuracy errors for low dc gain Here we will discuss frequency compensation techniques

3 Stability Problem INPUTx i x d A(s) x o LOAD x f β H CL ( s) A( s) A / A( s) A( s) s T ( s) Feedback forces x d to become smaller It takes time to detect x o and feedback to the input x d could be overcorrected (diverge and create instability) How to find the optimal (practical) x d will be based on frequency compensation techniques

4 Gain Margin & Phase Margin T (db) T <T(jω) -9º -8º -7º ϕ m Conceptual Gain Margin G M and Phase Margin ϕ m f x f -8º GM f x f -8º f (dec) f (dec) G M is the number of dbs by which T ( j 8) can increase until it becomes db G M log T j 8 Phase margin ϕ m is the number of degrees by which T jx can be reduced until it reaches π (-8º) m 8 T jx or T j 8 x m * ω x is the crossover frequency 4

5 At the crossover point, The non-ideal closed loop transfer function becomes Gain Margin & Phase Margin m j m x x e j T j T 8 m m ideal j ideal x ideal x x x x x CL j A e A j T A j T j A j A j A j H x sin cos /, sin cos Ideal m m Ideal x CL A A j H 5

6 Gain Margin & Phase Margin Observe that different ϕ m yield different errors. i.e. ϕ m 9º.77 6º. 45º. º.9 5º.8 º H CL j (oscillatory behavior) x In practical systems, ϕ m = 6º is required A worst case ϕ m = 45º for a typical lower limit For ϕ m < 6º, we have A jx Aideal indicating a peaked closedloop response. 6

7 A good Opamp design implies: (i) over as wide a band of frequencies as possible (ii) The zeroes of must be all in the left-hand plane These two conditions often conflict with each other. These trade-offs should be carefully considered. Let s consider a practical amplifier characterized with these poles, The characteristic equation becomes Why a dominant pole is required for a stable amplifier? 7 ) ( j A ) ( j A / / / ) ( s s s A s s s A s A ) ( A s s s s s s A s A

8 Critical Value of βa s s s A Note that βa is the critical parameter that determines the pole locations for a given α, α and α ( β ). Furthermore, when βa =, the roots are at -α, -α and -α. Therefore, for small βa, the roots should be in the left-hand plane (LHP). However, for βa >>, two of the roots might be forced to move to the righthand plane (RHP). This can be verified by applying Routh s stability criterion. Let us write the polynomial as b s bs bs b In order to have, in the above equation, left half plane roots, all the coefficients must be positive and satisfy b b bb 8

9 The condition for imaginary-axis roots become Now, for s=jω being a root, both real and imaginary parts must be zero. That is, Then the two roots are placed at Critical Value of βa 9 b b j b b b j b j b j b, b b b b or b b b b, b b j b b j p

10 Then Thus, the critical value of βa becomes Thus when βa becomes (βa ) C, the amplifier will oscillate at Also when βa > (βa ) C, the amplifier has RHP poles, therefore is unstable. Critical Value of βa b b b b C A C A b b OSC

11 Critical Value of βa Let us consider some numerical examples. Let A = 5, (i) Three equal poles α =α =α = 7 rad/s. 7 The amplifier oscillates at OSC 5 A 8, 8 / A 8 C C rad/s (ii), then the critical loop gains yields 4 4, C 4 4 A thus the amplifier is stable if C. A Since R / R R,. causes R / R 4, which means that for an inverting (non-inverting) configuration, the gain must be greater than -4 (5) to keep the amplifier stable.

12 Critical Value of βa (iii) Let us determine A C under the most stringent condition β=. Then from previous equation, A In order to have a large A, the poles must be widely separated. i.e. α <<α <<α, then the A inequality can be approximated as A To obtain a conservative A, let α =α, which yields A This inequality bounds the DC gain to provide a stable closed loop configuration.

13 Peaking in the frequency domain usually implies ringing in the time domain Normalized second-order all-pole (low pass) system Peaking and Ringing ) ( Q j s Q s s H j s j s

14 GP: Peak gain We have the error function, To find out the maximum value of E(jω), calculate the derivative of D(ω) and make it equal to zero For, use ω * in Eq. () and we get the peak gain Peaking and Ringing D Q j E ) ( 4 ) ( 4 Q D d d * Q OR Q / 4 j E Q Q GP () () 4

15 Peaking and Ringing OS: Overshoot Inverse Laplace transform b - domain : ( s a) s b Inverse Laplace time - doamin : e For a nd order all pole error function, the impulse response is H ( s) s j s Inverse Laplace h ( t) b s Q e at Consider the normalized step response of this system, b sin( bt) u( t) s a b b at sin v in bt a b u t ( t) ( t) Q 4Q y( t) t h( t) dt e sinbt, tan b at b a for v in ( t) u( t) ( t) dt () Use the damping factor to represent, a, b 5

16 Peaking and Ringing Usually, the damping factor ξ=/q is used to characterize a physical nd order system. Thus, we rewrite the normalized time-domain equation () at y( t) e sin b tan b a tan t bt e sin t For under damped case ξ < (Q >.5), the overshoot is the peak value of y(t). To find the peak value, we first calculate the derivative of y(t), and make it equal to. d d t y( t) e sin t t t e cos t, for small (4) 6

17 Peaking and Ringing tan tan cos sin ) ( t t e t e t y t d d t t To satisfy the equality above, we have In other words, the normalized step response y(t) in Eq.() achieves extreme values at time steps of n=,,,,,,..., n n t 7

18 Peaking and Ringing n= n= n= n= n=4 n=5 n=6 Peak Value Q=5 Time t 8

19 Peaking and Ringing sin ) (,, sin ) (,, sin ) (,, ) (,, e e t y t n e e t y t n e e t y t n t y t n Therefore, the global peak value of y(t) is achieved when n=. Thus, the overshoot is defined as Value Final Value Final Peak Value (%) e OS (5) 9

20 ϕ m : Phase Margin For a nd order all-pole error function Therefore, the loop gain is given by The cross-over frequency thus can be obtained Peaking and Ringing ) ( ) ( ) ( ) ( s Q s s T s T s T s E ) ( Q j T x x x

21 Peaking and Ringing Solve the equation and get the crossover frequency, / / x 4 Q Q And thus the phase margin is 4 8 cos m T j x cos Q Q Study the relationship between phase margin and gain peaking (Eq.) or overshoot (Eq.5), we have GP(6) GP(45).dB.4dB OS(6) 8.8% OS(45) % Q.8 Q.8

22 Peaking and Ringing Gain Peaking (db) 8 8 Q Factor Overshoot (%) Phase Margin (degree) Phase Margin (Degrees)

23 The Rate of Closure (ROC) One effective method of assessing stability for minimum phase systems from the magnitude Bode plots is by determining the ROC. The Rate of Closure (ROC) Determining the ROC is done by observing the slopes of and at their intersection point (cross-over frequency ) and deciding the magnitude of their difference. ROC Slope A Slope / The ROC is used to estimate the phase margin and therefore the stability (How?) f f x

24 The Rate of Closure (ROC) Observing a single-root transfer function H ( jf ) jf / f f f f f f / f Slope Slope Slope Empirical Equation H db/ dec H H db/ dec H 9 H db/ dec H 45 H 4. 5 Slope This correlation holds also if H(s) has more than one root, provided the roots are real negative, and well separated, say, at least a decade apart. H 4

25 The Rate of Closure (ROC) In a feedback system, suppose both A and /β have been graphed. 4.5 Thus, the ROC can be used to estimate the phase margin (Page 4) Cases ROC ROC ROC ROC m 8 T ( jf x ) db/ dec db/ dec 4dB/ dec 4dB/ dec m m m m

26 The Rate of Closure (ROC) A( s) s p A ol s p 8 6 Gain (db) 4 A ol ** /β * ** /β /β * db/dec ROC Stability ** 4dB/dec ROC Marginal Stability /β 4 K K K M * Frequency(Hz) 6

27 Stability in Constant-GBP OpAmp t Constant-GBP OpAmp (i.e. A( s) s j ) Unconditionally stable with frequency-independent feedback, or. (e.g. in a non-inverting or inverting amplifier, the feedback network contains only resistors) Stable for any β. In feedback systems, since now we have T A A, A 9 these circuits enjoy m 8 A jfx Typically, due to additional high-order poles in OpAmps, 6 m 9 j 7

28 Feedback Pole Feedback Pole Feedback network includes reactive elements Stability may no longer be unconditional jf f(db) jf / A /β f p A(jf x ) A pole f p (or a zero) of β becomes a zero f z (or a pole) for /β. /β β f p f z f x f t f (dec) β f t For the case fz f t The effect of a pole within the feedback loop 8

29 Feedback Pole Examine the error function,, Using OpAmp high-frequency approximation: E( s) f f j A f f f Refer to page 4, and we have s=jπf. The peak value of E(s) can be obtained. For Q >>, the approximate result is t t z f f f / x z ft, Q t f z 9

30 Feedback Pole The lower f z compared to β f t, the higher the Q and, hence, the more pronounced the peaking and ringing. Derive the phase margin T ( jf ) A( jf ) / ( jf ) 9 tan f / f x x f x f z f f As fz f t, T ( jf ) x 8 and ROC = 4dB/dec The circuit is on the verge of oscillation! z t x x z

31 Differentiator Feedback pole example: differentiator C R Assume constant-gbp OpAmp jf GB / j f jf A t / ZC Z R jf / C f z, Z C jc R S V i a Uncompensated Differentiator C R V o To stabilize the differentiator, add a series resistance Rs. V i a Vo Compensated Differentiator

32 Differentiator At low frequency, R S has little effect because R S << /jωc At high frequency, C acts as a short compared to R S, The feedback network becomes /β = +R/R S. A H(dB) Uncompensated A H(dB) Compensated A A /β /β f (f z ) f x f t f (dec) f (f z ) f x f t f (dec)

33 Differentiator Assume R S << R C, the series resistor R S introduces an extra pole frequency f e ZC R Z C jf jf / / f f z e Z C RS,, jc f x f t f z Choose R S R / ft / fz, we have f e f t f z T ( jf x ) A( jf 9 tan 5 x ) / ( jf ) f / f tan f / f x e x x z Therefore, ROC = db/dec, ϕ m 45º

34 Stray Input Capacitance Compensation All practical OpAmps exhibit stray input capacitance. The net capacitance C n of the inverting input toward ground is C n C C / C d C d is the diffrential Cap between input pins, C c / is the common-mode cap of each input to ground, and C ext is the external parasitic cap. C ext C f In the absence of C f, there s a pole in feedback R / R jf R // R ROC 4 db/dec (See page 9) C n R V i C n R a Vo 4

35 Stray Input Capacitance Compensation Solution: Introduce a feedback capacitance C f to create feedback phase lead. In the presence of we have, To have 45 (i.e. ROC= db/dec): Make the cross-over frequency exactly at H(dB) Since A A ϕ m º ϕ m 45º Solve the equation to get +R /R /β ϕ m 9º f z f p f t f (dec) 5

36 Stray Input Capacitance Compensation To have 9 (i.e. ROC= db/dec): Place exactly on the top of to cause a pole-zero cancellation Thus using simple algebra R (Neutral Compensation) H(dB) A A ϕ m º ϕ m 45º +R /R /β ϕ m 9º f z f p f t f (dec) 6

37 Capacitive-Load Isolation There re applications in which the external load is heavily capacitive. Load capacitance C L A new pole is formed with output resistance r o and C L Ignore loading by the feedback network The loaded gain is A loaded A jf / f p, f p r ocl ROC is increased and thus invite instability Solution: Add a small series resistance R S to decouple the output from C L 7

38 Capacitive-Load Isolation R R S r O R C f R r o R R C L 8

39 Uncompensated OpAmp The poles of the uncompensated OpAmp are located close together thus accumulating about 8 of phase shift before the -db crossover frequency. Unstable device, thus efforts must be done to stabilize it. Example of uncompensated OpAmps is 748, which is the uncompensated version of 74. They can be approximated as a three-pole system Three-pole OpAmp model 9

40 Stability of Uncompensated OpAmp With a frequency-independent feedback (i.e. curve is flat) around an uncompensated OpAmp, we have curve can be visualized as the curve with the line is the new -db axis For ROC db/dec 45 For db/dec ROC 4 db/dec 45 For ROC < 4 db/dec Three-pole open-loop response 4

41 Stability of Uncompensated OpAmp The uncompensated OpAmp provides only adequate phase margin only in high-gain applications (i.e. high ) To provide adequate phase margin in low-gain application, frequency compensation is needed. Internal compensation Achieved by changing External compensation Achieved by changing 4

42 Internal Frequency Compensation How to stabilize the circuit by modifying the open loop response? Dominant-Pole Compensation Shunt-Capacitance Compensation Miller Compensation Pole-Zero Compensation Feedforward Compensation 4

43 Dominant-Pole Compensation An additional pole at sufficiently low frequency is created to insure a roll-off rate of - db/dec all the way up to the crossover frequency. Cases: : ROC = db/dec 45 : ROC = db/dec 9 This technique causes a drastic gain reduction above 4

44 Dominant-Pole Compensation Numerical example: r d =,r o = g = ma/v, R = kω, g = ma/v, R =5 kω,, Find the required value of for 45 with For ϕ m =45º, we have Draw a straight line of slope - db/dec until it intercepts with the DC gain asymptote at point D and get. Thus: Requires EXTREMELY LARGE passive components 44

45 Shunt-Capacitance Compensation The dominant-pole technique adds a fourth pole Extra cost and less bandwidth. This technique rearranges the existing rather than creating a new pole. It decreases the first (dominant) pole to sufficiently low frequency to insure a roll-off rate of - db/dec all the way up to the crossover frequency. 45

46 Shunt-Capacitance Compensation Cases: : ROC = db/dec 45 : ROC = db/dec 9 Since is chosen to insure roll-off rate of - db/dec all the way up to the crossover frequency. Thus: 46

47 Shunt-Capacitance Compensation The first pole is decreased by adding an extra capacitance to the internal node causing it. Given the value of from the desired value of,we can find 47

48 Shunt-Capacitance Compensation Numerical example: r d =,r o = g = ma/v, R = kω g = ma/v, R =5 kω,, Find the required value of for 45 with For 45, Then, 59 EXTREMELY LARGE Unsuitable for monolithic fabrication 48

49 Miller s Theorem If A v is the voltage gain from node to, then a floating impedance Z F can be converted to two grounded impedances Z and Z : Miller Compensation v F F F A Z V V V Z Z Z V Z V V v F F F A Z V V V Z Z Z V Z V V [Liu] 49

50 Applying Miller s theorem to a floating capacitance connected between the input and output nodes of an amplifier. The floating capacitance is converted to two grounded capacitances at the input and output of the amplifier. The capacitance at the input node is larger than the original floating capacitance (Miller multiplication effect) Miller Compensation F v v F v F C A j A C j A Z Z F v v F v F C A j A C j A Z Z [Liu] 5

51 Miller Compensation This technique places a capacitor in the feedback path of one of the internal stages to take advantage of Miller multiplication of capacitors. The reflected capacitances due to and the DC voltage gain between and ( ) yields, and,, and, A low-frequency dominant pole can be created with a moderate capacitor value. 5

52 Miller Compensation Accurate transfer function Pole/zero locations RHP zero Dominant Pole Second Pole 5

53 Miller Compensation Right-half plane zero: The RHP zero is a result of the feedforward path through The circuit is no longer a minimum-phase system. It introduces excessive phase shift, thus reduces the phase margin. In bipolar OpAmps, it is usually at much higher frequency than the poles 5

54 Miller Compensation Pole Splitting Increasing lowers and raises The shift in eases the amount of shift required by Higher bandwidth Increasing above a certain limit makes stops to increase., 54

55 Miller Compensation Numerical example: r d =,r o =, g = ma/v, R = kω, g = ma/v, R =5 kω,, Find the required value of for 45 with From and, we can calculate 5.9 and.8 Assume is large 8. Since is the first non-dominant pole For 45, Then,.8 Much smaller than that of shuntcapacitance compensation Suitable for monolithic fabrication 55

56 Pole-Zero Compensation This technique uses a large compensation capacitor ( ) to lower the first pole. It also uses a small resistor ( ) to create a zero that cancels the second pole. The compensated response is then dominated by the lowered first pole up to Transfer function Pole/zero locations,, 56

57 Pole-Zero Compensation and lowers the dominant pole, creates a zero, and creates an additional pole Choose such that cancels The open loop gain now becomes - To have 45, the cross-over frequency should be - Since the compensated response is dominated by pole up to - Thus, 57

58 Pole-Zero Compensation Numerical example: r d =,r o =, g = ma/v, R = kω, g = ma/v, R =5 kω,, Find the required value of for 45 with g V d R C + V - C c R c g V R C + V - From and, we can calculate 5.9 and.8 For 45, Then, is chosen such that Since Ω 5.9 It will not affect the phase margin Relaxed compared to shuntcapacitance compensation, but still LARGE 58

59 Feedforward Compensation In multistage amplifiers, usually there is one stage that acts as a bandwidth bottleneck by contributing a substantial amount of phase shift in the vicinity of the cross-over frequency This technique creates a high-frequency bypass around the bottleneck stage to suppress its phase at, thus improving 59

60 Feedforward Compensation The bypass around the bottleneck stage is a high-pass function The compensated open-loop gain is At low frequency: The high low-frequency gain advantage of the uncompensated amplifier still hold. At high frequency: The dynamics are controlled only by Wider bandwidth & Lower phase shift 6

61 Summary of Internal Frequency Compensation Dominant-pole compensation: It creates an additional pole at sufficiently low frequency. It doesn t take advantage of the existing poles. It suffers from extremely low bandwidth. Shunt-capacitance compensation: It rearranges the existing poles rather than creating an additional pole. It moves the first pole to sufficiently low frequency. The value of the shunt capacitance is extremely large Extra cost 6

62 Summary of Internal Frequency Compensation Miller compensation: It takes advantage of Miller multiplicative effect of capacitors, thus requires moderate capacitance to move the first pole to sufficiently low frequency. It causes pole splitting, where the dominant pole is reduced and the first non-dominant pole is raised in frequency. 6

63 Summary of Internal Frequency Compensation Pole-zero compensation: Similar to shunt-capacitance technique, a large capacitor is used to shift the first pole to sufficiently low frequency. A small resistance is used to create a zero that cancels the first nondominant pole Feedforward Compensation It places a high frequency bypass around the bottleneck stage that contributes the most phase shift in the vicinity of 6

64 External Frequency Compensation How to stabilize the circuit by modifying its feedback factor? Reducing the Loop Gain Input-Lag Compensation Feedback-Lead Compensation 64

65 Reducing the Loop Gain This method shifts /β curve upwards until it intercepts the a curve at º, where is the desired phase margin. The shift is obtained by connecting resistance across the inputs. [Franco] Uncompensated Shifts the curve upwards to improve ϕ m 65

66 Reducing the Loop Gain R c is chosen to achieve the desired phase margin : Then, [Franco] 66

67 Reducing the Loop Gain Prices that we are paying for stability: Gain Error: The presence of reduces, thus resulting in a larger gain error. DC Noise Gain: The presence of causes an increased DC-noise gain which may result in an intolerable DC output error. THERE S NO FREE LUNCH! 67

68 Input-Lag Compensation The high DC-noise gain of the previous method can be overcome by placing a capacitance in series with. High frequencies: is short. curve is unchanged compared to the previous case. Low frequencies: is open [Franco], we now have much higher DC loop gain & much lower DC output error. 68

69 Input-Lag Compensation R c is chosen to achieve the desired phase margin : Then, [Franco] To avoid degrading, it is good practice to position the second breakpoint of curve a decade below. Then,. 69

70 Input-Lag Compensation Advantage(s): Lower DC-noise gain due to the presence of. It allows for higher slew rate compared with internal compensation techniques: Op-amp is spared from having to charge/discharge internal compensation capacitance. Disadvantage(s): Long settling tail because of the presence of pole-zero doublet of the feedback network ( ). Increased high-frequency noise in the vicinity of the cross-over frequency. Low closed-loop differential input impedance which may cause highfrequency input loading, ** is the open loop input impedance of the Op-Amp. 7

71 Feedback-Lead Compensation This technique uses a feedback capacitance to create phase lead in the feedback path. The phase lead is designed to be in the vicinity of the crossover frequency which is were is boosted. [Franco] 7

72 Feedback-Lead Compensation Analysis: where,, The phase-lag provided by is maximum at. The optimum value of, that maximizes the phase margin, is the one that makes this point at the crossover frequency. [Franco] The cross-over frequency can be obtained from Having, the optimum can be found 7

73 Feedback-Lead Compensation How much phase margin can we get? At the geometric mean of and, we have 9 tan The larger the value of, the greater the contribution of / to the phase margin. [Franco] E.g. 55 Thus, 55 - The phase margin is improved by 55 due to feedback-lead compensation. 7

74 Feedback-Lead Compensation Advantage(s): helps to counteract the effect of the input stray capacitance as we discussed beforehand. It provides better filtering for internally generated noise. Disadvantage(s): It doesn t have the slew-rate advantage of the input-lag compensation. 74

75 Decompensated OpAmps These OpAmps are compensated for unconditional stability only when used with above a specified value They provide a constant GBP only for They offer higher GBP and slew rate. Example The fully compensated LF56 OpAmp uses to provide 5 and for any V/V. The decompensated version of the same OpAmp, LF57, uses and provides and 5 but only for any 5 V/V. 75