Section I6: Stability of Feedback Amplifiers Weve seen that negative feedback, where a portion of the output signal is subtracted from the input signal, improves amplifier performance by reducing sensitivity to parameter variations. The subtraction is actually performed by feeding back the portion of the output signal so that it is 180o out of phase with the input signal. Although systems may be designed to perform such perfect subtraction, variations in phase shift outside of midrange frequencies results in less than perfect subtraction. In particular, an additional phase shift of 180o, resulting in a total phase shift of 360o (or, equivalently, 0o), changes negative feedback to positive feedback. If this happens, subtraction changes to addition the signal grows and the system may become unstable. Although a positive feedback system may be bounded, it no longer depends on just the input signal to provide an output indeed; it may not require an input at all. Amplifier stability depends only on the properties of the system and not the driving function. Therefore, if a system is unstable, any excitation (even thermal excitation) will cause the system to operate in an unstable manner. Conversely, if a system is stable, any bounded excitation causes a bounded response. It is therefore of utmost importance to make sure that a circuit is stable for all operating frequencies when designing amplifiers or amplification systems dont want to let the smoke out! Simply put, if the transient response of a system to an impulse input decays to a constant value as time approaches infinity, the system is stable. For example, if the response of a given system to an impulse input is

)sincos()( 21 tCtBeAeth tt ++= , (Equation 11.37) the system would be stable. The response illustrated in Equation 11.37 consists of two parts, a constant and a sinusoid, that are both multiplied by a decaying exponential. These exponential terms define a decay envelope that controls the overall system behavior. If, on the other hand, the response of the system to an impulse is

)sincos()( 211 tCtBeAethtt ++= ++ ,

(even if only one exponential was positive), we may be in for a world of hurt! Rather than decaying as in the function of Equation 11.37, the terms in the equation above are growing exponentially. Without some kind of boundary mechanism, this function would become completely unstable.

Im sure youve seen movies about this positive feedback phenomena bridges collapsing, buildings coming apart etc. well, in the world of

microelectronics, the least destructive thing you can hope for is some smoke and a bad smell!

Between the two extremes is marginally stable, or oscillatory, behavior. This occurs when the signal envelope is constant (i.e., the signal neither grows as for the unstable case nor decays as for the stable case). In terms of a sinusoidal signal (indicated in blue), illustrative figures for each case are given below.

When we discuss stability, it is convenient to plot the poles (and zeros) of the transfer function in the s-plane, as illustrated in the figure below (based on Figure 11.9 of your text). The s-plane is a two dimensional representation that allows us to plot the poles and/or zeros of a transfer function in terms of the real part (, the horizontal axis) and the imaginary part (j, on the

vertical axis). Poles or zeros may be purely real if they lie completely on the real axis or occur in complex conjugate pairs if they have any imaginary component. To create a complex conjugate pair, the sign of the imaginary portion is switched; i.e., if a pole (or zero) occurs at A+jB, the complex conjugate is at A-jB.

aplace transforms allow us to express time functions in term of the complex Lfrequency variable s and define a transfer function that describes the system response. The characteristic equation of a system is the denominator of the transfer function (given in Equation 11.3 for the generic closed loop feedback system) set equal to zero, or

0)()(1 =+ sHsG .

The roots of the characteristic equati system, determine

A system is stable if all roots of the characteristic equation life in the left

the right half plane (Quadrants I and IV), or

on, or poles of thethe nature of the system stability. There are three possibilities that define system stability:

half of the s-plane (Quadrants II and III). A system is unstable if o any of the roots lie in

o any multiple complex pairs of roots (double, triple, etc.) lie on the j

o iple real roots lie at the origin.

ts lies on the j axis, or in the left half-

A system is conditionally stable if all roots lie in the left half-plane only

he location of typical roots in the s-plane and the corresponding time

Root Location Form of Solution

axis, or any mult

A system is marginally stable if o any single pair of conjugate rooo a single root lies at the origin and all other roots lie

plane.

for some particular condition, set of conditions, or range of conditions, of the system parameters. These systems are stable when the system parameters are appropriate, but may become unstable with any variations.

Tfunctions are given in Table 11.1, reproduced below, where A and B are real constants:

Single roots along the negative real axis, e-1t+Be-2tsuch as 1 and 2, where 1 and 2 > 0 A

Single roots along the positive real axis, such as 1 and 2, where 1 and 2 >

Ae+1t+Be+2t

A single complex conjugate pair of roots at cost+Bsint j, on the j (imaginary) axis.

A

One pair of complex roots at -j in the -t(Acost+Bsint) left half plane ( > 0).

e

One pair of complex roots at -j in the +t(Acost+Bsint) right half plane ( < 0).

e

Double roots on the real axis in the left half plane -t(A+B) (i.e., assuming both roots at and > 0).

e

A single root at the origin. A A double root at the origin. A+Bt

s can be seen from the time responses in the table above, stable systems A

are those that have roots only in the left half-plane (the negative real axis) and are dominated by a decaying exponential. Roots that lie in the right half plane (the positive real axis) result in responses that increase with time and systems that are unsuitable for practical use. Single conjugate pairs of roots that lie completely on the imaginary axis, other than at the origin, result in an undamped sinusoidal (oscillatory) response. If all other roots lie in the left half-plane, except possibly a single root at the origin, this system is considered an oscillator. This is the limiting case (marginally stable) between stable and unstable systems. However, oscillators generally have some natural or introduced damping mechanism, since variations in device or

operational parameters may move poles on the imaginary axis into the right half-plane. System Stability and Frequency Response

s a designer, you have a burning desire to know whether you have any

o determine whether any roots of the

ve in the s-plane as illustrated by C1 in Figure 1.11a (below left) maps into a closed curve in the GH-plane as shown by C2

Asystem roots in the right half-plane, right? Well, it turns out that this is critical information, since any roots in the right half-plane indicate an unstable system and a design you dont want to be part of. It turns out that it doesnt have to be a mathematical torture ritual though amplitude and phase plots are often readily available from the device manufacturer and will provide all information needed. In addition, simulation software such as PSpice will generate frequency curves directly from the circuit. To extend the stability criteria from the s-plane to the frequency response curves, we need to use contour integration or the concept of mapping between planes. Tcharacteristic equation are in the right half-plane, we examine the contour (notice the direction of the path) as shown in Figure 11.10 and to the right. Since stability depends on whether any roots fall within the closed contour, R is made so large that all possible roots in the right half-plane are included (i.e., R). A plot of G(s)H(s) as s follows the contour of Figure 11.10 (where s=+j) yields the mapping of this contour located in the s-plane to a contou For example, a closed cur

r in the GH-plane.

1in Figure 11.11b (below right). For single-valued functions, a one-to-one correspondence exists between a point on C1 (the curve in the s-plane) and a point on C2 (the map in the GH-plane). If a point is moved along C1 in the direction of the arrow (clockwise), the mapped point moves along C2 in a direction that depends on the GH function.

OK, here comes the fun stuff

Let sj indicate a root of the characteristic equation, 1+G(s)H(s)=0. Also, let the curve C1 pass through the point sj in the s-plane. This means that, for s=sj, 1+G(sj)H(sj)=0, or G(sj)H(sj)=-1. So, the point sj in the s-plane maps to the point -1 in the GH-plane and the curve C2 passes through GH=-1.

Ready for more? The characteristic function for a closed-loop system may be written in factored form as shown below:

)())(()())((

)()(1 211pba

r

ssssssssssss

KsHsG ++++++=+ LL

, (Equation 11.38) where s1, -s2,, -sr are the zeros and sa, -sb, , -sp are the poles. Note: The representations in the figures that follow are not intended to indicate all poles and zeros. Remember that real poles/zeros may occur singly, but complex poles/zeros must occur in conjugate pairs. Also, both poles and zeros are shown by points, where each pole should be indicated by the symbol x and each zero by the symbol o. These figures are for discussion and illustration only. Once a value is assumed for s in Equation 11.38, each factor is a complex number. Vectors may be drawn from each of the fixed points (s1, s2, sa, sb, etc.) to a variable point in the s-plane, as shown in the figure to the

right. The vectors extend from the fixed points (shown as s1, s2, and s3 for the zeros and sa, sb, and sc for the poles in the figure above) to a variable point (s in the figure above). If we now allow the variable point s to move in a clockwise direction about a fixed point, illustrated by using the fixed point s2 (a zero of the function G(s)H(s)) in Figure 11.12 (shown to the right), the curve denoted C3 is generated after a complete revolution. The vector s+s2 will make one complete clockwise revolution because the contour encircles the zero s2. Since all other zeros and poles are external to the contour, none of the remaining vectors make a complete revolution (remember that, as s is moving about s2, the vectors from the other poles and zeros are also changing). The factor s+s2 in Equation 11.38 changes phase by 360o during the complete revolution of s around s2 and 1+G(s)H(s) also experiences a change in phase of 360o. Therefore, a vector representing 1+G(s)H(s) in the [1+GH]-plane (dont worry, we wont keep this plane long and we are going somewhere with this), would make one clockwise encirclement of the origin as shown in Figure 11.13a. The remaining poles and zeros contribute no net change to the phase of 1+G(s)H(s). Because the root we are looking at is a zero (i.e., makes the numerator of Equation 11.38 equal to zero), one clockwise rotation about s2 in the s-plane results in one clockwise encirclement of the origin in the [1+GH]-plane, as shown in the figure above. This behavior in the [1+GH]-plane is related to the GH plane as shown in Figure 11.13b and to the right. Specifically, one clockwise encirclement of the origin in the [1+GH]-plane is equivalent to one clockwise encirclement of the 1 (or 1+j0) point in the GH plane.

Well!

Now, lets suppose the closed contour in the s-plane is made to encircle both a zero (s2) and a pole (sa) in the clockwise direction as shown in Figure 11.14 and to the right. In this case, both the vector from s2 to s and that from sa to s will rotate through one complete clockwise revolution. The factor (s+s2) that corresponds to the zero s2 will contribute +360o to the change in phase of 1+G(s)H(s) since it is in the numerator of Equation 11.38. The factor (s+sa) that corresponds to the pole sa will contribute 360o to the change in phase since it is in the denominator. Therefore, the net change in phase of 1+G(s)H(s) is zero, and the resulting map to the GH plane does not encircle the 1 point, as illustrated in Figure 11.15 and to the right. We can keep adding in poles and zeros and keep track of the net phase of the characteristic equation. For example, if a closed contour in the s-plane is enlarged to include s1, s2, and sa, the net change in phase of 1+G(s)H(s) is +360o. A clockwise encirclement of a zero causes a clockwise encirclement of the 1 point in the GH plane. A clockwise encirclement of a pole causes a counterclockwise encirclement of the 1 point in the GH plane (because of the negative angle introduced by a pole). To summarize: Define

o nN as the number of clockwise encirclements of the 1 point o nR as the number of zeros located within the contour in the s-plane o nP as the number of poles located within the contour in the s-plane

A clockwise encirclement of a region in he s-plane results in nN=nR-nP clockwise encirclements of the 1 point in the GH plane.

o nN is positive (nR > nP). In this case, the 1 point is encircled in the same direction (clockwise) as the contour in the s-plane.

o nN is zero if nR=nP, which means that the 1 point is not encircled. o nN is negative if nR < nP and the 1 point is encircled in the direction

opposite to the contour in the s-plane; i.e., counterclockwise instead of clockwise.

To find out what the heck this actually meansread on

Bode Plots and System Stability As mentioned earlier, a closed loop system is unstable if the poles of the characteristic equation lie in the right half of the s-plane. If we expand the previous discussion and make the contour of Figure 11.10 large enough to include the entire right half of the s-plane and map the results to the GH-plane, information regarding the stability of the closed-loop system may be obtained. Specifically, the number of clockwise encirclements of the 1 point in the GH-plane provides this information, and the stability criterion may be stated as follows: If the loop gain function, G(s)H(s), is expressed as the ratio of two factored polynomials in the variable s and written in the form

)1()1)(1(

)1()1)(1()()( 21

pban

r

ssss

sssKsHsG

+++

+++=LL

, (Equation 11.39, Modified) as s travels a closed contour comprising the imaginary axis from -j to +j and then the right-hand semicircle from s=Rej/2 to s=Re-j/2 as R approaches infinity (i.e., the entire right half-plane), the polar plot of G(s)H(s) encircles the 1+j0 point in a clockwise direction nN times. nN is given by

PRN nnn = , (Equation 11.40) where nN is the number of clockwise encirclements of the 1 point (a negative nN corresponds to counterclockwise encirclements), nP is the number of poles of G(s)H(s) in the right half plane, and nR is the number of zeros which lie in the right half-plane. What this boils down to is that the 1 point is the key. We can relate this to the Bode plot format by noting that

)01(1)sin(cos111 jje j === . (Equation 11.42) This means that the 1 point occurs when the gain is 0dB at 180o phase shift. We can therefore determine system stability as follows: Examine the corrected Bode plot; i.e., after corrections have been made to the straight-line asymptotic plot. By inspection, the frequency at which the phase shift crosses 180o is going to determine system stability. If the dB gain at this frequency is Less than 0 dB, the system in stable. Equal to 0 dB, the system is marginally stable; i.e., we are operating

on the j axis and the behavior is oscillatory. Greater than 0 dB, the system is unstable.