Frequency response and stability of feedback amplifiers () ()

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1 Frequency response and stability of feedback amplifiers The stability problem Source X p + X n X - Σ X f Load B (s): open loop transfer function B(s): feedback transfer function f () s + () s () () s B s We assume the amplifier is direct coupled with dc gain and with poles & zeros in the high frequency band. We assume that at low frequency B(s) reduces to a constant value. (s)b(s) becomes a constant, which should be positive otherwise the feedback would not be negative. For S f ( ) + ( ) ( ) ( ) B 7//3

2 The loop gain ( ) B( ) is a complex number that can be represented by its magnitude and phase: jφ L B B e ( ) ( ) ( ) ( ) ( ) ( ) The manner in which the loop gain values with frequency determines the stability of a system, here a feedback amplifier. Consider the frequency at which the phase angle becomes 8 o i.e. ( 8 ) 8. Now the loop gain will be a real negative number and ( ) B( ) the feedback will be positive. φ Let us now examine 3 cases: Case - Stable system: If, B ( 8 ) ( 8 ) < Then f >, the amplifier will still be stable. Case Oscillator: B, ( ) ( ) 8 8 The amplifier will have an output for zero input oscillator with frequency of oscillation equal 8. For the oscillating case a non-linearity may be introduces into the circuit to hold the amplitude of the oscillations at some predetermined and fixed amplitude. Case 3 Unstable System B ( 8 ) ( 8 ) > The system is unstable. Oscillations will occur at the circuit output with increased magnitude of the output voltage will increase rapidly and without oscillations. f 7//3

3 The Nyquist plot The Nyquist Plot is a formalized approach to testing the Stabilityof a system. It is simply a Polar plot of Loop Gain with frequency used as parameter. If the intersection of the Nyquist plot with the negative real axis occurs as shown to the left of the plot (-,), we know that the magnitude of the loop gain at 8 is ( ) B( ) > and the amplifier will be unstable. If the intersection occurs to the right of the point (-,) the amplifier will be stable. It follows that, if the Nyquist plot encircles the point (-,) then the amplifier will be unstable. 7//3 3

4 Example: Consider a feedback amplifier for which the open-loop transfer function (s) is given by: () s 4 + s / ssuming that B is a constant and is independent of frequency.. Find the frequency ( 8 ) at which the phase shift is 8 o.. Show that the amplifier will be stable if the feedback factor B<B cr and unstable of B Bcr. 3. Find a value for B cr. 3 Solution:. ( ) + / 4 4 Thus φ 3tan ( / ), 8 t 8 3 tan / 6 4 φ thus ( ) rad/s. The feedback amplifier will be stable if at 8, B <. t boundary BB cr, thus B cr ( ) 8 B cr 3. / ( H ( 3) ) 3 /.8 7//3 4

5 Effect of feedback in the amplifier poles mplifier frequency response and stability are determined by its poles: For an amplifier or any other system to be stable its poles should lie in the left half of the s plane. pair of complex conjugate poles on the j axis gives rise to sustained sinusoidal oscillations. Poles in right half side of the s plane will give rise to growing oscillations Consider an amplifier with a poles pair at s ± If this amplifier is subjected to a disturbance such as that caused by closure of the power-supply switch, its transient response will contain terms of the form: V () [ ] t t t t t t t e e e e cos( t) + + This is a sinusoidal signal with an envelope n n n e t If the poles are in the left half of the s-plane, then will be negative and the oscillations will decay exponentially to zero. If the poles are in the right half, then will be positive and the oscillations increase exponentially. If the poles are on the j axis, is zero and oscillations are sustained at a fixed amplitude.. n 7//3 5

6 S plane Time S plane (a) Time S plane (b) Time Relationship between pole location and transient response (c) f () s + () s () () B s s The poles of f (s) are the zeros of + () s B() s, thus the feedback amplifier poles are obtained by solving the characterised equation: + s B s () () Consider first an amplifier with B independent of frequency. The single-pole open-loop transfer function characterized by: () s + s / p p real negative. 7//3 6

7 The closed loop transfer function is given by ( ) () ( ) / + B f s + s / p + B Feedback moves the pole along the negative real axis ( + B) pf p This is shown below: Note that at low frequency and f differ by log ( + B ) while at high frequencies the two curves coincide. p For >> p ( + B ) f () s () s no feedback s db log(+ β) S plane p f pf p (+ β) p pf (log scale) (a) (b) Effect of feedback on (a) the pole location, and (b) the frequency response of an amplifier having a single-pole open-loop response We can see from the diagram above that applying negative feedback to an amplifier extends its bandwidth at the expense of gain. Since the pole of the closed-loop amplifier never enters the right half of the s-plane, the single pole amplifier is stable for any value of B. the amplifier is unconditionally stable. This is for such an amplifier never exceeds 9 o. (8 o is necessary for the feedback to become positive). 7//3 7

8 mplifier with a two pole response Consider an amplifier whose open-loop transfer function is characterised by real poles: () s ( + s / )( + ) p s / p the closed loop poles are obtained from + B() s s + s( p + p ) + ( + B ) p p thus the closed loop poles are given by: s ( p + p ) ± ( p + p ) 4( + B ) p p s the loop gain B is increased from zero the poles are brought closer together as shown below, until a value of loop gain B is reached at which the poles become coincident. If the loop gain is further increased, the poles become complex conjugate and have along a vertical line as shown below: The diagram become blow shows the locus of the poles for increasing loop gain this plot is called a root-locus diagram where root refers to root of the characteristic equation. From the diagram we see that this amplifier is unconditionally stable. The maximum phase shift of (s) is in this case 8 o (9 o /pole) where φ 8 is reached for. Thus there is no finite frequency at which the phase shift reaches 8 o. + p p - p - p S plane 7//3 8

9 The characteristic equation of a second order network can be written as s s + + Q - pole frequency Q pole Q factor The poles are complex id Q>.5. geometric interpretation for is the radial distance of the poles from the axis. Poles on the axis have Q. By comparing the p, p quadratic to s and the s equation above: ( + B ) ( ) o p p Q p + p S plane Q The normalised magnitude response V s frequency for a second order system is shown below: the response is maximally flat for Q.77 (poles at 45 o angles). 7//3 9

10 Exercise : n amplifier with a low frequency gain of and poles at 4 and 6 rad/s is incorporated in a negative feedback loop with feedback factor B. For what value of B of the poles of the closed loop amplifier coincide? What is the corresponding Q of the resulting second-order system? For what value of B is s maximally flat response achieved? What is the lowfrequency closed loop gain in the maximally flat case? Solution: From s ( p + p ) ± ( p + p ) 4( + B ) p p the poles coincide at the value of B, which makes ( ) ( + ) + p 4 p p p B 4 substituting o, p and p 4 6 ( + ) 4( + B ) B.45 and Q.5 maximally flat response is obtained when Q.77 ( + B ) p p.77 p + B.5 p in this case the low frequency gain is.96 V/V + B //3

11 Exercise: Consider the positive feedback circuit shown in Fig(a) below. Find the loop transmission L(s) and the characteristic equation. Sketch a rootlocus diagram for varying k, find k for a maximally flat response, and for circuit oscillation. K.586 Q.77 K3 Q K Q/3 45 s plane K Q.5 45 K.586 Q.77 K3 Q To obtain the loop transmission we short circuit the signal source and break loop at the amplifier input. We then apply a test voltage V t and find the returned voltage V r as indicated in Fig(b) above. The loop transmission is given by 7//3

12 Vr L( s) ( s) B( s) k kt( s) Vt Vr s(/ RC) T( s) Vt s + s(3/ RC) + (/ RC) k s( ) thus L( s) RC s + s(3/ RC) + (/ RC) The characteristic eqn. is: + L ( s) 3 k i.e. s + s + s RC RC RC 3 k s + s + RC RC Hence from s + s + Q, Q RC 3 k For k the poles have Q/3 and are (s)b(s) located on the negative real axis. s k increased the poles are brought closer together and eventually coincide (Q.5, k) Further increasing k results in the poles becoming complex and conjugate. The root locus is then a circle because the radial distance remains constant independent of k. The maximally flat response is obtained when Q.77 (with k.586). the poles are at 45 o angles. The poles cross the axis into the right half of the s plane at the valu of k that results in Q, that is at k3. For k 3 the circuit becomes unstable. This is a nd order system but in this case the amplifier has positive gain k, and a feedback network whose transfer feedback is frequency dependent T(s). Thus the feedback is positive and the circuit oscillates at the frequency for which the phase T(s) is zero. 7//3