Feedback: 8.8 The Stability Problem

1

Feedback: 8.8 The Stability Problem

Microelectronic Circuits - Fifth Edition Sedra/Smith 2Copyright 2004 by Oxford University Press, Inc.

Figure 8.1 General structure of the feedback amplifier. This is a signal-flow diagram, and the quantities x represent either voltage or current signals.

1f

A sA s

A s s

1f

A jA j

A j j

8.8.1 Transfer Function of the Feedback Amplifier


jL j A j j A j j e

Suppose that there is a frequency, , for which 180 180

180 180 1j je e 180

1f

A jA j

A j j

Then


Loop Gain:


1f

A jA j

A j j

If , then 1A j j fA j A j

(Armstrong’s “regeneration”)

If , then 1A j j fA j (Armstrong’s “oscillation”)



8.9.1 Stability and Pole Location

A quick review


Figure 8.29 Relationship between pole location and transient response.

2 coso n n ot j t j t tnv t e e e e t


8.9.2 Poles of the Feedback Amplifier

Recall the closed loop transfer function (transfer function with feedback):

1f

A sA s

A s s

Poles:

1 0A s s


8.9.3 Amplifier with a Single-Pole Response

Simple example: suppose an amplifier without feedback has a single pole at :

0

1 P

AA s

s

With negative feedback, we saw earlier that, assuming constant, the amplifier gain is:

0 0

0

1

1 1fP

A AA s

s A

Ps


0 0

0

1

1 1fP

A AA s

s A

The feedback has moved the pole along the axis from to .P 01P A



Let . When , 01P A

0 0 0 0

0 0

1 1

1 1 1fP P

A A A AA j

j A j A

s j

0 Pf

AA j

j

0 0 0

1P

P P

A A AA j

j j j



Summary: when , 01P A

0 Pf

AA j A j

j



Figure 8.30 Effect of feedback on (a) the pole location and (b) the frequency response of an amplifier having a single-pole open-loop response.



Figure 8.36 Bode plot for the loop gain A illustrating the definitions of the gain and phase margins.

1; 0dB

A A

180 Instability:

1020logdB

A A j j

AND

8.10.1 Gain and Phase Margins


8.10.2 Effect of Phase Margin on Closed-Loop Response

Typical phase margin design value:45

Value of phase margin affects shape of closed- loop gain versus frequency plot.



Example: suppose0;

0 1A

0 0 0

0 0

1

1 1f

A j A AA j

A j A A

00;A j A real

Closed loop gain at low frequencies:



Suppose

1 1A j for some frequency 1 .

1 1

1 1

j j jA j A j e e e

where 0

1

1 jA j e

. Thus, we can write



phase margin = 180

phase margin = 180

180 phase margin



Closed loop gain at frequency :1

1

1 1

1

1;

1j

f

A jA j A j e

A j

1

11

1 11

jj

f jj

ee

A jee



Magnitude of the closed loop gain at frequency :1

1

1 1 1 1

1 1 1

jj

f j j j

eeA j

e e e

For a phase margin of 45

180 phase margin = 180 45 135



Magnitude of the closed loop gain at frequency :1

1 135

1 1 1 1 1 1

0.76541 1f j j

A je e

1

1.31fA j

Thus the gain peaks at 131% of its value at low frequencies with a phase margin of 45.


Af (omega1) = factor / beta

0

2

4

6

8

10

12

14

0 50 100 150 200 250 300 350 400

phase margin, degrees



Construct a Bode plot for rather than for .AA

10 1020log 20logdB

A A A

10 1020log 20logdB

A A

10 10

120log 20log

dBA A

8.10.3 An Alternative Approach for Investigating Stability (Bode)


Example:

5

5 6 7

10

1 10 1 10 1 10A

j f j f j f

1 5 1 6 1 7tan 10 tan 10 tan 10f f f

8.10.3 An Alternative Approach for Investigating Stability (Bode)


Figure 8.37 Stability analysis using Bode plot of |A|.

8.10.3 An Alternative Approach for Investigating Stability (Bode)8.10.3 An Alternative Approach for Investigating Stability (Bode)

50

60

85

1020log 1f dBA

55.62 10

31.00 10

33.15 10


8.10.3 An Alternative Approach for Investigating Stability (Bode)• Note that instability occurs for smaller

values of .

• Result that we will not prove:– If the horizontal line intersects

the curve on a segment for which the slope is –20 dB/decade, then the phase margin will be a minimum of 45º.

fA

1020log 1 1020log A


8.11 Frequency Compensation

• Frequency compensation modifies the open-loop transfer function of an amplifier (with three or more poles) so that the closed-loop transfer function is stable for any value chosen for the closed-loop gain.

A s

fA s


8.11 Frequency Compensation: Theory• The simplest method of frequency

compensation modifies the original open-loop transfer function, , by introducing a new low frequency pole at to form a new open-loop transfer function,

, which has a slope of –20 dB/decade at the intersection of the curve and the curve. 1020log 1

1020 log 'A j

A j

'A j

D


Figure 8.38 Frequency compensation for = 102. The response labeled A is obtained by introducing an additional pole at fD.

8.11 Frequency Compensation: Example:

Stability for closed-loop gains of 40 dB or higher by adding a new pole at .D


Figure 8.38 Frequency compensation for = 102. The response labeled A is obtained by introducing an additional pole at fD.

8.11 Frequency Compensation: Difficulty:

Stability achieved, but high open-loop gain, and hence the benefits of negative feedback, only for: 210f Hz


8.11.3 Miller Compensation and Pole Splitting

The Miller effect allows shiftingto and shifting to a higher frequency. Stability and high open-loop gain for

1Pf

'Df2Pf

310f HzFigure 8.38 Frequency compensation for = 102. The response labeled A is obtained by introducing an additional pole at fD. The A response is obtained by moving the original low-frequency pole to f D.


Figure 8.40 (a) A gain stage in a multistage amplifier with a compensating capacitor connected in the feedback path and (b) an equivalent circuit. Note that although a BJT is shown, the analysis applies equally well to the MOSFET case.




Node equations:

11

0Bi B f B C

VI s C V s C V V

R

22

0Cf C B m B C

Vs C V V g V s C V

R

B:

C:

Transresistance amplifier:



Collect terms:

11

0Bi B f B C

VI s C V s C V V

R

22

0Cf C B m B C

Vs C V V g V s C V

R

B:

C:

11

1B f C f iV sC sC V sC I

R

22

10B f m C fV s C g V s C s C

R



MATLAB: %Declare symbolic variables

syms s R1 R2 C1 C2 Cf gm Ii Vb Vc Vo A b x

%Define equations: Ax = b

A=[1/R1+s*(C1+Cf) -s*Cf; gm-s*Cf 1/R2+s*(Cf+C2)];

b=[Ii; 0];

%Solve the equations:

x=A\b;

Vo=x(2);

Vo=simplify(Vo)



MATLAB:Vo = R1*Ii*R2*(-gm+s*Cf)/

(s*R1*Cf*R2*gm+1+s*R1*C1+s*R1*Cf+s*R2*Cf+s^2*R2*Cf*R1*C1+s*R2*C2+s^2*R2*C2*R1*C1+s^2*R2*C2*R1*Cf)

1 221 1 1 2 2 1 2 1 2 1 2 1 2 1 2

sC g R RV f moI s C R C R C g R R R R s C C C C C R Ri f m f

21 1 1 2 2 1 2 1 2 1 2 1 2 1 2D s s C R C R C g R R R R s C C C C C R Rf m f




(3 capacitors, 2 poles? Miller’s Theorem.)

21 1

1 1 1' ' ' ' ' '

1 2 1 2 1 2

s s sD s s

P P P P P P

Suppose one pole is dominant: ' '1 2P P

2

1' ' '

1 1 2

s sD s

P P P




By comparison:

2

1' ' '

1 1 2

s sD s

P P P

1'

11 1 2 2 1 2 1 2

P C R C R C g R R R Rf m

'ln ln1 1 1 2 2 1 2 1 2C R C R C g R R R RP f m

Prepare for logarithmic differentiation:



'ln ln1 1 1 2 2 1 2 1 2C R C R C g R R R RP f m

Logarithmic differentiation:

'

1 1 2 1 21 0'

1 1 2 2 1 2 1 21

g R R R Rd mPdC C R C R C g R R R Rf f mP

Thus, , so the Miller capacitor lowers the frequency of the lower frequency pole.

'1

0d PdC f


8.11.3 Miller Compensation and Pole SplittingWhat about ?

By comparison:

'2P


2

1' ' '

1 1 2

s sD s

P P P

11 2 1 2 1 2' '

1 2

C C C C C R RfP P

1'

2 '1 1 2 1 2 1 2

PC C C C C R RP f



Prepare for logarithmic differentiation:

1'2 '

1 1 2 1 2 1 2P C C C C C R RP f

' 'ln ln ln ln2 1 1 2 1 2 1 2C C C C C R RP P f

' '

1 12 1 1 2' '

1 2 1 22 1

d d C CP PdC dC C C C C Cf f fP P

'

1 1 2 1 22 1 2'

1 2 1 21 1 2 2 1 2 1 22

g R R R Rd C CmPdC C C C C CC R C R C g R R R Rf ff mP



'

1 1 2 1 22 1 2'

1 2 1 21 1 2 2 1 2 1 22

g R R R Rd C CmPdC C C C C CC R C R C g R R R Rf ff mP

'1 2 1 2 1 22

'1 2 1 21 1 2 2 1 2 1 22

C C g R R R R C C Cdf f m fPdC C C C C CC R C R C g R R R Rf ff mP

Note:

1 2 1 21

1 1 2 2 1 2 1 2

C g R R R Rf m

C R C R C g R R R Rf m

11 21

1 1 21 2 1 2 11 2

C C CfC CC C C C Cf

C C Cf



'1 2 1 2 1 22

'1 2 1 21 1 2 2 1 2 1 22

C C g R R R R C C Cdf f m fPdC C C C C CC R C R C g R R R Rf ff mP

For the Miller effect to be large, :2g Rm

'1 22 1 0

'1 2 1 22

C C C Cdf fPdC C C C C Cf fP

'2 0

d PdC f



The Miller effect can split the two poles, pushing the frequency of the lower one lower and pushing the higher frequency higher.

Shifting the lower frequency pole gives stability without the addition of an extra pole.


8.11.3 Miller Compensation and Pole SplittingShifting the higher frequency pole shifts the – 20 dB/decade segment to the right and thus gives broader bandwidth.

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