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Ecen 457 (Ess) ECEN 457 (ESS) Op-Amps Stability and Frequency Compensation Techniques Analog & Mixed-Signal Center Texas A&M University Reference: Sergio Franco, “Design with Operational Amplifiers and Analog Integrated Circuits” 4th Edition , Chapter 8, 2015 Stability of Linear Systems Harold S. Black, 1927 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 2 Stability Problem LOAD INPUT xd xo A(s) xi xf β A(s) 1/ A(s) H (s) CL 1 1 As 1 1T (s) A(s) Feedback forces xd to become smaller It takes time to detect xo and feedback to the input xd could be overcorrected (diverge and create instability) How to find the optimal (practical) xd will be based on frequency compensation techniques 3 Gain Margin & Phase Margin |T| (dB) G is the number of dBs by T0 M which | T ( j 180 ) | can increase f x f-180º f (dec) until it becomes 0 dB GM 1 GM 20log |T j180 | <T(jω) 0 fx f-180º f (dec) Phase margin ϕm is the number T j -90º of degrees by which x ϕ can be reduced until it reaches -180º m –π (-180º) -270º m 180 T jx Conceptual Gain Margin GM or and Phase Margin ϕ m T jx m 180 * ωx is the crossover frequency 4 Gain Margin & Phase Margin At the crossover point, jm T jx 1T jx 1m 180 e The non-ideal closed loop transfer function becomes A jx A jx Aideal H CL jx 1 Ajx 1T jx 11/T jx A A ideal ideal jx 1 e 1 cosm j sinm 1 1 H CL jx AIdeal , AIdeal 2 1 cosm sin m 5 Gain Margin & Phase Margin Observe that different ϕm yield different errors. i.e. H j • In practical systems, ϕm CL x ϕm = 60º is required 90º 0.707 60º 1.00 • A worst case ϕm = 45º for a typical lower limit 45º 1.31 • For ϕ < 60º, we have 30º 1.93 m A jx Aideal 15º 3.83 indicating a peaked closed- ∞ (oscillatory 0º loop response. behavior) 6 Why a dominant pole is required for a stable amplifier? A good Op amp design implies: (i) A ( j ) 1 over as wide a band of frequencies as possible (ii) The zeroes of A ( j ) 1 0 must be all in the left-hand plane Note that A( j) 0 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, A A A(s) 0 0 1 2 3 1 s /1 1 s / 2 1 s /3 s 1 s 2 s 3 The characteristic equation becomes A A(s) 1 0 1 2 3 1 0 s 1 s 2 s 3 s 1 s 2 s 3 A01 23 0 7 Critical Value of βA0 3 2 s s 1 2 1 s1 2 13 23 1 23 1 A0 0 Note that βA0 is the critical parameter that determines the pole locations for a given α1, α2 and α3 (0≤ β ≤1). Furthermore, when βA0 = 0, the roots are at -α1, -α2 and -α3. Therefore, for small βA0, the roots should be in the left-hand plane (LHP). However, for βA0>>1, two of the roots might be forced to move to the right- hand plane (RHP). This can be verified by applying Routh’s stability criterion. Let us write the polynomial as 3 2 b3s b2s b1s b0 0 In order to have, in the above equation, left half plane roots, all the coefficients must be positive and satisfy b2b1 b3b0 0 8 Critical Value of βA0 The condition for imaginary-axis roots become 3 2 b3 j b2 j b1 j b0 0 2 2 b0 b2 j b1 b3 0 Now, for s=jω being a root, both real and imaginary parts must be zero. That is, 2 2 b0 b2 0, b1 b3 0 or b3b0 b1b2 Then the two roots are placed at b b j 0 j 1 p2,3 b2 b3 9 Critical Value of βA0 b2b1 b0 b3 Then 1 231 A0 C 1 2 3 1 2 13 23 Thus, the critical value of βA0 becomes 1 1 2 2 3 3 A0 C 2 2 3 1 3 1 2 Thus when βA0 becomes (βA0)C, the amplifier will oscillate at b1 OSC 1 2 13 23 b3 Also when βA0 > (βA0)C, the amplifier has RHP poles, therefore is unstable. 10 Critical Value of βA0 5 Let us consider some numerical examples. Let A0 = 10 , 7 (i) Three equal poles α1=α2=α3=10 rad/s. 7 The amplifier oscillates at OSC 1 3 10 3rad/s 5 A0 C 8, C 8/ A0 810 (ii) 2 3 , then the critical loop gains yields 1 104 104 4 2 4 210 A0 C 2 210 , thus the amplifier is stable if C 0.2 1 A0 Since R 1 / R 1 R 2 , 0 . 2 causes R 2 / R 1 4 , which means that for an inverting (non-inverting) configuration, the gain must be greater than -4 (5) to keep the amplifier stable. 11 Critical Value of βA0 (iii) Let us determine A0C under the most stringent condition β=1. Then from previous equation, 1 1 2 2 3 3 A0 2 2 3 1 3 1 2 In order to have a large A0, the poles must be widely separated. i.e. α1<<α2<<α3, then the A0 inequality can be approximated as 2 3 A0 1 1 To obtain a conservative A0, let α2=α3, which yields 2 A0 2 1 This inequality bounds the DC gain to provide a stable closed loop configuration. 12 Peaking and Ringing Peaking in the frequency domain usually implies ringing in the time domain Normalized second-order all-pole (low pass) system 2 1 H (s) 0 s j 2 s 2 0 s 2 1 0 1 2 j Q s j 0 Q 0 13 Peaking and Ringing GP: Peak gain – We have the error function, 1 1 E( j) 2 (1) 2 1 2 D 1 2 2 0 Q 0 To find out the maximum value of |E(jω)|, calculate the derivative of D(ω) and make it equal to zero 3 d 1 4 2 1 2 D() 2 2 0 1 OR 0 2 2 4 * 2Q2 0 d Q 0 0 For Q 1 / 2 , use ω* in Eq. (1) and we get the peak gain 2Q2 GP Ej0 1 (2) 4Q2 1 14 Peaking and Ringing OS: Overshoot – Inverse Laplace transform b Inverse s - domain : Laplace time - doamin : eat sinbt u t (s a)2 b2 For a 2nd order all pole error function, the impulse response is 2 2 b v (t) (t) H (s) 0 0 in s j 2 2 2 2 0 b s a b 0 s s 0 a Q 2Q 2 Inverse 1 Laplace h(t) 0 eat sin(bt)u(t) b 1 b 4Q2 0 Consider the normalized step response of this system, t 0 at 1 b y(t) h(t)dt 1 e sinbt , tan for vin (t) u(t) (t)dt (3) 0 b a 2 Use the damping factor to represent, a 0 ,b 1 0 15 Peaking and Ringing Usually, the damping factor ξ=1/2Q is used to characterize a physical 2nd order system. Thus, we rewrite the normalized time-domain equation (3) 0 at 1 0t 2 y(t) 1 e sinbt 1 e sin 1 0t b 1 2 (4) b 1 2 1 2 tan1 tan1 , for small a For under damped case ξ < 1 (Q > 0.5), 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 0. d y(t) e0t sin 1 2 t 2 0 d0t 1 0t 2 e cos 1 0t 16 Peaking and Ringing d y(t) 0 d0t 0t 2 0t 2 e sin 1 0t e cos 1 0t 1 2 2 2 1 tan 1 0t tan To satisfy the equality above, we have n 0t ,n 0,1,2,... 1 2 In other words, the normalized step response y(t) in Eq.(3) achieves extreme values at time steps of n=0, 1, 2, … 17 Peaking and Ringing n=0 n=1 n=2 n=3 n=4 n=5 n=6 Peak Q=5 Value Time t 18 Peaking and Ringing n 0, 0t 0, y(t) 0 1 1 2 1 2 n 1, 0t , y(t) 1 e sin 1 e 1 2 1 2 2 2 2 1 1 2 1 2 n 2, 0t , y(t) 1 e sin2 1 e 1 2 1 2 3 3 3 1 1 2 1 2 n 3, 0t , y(t) 1 e sin3 1 e 1 2 1 2 Therefore, the global peak value of y(t) is achieved when n=1. Thus, the overshoot is defined as Peak Value Final Value 2 OS(%) 100 100e 1 (5) Final Value 19 Peaking and Ringing ϕm : Phase Margin For a 2nd order all-pole error function T (s) 1 1 E(s) 1T (s) 11 T (s) s2 1 s 2 1 0 Q 0 Therefore, the loop gain is given by 1 1 The cross-over frequency thus can be obtained 1 2 2 2 T ( j ) x x 1 x 2 0 0Q 20 Peaking and Ringing Solve the equation and get the crossover frequency, 1/ 2 1 1 1/ 2 1 4 4 1 2 2 x 0 4 2 0 4Q 2Q And thus the phase margin is 1 1 180 T j cos1 1 cos1 4 4 1 2 2 m x 4 2 4Q 2Q Study the relationship between phase margin and gain peaking (Eq.2) or overshoot (Eq.5), we have GP(60) 0.3dB OS(60) 8.8% Q 0.82 GP(45) 2.4dB OS(45) 23% Q 1.18 21 Peaking and Ringing 30 30 100 90 24 24 80 70 18 18 60 50 Q Factor 12 12 40 Gain Peaking (dB) Overshoot (%) Overshoot 30 6 6 20 10 0 0 0 0 10 20 30 40 50 60 0 8 16 24 32 40 48 56 64 Phase Margin (degree) Phase Margin (Degrees) 22 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.
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