feedback basics - penn engineering · 2008. 11. 19. · feedback block diagram point of view a β-+...

ESE319 Introduction to Microelectronics

12008 Kenneth R. Laker (based on P. V. Lopresti 2006) updated 29Oct08 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

F

Operational Amplifier From Another View



Feedback Block Diagram Point of View

A

β

-+vi vo

vo=A

1AFvi

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

G=vovi= A1A F

The quantity ' loop-gain and G = 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=vovi

A

β

-+vi vo

F

vf

F=v fvo

G=vovi≈1F

|v f=0



Feedback in the Emitter FollowerEmitter current equation:

ie=1

RsT1

r eREvi

Create an “artificial” feedbackequation, multiply by :

ie=

T1Rs

1T1Rs

r eRE vi=

A1AF

vi

T1/Rs

Note: , i.e. the BJT “beta”. T=

ib

ie

T ib



Emitter Follower Emitter Current

ie=

T1Rs

1T1Rs

r eRE vi=

A1AF

vi

The forward gain term, A:

A=T1Rs

The feedback term, :

F=r eRE

If the “loop gain” is large,

ie≈1 Fvi=

1r eRE

vi

F

AF≫1

ib ib

ie

T

Dependence on is eliminated.T



Loop Gain Sensitivities For:Forward gain:dGdA

=1

1A F 2

dGG

=1

1A F 2

1AFA

dA

11AF

dAA

0

Feedback gain:dGd F

=−A2

1A F 2

dGG

=−A2

1AF 2

1AFA

d F

−A F1A F

d F F

−d F F

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

G= A1AF

F

==AF∞

AF∞



Feedback - One-pole A(s)

Consider the case where:

As=a K0sa

A

β

-+vi vov

f

F

G=vovi=

A1AF

G s ,F =

a K0sa

1a F K01sa

=a K0

sa 1F K0

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 G s ,F

stable for all ΓF!



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

a K0sa

G s=a K 0

sa 1F K0=N sD s

GBWOL=GBWCL=a K0

a

Kω

Gain - dB

- 20 dB/dec

20 log10K0

1F K0

a 1F K00

- 20 dB/dec

20 log10K0

frequency (ω)

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



One-pole Feedback - Root LocusG s=

a K 0sa 1F K0

= N sD s

As=a K0sa

D s=sa 1F K0=saa F K0=0Pole of G(s):

saa F K0=0⇒a F K0sa

=−1=1∗e j± 2k

a F K0∣sa∣

=1 −sa =± 2kand

x

si

|si+a |

-a

s ia

-ax

12

s1a =0≠± 2 k1

s2a =2

∞F K0≥ 0 Not allowed

s1s2

j

js= j



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

As=K sa sb sc

For convenient Bode plotting we can rewrite A(s) as:

As=K abc

sa1

sb1sc1

=K0 sa1

sb1sc1

abc

K0=Kabc

where

dc gain



Frequency-Dependent FeedbackA sketch of the Bode plot would look something like:

Sketching the frequency response is “easy,” if the polynomialis factored into its roots!

As=K0 sa1

sb1 s

c1

abca b c

Kω

20 log10

|A(s)| - dB

+ 20 dB/dec

- 20 dB/dec

open-loop (OL)

20 log10K0



Frequency-Dependent FeedbackIf the frequency-dependent quantity, A(s), is embedded in a feedback loop, the poles are not so easily factored:

As=K0sa

sbsc

G s= As1 F As

G s=K0

sa sbsc

1F K0sa

sbsc

=K0sa

sbsc F K0 sa

open-loop (OL)

closed-loop (CL)

1. zeros of G(s) = zeros of A(s) and are independent of feedback2. poles of G(s) ≠ poles of A(s) are a function of the feedback F

F

frequency dependent feedback



Frequency-Dependent Feedback – Root Locus

Rationalizing this expression leads to:G s=K 0

sa sbscF K0sa

=N sD s

The numerator is factored, but the denominator is not. Wehave a new quadratic D(s) for G(s). It could be factoredusing the quadratic formula, but we will do it another way!Factoring a polynomial is equivalent to finding its roots, i.e.the values of “s” that make the polynomial equal zero.

D s= sbsc F K0 sa =0

G s=K0

sa sbsc

1F K0sa

sbsc



Frequency-Dependent Feedback – Root LocussbscF K0 sa =0

Since the poles of G(s) will not equal any of the poles of A(s), we can divide by (s + b) (s + c) and obtain:

1F K0sa

sbsc=0

The “loop-gain” terms, and K0, are positive real numbers, sofor a root, say s = -r

1 to exist, the value of the frequency-dependent

terms must be real and negative when evaluated at s = -r1!

F K0 sa

sbsc=−1

F




F K0 sa

sbsc=−1=1∗e j±2k

Working with the complex numbers in polar form:

F K0∣sa∣

∣sb∣∣sc∣=1

⇒∣sa∣

∣sb∣∣sc∣=

1F K0

sia −sib−sic=± 2k

we add the angles for zeros to the point “s

i” in the complex plane and subtract

the angles for poles.

and

where k = 0, 1, ...




A sketch of the operations looks like:

.

-a-b-c

si

|si+c |

s ic s ibs ia

x x o

|si+b |

|si+a |

∣s ia∣∣sib∣∣s ic∣

=1

F K0

s ia −sib− sic=± 2k

Move point si until the phase angle relation is satisfied.

The only values of si that satisfy the phase angle relationship

lie on the negative - axis.

Root Locus Protocol

j

s= j



Frequency-Dependent Feedback - Root LocusOnly D(s) root locations where the angles with respect to open-loop pole/zero locations of A(s) are odd multiples of ±180o are candidates.

1. All roots of D(s) must lie along the “green” loci.

tot= sa − sb−sc=± 2kD s = sb sc F K0sa =0

∞F ≥0 -a-b-cx x o

13 2

4

j

0≤ F∞ s1s3s4

s2

G(s) stable for all ΓF!

3. If s2 = -a there is a zero/pole cancellation2. Bandwidth of G(s) > bandwidth of A(s)



Frequency-Dependent Feedback – Root LocusNow, to find the gain required for a given root or pole of G(s), measure the distances from each critical point and form the ratio:

F K0r=∣sb∣∣sc∣

∣sa∣ s=−sr

The only allowed poles {e.g. “sr”} of G(s) are on the “green” loci.

Where s = - sr is on the root-locus

∣sa∣∣sb∣∣sc∣

= 1F K0

⇒ |∣−s rc∣

-a-b-cx x o

∣−s rb∣ ∣−s ra∣

l“s

r”

F K 0

∣−s rc∣



Root Locus Two-pole A(s)

As=K0bc

sb sc

G s=K0bc

sbscF K 0bc=N sD s b c

Kω

20 log10

|A(s)| - dB

- 20 dB/dec

- 40 dB/dec

-b-cX X

∞F ≥ 0

20 log10K0

jcomplex conjugate poles of G(s)

G(s) stable for all ΓF!

Note: no zero at s = -a



Root Locus Three-pole A(s)

As=K0bcd

sb scsd G s=

K 0bcdsbscsd F K 0bcd

=N sD s

-b-cX XX-d

G(s) NOT stable for all ΓF!

j



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 example, we noted that increasing feedback increases the closed-loop bandwidth – i.e. the low and high frequency break points moved in opposite directions as ΓF increases. Higher ΓF , as a trade-off, reducesthe closed-loop mid-band gain.



GM

oscillation or instability

G j=Vo jV i j

=A j

1A jF

F

F

F

F

F

F

loop-gain

mutually consistent stability conditions



G j

∣G j∣

F

PM50o



∣G j∣

70oPM55o



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 1/Γ

F separately.20 log∣A jF∣=20 log∣A j∣−20 log

1F

A jF≫1⇒G j=A j

1A jF≈1F

20 log∣A j∣=1=20 log1F

⇒∣A j1 F∣=1

RECALL:

when

closed-loop gain

20 logF=−20 log1F



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

-20 dB /dec

-40 dB /dec

-60 dB /dec

20 log∣AF∣20log 1/ΓF = 85 dB

20log 1/ΓF = 60 dB

20log 1/ΓF = 50 dB

dB

100

80

60

40

20

0000

-90o

-180o

-270o

-108o 72o PM

stable

zero PM & GM

unstable

f 180

25 dB GM

f 1

f1

f1

PM < 0o

∣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)



Alternative Stability Analysis - cont.

“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

arg A jf 1−135o

=>

∣A jf ∣

arg A jf



Frequency Compensation

frequency compensation – modifying the open-loop A(s) so that the closed-loop G(s) is stable for any desired value of |G(jf)|.

desired implementation - minimum on-chip or external components.

Ex: LM 741 op amp is frequency compensated to be stable with 60o PM for |G(jf)| = 1/ΓF = 1.



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

dB

100

80

60

40

20

0000

-90o

-180o

-270o

∣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

∣Acomp jf ∣ 20log 1/ΓF1 = 85 dB

20log 1/ΓF2

= 45 dB

Acomp jf ≈105

1 jf /1031 jf /1061 jf /107

f (Hz)

C comp



Frequency Compensation Using Miller EffectVCC Ccomp

C1 C2R1 R2V gmVI i

I i

Vo

Vo

C1=CC 1gmR2R1=RB∥r

RC

R2=RC∥r o∥Rin2Rin2

C2

VoI i=

sC comp−gmR1R21s [C1R1C2R2C compgmR1R2R1R2]s

2[C1C 2C compC 1C2]R1R2

.=sC comp−gmR1R2

1 sp1c

1 sp2c

=

sC comp−gmR1R2

1s 1 p1c

1 p2c

s2

p1cp2c

where p2c≫p1c

p1=1R1C1

p2=1R2C2

BJT Miller Capacitance

with Comp = 0

C comp



Compensation Using Miller Effect - cont.VoI i=

sC comp−gmR1R2

1s 1 p1c

1 p2c

s2

p1cp2c

≈sC comp−gmR1R2

1s 1p1c

s2 1p1c p2c

p2c≫p1cIf

VoI i=

sC comp−gmR1R21s [C1R1C 2R2C compgmR1R2R1R2]s

2[C1C2C compC1C 2]R1R2

and

p1c=1

C1R1C2R2C compgmR1R2R1R2≈ 1C compgmR1R2

p2c= p1c p2c p1c

=C1R1C2R2C compgmR1R2R1R2

[C1C2C compC1C 2]R1R2≈

C comp gmC1C2C compC1C 2

≈gm

C1C 2C comp≫C 2

pole-splitting

assumption

Compensation Miller Capacitance

If



Compensation Using Miller Effect - cont.f p1c=

p1c2

≈1

2 [C compgmR1R2]f p2c=

p2c2

≈gm

2 [C1C2]

A jf = 105

1 jf /1051 jf /1061 jf /107= 10

5

1 jff p11 jff p2

1 jff p3

Using Miller effect compensation: fp1

-> fp1c

fp2c

>> fp2

Also fp3

is unaffected by the compensation => fp3c

= fp3

Acomp jf =105

1 jf /1021 jf / f p2c1 jf /107= 10

5

1 jff p1c1 jff p2c

1 jff p3c

(pole-splitting)



Miller Compensation Example

Design Objective: determine Ccomp s.t. fp1c = 102 Hz

Given: A jf =105

1 jf /1051 jf /1061 jf /107= 10

5

1 jff p11 jff p2

1 jff p3

f p2c=gm

2C1C2≈60MHz

Acomp jf =105

1 jf /1021 jf / f p2c1 jf /107= 10

5

1 jff p1c1 jff p2c

1 jff p3c

where C1 = 100 pF, C

2 = 5 pF, g

m = 40 mS, R

1 = 100/2π kΩ and R

2 = 100/π kΩ

f p1c=100 Hz≈1

2 [C compgmR1R2]⇒C comp=78.5 pF



Compensation Using Miller Effect - cont.dB

100

80

60

40

20

000

∣A jf ∣

102 104 106 108

-20 dB /dec

-40 dB /dec

-60 dB /dec

-60 dB /dec -40 dB /dec

-20 dB /dec

∣Acomp jf ∣

20log 1/ΓF2

= 0 dBf (Hz)

0

-90o

-180o

-270o

108106104

1-pole roll-off !

Acomp jf ≈105

1 jf /1021 jf /6∗1071 jf /107

PM45o for∣G jf ∣= 1F≥1



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.

We next will use feedback to convert a well-behaved stablecircuit into an oscillator. Sometimes, because of parasitics,feedback in amplifiers turns them into unexpected oscillators.

The root-locus method was used to help us understand how feedback can create the conditions for oscillation and instability.

feedback basics - penn engineering · 2008. 11. 19. · feedback block diagram point of view a β-+...

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