Microelectronic Circuits II. Ch 9 : Feedback

Transcription

1 Microelectronic Circuits II Ch 9 : Feedback 9.9 Determining the Loop Gain 9.0 The Stability problem 9. Effect on Feedback on the Amplifier Poles 9.2 Stability study using Bode plots 9.3 Frequency Compensation CNU EE 9.2-

2 Determining the Loop Gain An alternative approach for finding Loop Gain, Aβ. Let external source x s to be zero. 2. Open the feedback loop by breaking the connection of x o to the feedback network 3. Apply test signal x t and then, the signal at the output of the feedback network is x f =βx t 4. The signal at the input of the basic amplifier is x i =-βx t 5. The signal at the output of the amplifier, where the loop was broken, is x o =-Aβx t 6. Loop gain Aβ = negative of the ratio of the returned signal to the applied test signal; Aβ=-x o /x t CNU EE 9.2-2

3 Determining the Loop Gain Condition in breaking the feedback of a practical amplifier circuit - The condition that Existed prior to breaking the loop does not change à terminate the loop where it is opened with an impedance equal to that seen before the loop was broken - Break the loop at XX - Apply V t to the left of XX - Terminals at the right of XX should be loaded with Z t - Z t : impedance previously seen looking to the left of XX - Loop gain Ab : Ab = - V V r t - In some cases, it may be convenient to determine Ab by applying a test circuit I t and finding the returned current signal I r à Ab = I r / I t CNU EE 9.2-3

4 Determining the Loop Gain An alternative equivalent method for determining Aβ ( Rosenstark, 986). The loop is broken at a convenient point 2. Determine the open-circuit transfer function T oc (c) 3. Determine the short-circuit transfer function T sc (d) 4. T oc and T sc are combined and then loop gain Ab : Ab = - æ ç + è T oc T sc This method is useful when it is not easy to determine the termination impedance Z t ö ø CNU EE 9.2-4

5 Determining the Loop Gain Example of determining the loop gain - Place to break the loop = input terminals of the op amp - V t is applied to the right-hand-side terminals - R id terminates the left-hand-side terminals Returned voltage V r V r = -mv { RL //[ R2 + R ( Rid + R) ]} { R //[ R + R ( R + R) ]} + L 2 id r o [ R ( Rid + R) ] Rid [ R ( R + R) ] + R R + R id CNU EE id V L = Ab = - V V r r = - t V Loop Transmission

6 Determining the Loop Gain Equivalence of Circuits from a Feedback-Loop Point of View - The Poles of a circuit are independent of the external excitation - The poles, or the natural modes, can be determined by setting the external excitation to zero - The poles of a feedback amplifier depend only on the feedback loop à the characteristic equation (whose roots are poles) is completely determined by the loop gain à a given feedback loop may be used to generate a number of circuits having the same poles but different transmission zeros à The closed loop-gain & the transmission zeros depend on how & where the input signal is injected into the loop - Noninverting op-amp circuit & inverting op-amp circuit have the same feedback loop - Two or more circuits are equivalent from a feedback-loop point of view à stability is a function of the loop à perform the stability analysis only once for a given loop à concept of loop equivalence in the synthesis of active filters CNU EE 9.2-6

7 Transfer function of the feedback amplifier Closed loop transfer function : What happens at higher frequencies? For s=jω, (at HF) Stability Problem A(s) : open-loop transfer function direct-coupled with constant dc gain A 0, pole & zeros occurring in the high-frequency band b(s) : feedback transfer function positive constant at low-frequencies à Loop gain A(s) b(s) is positive at low freq. Loop gain : Loop gain varies with frequency that determines the stability or instability of the feedback amplifier At the frequency w 80 at which the phase angle φ(w) = 80 o - Loop gain A(jw)b(jw) will be a real number with negative sign à the feedback will become positive. - If the magnitude of loop gain at ω=80 is less than unity, à A f is larger than A but the loop is still stable. - If the magnitude of loop gain at ω=80 is equal to unity, à A f becomes infinite and the amplifier will have an output for zero input è Oscillator à external input x s =0, x i generated by any disturbances à X i sin(w 80 t) at w= w 80 ( w ) b ( w ) X = A j j X = - X f i i à X f is multiplied by - in the summer block à X i is sustained - If the magnitude of loop gain at ω=80 is greater than unity, the amplifier will also oscillate and the oscillations will grow in amplitude until some nonlinearity reduces the magnitude of the loop gain to exactly unity CNU EE 9.2-7

8 Nyquist Plot - polar plot of loop gain with frequency used as a parameter, testing for stability - radial distance= Ab, angle=phase angle f - The Nyquist plot intersects the negative real axis at the frequency w 80 - If this intersection occurs to the left of the point (-,0), the magnitude of loop gain at w 80 >> & the amplifier will be unstable - If this intersection occurs to the right of the point (-,0), the amplifier will be stable Nyquist criterion - If the Nyquist plot encircles the point (-, 0), then the amplifier will be unstable. Stability Problem CNU EE 9.2-8

9 Effect of Feedback on the Amplifier Poles u Stability and pole location - Amplifier frequency response and stability are determined directly by its poles - An Amplifier w/ a pole pair at s=s o +/- jw o à transient response s 0 ( ) t + j wn t - j wn t s 0 t = e ée + e ù = 2e t cos( w t ) ë Sinusoidal signal with an envelop : - s o is negative : poles should be in the left half of the s plane à oscillations decay exponentially toward zero à Stable system - s o is positive : poles in the right half of the s plane à oscillation grow exponentially (until some nonlinearity limits their growth) - s o is zero : poles on the jw axis à Sustained sinusoidal oscillation - The existence of any right-half-plane poles results in instability û n e s ot CNU EE 9.2-9

10 Effect of Feedback on the Amplifier Poles Poles of the feedback amplifier - the poles of the feedback amplifier are zeros of the + A(s)b(s) - The feedback amplifier poles are obtained by solving ( ) b ( s) + A s = 0 à Characteristic equation of the feedback loop à Applying feedback to an amplifier changes the location of the poles How feedback affects the amplifier poles? - Assumption. open-loop amplifier has real poles and no finite zeros (i.e, all the zeros are at s=infinite) 2. feedback factor b is independent of frequency CNU EE 9.2-0

11 Effect of Feedback on the Amplifier Poles Amplifier with a single-pole response - open-loop transfer function with a single pole - closed-loop transfer function - feedback moves the pole along the negative real axis to a frequency w pf - at low frequency, difference between A and A f is 20log(+A o b) - for frequencies w >> w p (+A o b) : A Aow p ( s)» A( s) s f» - Extension of bandwidth at the expense of a reduction in gain - single-pole amplifier is stable for any value of b à unconditionally stable ( ) w w b Pf = P + A0 Phase lag associated with a single-pole response can never be greater than 90 o àloop gain never achieves the 80 o phase shift required for the feedback to become positive CNU EE 9.2-

12 Effect of Feedback on the Amplifier Poles Amplifier with two-pole response Closed-loop poles from +A(s)β =0 ( ) ( ) 2 s s wp wp 2 A0 b wpw P = 0 - root-locus diagram shows the locus of the poles for increasing loop gain à poles = roots of the characteristic equation - Unconditionally stable : maximum phase shift is 80 o (90 o per pole) at w=infinite à there is no finite frequency at which the phase shift reaches 80 o As A 0 β is increased from zero, two poles get closer, then coincident and become complex conjugate moving along a vertical line. CNU EE 9.2-2

13 Effect of Feedback on the Amplifier Poles Amplifier with two-pole response - The characteristic equation of a second-order network : à w o : pole frequency, Q : pole Q factor à w o : radial distance of the poles from the origin à Q : distance of the poles from the jw axis, poles on jw when Q = inf. à if Q > 0.5 (z < ), poles are complex ( + A0 b ) w pw p2 - Q factor for the poles of the feedback amplifier Q = w + w ( ) p p2 - No peaking for Q < Q=0.707 (poles at 45 o angle) à maximally flat response CNU EE s 2 z = + 2zw s + w 2Q 0 0 s = zw = w0 2Q 2 0 s = -s ± jw w = w d d 0 -z 2

14 Effect of Feedback on the Amplifier Poles Amplifier with three or more poles Root-locus diagram for an amplifier with three poles. - Increasing the loop gain from zero moves the highest-frequency pole outward while the two order poles are brought closer together - As A 0 β is increased further, two poles become coincident and then become complex and conjugate - Pair of complex-conjugate poles enters the right-half of the s plane, thus causing the amplifier to become unstable -An amplifier w/ three poles has a phase shift that reaches -270 o as w approaches infinity à there exists a finite frequency, w 80, at which the loop gain has 80 o phase shift For a given A 0, there exists maximum value of β above which the amplifier becomes unstable. There exists a minimum value for the closed-loop gain A f0 below which the amplifier become unstable To obtain lower values of closed-loop gain à alter the loop transfer function L(s) : frequency compensation CNU EE 9.2-4

15 Gain and Phase margins Stability study using Bode Plots Bode plot for Ab - Magnitude of loop gain at w 80 < unity (negative db) à stable - Gain margin : difference between the value of Ab at w 80 and unity [decibel] à the amount by which the loop gain can be increased while stability is maintained - Phase angle at the frequency for which Ab = < 80 o à stable - Phase margin : difference between the phase angle at the frequency for Ab = and 80 o CNU EE 9.2-5

16 Effect of Phase Margin on Closed-Loop Response - Feedback amplifier à a phase margin of at least 45 o - The amount of phase margin has a profound effect on the shape of the closed-loop gain response Assumption : a large low-frequency loop gain A o b >> à the closed-loop gain at low frequencies ~ /b - Denoting the frequency at which the magnitude of loop gain is unity by w - At w, the closed-loop gain : - Magnitude of the gain at w : Stability study using Bode Plot A ( jw ) b = e A( - jq jw ) - For a phase margin of 45 o, q = 35 o : where q = 80 o - phase margin ( b ) - jq Af ( jw) = = - jq + A( jw) b + e A f ( jw) A f e b = + e - The gain peaks by a factor of.3 above the low-frequency value of /b - This peaking increases as the phase margin is reduced, eventually reaching when the phase margin is zero. Zero phase margin implies that the amplifier can sustain oscillation [poles on the jw axis; Nyquist plot passing through (-, 0)] CNU EE jq ( jw ) =.3 b

17 Stability study using Bode Plot Alternative approach for investigating stability - Stability investigation by constructing Bode plots for the gain loop Ab à tedious & time-consuming process - Simpler approach. Bode plot for the open-loop gain A(jw) 2. plot 20 log(/b) as a horizontal straight line on the same plane used for 20 log A 3. difference between the two curves à loop gain (db) (assume b = constant) 4. study stability by examining the difference between the two plots - Open-loop transfer function with three poles which are widely separated à 0. MHz, MHz and 0 MHz à Phase : -45 o at the st pole frequency, -35 o at the 2 nd, and -225 o at the 3 rd pole frequency à The frequency at which the phase of A(jw) is -80 o lies on the -40dB/decade segment A = ( + jf / 0 ) ( + jf / 0 ) ( + jf / 0 ) 5 ( ) ( ) ( ) Open-loop gain f = - étan f / 0 + tan f / 0 + tan f / 0 7 ù Phase ë û CNU EE 9.2-7

18 Stability study using Bode Plot Alternative approach for investigating stability - f 80 at which phase angle is 80 o = 3.2x0 6 Hz - line (a) : closed-loop gain of 83.6 db 20 log(/b)=85db à b=5.623x0-5 à Loop gain = difference between A curve & /b line à the point of intersection X = freq. at Ab = à 5.6x0 5 Hz, -08 o à phase margin of 72 o, gain margin of 25 db - Line (b) : closed-loop gain of 50 db à A 0 = 00dB, 20 log(a 0 b) ~ 50dB à 20 log(/b) ~ 50dB à the point of intersection X 2 = phase > 80 o à unstable - Minimum value of 20 log(/b) = 60 db - The closed-loop amplifier will be stable if the 20 log(/b) line intersects the 20 log A curve at a point on the -20-dB/decade segment à phase margin of at least 45 o - At the intersection of 20 log(/ b(jw) ) and 20 log A(jw) the difference of slopes (called the rate of closure) should not exceed 20-dB/decade CNU EE 9.2-8

19 Frequency compensation Theory The simplest method of frequency compensation : introduce a new pole in the function A(s) at a sufficiently low frequency, f D à modified open-loop gain A (s) intersects the 20 log(/ b ) curve with a slope difference of 20 db/decade Compensation of A(s) such that b=0-2 will be stable - Locate point Y at the freq. of st pole, f P - Draw a line w/ -20-dB/decade slope à intersect the dc gain line at point Y à New pole at f D à compensated openloop response A (s) w/ 4 poles at f D, f P, f P2, f P3 à 20 log(/b) line intersects 20 log A curve at point Y on the -20-dB/decade segment à closed-loop amplifier w/ the b is stable but the loop gain is drastically reduced at most freq. Elimination of f P by shifting from f= f P to f= f D - Start from point Z (at the freq. of 2 nd pole) & draw the line ZZ à open-loop curve A (s) = higher gain than A (s) - Make the pole f D dominant & eliminate the need for introducing an additional lowfrequency pole CNU EE 9.2-9

20 Frequency compensation Implementation - Each stage of multistage amplifier is responsible for one or more of the transfer-function pole - First pole f P is introduced at the interface between the two cascaded differential stages (a) Two cascaded gain stages of a multistage amplifier (b) Equivalent circuit for the interface between the two stages - I x : output signal current of Q Q 2 stage - R x & C x : total resistance & capacitance between B and B - pole f P : f P = 2pC R (c) Compensating capacitor C C between B & B - desired pole f D : f - C C shifts the pole from f P to f D determined by Z - C C changes location of the other poles - Quite large value of C C à feedback path of the amplifier à Miller effect à much larger effective capacitor CNU EE x x ' D = 2p ( C x + CC ) Rx

21 Frequency compensation Miller compensation and Pole splitting - Common-emitter amplifier w/ compensating capacitor C f in the feedback path - R, C : Total R & C between B & ground; R 2, C 2 : Total R & C between C & ground - C includes Miller component due to C m ; I i : output signal current of the preceding stage - In the absence of C f : f P 2pC R = f P 2 = 2pC R 2 2 CNU EE 9.2-2

22 Miller compensation and Pole splitting - With C f present : - Zero is at a much higher frequency than the dominant pole à neglected - Dominator polynomial D(s) : w P, w P2 : new 2 poles Frequency compensation ( f - m ) V sc g R R o 2 = 2 I + s é ëc R + C R + C ( g R R + R + R ) ù û + s é ëc C + C ( C + C ) ù û R R i 2 2 f m 2 2 f 2 2 æ s D( s) = ç + è w 2 s s - If w P is dominant; w P << w P2 D( s)» + / / à/ w + w w / P öæ ç s + øè w P / P2 ö æ = + sç ø è w P P2 / P + w / P2 ö 2 s + / ø wpw / P2 ' w P = C R + C R + C g R R + R + R ( ) 2 2 f m Approximated two new poles : / w P» g m R C 2 f R w g C / m f P 2» CC 2 + C f C ( C + 2 ) - C f à w P & w P2 == pole splitting - Increase in w P2 à move point Z further to the right à higher compensated open-loop gain - w P : C f is multiplied by the Miller-effect factor g m R 2 à much larger capacitance g m R 2 C f == required value of C f is much smaller than that of C C CNU EE