Transcript Document
Chapter 14 Feedback, Stability and Oscillators
Microelectronic Circuit Design
Richard C. Jaeger Travis N. Blalock
2 Microelettronica – Circuiti integrati analogici 2/ed
Richard C. Jaeger, Travis N. Blalock
Copyright © 2005 – The McGraw-Hill Companies srl
Chapter Goals
• Review concepts of negative and positive feedback.
• Develop 2-port approach to analysis of negative feedback amplifiers.
• Understand topologies and characteristics of series-shunt, shunt-shunt, shunt-series and series-series feedback configurations.
• Discuss common errors that occur in applying 2-port feedback theory.
• Discuss effects of feedback on frequency response and feedback amplifier stability and interpret stability in in terms of Nyquist and Bode plots.
• Use SPICE ac and transfer function analyses on feedback amplifiers.
• Determine loop-gain of closed-loop amplifiers using SPICE simulation or measurement.
• Discuss Barkhausen criteria for oscillation and amplitude stabilization • Understand basic
RC, LC
and crystal oscillator circuits and present
LCR
model of quartz crystal.
2 Microelettronica – Circuiti integrati analogici 2/ed
Richard C. Jaeger, Travis N. Blalock
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Feedback Effects
• • • •
Gain Stability
: Feedback reduces sensitivity of gain to variations in values of transistor parameters and circuit elements.
Input and Output Impedances
amplifier.
: Feedback can increase or decrease input and output resistances of an
Bandwidth
: Bandwidth of amplifier can be extended using feedback.
Nonlinear Distortion:
nonlinear distortion.eg: removal of dead zone in class B amplifiers Feedback reduces effects of
2 Microelettronica – Circuiti integrati analogici 2/ed
Richard C. Jaeger, Travis N. Blalock
Copyright © 2005 – The McGraw-Hill Companies srl
• •
Classic Feedback Systems
V o (
s
) V d (
s
)
A
(
s
) V f (
s
) V o (
s
) b (
s
)
A v
V o (
s
) V i (
s
) 1
A
A
b
A(s)
= transfer function of open-loop amplifier or open-loop gain.
b
(s)
= transfer function of feedback network.
V d (
s
) V i (
s
) V f (
s
)
T(s)
= loop gain
A
1
T
For negative feedback:
T(s)
> 0 For positive feedback:
T(s)
< 0
2 Microelettronica – Circuiti integrati analogici 2/ed
Richard C. Jaeger, Travis N. Blalock
Copyright © 2005 – The McGraw-Hill Companies srl
Voltage Amplifiers: Series-Shunt Feedback (Voltage Gain Calculation)
v 1 i 2
h h
T 11 i 1 A 21 i 1
h h
F 12 T 22 v v 2 2 0 v i
h
A ( 21
R
i 1
I
h
(
h
T 11 T 22 )i 1
h G L
F 12 ) v 2 v 2 i A 2
h h
A 11 i 1 A 21 i 1
h h
A 12 A 22 v 2 v 2
A v
v 2 v i
h
A 21
h
F 12 (
R I h
A 21
h
T 11 ) (
h
T 22
G L
) 1
A A
b i F 2
h h
F 11 F 21 i i 1 1
h h
F 12 F 22 v 2 v 2
h
F 12
h
A 12
A
(
R I
h h
T 11 A 21 ) (
h
T 22
G L
) b
h
F 12
2 Microelettronica – Circuiti integrati analogici 2/ed
Richard C. Jaeger, Travis N. Blalock
Copyright © 2005 – The McGraw-Hill Companies srl
Voltage Amplifiers: Series-Shunt Feedback (Two-Port Representation)
• Gain of amplifier should include effects of , , • Required
h
11
h
F
R I
and
R L
.
-parameters are found from their individual definitions.
• Two-port representation of the amplifier is as shown
2 Microelettronica – Circuiti integrati analogici 2/ed
Richard C. Jaeger, Travis N. Blalock
Copyright © 2005 – The McGraw-Hill Companies srl
Voltage Amplifiers: Series-Shunt Feedback (Input and Output
v i (
R I
h
T 11 )i 1
h
F 12 ( -
h
T 22
h
A 21
G L
) i 1 v
R
in
R
in i 1
R
i A in ( ( 1
R I
A
b
h
T 11 )(1 )
A
b ) Series feedback at a port increases input resistance at that port.
v 1 For output resistance: i x 0
R I
i 1 ( v 2 v x
h R
A 21
I
i 1
h
( T 11
h
T 22 )i 1 i 2
G L
)
h
F 12 i v x x v x
G L
v 2
R
out v x i x 1
h
T 22 1
G L A
b resistance at that port.
R
out
R
1 A out
A
b Shunt feedback at a port reduces
2 Microelettronica – Circuiti integrati analogici 2/ed
Richard C. Jaeger, Travis N. Blalock
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Voltage Amplifiers: Series-Shunt Feedback (Example)
• •
Problem:
Find resistances.
A,
b
,
closed loop gain, input and output
Given data:
R
1 =10 k W ,
R
2 =91 k W,
R id
=25 k W,
R o
=1 k W, A 10 4 .
•
Analysis:
Richard C. Jaeger, Travis N. Blalock
h
F 11 v 1 i 1 v 2 0
R
1
R
2 9 .
01 kΩ
h
F 22 i 2 v 2 i 1 0 1
R
1
R
2 1 101 kΩ
h
F 12 v 1 v 2 i 1 0
R
1
R
1
R
2 0 .
0990
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Voltage Amplifiers: Series-Shunt Feedback (Example contd.)
A
v o v i 1 kΩ 25 25 kΩ kΩ 9 .
01 kΩ ( 10 4 ) 1 .
1 .
96 kΩ 96 kΩ 1 .
00 kΩ 4730
A v
1
A A
b 1 4730 4730 ( 0 .
0990 ) 10 .
1
R
in
R
out
R
A in ( 1 1
R
A out
A
b
A
b ) 16.4M
W 1 .
41 W
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Transresistance Amplifiers: Shunt-Shunt Feedback (Voltage Gain Calculation)
i 2 i 1
y y
T 11 A 21 v v 1 1
y y
F 12 T 22 v v 2 2 0 i i
y
(
G I
A 21 v
y
T 11 ( 1
y
)v 1 T 22
y
F 12
G L
) v v 2 2 i A 1 i A 2
y
A 11 v 1
y
A 21 v 1
y
A 12 v 2
y
A 22 v 2
y
T
ij
y
A
ij
y
F
ij A tr
v 2 i i
y
F 21
y
A 12 (
G I y
F 21
y
T 11 ) (
y
T 22
G L
) 1
A A
b i F 1 i F 2
y
F 11 v 1
y
F 21 v 1
y
F 12 v 2
y
F 22 v 2
y
F 12
y
A 12
A
(
G I
y y
T 11 ) ( A 21
y
T 22
G L
) b
y
F 12
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Richard C. Jaeger, Travis N. Blalock
Copyright © 2005 – The McGraw-Hill Companies srl
Transresitance Amplifiers: Shunt-Shunt Feedback (Two-Port Representation)
• Gain of amplifier should include effects of , , • Required
y
11
y
F
R I
and
R L
.
-parameters are found from their individual definitions.
• Two-port representation of the amplifier is as shown.
2 Microelettronica – Circuiti integrati analogici 2/ed
Richard C. Jaeger, Travis N. Blalock
Copyright © 2005 – The McGraw-Hill Companies srl
i i
Transresistance Amplifiers: Shunt-Shunt Feedback (Input and Output Resistances)
(
G I y
11 )v 1
y
12 (
y
T 22
y
A 21
G L
) v 1
R
in v 1 i 1
G I
(1
A h
T 11 b )
R
( 1 A in
A
b i Shunt feedback at a port reduces ) resistance at that port.
For output resistance: i 1
G I
v 1 i 2 i x
G L
v 2 i 0 x
y
(
G I
A 21 v 1 (
y
T 22
y
T 11 )v 1
y G
F 12
L
) v v x x
R
out v x i x 1
y
T 22 1
G L A
b
R
out
R
1 A out
A
b Resistance at output port is reduced due to shunt feedback.
2 Microelettronica – Circuiti integrati analogici 2/ed
Richard C. Jaeger, Travis N. Blalock
Copyright © 2005 – The McGraw-Hill Companies srl
• • •
Transresistance Amplifiers: Shunt-Shunt Feedback (Example)
Problem:
resistances.
Find
A,
b
,
closed-loop gain, input and output
r o
50 V 1 .
35 V 0 .
977 mA 52 .
6 k W
Given data:
V A
= 50 V, b
F
= 150
Analysis:
From dc equivalent circuit,
I C
V CE R V CC C
V CC
R C
V
( b
I BE
R F F C
I
B
0 .
970 mA )
R C
1 .
35 V i 2 1
R F
1
R F g m
40 ( 0 .
977 mA) 39.1mS
y
F 11
y
F 22 i v 1 1 v 2 v 2 i 2 0 0 10 5 S 10 5 S
r
1
g m
3 .
84 k W
y
F 12 i v 1 2 v 1 0 1
R F
10 5 S
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Richard C. Jaeger, Travis N. Blalock
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Transresistance Amplifiers: Shunt-Shunt Feedback (Example contd.)
i b v o
4 .
76 kΩ
i i
4 .
76 kΩ
r
b
o i b
( 1 .
41 kΩ
r o
)
A
v o i i 4 .
76 4 .
kΩ 76 kΩ 3 .
84 kΩ ( 150 ) 1 .
41 kΩ 52 .
6 kΩ
A tr
1
A A
b 114 kΩ 1 114 kΩ ( 0 .
01 mS ) 53 .
3 kΩ 114 kΩ
R
in
R
A in ( 1
A
b )
R I R F r
( 1
A
b ) 995 W
R
out 1
R
A out
A
b
R L R F
1
A R C
b
r o
640 W
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Richard C. Jaeger, Travis N. Blalock
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Current Amplifiers: Shunt-Series Feedback (Voltage Gain Calculation)
v i 1 2
g g
T 11 A 21 v 1 v 1
g g
F 12 T 22 i i 2 2 i i 0 (
g G I
A 21 v 1
g
T 11 (
g
)v 1 T 22
g
F 12 i 2
R L
) i 2 i 1 v 2 i A 1 i F 1
g
T
ij
g
A
ij
g
F
ij A i
i 2 i i
g
A 21
g
F 12 (
G I g
A 21
g
T 11 ) (
g
T 22
R L
) 1
A A
b
A
(
G I
g
A 21
g
T 11 ) (
g
T 22
R L
) b
g
F 12
g
F 12
g
A 12
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Richard C. Jaeger, Travis N. Blalock
Copyright © 2005 – The McGraw-Hill Companies srl
CurrentAmplifiers: Shunt-Series Feedback (Two-Port Representation)
• Gain of amplifier should include effects of , , • Required
g
11
g
F
R I
and
R L
.
-parameters are found from their individual definitions.
• Two-port representation of the amplifier is as shown
2 Microelettronica – Circuiti integrati analogici 2/ed
Richard C. Jaeger, Travis N. Blalock
Copyright © 2005 – The McGraw-Hill Companies srl
i i
Current Amplifiers: Shunt-Series Feedback (Input and Output Resistances)
(
G I g
T 11 )v 1
g
F 12 (
g
T
g
22 A 21
R L
) v 1 1
R
in v 1 i i
G I
(1
A g
T 11 b ) ( 1
R
A in
A
b ) Shunt feedback at a port decreases resistance at that port.
i For output resistance: 1
G I
v 1 v 2 v x
R L
i 2 v x 0
g
(
G I
A 21 v 1
g
T 11 (
g
)v 1 T 22
R L g
F 12 i 2 ) i 2
R
out
R
A out v x i 2 1
A
b
g
T 22
R L
1
A
b Series feedback at output port increases resistance at that port.
2 Microelettronica – Circuiti integrati analogici 2/ed
Richard C. Jaeger, Travis N. Blalock
Copyright © 2005 – The McGraw-Hill Companies srl
Transconductance Amplifiers: Series-Series Feedback (Voltage Gain Calculation)
v 2 v 1
z z
T 11 T 21 i 1 i 1
z z
T 12 T 22 i i 2 2 0 v i
z
(
R S
T 21 i 1 (
z z
T 11 T )i 1 22
z
F 12
R L
) i 2 i 2 v 1 v 2
A tc
i 2 v i
z
A 21
z
F 12 (
R I z
A 21
z
T 11 ) (
z
T 22
R L
) 1
A A
b
A
(
R I
z
A 21
z
T 11 ) (
z
T 22
R L
) b
z
F 12
z
F 12
z
A 12
2 Microelettronica – Circuiti integrati analogici 2/ed
Richard C. Jaeger, Travis N. Blalock
Copyright © 2005 – The McGraw-Hill Companies srl
Transconductance Amplifiers: Series-Series Feedback (Input and Output Resistances)
R
in
R
A in v i i 1 1
z
T 11
A
b
R I
1
A
b
R
out
R
A out v x i 2 1
A
b
z
T 22
R L
1
A
b • Gain of amplifier should include effects of , , • Required
g
11
z
F
R I
and
R L
.
-parameters are found from their individual definitions.
• Two-port representation of the amplifier is as shown Series feedback at input and output port increases resistance at both ports.
2 Microelettronica – Circuiti integrati analogici 2/ed
Richard C. Jaeger, Travis N. Blalock
Copyright © 2005 – The McGraw-Hill Companies srl
• • •
Erroneous Application of 2-port Feedback Theory
Problem:
Find
A,
b
,
closed-loop gain, input and output resistances.
Given data:
V
REF = 5 V, b
o
= 100,
V A
= 50 V,
A o
= 10,000,
R id
=25 k W ,
R o
=0
Analysis:
The circuit is redrawn to identify amplifier and feedback networks and appropriate 2-port parameters of feedback network are found.
This case seems to use series-series feedback.
i e
is sampled by feedback network instead of
i o .
This assumption is made since a
o
is approximately 1.
2 Microelettronica – Circuiti integrati analogici 2/ed
Richard C. Jaeger, Travis N. Blalock
Copyright © 2005 – The McGraw-Hill Companies srl
Erroneous Application of 2-port Feedback Theory (contd.)
Z
-parameters are found as shown. From
A-
circuit,
I E
=1 mA
R
in
R
A in 1
A
b (
R id
R
) 1
A
b
r
i o A tc
1000 ( 0 .
025 V) 1 mA
v i i o v i
R R id
id A
1
A R
b 2 .
5 k W
r o
50 V 1mA
A o r
( b b
o o
1 )
R
1 .
64 S 1 1 .
64 S(5k W ) 50 k W 0 .
200 mS 246 MΩ
R
in
R
A in 1
A
b 27 .
7 GΩ
2 Microelettronica – Circuiti integrati analogici 2/ed
Richard C. Jaeger, Travis N. Blalock
Copyright © 2005 – The McGraw-Hill Companies srl
Erroneous Application of 2-port Feedback Theory (contd.)
Results for
R
out are in error because output of op amp is referenced to ground, base current of BJT is lost from output port and feedback loop and
R
out is limited to
R
out b
o r o
SPICE analyses confirm results for
A tc R
in , but results for
R
out are in error. For and
A tc
and
R
in , amplifier can be properly modeled as a series-shunt feedback amplifier, as collector of
Q
1 can be directly connected to ground for calculations and a valid 2-port representation exists as shown.
3 and 4 are not valid terminals as current entering 3 is not same as that exiting 4. Amplifier can’t be reduced to a 2-port.
2 Microelettronica – Circuiti integrati analogici 2/ed
Richard C. Jaeger, Travis N. Blalock
Copyright © 2005 – The McGraw-Hill Companies srl
• • •
Analysis of Shunt-Series Feedback Pair
Problem:
Find
A,
b
,
closed-loop gain, input and output resistances.
Given data:
b
o
= 100,
V A
= 100 V, Q-point for
Q
1 :(0.66 mA, 2.3 V), Q-point for
Q
2 :(1.6 mA, 7.5 V)
Analysis:
The circuit is redrawn to identify amplifier and feedback networks and appropriate 2-port parameters of feedback network are found.
Shunt-shunt transresistance configuration is used.
2 Microelettronica – Circuiti integrati analogici 2/ed
Richard C. Jaeger, Travis N. Blalock
Copyright © 2005 – The McGraw-Hill Companies srl
Analysis of Shunt-Series Feedback Pair (contd.)
Small signal parameters are found from given Q-points.
For
Q
1 ,
r
=3.79 k W ,
r
o = 155 k W .
For
Q
2 ,
r
=1.56 k W ,
r
o = 64.8 k W .
v th
R th
i i 4 .
94
R
10 kΩ
B R B
r
10 5 i i
r o
1 1 b
o
1 (
r o
1 8 .
88 kΩ
R C
) v 2 v th 8 .
88 kΩ ( b
o
2
r
2 1 )( 0 .
901 kΩ ) ( b
o
2 1 )( 0 .
901 kΩ )
2 Microelettronica – Circuiti integrati analogici 2/ed
Richard C. Jaeger, Travis N. Blalock
Copyright © 2005 – The McGraw-Hill Companies srl
Analysis of Shunt-Series Feedback Pair (contd.)
v
A
i i 2 4 .
43 10 5 W b
y
F 12 1 9100 S
A tr
1
A A
b 8910 W
R
in
R
( 1 A in
A
b )
R
(
B
1
r A
b 1 ) 42 .
5 W
R
out (
R
1 A out
A
b ) 1 .
86 W Closed-loop current gain is given by:
A i
i o i i
a
o i e i i
a
o v 2
901 W
i i
a
o
901 W
A tr
9 .
79
2 Microelettronica – Circuiti integrati analogici 2/ed
Richard C. Jaeger, Travis N. Blalock
Copyright © 2005 – The McGraw-Hill Companies srl
Direct Calculation of Loop Gain
Example: • • Original input source is set to zero.
Test source is inserted at the point where feedback loop is broken.
v r
T
A
b b v o b v r v x
A
( 0 v x ) b
A
v x
R
3
R
2
R
1 is added for proper termination of feedback loop.
v r v o
R
2
R
1
R
1
R
2
R
1
R
1
A
v x
R R id id
R
3
T
v r v x
A
R
2
R
1
R
1
R R id id
R
3
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Richard C. Jaeger, Travis N. Blalock
Copyright © 2005 – The McGraw-Hill Companies srl
Calculation of Loop Gain using Successive Voltage and Current Injection
Current injection:
Current
Voltage injection:
Voltage source
v X
is inserted at arbitrary point P in circuit.
v 1 v 2 1 v 1 b
A
b v x v x b where ( 1 1
A A
b b ) b v x
R A R
A R B T v
v 2 v 1 1
A
b b b
T
1
R B R A
R R B A
source
i X
i 1 v x
R A
is inserted again at P. i 2 v x
A R B
v x v x
R B A T i
T
i 2 i 1 1 2
T v T i
T v
A T i R
1 /
R A
1
B
As
A
b
= T
R R B A R R B A
1
R T
1 1
T v T i B R A
2 Microelettronica – Circuiti integrati analogici 2/ed
Richard C. Jaeger, Travis N. Blalock
Copyright © 2005 – The McGraw-Hill Companies srl
Simplifications to Successive Voltage and Current Injection Method
• Technique is valid even if source resistances with
v X
analysis.
and
i X
are included in • If at P, and
T R B
=
T v
is zero or
R A
is infinite,
T
can be found by only one measurement . In ideal op amp, such point exists at op amp input.
• If at P,
R B
output.
is zero,
T
=
T v
. In ideal op amp, such point exists at op amp • If
R A
= 0 or
R B
is infinite, • In practice, if
R B
>>
R A
or
T R
=
A T I
>> .
R B
, the simplified expressions can be used.
2 Microelettronica – Circuiti integrati analogici 2/ed
Richard C. Jaeger, Travis N. Blalock
Copyright © 2005 – The McGraw-Hill Companies srl
Blackman’s Theorem
• First we select ports where resistance is to be calculated.
• Next we select one controlled source in the amplifier’s equivalent circuit and use it to disable the feedback loop and also as reference to find
T SC
and
T OC
.
R CL
R D
1 1
T SC T OC R CL
= resistance of closed-loop amplifier looking into one of its ports (any terminal pair)
R D
= resistance looking into same pair of terminals with feedback loop disabled.
T SC T OC
= Loop gain with a short-circuit applied to selected port = Loop gain with same port open-circuited.
2 Microelettronica – Circuiti integrati analogici 2/ed
Richard C. Jaeger, Travis N. Blalock
Copyright © 2005 – The McGraw-Hill Companies srl
Blackman’s Theorem (Example 1)
Problem:
Find input and output resistances.
Given data:V REF =5 V,
R
=5 k W b
o
=100,
V A
=50 V,
A o
=10,000,
R id
=25 k W ,
R o
=0
Assumptions
: Q-point is known,
g m
= 0.04 S,
r
=25 k W ,
r o
=25 k W .
For output resistance:
T SC T OC R
out
R D
A o
1
A
v o
1
v r o
1 3 .
18 MΩ 1
A
o A
1
r o
b
o r
r
( b ( ( ( (
R o
9940
R
b
R
/
R id o
(
R id
1 )(
R R id
1 1 )
R
) )( )
R
R id R id R
) 6350 3 .
18 MΩ
r o id
6350 )
r o
) 9940 5 .
06 MΩ For input resistance:
R
in
R
D
R id
25 kΩ (
R g
1
m
) 25 k Ω 1 9940 / 1 249 MΩ
2 Microelettronica – Circuiti integrati analogici 2/ed
Richard C. Jaeger, Travis N. Blalock
Copyright © 2005 – The McGraw-Hill Companies srl
Blackman’s Theorem (Example 2)
Problem:
Find input and output resistances.
Given data:
k W ,
r
o b
o
= 100, = 64.8 k W .
V A
= 100 V, Q-point for
Q
1 :(0.66 mA, 2.3 V), Q-point for
Q
2 :(1.6 mA, 7.5 V). For
Q
1 ,
r
=3.79 k W ,
r
o = 155 k W , For
Q
2 ,
r
=1.56
2 Microelettronica – Circuiti integrati analogici 2/ed
Richard C. Jaeger, Travis N. Blalock
Copyright © 2005 – The McGraw-Hill Companies srl
Blackman’s Theorem (Example 2 contd.)
For output resistance:
T SC
b ( 9 .
1 kΩ b 10 kΩ 3 .
79
R D
kΩ .
1 .
( 56 kΩ
o
( 1 ) 0 .
901 kΩ
o
1 ) 0 .
901 kΩ
y f
)( 12
r o
2
g m
1 ) 1 48 .
79 kΩ 7
r
2 ( 10 b
o R C
( 1 /
r o y f
1 ) 22 ( ) 1 / kΩ 1 .
56 kΩ
y f
22 ) ( b
o
321 kΩ 1 ) 0 .
901 kΩ
T OC
( 2 .
11 kΩΩ)(4 0.66mA)(1.
93kΩ3 0.901kΩ 1.56kΩ 0.901kΩ
R
out 321 kΩ 1 48 .
7 / 1 4 .
33 2 .
99 MΩ For input resistance:
R D
10 kΩ 9 .
1 kΩ r 1 2 .
11 kΩ 1 9.1kΩ 4 .
33
R
in 2 .
11 kΩ 1 0 /( 1 48 .
7 ) 42 .
5 Ω
2 Microelettronica – Circuiti integrati analogici 2/ed
Richard C. Jaeger, Travis N. Blalock
Copyright © 2005 – The McGraw-Hill Companies srl
Blackman’s Theorem (Example 3)
• •
Problem:
disabled by Find expression for output resistance.
Analysis:
Feedback loop is to zero.
Assuming
R D
r o
3
g
1 =
r o
2
g
b
o r
= 3 ( 1 3 /
g
g m
( 1 / and m
f
1 )
g m
1 ) >> b o
r o
3
i
>>1.
i
i e
T SC
R
out ( b
o
3 1 )
i b
Next,
o
1
i r o
1 b 1
o
1 ( 1 b 2
o
b m
o f
1 ) 1 is set to 1 1
T OC
b
o
2
r o
( b
o
1 )
i e i
1 b m
f
1 1
2 Microelettronica – Circuiti integrati analogici 2/ed
Richard C. Jaeger, Travis N. Blalock
Copyright © 2005 – The McGraw-Hill Companies srl
Use of Feedback to Control Frequency Response
A v
s
2
H
L
( 1
H A o
F L
BW
F
1
L
A o H
b ( 1
A o F H
b
A o
) b ( 1 )
H A o
H s
b ( 1 )
s
A o
b )
A v
1
A
A
b
L
H
Upper and lower cutoff frequencies as well where
A
(
s
A o
L
)(
H s
s
H
) as bandwidth of amplifier are improved, gain is stabilized at GBW
A
mid
A
mid BW
F
1
A A A o o o
b
H
1 b
2 Microelettronica – Circuiti integrati analogici 2/ed
Richard C. Jaeger, Travis N. Blalock
Copyright © 2005 – The McGraw-Hill Companies srl
Use of Nyquist Plot to Determine Stability
• If gain of amplifier is greater than or equal to 1 at the frequency where feedback is positive, instability can arise.
A v
A
• Poles are at frequencies where
T(s)
• In Nyquist plots, each value of
s
in =-1.
s
-plane has corresponding value of
T(s).
• Values of
s
on
j
some value of
s
axis are plotted.
• If -1 point is enclosed by boundary, there is for which
T(s)
=-1, pole exists in RHP and amplifier is unstable.
• If -1 point lies in outside interior of Nyquist plot, all poles of closed-loop amplifier are in LHP and amplifier is stable.
2 Microelettronica – Circuiti integrati analogici 2/ed
Richard C. Jaeger, Travis N. Blalock
Copyright © 2005 – The McGraw-Hill Companies srl
First-Order Systems
T
(
j
)
j T o
1 At dc,
T
(0) =
T o
, but for >>1,
T
(
j
)
j T o
As monotonically approaches zero and phase asymptotically approaches -90 0 .
For a simple low-pass amplifier,
T
s A o
o o
b
s T o
o
It can also represent a single-pole As b changes, value of
T
(0) =
T o
regardless of value of
T o
.
is scaled but as
T
(0) changes, radius of circle changes, but it can never enclose the -1 point, so amplifier is stable op amp with resistive feedback
2 Microelettronica – Circuiti integrati analogici 2/ed
Richard C. Jaeger, Travis N. Blalock
Copyright © 2005 – The McGraw-Hill Companies srl
T
Second-Order Systems
In given example,
T
(
j
)
14 j
1 2
T
(0) =14, but, for high frequencies
A o
1
s
1
1
s
2
b
T o
1
s
1
1
s
2
As
T
(
j
) (
j
) 2
14
2
14
2 increases, magnitude monotonically decreases from 14 towards zero and phase asymptotically approaches -180 0 The transfer function can never enclose the -1 point but can come arbitrarily close to it.
2 Microelettronica – Circuiti integrati analogici 2/ed
Richard C. Jaeger, Travis N. Blalock
Copyright © 2005 – The McGraw-Hill Companies srl
Phase Margin
Phase Margin
is the maximum increase in phase shift that can be tolerated before system becomes unstable.
m
T
Where
T
( (
j
1 )
j
1 ) ( 180 ) 1 180
T
(
j
1 ) First we determine frequency for which magnitude of loop gain is unity, corresponding to intersection of Nyquist plot with unit circle shown and then determine phase shift at this frequency. Difference between this angle and -180 0 is phase margin.
Small phase margin causes excessive peaking in closed-loop frequency response and ringing in step response.
2 Microelettronica – Circuiti integrati analogici 2/ed
Richard C. Jaeger, Travis N. Blalock
Copyright © 2005 – The McGraw-Hill Companies srl
Third-Order Systems
T
1
s
1
1
T o
s 2
1
s
3
In given example,
T
(
s
)
14 s
3
s
2 3
s
2
T
(0) = 7, but, for high frequencies As
T
(
j
) (
j
) 3
14
3
j 14
3 increases, polar plot asymptotically approaches zero along positive imaginary axis and plot can enclose the -1 point under any circumstances and system is unstable.
2 Microelettronica – Circuiti integrati analogici 2/ed
Richard C. Jaeger, Travis N. Blalock
Copyright © 2005 – The McGraw-Hill Companies srl
Gain Margin
GM 1
T
(
j
180 ) Where
T
(
j
180 ) GM dB 180 20 log( GM)
Gain Margin
is the reciprocal of magnitude of
T
(
j
) evaluated at frequency for which phase shift is 180 0 .
If magnitude of
T
(
j
) is increased by a factor equal to or exceeding gain margin, then closed-loop system becomes unstable, because Nyquist plot then encloses -1 point.
2 Microelettronica – Circuiti integrati analogici 2/ed
Richard C. Jaeger, Travis N. Blalock
Copyright © 2005 – The McGraw-Hill Companies srl
Bode Plots
2 Microelettronica – Circuiti integrati analogici 2/ed
Richard C. Jaeger, Travis N. Blalock
A
b 2 10 19
s
10 5
s
10 6
s
10 7 At 1.2e+6 rad/s, magnitude of loop gain is unity and corresponding phase shift is 145 0 , and phase margin is given by 180 0 - 145 0 = 35 0 .
Amplifier can tolerate additional phase shift of 35 0 before it becomes unstable.
At 3.2e+6 rad/s, phase shift is exactly 180 0 and corresponding magnitude of loop gain is -17 dB, and phase margin is given by 17 dB.
Gain of amplifier must increase by 17 dB before amplifier becomes unstable.
Copyright © 2005 – The McGraw-Hill Companies srl
Use of Bode plot to Determine Stability
Frequency at which curves corresponding to magnitudes of open-loop gain and reciprocal of feedback factor intersect is the point at which loop gain is unity, phase margin is found from phase plot.
A
b
s
10 5
s
2 10 24 3 10 6
s
10 8 20 log
A
20 log
A
20 log 1 b Assuming feedback is independent of frequency, For 1/ b =80 dB,
m
=85 0 , amplifier is stable.
For 1/ b =50 dB,
m
=15 0 , amplifier is stable, but with significant overshoot and ringing in its step response.
For 1/ b =0 dB,
m
= -45 0 , amplifier is unstable
2 Microelettronica – Circuiti integrati analogici 2/ed
Richard C. Jaeger, Travis N. Blalock
Copyright © 2005 – The McGraw-Hill Companies srl
• •
Operational Amplifier Compensation Example
Problem:
Find value of compensation capacitor for
m
=70 0 .
Given data:
R C
1 =3.3 k W ,
R C
2 =12 k W ,SPICE parameters-BF=100, VAF=75 V, IS=0.1 fA, RB=250 W , TF=0.75 ns, CJC= 2 pF.
Assumptions:
Dominant pole is set by
C C
and
pnp
C-E stage.
R Z
is included to remove zero associated with
C C
,
pnp
and
npn
transistors are identical, quiescent value of
V o
=0, VJC=0.75 V, MJC=0.33.
Q 4
and
Q 5
are in parallel, small signal resistances of diode-connected
Q 7
neglected.
and
Q 8
can be
2 Microelettronica – Circuiti integrati analogici 2/ed
Richard C. Jaeger, Travis N. Blalock
Copyright © 2005 – The McGraw-Hill Companies srl
•
Operational Amplifier Compensation Example (contd.)
Analysis:
For
V o I C
1
=
I C
2
=
250 m A. =0,voltage across
R C
2 =12-0.75=11.3 V and
I C
3 =11.3V/12 k W =938 m A.
Q 4
and
Q 5
and
Q 8
mirror currents in , so, m A. For
V o V CE5
=12 V,
V CE3
=0,
V CE2 I C
4 =0,
=
I C
5
V CE4 Q
=
938 =12 V,
7 V I
=11.3 V. For 11.9 V =12.8 V,
V CE1
=12-3300(0.25 mA)+0.75= Small signal parameters are found using their respective formulae.
A v
1
A v
0 .
01 ( 696 kΩ 2 2
g m
2 (
r o
3 3 .
3 kΩ
R C
2
g
0 .
0375 ( 92 kΩ 338
A v
3 0 .
974
m
1 2
r
2 4
A v
( ( 2
r o
1 b
o
( b
R C
1 12 kΩ 1 )
o
4
R L
A v
1
A v
2 1 )
r
3 .
07 kΩ )
r
2 3 )
A v
3 4
R L
2 ( b
o
4 3 .
09 kΩ 3090 2 2610 7 .
93 1 )
R L
( 117 ) 500 W ) ( 117 ) 500 ( 117 ) 500
2 Microelettronica – Circuiti integrati analogici 2/ed
Richard C. Jaeger, Travis N. Blalock
Copyright © 2005 – The McGraw-Hill Companies srl
Operational Amplifier Compensation Example (contd.)
Input stage pole: At
f T
,dominant pole due to
C C f
H f
H
2 1 82 .
5 MHz
r
59 .
2 MHz
C
C
m 2 1
g m
2
R C
R C r x
20 contributes phase shift of 90 0 . For
m
=70 0 , other 2 poles can contribute more phase margin of 20 0 .
Emitter Follower pole:
Q 4
in parallel, composite parameters are-
g C m
m =0.02 S,
r
=1.60 pF,
x R
=125 W ,
C
th
=1/
g m3
=56.2 pF, =267 W .
1 1 2 (
R th
r x
R L
) 1
C
and
Q 5 g m R L
C
are m
R th
f T
C C r x
tan 1 12 .
2 MHz
C
m 3
f T
59 .
2 MHz
G
m
1
T
R Z
=1/
g m3
g m
2 1 tan 1 2 1
f T
=27.5 W
f T
82 .
5 MHz 65 pF
f
remove zero associated with
C C.
B
f A T o
12 .
2 MHz 2610 4.67kHz
is included to
2 Microelettronica – Circuiti integrati analogici 2/ed
Richard C. Jaeger, Travis N. Blalock
Copyright © 2005 – The McGraw-Hill Companies srl
Barkhausen’s Criteria for Oscillation.
• For sinusoidal oscillations, 1
T
0
T
1 • • For sinusoidal oscillator, poles of closed-loop amplifier should be at frequency
0
on
j
axis.
Use positive feedback through frequency-selective feedback network to ensure sustained oscillation at
0
.
A v
1
A A s s
b 1
A T s
• • •
T
Barkhausen’s criteria state
T j
1 0 Or even multiples of 360 0 Phase shift around feedback loop should be zero degrees and magnitude of loop gain must be unity.
Loop gain greater than unity causes distorted oscillations.
2 Microelettronica – Circuiti integrati analogici 2/ed
Richard C. Jaeger, Travis N. Blalock
Copyright © 2005 – The McGraw-Hill Companies srl
Oscillators with Frequency-Selective
RC
Networks: Wien-Bridge Oscillator
T(s)
V o (
s
) V o V I (
s
) (
s
) V 1 (
s
) ( 1
Z
1 (
Z s
) 2 (
s Z
) (
s
) 2
sRCG
2
R
2
C
2 ) 3
sRC
1 At
0 T(j
=1/
RC o )
G T(j
o )
T(j G
3 o 3 )
0 This oscillator is used for frequencies upto few MHz, limited primarily by characteristics of amplifier
.
2 Microelettronica – Circuiti integrati analogici 2/ed
Richard C. Jaeger, Travis N. Blalock
Copyright © 2005 – The McGraw-Hill Companies srl
Oscillators with Frequency-Selective
RC
Networks: Phase-Shift Oscillator
V V o ( 2 (
s
)
s
)
sCR
1
T
(
s
) V V o o ' ( (
s
)
s
) 3
s
2
s
3
C
3
R
2
R
1
R
2
C
2 4
sRC
1 ( 1
o
2
o
1 3
RC
At
0 T(j
o )
o
2
C
2
RR
1
4
2
C
1 12
R
1
R
sC
V o ' 0 (
s
) (2sC
sC
G) (2sC
sC
G) V V 1 ( 2 (
s
)
s
)
2 Microelettronica – Circuiti integrati analogici 2/ed
Richard C. Jaeger, Travis N. Blalock
Copyright © 2005 – The McGraw-Hill Companies srl
Amplitude Stabilization
• Loop gain of oscillator changes due to power supply voltage, component value or temperature changes.
• If loop gain is too small, desired oscillation decays and if it is too large, waveform is distorted.
• Amplitude stabilization or gain control is used to automatically control loop gain and place poles exactly on
j
axis.
• At power on, loop gain is larger than that required for oscillation.As oscillation builds up, gain is reduced o minimum required to sustain oscillations.
2 Microelettronica – Circuiti integrati analogici 2/ed
Richard C. Jaeger, Travis N. Blalock
Copyright © 2005 – The McGraw-Hill Companies srl
Amplitude Stabilization in
RC
Oscillators: Method 1
R
1 is replaced by a lamp. Small-signal resistance of lamp depends on temperature of bulb filament.
If amplitude is large, current is large, resistance of lamp increases, gain is reduced. If amplitude is small, lamp cools, resistance decreases, loop gain increases. Thermal time constant of bulb averages signal current and amplitude is stabilized.
2 Microelettronica – Circuiti integrati analogici 2/ed
Richard C. Jaeger, Travis N. Blalock
Copyright © 2005 – The McGraw-Hill Companies srl
Amplitude Stabilization in
RC
Oscillators: Method 2
R
2
R
3
R
2
R
3
R
4 slightly >3 ensuring oscillation, but, when one diode is on, gain is reduced to slightly<3.
i v o
R
3 v 1 v o v 1
R
4
V D
v 1 v 3 o 1 For positive signal at voltage across on voltage.
R 4 R 3
loop gain is reduced.
v D o 2
,
D 1
turns on as exceeds diode turn is in parallel with similarly at negative signal peak.
R
functions
3
, v o 2
R
2
R
1 shift oscillators.
3
V
1
D R
4
R
3
R R
4 1
R
2
R
1 Same method can also be used in phase 2
R
2
R
1
2 Microelettronica – Circuiti integrati analogici 2/ed
Richard C. Jaeger, Travis N. Blalock
Copyright © 2005 – The McGraw-Hill Companies srl
LC
Oscillators: Colpitts Oscillator
0 0
s
(
C
3
s
(
C GD C
3 )
g m
1 / )
sL
s
(
C
1
C
3
sC
3 )
g m
G
V g V s ( (
s s
) )
g s
2
m
sL C
1
G C
3 (
C
1
C GD
C
3
L
) (
C
1
C
3 )
s
(
C GD
C
3 )
G
=0, collect real and imaginary parts and set
GC
3
G
1 /(
R S r o
)
C
3
C
2
C GS
them to zero.
o
1
C TC
C GD
C C
1
C
1 3
C
3
0
oscillation with amplitude stabilization.
2 Microelettronica – Circuiti integrati analogici 2/ed
Richard C. Jaeger, Travis N. Blalock
Copyright © 2005 – The McGraw-Hill Companies srl
LC
Oscillators: Hartley Oscillator
G-S and G-D capacitances are neglected, assume no mutual coupling between inductors.
0 0 (
sC
1 / 1 /
sL
2 )
sL
2
g m
( 1 /
sL
1 ) ( 1 / 1 /
sL sL
2 ) 2
g m
g o
V g V s ( (
s
)
s
)
sC g m
g o
g m sL
2 1
s
2
L
1
L
2
C
1
L
1 1
L
2 =0, collect real and imaginary parts and set them to zero.
o
1
C
(
L
1
L
2 ) At
0
m
f
L L
1 2 Generally more gain is used to ensure oscillation with amplitude stabilization.
2 Microelettronica – Circuiti integrati analogici 2/ed
Richard C. Jaeger, Travis N. Blalock
Copyright © 2005 – The McGraw-Hill Companies srl
Amplitude Stabilization in
LC
Oscillators
• Inherent nonlinear characteristics of transistors are used to limit oscillation amplitude. Eg: rectification by JFET gate diode or BJT base-emitter diode.
• In MOS version, diode and
R G
form rectifier to establish negative bias on gate, capacitors act as rectifier filter.
• Practically, onset of oscillation is accompanied by slight shift in Q-point values as oscillator adjusts to limit amplitude.
2 Microelettronica – Circuiti integrati analogici 2/ed
Richard C. Jaeger, Travis N. Blalock
Copyright © 2005 – The McGraw-Hill Companies srl
Crystal Oscillators
Z C
Z Z P P
Z Z S S
1
sC P
s
2
s
2
s R L s R L
1
LC
1
S LC T
Crystal: A piezoelectric device that vibrates is response to electrical stimulus, can be modeled electrically by a very high (>10,000) resonant circuit.
Q C T
C C P P
C S C S L
,
C S
,
R
represent intrinsic series resonance path through crystal.
C P
is package capacitance. Equivalent impedance has series resonance where
C S
resonates with parallel resonance where
L L
and resonates with Below
S
and above
P
, crystal appears capacitive, between
S
and
P
it exhibits inductive reactance.
series combination of
C S
and
C P
.
2 Microelettronica – Circuiti integrati analogici 2/ed
Richard C. Jaeger, Travis N. Blalock
Copyright © 2005 – The McGraw-Hill Companies srl
• •
Crystal Oscillators: Example
Problem:
Find equivalent circuit elements for crystal with given parameters.
Given data:
f
S =5 MHz,
Q
=20,000
R
=50 W ,
C P
=5 pF
Analysis:
L
RQ
S C S
S
1 2 50 ( 20 , 000 ) 2
L
( 5 10 6 ) 10 7 1 31 .
8 mH 2 ( 0 .
0318 ) 31 .
8 fF
f P
2
L
5.02MHz
1
C C P P
C S C S
2 1 ( 31 .
8 mH)( 31 .
6 fF)
2 Microelettronica – Circuiti integrati analogici 2/ed
Richard C. Jaeger, Travis N. Blalock
Copyright © 2005 – The McGraw-Hill Companies srl
Crystal Oscillators: Topologies
Colpitts Crystal Oscillator Crystal Oscillator using BJT Crystal Oscillator using JFET Crystal Oscillator using CMOS inverter as gain element.
2 Microelettronica – Circuiti integrati analogici 2/ed
Richard C. Jaeger, Travis N. Blalock
Copyright © 2005 – The McGraw-Hill Companies srl