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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
ECEN 326 Electronic Circuits - Aydın I. Karsılayan - Feedback 2
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
ECEN 326 Electronic Circuits - Aydın I. Karsılayan - Feedback 3
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
ECEN 326 Electronic Circuits - Aydın I. Karsılayan - Feedback 4
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
ECEN 326 Electronic Circuits - Aydın I. Karsılayan - Feedback 5
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
ECEN 326 Electronic Circuits - Aydın I. Karsılayan - Feedback 6
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
ECEN 326 Electronic Circuits - Aydın I. Karsılayan - Feedback 7
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)
ECEN 326 Electronic Circuits - Aydın I. Karsılayan - Feedback 8
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
ECEN 326 Electronic Circuits - Aydın I. Karsılayan - Feedback 9
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 |
ECEN 326 Electronic Circuits - Aydın I. Karsılayan - Feedback 10
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
ECEN 326 Electronic Circuits - Aydın I. Karsılayan - Feedback 11
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
ECEN 326 Electronic Circuits - Aydın I. Karsılayan - Feedback 12
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
ECEN 326 Electronic Circuits - Aydın I. Karsılayan - Feedback 13
Series-ShuntSeries-Shunt Example
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
ECEN 326 Electronic Circuits - Aydın I. Karsılayan - Feedback 14
Series-ShuntSeries-Shunt Example
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
ECEN 326 Electronic Circuits - Aydın I. Karsılayan - Feedback 15
Series-ShuntSeries-Shunt Example
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
ECEN 326 Electronic Circuits - Aydın I. Karsılayan - Feedback 16
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
ECEN 326 Electronic Circuits - Aydın I. Karsılayan - Feedback 17
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 |
ECEN 326 Electronic Circuits - Aydın I. Karsılayan - Feedback 18
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
ECEN 326 Electronic Circuits - Aydın I. Karsılayan - Feedback 19
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
ECEN 326 Electronic Circuits - Aydın I. Karsılayan - Feedback 20
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
ECEN 326 Electronic Circuits - Aydın I. Karsılayan - Feedback 21
Shunt-ShuntShunt-Shunt Example
Differential-mode AC half circuit:
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
ECEN 326 Electronic Circuits - Aydın I. Karsılayan - Feedback 22
Shunt-ShuntShunt-Shunt Example
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
ECEN 326 Electronic Circuits - Aydın I. Karsılayan - Feedback 23
Shunt-ShuntShunt-Shunt Example
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
ECEN 326 Electronic Circuits - Aydın I. Karsılayan - Feedback 24
Shunt-ShuntShunt-Shunt Example
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
ECEN 326 Electronic Circuits - Aydın I. Karsılayan - Feedback 25
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)
ECEN 326 Electronic Circuits - Aydın I. Karsılayan - Feedback 26
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 |
ECEN 326 Electronic Circuits - Aydın I. Karsılayan - Feedback 27
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
ECEN 326 Electronic Circuits - Aydın I. Karsılayan - Feedback 28
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
ECEN 326 Electronic Circuits - Aydın I. Karsılayan - Feedback 29
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
ECEN 326 Electronic Circuits - Aydın I. Karsılayan - Feedback 30
Shunt-SeriesShunt-Series Example
Differential-mode AC half circuit:
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
ECEN 326 Electronic Circuits - Aydın I. Karsılayan - Feedback 31
Shunt-SeriesShunt-Series Example
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 = ∞
ECEN 326 Electronic Circuits - Aydın I. Karsılayan - Feedback 32
Shunt-SeriesShunt-Series Example
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 = ∞
ECEN 326 Electronic Circuits - Aydın I. Karsılayan - Feedback 33
Shunt-SeriesShunt-Series Example
Voltage-mode equivalent circuit:
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
ECEN 326 Electronic Circuits - Aydın I. Karsılayan - Feedback 34
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)
ECEN 326 Electronic Circuits - Aydın I. Karsılayan - Feedback 35
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 |
ECEN 326 Electronic Circuits - Aydın I. Karsılayan - Feedback 36
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
ECEN 326 Electronic Circuits - Aydın I. Karsılayan - Feedback 37
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
ECEN 326 Electronic Circuits - Aydın I. Karsılayan - Feedback 38
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
ECEN 326 Electronic Circuits - Aydın I. Karsılayan - Feedback 39
Series-SeriesSeries-Series Example
Differential-mode AC half circuit:
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
ECEN 326 Electronic Circuits - Aydın I. Karsılayan - Feedback 40
Series-SeriesSeries-Series Example
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 = ∞
ECEN 326 Electronic Circuits - Aydın I. Karsılayan - Feedback 41
Series-SeriesSeries-Series Example
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 = ∞
ECEN 326 Electronic Circuits - Aydın I. Karsılayan - Feedback 42
Series-SeriesSeries-Series Example
Voltage-mode equivalent circuit:
vo
15k
vi vi viAv
vo = −(15k)(0.032vi) = Avvi ⇒ Av = −480
Adm = −480
Rid = ∞Rod = 2 × 15k = 30 kΩ
ECEN 326 Electronic Circuits - Aydın I. Karsılayan - Feedback 43