Feedback Basics - University of Pennsylvania

ESE319 Introduction to Microelectronics

12008 Kenneth R. Laker updated 15Nov09 KRL

Feedback Basics● Stability● Feedback concept● Feedback in emitter follower● One-pole feedback and root locus● Frequency dependent feedback and root locus● Gain and phase margins● Conditions for closed loop stability● Frequency compensation



F

F

F

F

F

Operational Amplifier From Another View

ΓF

vf

vovi A

AA

A

A

A A

AA



Feedback Block Diagram Point of View

A

β

-+vi vo

vo=A

1AFv i

Fundamental feedback equation

Fundamental block diagram We can split the idealop amp and resistivedivider into 2 separateblocks because theydo not interact with, or“load,” each other.

1. The op amp output isan ideal voltage source.

2. The op amp input draws no current.

F

vf



Comments on the Feedback Equation

Acl=vo

v i= A1AF

The quantity loop-gain and A

cl = closed-loop gain.

AF≫1(and the system is stable – a topic to be discussed later!) the closed-loop gain approximates and is independent of “A” !

AF

1/F

The quantity A = open-loop gain and = feedback gain.

If the loop gain is much greater than one, i.e.

F

A=vo

vi

A

β

-+vi vo

F

vf

F=v f

vo

Av−cl=vo

vi≈ 1F

|v f =0



Feedback in the Emitter FollowerEmitter current equation:

ie=1

Rs

1r eRE

v i

Create an “artificial” feedback equation, multiply numerator & denominator by :

vo=

RE 1Rs

1RE 1

R s

reRE RE

v i=A

1AFvi

1/Rsib

ie

ib

vo=RE ie=RE

R s

1reRE

vi



Emitter Follower Emitter CurrentThe forward gain open-loop term, A:

A=RE 1

Rs

The feedback term, :

F=reRE

RE

If the “loop gain” is large,

vo≈1F

vi=RE

r eREv i

F

AF≫1

ib i

b

ie

Dependence on is eliminated.

vo=

RE 1Rs

1RE 1

R s

reRE RE

v i=A

1AFvi

AF=RE 1

Rs

reRE

RE=1

r eRE

Rs≫1



Loop Gain Sensitivities For:Open-loop gain:dAcl

dA= 1

1AF 2

dAcl

Acl=

1

1AF 21AF

AdA

dAcl

Acl= 1

1AF dAA

0

Feedback gain:dAcl

d F=

−A2

1AF 2

dAcl

Acl=

−A2

1AF 21AF

Ad F

dAcl

Acl=

−AF

1AF d F

F−

d F

F

High loop gain makes system insensitive to A, but sensitive to !

Acl=A

1AF

F

AF ∞ AF ∞



Feedback - One-pole A(s)

Consider the case where:

As=a K 0

sa

A

β

-+vi vov

f

F

Acl=vo

v i=

A1AF

Acl s ,F =

a K 0

sa

1F a K 01

sa

=a K 0

saF a K 0

open-loop pole

closed-loop pole

Where and K0 are positive real quantities.F

X-a

j

s= j

F=0

∞F

Root Locus Plot for Acl s ,F

stable for all ΓF!

pole(s) of Acl(s) are the roots of 1 + AΓF = 0, or

AsF=−1=1∗e j± 2k for k = 0, 1, ...



Feedback - One-pole A(s) cont.As=

a K 0

saAcl s=

a K 0

saF a K 0= N s

D s

GBW OL=GBW CL=a K 0

a

Kω

Gain - dB

- 20 dB/dec

20 log10K 0

1F K 0

a1F K 00

- 20 dB/dec

20 log10 K 0

frequency (ω)

open-loop (OL) closed-loop (CL)



One-pole Feedback - Root LocusAcl s=

a K 0

saF a K 0=

N sD sAs=

a K 0

saD s=saF a K 0=0Pole of A

cl(s):

saF a K 0=0⇒F a K 0

sa=−1=1∗e j ±2k

F1 a K 0

∣r1a∣=1−sr1 =± 2k and

x

r1

|r1+a|

-a

r1a

-ax

12

r1a=0≠± 2k1

r1a=

∞F K0≥ 0 Not allowed

r1r

1

j

js= j

2

for k = 0, 1, ...

AsF=−1=1∗e j± 2k

r1 is a root of D(s) = 0 or 1 + A(s)Γ

F = 0 iff

or s=−aF a K 0

for p.r. ΓF1



Frequency-Dependent FeedbackConsider the case where the open-loop gain is:

As=K sasb sc abc

dc Gain: Ass=0=K 0a0b0c

=K abc

=K 0

A sketch of the Bode plot would look something like:

a b c

Kω

20 log10

|A(s)| - dB

+ 20 dB/dec

- 20 dB/dec

open-loop (OL)

20 log10 K 0

As=K sasb sc

abc



Frequency-Dependent Feedback



Acl s =K 0

sasb sc

1F K 0sa

sbsc

=K 0sa

sb scF K 0sa

-a-b-cx x o

j

What's the Root Locus for Acl(s)?

Is Acl(s) Stable for all Γ

F?

abc





F?



-a-b-cx x ol

“ri”

j

Acl s =K 0

sasb sc

1F K 0sa

sbsc

=K 0sa

sb scF K 0sa

Acl(s) IS stable for all ΓF!



F?

1. zeros of Acl(s) = zeros of A(s) and are independent of feedback

2. poles of Acl(s) ≠ poles of A(s) are a function of the feedback

abc





F?

As=K 0bc

sb sc

Acl s=K 0bc

sbscF K 0bc=

N sD s

b c

Kω

20 log10

|A(s)| - dB

- 20 dB/dec

- 40 dB/dec

-b-cX X

20 log10 K 0

j

two-pole bc





F?

As=K 0bc

sb scAcl s=

K 0bcsbscF K 0bc

=N sD s

-b-cX X

∞F ≥ 0

jcomplex conjugate poles of A

cl(s)

Av(s) IS stable for all ΓF!

bc



As=K 0bcd

sb scsd Acl s=

K 0bcdsbscsd F K 0bcd

= N sD s

-b-cX XX-d

j



F?

bcd





F?

As=K 0bcd

sb scsd Acl s=

K 0bcdsbscsd F K 0bcd

= N sD s

-b-cX XX-d

Acl(s) NOT stable for all ΓF!

jbcd



The Root Locus MethodThis graphical method for finding the roots of a polynomialis known as the root locus method. It was developed beforecomputers were available. It is still used because it givesvaluable insight into the behavior of feedback systems asthe loop gain is varied. Matlab (control systems toolbox) will plot root loci.

In the frequency-dependent feedback (two-pole & one-zero) example, we noted that increasing feedback increases the CL bandwidth – i.e. the low & high frequency break points moved in opposite directions as ΓF increases. Higher ΓF , as a trade-off, reduces the closed-loop mid-band gain.



GM

oscillation or instability

Acl j=V o jV i j

=A j

1A j F

F

F

F

F

F

F

loop-gain

mutually consistent stability conditions

20log∣A j∣F

A jF=1 e j±2k

∣A j180∣F1

20 log∣A j180∣F0dB

j1=arg [A j1F ]−180o

GM=0dB−20 log10∣A j180∣F



∣Acl j ∣

F

PM50o

(maximally flat response)

-12 dB/octave

ω (log scale)

Nor

mal

ized

gai

n (d

B)Acl j=

vo

v i=

A j1A jF



70oPM55o

∣Acl j ∣

-12 dB/octave

(maximally flat response)

Nor

mal

ized

gai

n (d

B)



Alternative Stability Analysis1. Investigating stability for a variety of feedback gains Γ

F by

constructing Bode plots for the loop-gain A(jω)ΓF can be tedious

and time consuming.

2. A simpler approach involves constructing Bode plots for A(jω) and Γ

F (or 1/Γ

F) separately.

20 log∣A jF∣=20 log∣A j∣−20 log 1F

A jF≫1⇒ Acl j=A j

1A jF≈1F

20 log∣A j∣=1=20 log 1

F⇒20 log∣A j1F∣=0dB⇒∣A j1F∣=1

RECALL:

when

closed-loop gain

20 logF=−20 log 1F



Alternative Stability Analysis - cont.

stable iff20log 1/Γ

F > 60 dB

or20log Γ

F < -60 dB

orΓ

F < 0.001

.=20 log∣A jf ∣−20 log 1F

20log 1/ΓF = 60 dB

dB

100

80

60

40

20

0000

-90o

-180o

-270o

f1

20log∣A jf ∣

arg A jf 102 104 106 108

108106104102

A jf = 105

1 jf /1051 jf /1061 jf /107

20 log∣A jf F∣

f (Hz)

f (Hz)

zero PM & GM

f180

20 log∣A jf 1F∣=0

-40 dB /dec

-60 dB /dec

-20 dB /dec

f1 = f

180

=> f1 < f

180

20 log∣A jf F∣



f180

f1

PM = GM = 0

20log 1/ΓF = 85 dB

f1 < f

180GM =25 dB

20 log∣A j f 1F∣=0

20 log∣A jf F∣=20 log∣A jf ∣−20 log 1F

A jf = 105

1 jf /1051 jf /1061 jf /107

jf 1=arg [A jf 1F ]−180o

φ(jf1) = -108o

PM =φ(jf1) + 180o = 72o stable!

20log∣A jf ∣



PM = GM = 0

A jf = 105

1 jf /1051 jf /1061 jf /107


-40 dB /dec

20log 1/ΓF = 50 dB 20 log∣A j f 1F∣=0

f180

GMGM = - 20 dB

φ(jf1) = -210o PM =φ(jf1) + 180o = -30o unstable!

20log∣A jf ∣




stable iff20log 1/Γ

F > 60 dB

or20log Γ

F < -60 dB

orΓ

F < 0.001

-40 dB /dec

20 log∣AF∣

dB

100

80

60

40

20

0000

-90o

-180o

-270o

72o PM

stable

zero PM & GM 25 dB GM

20 log∣A jf ∣

arg A jf 102 104 106 108

108106104102

f (Hz)

f (Hz)

f1

φ(jf1) = -108o

-20 dB /dec

f180

A jf = 105

1 jf /1051 jf /1061 jf /107


-60 dB /dec




“Rule of Thumb” – Closed -loop amplifier will be stable if the 20log1/ΓF line intersects the 20log|A(j f)| curve on the -20 dB/dec segment.

Using “Rule of Thumb”:

PM45o=>

20log∣A jf ∣-20 dB /dec

-40 dB /dec

-60 dB /dec

stable

unstable

unstable

A jf = 105

1 jf /1051 jf /1061 jf /107


f180

f1

arg (A(jf1) ΓF) arg (A(jf1) ΓF) > - 135o



Rate of Closure (RC) RC=slope A j2 f dB/dec−slope 1

F

dB /dec

RECALL:“Rule of Thumb” – Closed-loop amplifier will be stable if 20log1/ΓF line intersects 20log|A(j f)| curve on the -20 dB/dec segment.

“Equivalent Rule of Thumb” – Closed-loop amplifier will be stable if

RC ≥−20dB /dec

RC = - 20 dB/dec – 0 dB/dec = - 20 dB/dec

RC = - 20 dB/dec – 0 dB/dec = - 40 dB/decRC = - 20 dB/dec – 0 dB/dec = - 40 dB/dec

argPM = 72o

100

20log∣A jf ∣



Frequency Dependent Feedback

100

80

60

40

20

01 10 100 1k 10k 100k 1M 10M

Frequency (Hz)

dB A(j2πf)

1/Γ1(j2πf)

1/Γ2

1/Γ3

1/Γ4(j2πf)

RC=slope A j2 f dB/dec−slope 1F

dB /dec

RC1=−20dB /dec−20dB /dec =−40dB /dec

RC 2=−20 dB /dec−0dB /dec =−20 dB /dec

RC3 = - 40 dB/dec – (0 dB/dec) = - 40 dB/dec

RC4 = - 40 dB/dec – (-20 dB/dec) = - 20 dB/dec

“Equivalent Rule of Thumb” – Closed-loop amplifier will be stable if

RC ≥−20dB /dec

-20 dB /dec



Frequency Compensation

frequency compensation – modifying the open-loop A(s) so that the closed-loop A

cl(s) is stable for any desired value of |A

cl(jf)| by extending the -20 dB/dec segment.

desired implementation - minimum on-chip or external components.

Ex: LM 741 op amp is frequency compensated to be stable with 60o PM for 20 log |A

cl(jf)| = 20 log 1/ΓF = 0 dB.

1/ΓF1/ΓF

1/ΓF

1/ΓF

20log∣A jf ∣



Compensation – What if 1st pole is shifted lower?VCC

dB

100

80

60

40

20

0000

-90o

-180o

-270o

20log∣A jf ∣

102 104 106 108

108106104

-20 dB /dec

-40 dB /dec

-60 dB /dec

-60 dB /dec

-40 dB /dec

-20 dB /dec

20log∣Acomp jf ∣ 20log 1/ΓF1

= 85 dB

20log 1/ΓF2

= 45 dB

Acomp jf ≈ 105

1 jf /1031 jf /1061 jf /107

f (Hz)



Frequency Compensation Using Miller EffectVCC C comp

C1 C2R1 R2V gm V I i

I i

V o

V o

C1=CC 1g m R2

R1=RB∥r

RC

R2=RC∥r o∥Rin2Rin2

C2

V o

I i=

s C comp−g mR1 R21s [C1 R1C2 R2C comp g m R1R2R1R2]s2[C1C 2C comp C1C2]R1 R2

.=sC comp−gmR1 R2

1 sp1c

1 sp2c

=

sC comp−gmR1 R2

1s 1 p1c

1 p2c

s2

p1cp2c

where p2c≫p1c

p1=1

R1C1

p2=1

R2C2

BJT Miller Capacitance

with Comp = 0

C comp



Compensation Using Miller Effect - cont.V o

I i=

sC comp−gmR1 R2

1s 1 p1c

1 p2c

s2

p1cp2c

≈sC comp−gmR1 R2

1s 1p1c

s2 1p1c p2c

p2c≫p1cIf

V o

I i=

s C comp−gmR1 R21s [C1 R1C2 R2C comp gm R1R2R1R2]s2[C 1C 2C comp C1C2]R1 R2

and

p1c=1

C1 R1C2 R2C comp g m R1R2R1R2≈ 1

C comp gm R2R1

p2c= p1c p2c

p1c=

C1 R1C2 R2C comp gm R1R2R1R2[C1C2C comp C1C 2]R1R2

≈C comp g m

C comp C1C2≈

g m

C1C2

Ccomp C1C2≫C1C2

pole-splitting

assumption

Compensation Miller Capacitance

If



Compensation Using Miller Effect - cont.

f p1c= p1c

2≈

12 [C comp gm R1 R2]

f p2c= p2c

2≈

gm

2 [C1C2]

A jf = 105

1 jf105

1 jf106

1 jf107

= 105

1 jff p1

1 jff p2

1 jff p3

Using Miller effect compensation: fp1

-> fp1c

<< fp1

and fp2

-> fp2c

>> fp2

Also fp3

, determined by another stage, is unaffected by the compensation => fp3c

= fp3

Acomp jf = 105

1 jff p1c

1 jff p2c

1 jff p3

(pole-splitting)

f p1=p1

2=

1R1C1

f p2=p2

2=

1R2C 2



Miller Compensation Example

Design Objective: determine Ccomp s.t. fp1c = 100 Hz and compute fp2c.

Given:A jf = 105

1 jfR1C1

1 jfR2C 2

1 jf107

=

105

1 jff p1

1 jff p2

1 jf107

=

105

1 jf105

1 jf106

1 jf107

f p2c=gm

2C1C2≈61MHz

Acomp jf = 105

1 jf102

1 jff p2c

1 jf107

= 105

1 jff p1c

1 jff p2c

1 jf107

where C1 = 100 pF, C

2 = 5 pF, g

m = 40 mS, R

1 = 100/2π kΩ and R

2 = 200/2π kΩ

f p1c=100Hz≈ 12 [C comp gm R1 R2]

=22

2 [C comp 40x10−31052x105]

⇒C comp=78.5 pF



Compensation Using Miller Effect - cont.dB

100

80

60

40

20

000

20log∣A jf ∣

102 104 106 108

-20 dB /dec

-40 dB /dec

-60 dB /dec

-60 dB /dec -40 dB /dec

-20 dB /dec20log∣Acomp jf ∣

20log 1/ΓF = 0 dBf (Hz)

0

-90o

-180o

-270o

108106104

1-pole roll-off !

Acomp jf ≈ 105

1 jf102

1 jf6.1∗107

1 jf107

PM45o for∣Acl jf ∣= 1F≥1

A jf = 105

1 jf105

1 jf106

1 jf107

Acomp jf



SummaryFeedback has many desirable features, but it can createunexpected - undesired results if the full frequency-dependent nature (phase-shift with frequency) of the feedback circuit is not taken into account.

Feedback can be used to convert a well-behaved stable circuitinto an oscillator. Sometimes, due to parasitics, feedback inamplifiers results in unexpected oscillations.

The root-locus method was used to show how feedback cancreate the conditions for oscillation and instability.

Feedback Basics - University of Pennsylvania

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