ECEN 326 Electronic Circuits - Texas A&M Universityspalermo/ecen326/feedback.pdf · Ideal Feedback...

ECEN 326

Electronic Circuits

Feedback

Dr. Aydın Ilker Karsılayan

Texas A&M University

Department of Electrical and Computer Engineering

Ideal Feedback ConfigurationIdeal Feedback Configuration

ε

Sfb

SoSiS

Feedback Network

Basic Amplifier

a

f

Σ

So

Si= A =

a

1 + T, T = af

Sε

Si=

1

1 + T,

Sfb

Si=

T

1 + T

ECEN 326 Electronic Circuits - Aydın I. Karsılayan - Feedback 1

Effect of FeedbackEffect of Feedback Gain sensitivity

Gain of the basic amplifier (a) usually depends on temperature,

operating conditions and process parameters.

dA

da=

(1 + af) − af

(1 + af)2=

1

(1 + af)2

δA =δa

(1 + af)2⇒

δA

A=

1 + af

a

δa

(1 + af)2

δA

A=

δa

a1 + af

=

δa

a1 + T


Effect of FeedbackEffect of Feedback Distortion

So

Sε

a3

a1

a3

a2

a2

So1a1

So2So1

So

Si

A3

A1

A2

A2

A3

So1a1

(1+a1f)

So2So1

A1 =a1

1 + a1f≈

1

f, A2 =

a2

1 + a2f≈

1

f


Two-Port NetworksTwo-Port Networks y parameters

i2

2vy21v1

y22

i1

v1y12v2

y11

1

Two−port network

1

i1 = y11v1 + y12v2

i2 = y21v1 + y22v2

y11 =i1

v1

∣∣∣∣∣∣∣∣v2=0, y12 =

i1

v2

∣∣∣∣∣∣∣∣v1=0

y21 =i2

v1

∣∣∣∣∣∣∣∣v2=0, y22 =

i2

v2

∣∣∣∣∣∣∣∣v1=0


Two-Port NetworksTwo-Port Networks z parameters

i2i1

Two−port network

2v1 v2

12

z

iz z21i1

z11 22

v1 = z11i1 + z12i2

v2 = z21i1 + z22i2

z11 =v1

i1

∣∣∣∣∣∣∣∣i2=0, z12 =

v1

i2

∣∣∣∣∣∣∣∣i1=0

z21 =v2

i1

∣∣∣∣∣∣∣∣i2=0, z22 =

v2

i2

∣∣∣∣∣∣∣∣i1=0


Two-Port NetworksTwo-Port Networks h parameters

i1 1vh11

v2 h

i2

222

h21i1h12v1

Two−port network

v1 = h11i1 + h12v2

i2 = h21i1 + h22v2

h11 =v1

i1

∣∣∣∣∣∣∣∣v2=0, h12 =

v1

v2

∣∣∣∣∣∣∣∣i1=0

h21 =i2

i1

∣∣∣∣∣∣∣∣v2=0, h22 =

i2

v2

∣∣∣∣∣∣∣∣i1=0


Two-Port NetworksTwo-Port Networks g parameters

i1

v11

i2gv2

21

g22

i2g12 v1

g11

Two−port network

i1 = g11v1 + g12i2

v2 = g21v1 + g22i2

g11 =i1

v1

∣∣∣∣∣∣∣∣i2=0, g12 =

i1

i2

∣∣∣∣∣∣∣∣v1=0

g21 =v2

v1

∣∣∣∣∣∣∣∣i2=0, g22 =

v2

i2

∣∣∣∣∣∣∣∣v1=0


Feedback ConfigurationsFeedback Configurations

According to whether the output signal So is a current or a

voltage and whether the feedback signal Sfb is a current or a

voltage, four feedback configurations exist:

Series-Shunt Feedback

Sfb = vfb (voltage), So = vo (voltage)

Shunt-Shunt Feedback

Sfb = ifb (current), So = vo (voltage)

Shunt-Series Feedback

Sfb = ifb (current), So = io (current)

Series-Series Feedback

Sfb = vfb (voltage), So = io (current)


Series-ShuntSeries-Shunt Ideal

vε

zo

voav

i

z ε

o

v

i

i

fv

i

Basic amplifier

Feedback network

vo

vi=

a

1 + afZi = (1 + af)zi Zo =

zo

1 + af


Series-ShuntSeries-Shunt h-parameter representation

voh12a v

h v

o

h11f

h11a

ii

i

ii

12f

v

o i

zS

s

h22f

1

h22a

1yL

1

h21f

h21a

Feedback network

Basic amplifier

|h12f | |h12a| |h21a| |h21f |


Series-ShuntSeries-Shunt h-parameter representation - simplified

ii v

12f vo

i o

h

vs

z

ii

Basic amplifier

yo

1h21a

Feedback network

zi = zS + h11a + h11f

yo = yL + h22a + h22f


Series-ShuntSeries-Shunt h-parameter representation - simplified

vε

zo

v

av o

ii

z ε

s

h

v

vo

i

12f

Basic amplifier

Feedback network

a = −h21a

ziyof = h12f zo =

1

yo


Series-ShuntSeries-Shunt Example

4.6k

3k

1k

50Vi1 Vo1 Vo2Q1

Q2

Q3

Q4

Q5

Q6

Q7Q9

Q11

Q12

Q8 Q10

Q13

Vi2

+3V

−3V

50 3k5k

1k

500

5k

Vi1,dc = Vi2,dc = 0, VA = ∞, VT = 25mV, β = 100



Differential-mode AC half circuit:

1k

3k

Q1

Q2

Q3vi

vo

Feedback network

2505k

50

Input: Voltage

Output: Voltage

IC1 = 1.3 mA

IC2 = 0.5 mA

IC3 = 0.7 mA

h-parameter representation of the feedback network:

2 5.25k

5k

250

238

0.0476 v

v1 v2i1 i2 v2

i2v1i1



Q1

Q2

Q3vi

5.25k

3kvo

0.0476 vo

Ideal feedback

50

238

1k

f = 0.0476

3k

Q1

Q2

Q3

1k

vε

zi

vo

zo

50

238 5.25k

a =vo

vε= 192

zi = 26 kΩ

zo = 64.6 Ω



vε

0.0476 vo

vi

64.6

26k 192 vε vo a = 192

f = 0.0476

zi = 26 kΩ

zo = 64.6 Ω

vivi 18.9 vi

6.37vo

263.6k A =vo

vi=

a

1+af= 18.9

Zi = zi(1+af) = 263.6 kΩ Zo =zo

1+af= 6.37 Ω

Adm = 18.9

Rid = 2 × 263.6k = 527.2 kΩ

Rod = 2 × 6.37 = 12.7 Ω


Shunt-ShuntShunt-Shunt Ideal

zo

oi

o

i

vv z

fv

i aiε

ε

ii

Feedback network

Basic amplifier

vo

ii=

a

1 + afZi =

zi

1 + afZo =

zo

1 + af


Shunt-ShuntShunt-Shunt y-parameter representation

i21f v

21a vi vo12av oi

12f

y

o

is y v

y v y

y11a

1

y11f

1y22f

1

y22a

1yS

1yL

1

Basic amplifier

Feedback network

|y12f | |y12a| |y21a| |y21f |


Shunt-ShuntShunt-Shunt y-parameter representation - simplified

y21a vi

y12f o

vi

v

vois

Feedback network

Basic amplifier

1yi

1yo

yi = yS + y11a + y11f

yo = yL + y22a + y22f


Shunt-ShuntShunt-Shunt y-parameter representation - simplified

zo

i i

i

vvi ozs ai

ε

y12f vo

ε

Feedback network

Basic amplifier

a = −y21a

yiyof = y12f zi =

1

yizo =

1

yo


Shunt-ShuntShunt-Shunt Example

5k

6

Q5 Q4

Q13

2.3k

1k 1k

Q

Vo2

Q9 Q10

Q2

Q3

Q1

Vo1

Q8Q7

Vi1 Vi2

Q12

Q11

+3V

−3V

100

1k 1k

100

Vi1,dc = Vi2,dc = 0, VA = ∞, VT = 25mV, β = 100




2.5k1k

Q1vi Q2

Q3

vo

Feedback network

100

1k Input: Current

Output: Voltage

IC1 = 1 mA

IC2 = 1 mA

IC3 = 2 mA

y-parameter representation of the feedback network:

v2

10001k

1k1k

i1 i2

v2v1

v1i1 v2

i2



Q1 Q2

Q3

vo

Ideal feedback

100vi

vo1k

10001k 2.5k

1k

f = −0.001

Q1 Q2

Q3

iε

zi

714.2

vo

zo

100 1k

1ka =

vo

iε= −3403

zi = 87.7 Ω

zo = 21.7 Ω



iε

ii

−0.001 vo

−3403 iε

21.7

87.7 vo

ii =vi

100

a = −3403

f = −0.001

zi = 87.7 Ω

zo = 21.7 Ω

ii

ii 4.9vo

−773 ii19.9 A =vo

ii=

a

1+af= −773

Zi =zi

1+af= 19.9 Ω Zo =

zo

1+af= 4.9 Ω



Voltage-mode equivalent circuit:

vx vxAvvi

4.9vo

Ri

100

100 ‖ Ri = Zi ⇒1

100+

1

Ri=

1

Zi⇒ Ri = 24.8 Ω

ii =vi

100, vo = −773ii ⇒

vo

vi= −7.73

vo = −7.73vi = AvRi

Ri + 100vi ⇒ Av = −38.9

Adm = −7.73

Rid = 2 × (100 + 24.8) = 249.6 Ω

Rod = 2 × 4.9 = 9.8 Ω


Shunt-SeriesShunt-Series Ideal

zi

iε

v

i

ii aiεi

o

z

fio

o

Basic amplifier

Feedback network

io

ii=

a

1 + afZi =

zi

1 + afZo = zo(1 + af)


Shunt-SeriesShunt-Series g-parameter representation

vi21f

v g ii

g22f

g22a

zL

s 21a

o

i v

io

i

i

g

oyS

1g11a

1

g11f

1

Feedback network

Basic amplifier

g12a

g12f

|g12f | |g12a| |g21a| |g21f |


Shunt-SeriesShunt-Series g-parameter representation - simplified

vi g i

iozo

21a vis

og12f i

Feedback network

yi

1

Basic amplifier

yi = yS + g11a + g11f

zo = zL + g22a + g22f


Shunt-SeriesShunt-Series g-parameter representation - simplified

zi

iε

vis i

og

ai z

i

io

12f

ε o

Basic amplifier

Feedback network

a = −g21a

yizof = g12f zi =

1

yi


Shunt-SeriesShunt-Series Example

20k

50k50k

20k

Vi1 Vi2

Vo1 Vo2M1

M2

M8

M7M5

M3

M9

M4

M6

20/1

40/1

50/150/1 50/1

20/1

30/1 30/1

40/1100µ

−3V

+3V

1k 1k

2k5k2k

k′n = 190 µA/V2, k′

p = 110 µA/V2, λn = λp = 0

Vtn = 1.5 V, Vtp = −1.6 V, χ = 0




1k

2k

voM1

M2

vi

50k

Feedback network

20k

2.5k

Input: Current

Output: Current

ID1 = 50 µA

ID2 = 100 µA

g-parameter representation of the feedback network:

1.11k

2k

2.5k 4.5k

v2i2v1 v2

i1 i2 v1i1

−0.55 i2



voM1

M2

io−0.55 io

50k

Ideal feedback

vi1k

4.5k

1.11k

20k

f = −0.55

M1

M2

1.11k

io

zovo

1kiε

zi50k

4.5k20k

a =io

iε= −11.6

zi = 818.2 Ω

zo = ∞



iε

ii

io

818.2

20k

−11.6 iε

−0.55 io

ii =vi

1k

a = −11.6

f = −0.55

zi = 818.2 Ω

zo = ∞

ii

ii 20k

voio

−1.57 ii110.9 A =io

ii=

a

1+af= −1.57

Zi =zi

1+af= 110.9 Ω Zo = ∞




vx vxAvRivi

20kvo

1k

1k ‖ Ri = Zi ⇒1

1k+

1

Ri=

1

Zi⇒ Ri = 124.7 Ω

ii =vi

1k, vo = −(20k)(−1.57ii) ⇒

vo

vi= 31.4

vo = 31.4vi = AvRi

Ri + 1kvi ⇒ Av = 283.2

Adm = 31.4

Rid = 2 × (1k + 124.7) = 2.25 kΩ

Rod = 2 × 20k = 40 kΩ


Series-SeriesSeries-Series Ideal

avε

i

ε z

vi

fio

i

i

io

zov

Basic amplifier

Feedback network

io

vi=

a

1 + afZi = zi(1 + af) Zo = zo(1 + af)


Series-SeriesSeries-Series z-parameter representation

z11a

ii

ii

i

S zL

iz

z

z

o

io

io

vs

11f 22f

22a

i

z

z21f

z21a

Feedback network

z12f

z12a

Basic amplifier

|z12f | |z12a| |z21a| |z21f |


Series-SeriesSeries-Series z-parameter representation - simplified

io

io

z i

zo

v

ii

i

s

iz

z12f

Feedback network

Basic amplifier

21a

zi = yS + z11a + z11f

zo = zL + z22a + z22f


Series-SeriesSeries-Series z-parameter representation - simplified

ε

ii

ziv

i

avε zo

o

o

z12f

s

i

v

Basic amplifier

Feedback network

a = −z21a

zizof = z12f


Series-SeriesSeries-Series Example

M7

M4

M6M3

M2

M1

M13

M5Vo1

30k

500

1k30k

Vo2

100µ

M11

M12

30/1

25/1

Vi2Vi1

M8 M9 M1030/130/1 30/130/1

10/1

10/110/1

20/1

10/1

50/1

−3V

+3V

20/1

25k25k

15k 15k

5k5k

k′n = 190 µA/V2, k′

p = 110 µA/V2, λn = λp = 0

Vtn = 1.5 V, Vtp = −1.6 V, χ = 0




15k

30k

M2

M1

500

vi

M3

vo

io

5k

Feedback network

25k

250Input: Voltage

Output: Current

ID1 = 100 µA

ID2 = 100 µA

ID3 = 100 µA

z-parameter representation of the feedback network:

i1

239

i2v1 v2

250 5005k

v1i1

21.7 i2

v2i2

456



30k

M2

M1vi

M3

Ideal feedback

15k

vo

456

io21.7 io239

25k

f = 21.7

15k

M2

M3M1

239

zi

456

30k

vo

zo

iovε

25k

a =io

vε= 0.106

zi = ∞zo = ∞



0.106 vε

21.7 io

vε

15kvi

io

a = 0.106

f = 21.7

zi = ∞zo = ∞

0.032 vi 15kvi

vo

vi A =io

vi=

a

1+af= 0.032

Zi = ∞ Zo = ∞




vo

15k

vi vi viAv

vo = −(15k)(0.032vi) = Avvi ⇒ Av = −480

Adm = −480

Rid = ∞Rod = 2 × 15k = 30 kΩ


ECEN 326 Electronic Circuits - Texas A&M Universityspalermo/ecen326/feedback.pdf · Ideal Feedback...

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Transcript of ECEN 326 Electronic Circuits - Texas A&M Universityspalermo/ecen326/feedback.pdf · Ideal Feedback...