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.
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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: Feedback can increase
or decrease input and output resistances of an
amplifier.
• Bandwidth: Bandwidth of amplifier can be extended
using feedback.
• Nonlinear Distortion: Feedback reduces effects of
nonlinear distortion.eg: removal of dead zone in classB amplifiers
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Classic Feedback Systems
Vo(s)  V (s) A(s)
d
V (s)  Vo(s)b (s)
f
V ( s)
A s
As
Av s   o 
  
V (s) 1 A sb  s 1 T  s
i
• A(s) = transfer function of
open-loop amplifier or
open-loop gain.
• b(s) = transfer function of
feedback network.
T(s) = loop gain
For negative feedback: T(s) > 0
For positive feedback: T(s) < 0
V ( s)  V ( s)  V ( s)
d
i
f
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Voltage Amplifiers: Series-Shunt
Feedback (Voltage Gain Calculation)
v  hT i  h F v  v  ( R  hT )i  h F v
I 11 1 12 2
1 11 1 12 2
i
i  h A i  hT v  0  h A i  (hT  G )v
21 1
22 L 2
2 21 1 22 2
A
h
A
21
Av  2 

v h A h F  ( R  hT )(hT  G ) 1 Ab
21 12
I 11 22 L
i
v
A i  hA v
v1A  h11
1 12 2
A i  hA v
i2A  h 21
1 22 2
F i  hF v
v1F  h11
1 12 2
F i  hF v
i2F  h21
1 22 2
A
h 21
A 
( R  hT )(hT  G )
I 11 22 L
b  hF
12
A  h F h F  h A
T  hA  hF
h
hij
and
21
21, 12
12
ij ij
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Voltage Amplifiers: Series-Shunt Feedback
(Two-Port Representation)
• Gain of amplifier should include
effects of h F , h F , RI and RL.
11 22
• Required h-parameters are found
from their individual definitions.
• Two-port representation of the
amplifier is as shown
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Voltage Amplifiers: Series-Shunt
Feedback (Input and Output
- hA
Resistances)
21 i
v  ( R  T )i  F
I h11 1 h12 ( T  G ) 1
h 22 L
v
R  i  ( R  hT )(1 Ab )
I 11
in i
1
R  R A (1 Ab )
in
in
Series feedback at a port increases
input resistance at that port.
i
For output resistance:
v   R i v  vx i  ix  G v
L 2
I1 2
2
1
ix  h A i  (hT  G )vx
21 1
22 L
0  ( R  hT )i  h F vx
I 11 1 12
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













1
T
A
v x h 22  GL
Rout
Rout 

Rout 
ix
1  Ab
1 Ab
Shunt feedback at a port reduces
resistance at that port.
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Voltage Amplifiers: Series-Shunt
Feedback (Example)
• Problem: Find A, b, closedloop gain, input and output
resistances.
• Given data: R1=10 kW, R2=91
kW, Rid=25 kW, Ro=1 kW,
A104.
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• Analysis:
Richard C. Jaeger, Travis N. Blalock
v
F
 R R 9.01kΩ
h11  1
1 2
i
1 v 0
2
i
1
1
F
2



h 22
v
R R
101kΩ
2 i 0
1
2
1
v
R
F
1
1  0.0990

h12 
v
R R
2 i 0
1
2
1
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Voltage Amplifiers: Series-Shunt
Feedback (Example contd.)
v
25kΩ
1.96kΩ
A o 
(104)
 4730
v 1kΩ  25kΩ  9.01kΩ
1.96kΩ 1.00kΩ
i
A
4730
Av 

10.1
1 Ab 1 4730(0.0990)
R  R A (1 Ab ) 16.4MW
in
in
A
R out
Rout 
1.41W
1 Ab
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Transresistance Amplifiers: Shunt-Shunt
Feedback (Voltage Gain Calculation)
i  y T v  y F v  i  (G  y T )v  y F v
I 11 1 12 2
1 11 1 12 2 i
i  y A v  y T v  0  y A v  ( y T  G )v
21 1
22 L 2
2 21 1 22 2
yF
A
21
Atr  2 

y F y A  (G  y T )( y T  G ) 1 Ab
i
21 12
I 11 22 L
i
v
A v  yA v
i1A  y11
1 12 2
A v  yA v
i2A  y 21
1 22 2
F v  yF v
i1F  y11
1 12 2
F v  yF v
i2F  y 21
1 22 2
yA
21
A 
(G  y T )( y T  G )
I 11 22 L
b  yF
12
y T  y A  y F and y A  y F , y F  y A
21
21 12
12
ij ij ij
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Transresitance Amplifiers: Shunt-Shunt
Feedback (Two-Port Representation)
• Gain of amplifier should include
effects of y F , y F , RI and RL.
11 22
• Required y-parameters are found
from their individual definitions.
• Two-port representation of the
amplifier is as shown.
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Richard C. Jaeger, Travis N. Blalock
Copyright © 2005 – The McGraw-Hill Companies srl
Transresistance Amplifiers: Shunt-Shunt
Feedback (Input and Output
- yA
Resistances)
21 v
i  (G  y T )v  y F
i
I
11 1







12 ( y T  G ) 1
22
L


1


T
A

G

v
h

R
11

in
R  1 I
in i
(1 Ab ) (1 Ab )
i
Shunt feedback at a port reduces
resistance at that port.
For output resistance:
i  G v
i  ix  G v
I 1
L 2
1
2
ix  y A v  ( y T  G )vx
21 1
22 L
0  (G  y T )v  y F vx
I 11 1 12
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






1
T
v x y 22  GL
Rout 

ix
1  Ab







A
Rout
Rout 
1 Ab
Resistance at output port is reduced
due to shunt feedback.
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Transresistance Amplifiers: Shunt-Shunt
Feedback (Example)
• Problem: Find A, b, closed-loop
gain, input and output
resistances.
• Given data: VA= 50 V, bF= 150
• Analysis: From dc equivalent
circuit,
V V
I  CC BE  0.970mA
C
R R
R  C F
C
b
F
V V  ( I  I )R  1.35V
CE CC C B C
gm  40(0.977mA)  39.1mS
1
r 
 3.84kW
gm
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ro 
50V 1.35V
 52.6kW
0.977mA
i
1
F
y  1

 10- 5S
11 v
R
F
1 v 0
2
i
1
yF  2

 10- 5S
22 v
R
F
2 i 0
2
i
1
yF  1

 10- 5S
12 v
R
F
2 v 0
1
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Transresistance Amplifiers: Shunt-Shunt
Feedback (Example contd.)
4.76kΩ
i i
b i 4.76kΩ  r

vo   boi (1.41kΩ ro )
b
A 
vo
i
i

4.76kΩ
(150)1.41kΩ 52.6kΩ   114kΩ


4.76kΩ  3.84kΩ
A
114kΩ
Atr 

 53.3kΩ
1 Ab 1114kΩ(0.01mS)


R  R A (1 Ab )   R R r (1 Ab )  995W
in
in
F 
 I
R R R ro
A
R out
Rout 
 L F C  640W
1 Ab
1 Ab
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Richard C. Jaeger, Travis N. Blalock
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Current Amplifiers: Shunt-Series
Feedback (Voltage Gain Calculation)
i  g T v  g F i  i  (G  g T )v  g F i
I 11 1 12 2
1 11 1 12 2 i
v  g A v  g T i  0  g A v  ( g T  R )i
21 1
22 L 2
2 21 1 22 2
gA
A
21
A 2

i i g A g F  (G  g T )( g T  R ) 1 Ab
21 12
I 11 22 L
i
i
i  iA  i F
1 1 1
v  vA  vF
2
2 2
gA
21
A 
(G  g T )( g T  R )
I 11 22 L
b  gF
12
g T  g A  g F and g A  g F , g F  g A
ij
ij
ij
12
21
21 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 g F , g F , RI and RL.
11 22
• Required g-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
Current Amplifiers: Shunt-Series
Feedback (Input and Output
- gA
Resistances)
21 v
i  (G  g T )v  g F
i
I
11 1







12 ( g T  R ) 1
L
 22

1


T
A

g
G

v

R
11

in
R  1 I
in i
(1 Ab ) (1 Ab )
i
Shunt feedback at a port decreases
resistance at that port.
For output resistance:
i  G v
v  vx  R i
I 1
L2
1
2
vx  g A v  ( g T  R )i
21 1
22 L 2
0  (G  g T )v  g F i
I 11 1 12 2
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vx
Rout 
  g T  R 1 Ab 
L
 22
i
2
A 1 Ab 
 Rout


Series feedback at output port
increases resistance at that port.
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Transconductance Amplifiers: Series-Series
Feedback (Voltage Gain Calculation)
v  z T i  z T i  v  ( R  z T )i  z F i
1 11 1 12 2
S 11 1 12 2
i
v  z T i  z T i  0  z T i  ( z T  R )i
21 1
22 L 2
2 21 1 22 2
A
z
A
21
Atc  2 

v z A z F  ( R  z T )( z T  R ) 1 Ab
I 11 22 L
i 21 12
i
v  vA  vF
1
1 1
v  vA  vF
2
2 2
T  zA  zF
zij
ij ij and
A
z 21
A
( R  z T )( z T  R )
I 11 22 L
b  zF
12
A  z F z F  z A
z 21
21, 12
12
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Richard C. Jaeger, Travis N. Blalock
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Transconductance Amplifiers: Series-Series
Feedback (Input and Output Resistances)
v
R  i   z T  R 1 Ab 
in i  11 I 
1
 RA 1 Ab 
in
vx
• Gain of amplifier should include
effects of z F , z F , RI and RL.
11 22
• Required g-parameters are found
from their individual definitions.
• Two-port representation of the
amplifier is as shown
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Richard C. Jaeger, Travis N. Blalock
Rout 
  z T  R 1 Ab 
L
 22
i
2
A 1 Ab 
 Rout


Series feedback at input and output
port increases resistance at both ports.
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•
•
•
Erroneous Application of 2-port
Feedback Theory
Problem: Find A, b, closed-loop gain, input and output resistances.
Given data: VREF = 5 V, bo= 100, VA = 50 V, Ao = 10,000, Rid =25 kW, Ro =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.
ie is sampled by feedback network instead of io. This assumption is
made since ao 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 Acircuit, IE=1 mA
50V
1000(0.025V)
 50kW
 2.5kW ro 
1mA
1mA
R
bo
id A
io  v
i R  R o r  ( b 1)R

o
id
io
A
1.64S
Atc  

 0.200mS
v 1 Ab 11.64S(5kW)
i
R  RA 1 Ab   (R  R)1 Ab   246MΩ
R  RA 1 Ab   27.7GΩ
in
in
in
id
in
r 
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Erroneous Application of 2-port
Feedback Theory (contd.) Results for R
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 Rout is limited to Rout  boro
out
3 and 4 are not valid terminals as
current entering 3 is not same as that
SPICE analyses confirm results for Atc and exiting 4. Amplifier can’t be reduced
Rin, but results for Rout are in error. For Atc to a 2-port.
and Rin, amplifier can be properly modeled
as a series-shunt feedback amplifier, as
collector of Q1 can be directly connected to
ground for calculations and a valid 2-port
representation exists as shown.
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Richard C. Jaeger, Travis N. Blalock
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Analysis of Shunt-Series Feedback Pair
•
•
•
Problem: Find A, b, closed-loop gain, input and output resistances.
Given data: bo= 100, VA = 100 V, Q-point for Q1:(0.66 mA, 2.3 V), Q-point
for Q2:(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.
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Richard C. Jaeger, Travis N. Blalock
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Analysis of Shunt-Series Feedback Pair
(contd.)
Small signal parameters are found
from given Q-points.
For Q1, r=3.79 kW, ro = 155 kW.
For Q2, r =1.56 kW, ro = 64.8 kW.
R
B b (r R )
v  i
o1 o1 C
th
i R r
B 1
 4.94 105i
i
R  10kΩ r  8.88kΩ
o1
th
(b
1)(0.901kΩ)
o
2
v v 
2 th 8.88kΩ  r  ( b 1)(0.901kΩ)
2
o2
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Analysis of Shunt-Series Feedback Pair
(contd.)
v
A  2  4.43105W
b  yF   1 S
12 9100
i
i
A
Atr 
 8910W
1 Ab
A
R r
Rin
R 
 B  1  42.5W
in (1 Ab ) (1 Ab )
A
R out
Rout 
 1.86W
(1 Ab )
Closed-loop current gain is given by:
v
2
io aoie ao 901W ao
A 


Atr  9.79
i i
i
i
901W
i
i
i
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Richard C. Jaeger, Travis N. Blalock
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Direct Calculation of Loop Gain
Example:
R  R R is added for proper
3
2 1
• Original input source is set to zero. termination
of feedback loop.


• Test source is inserted at the point


R
R
R 


id 
1 
1   Av
where feedback loop is broken.
v r  vo
x R R 
R  R  R  R 
v r  bvo  bA(0  v x )   bAv x
2 1  2 1 
id 3 



vr
A



R
R
v
T  




id
1 
T   r  A

b
vx

 R  R  R
vx

R



2
1
3

 id

2 Microelettronica – Circuiti integrati analogici 2/ed
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
source iX is inserted again at P.
v x  Av x
vx
1 A
i 
 vx
i 
2
1 R
R
R
B
B
A


1 A R


i
R
R
Voltage injection: Voltage source vX is



B  B  T 1 B 
T  2
i i
inserted at arbitrary point P in circuit.
R

1/ R
R 
A

R
1
A
A

b v
A
v 
b

where
T T 1
R 1 T
1 1 Ab x
R  R T  v i
v
B
B
A
R
b
 (1 Ab )
2

T

T
A 1 Ti
v i
v  v  vx 
vx
2
1
1 Ab


v 1 Ab  b
R  R
Tv   2 
 T 1 B   B As Ab = T

v
b
R  RA

1
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 vX and iX are included in
analysis.
• If at P, RB is zero or RA is infinite, T can be found by only one measurement
and T = Tv . In ideal op amp, such point exists at op amp input.
• If at P, RB is zero, T = Tv . In ideal op amp, such point exists at op amp
output.
• If RA = 0 or RB is infinite, T = TI .
• In practice, if RB >> RA or RA >> RB, 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 TSC and TOC.
1 T
SC
R R
D 1 T
CL
OC
RCL = resistance of closed-loop amplifier looking into one of
its ports (any terminal pair)
RD = resistance looking into same pair of terminals with
feedback loop disabled.
TSC = Loop gain with a short-circuit applied to selected port
TOC = 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:VREF =5 V, R =5 kW bo=100, VA=50 V, Ao=10,000, Rid=25 kW, Ro=0
Assumptions: Q-point is known, gm = 0.04 S, r =25 kW, ro =25 kW.




b
(
R
R
)

For output resistance:
o
id 

R  ro 1
 3.18MΩ


D
r  ( R R ) 


id 

Aov
( bo 1)( R R ro )
id
1
T

 Ao
 9940
SC
1
r  ( bo 1)( R R ro )
id
(R R )
id
T
 Aov  Ao
 6350
1
OC
r  ( R R )
id
Rout  3.18MΩ 1 9940 /1 6350   5.06MΩ
R  R  (R g1 )  25kΩ
For input resistance:
D
id
m
R  25kΩ 1 9940 /1  249MΩ
in
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:bo= 100, VA = 100 V, Q-point for Q1:(0.66 mA, 2.3 V), Q-point
for Q2:(1.6 mA, 7.5 V). For Q1, r=3.79 kW, ro = 155 kW, For Q2, r =1.56
kW, ro = 64.8 kW.
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.)









bo (1/ y f )

22
For output resistance: R  r 1
  321kΩ
D o2
f
r  ( R r )  (1/ y ) 
22 
2
C o1
T
 (9.1kΩ 10kΩ 3.79kΩ)( g ) 79kΩ 10kΩ 1.56kΩ  ( bo 1)0.901kΩ
m1 

SC
( bo 1)0.901kΩ
f  48.7
.
y12
1.56kΩ  ( bo 1)0.901kΩ


0.901kΩ
1 

T
 (2.11kΩΩ)(4 0.66mA)(1.93kΩ3

  4.33


OC
 1.56kΩ  0.901kΩ  9.1kΩ 
Rout  321kΩ 1 48.7  /1 4.33   2.99MΩ
For input resistance:
R 10kΩ 9.1kΩ r  2.11kΩ
D
1
R  2.11kΩ 1 0 /(1 48.7)   42.5Ω
in
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: Find expression for output resistance.
•
Analysis: Feedback loop is
disabled by
Assuming gm1= gm2 = gm3 and mf >> bo >>1.
setting
reference source
i


b (1/ g )


o3
m1
r
R  r 1
to zero.
D o3
r  r  (1/ g )  o3


o2
3
m1


( bo 1)
i  ie  ( b 1)i  
 ( bo 1)
o3
b
2 bo  1
1
m
f
Next, i is set to 1
ie
b 1
T
  1 mo  1
T
 bo 1
SC
OC i
f
 1 b 1
b
r
o   o o
Rout  ro 
11 
2


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
Av s 
where
As
1 Asb  s
Ao s
H
As  
(s   )(s   )
L
H
2 Microelettronica – Circuiti integrati analogici 2/ed
Richard C. Jaeger, Travis N. Blalock
Ao s
H
Av s 
s2    (1 Ao b )s  
H
L H
 L

Assuming  (1 Aob )  ,
H
L

F
L
  (1 Aob )
F
L 
H
H
1 Ao b
BW   (1 Ao b )
F
H
Upper and lower cutoff frequencies as well
as bandwidth of amplifier are improved,
Ao
1
gain is stabilized at A


mid 1 A b b
o
GBW  A
 BW  Ao
F
H
mid
Copyright © 2005 – The McGraw-Hill Companies srl
Use of Nyquist Plot to Determine
Stability
• If gain of amplifier is greater than or equal to
•
•
•
•
•
2 Microelettronica – Circuiti integrati analogici 2/ed
Richard C. Jaeger, Travis N. Blalock
1 at the frequency where feedback is positive,
instability can arise.
As
Av s   
1 T where
s
Poles are at frequencies
T(s)=-1.
In Nyquist plots, each value of s in s-plane
has corresponding value of T(s).
Values of s on j axis are plotted.
If -1 point is enclosed by boundary, there is
some value of s 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.
Copyright © 2005 – The McGraw-Hill Companies srl
First-Order Systems
To
j 1
At dc, T(0) = To, but for  >>1,
T ( j ) 
To
T ( j )   j

As  increases, magnitude
monotonically approaches zero and
phase asymptotically approaches -900.
As b changes, value of T(0) = To is
For a simple low-pass amplifier, scaled but as T(0) changes, radius of
Aoo
To
circle changes, but it can never enclose


Ts
b
the -1 point, so amplifier is stable
s  o
s  o
It can also represent a single-pole regardless of value of To.
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
Second-Order Systems
In given example, T ( j ) 
14
2
j 1
T(0) =14, but, for high frequencies


14
14
T ( j )  ( j)2

2 2
T s  







Ao
1
s

1







1
To
b 

s

2












1
2 Microelettronica – Circuiti integrati analogici 2/ed
Richard C. Jaeger, Travis N. Blalock
s

1







1
s

2







As  increases, magnitude
monotonically decreases from 14
towards zero and phase asymptotically
approaches -1800 The transfer function
can never enclose the -1 point but can
come arbitrarily close to it.
Copyright © 2005 – The McGraw-Hill Companies srl
Phase Margin
m  T ( j )  (180)  180  T ( j )
1
Where T ( j1) 1
Phase Margin is the maximum
increase in phase shift that can be
tolerated before system becomes
unstable.
2 Microelettronica – Circuiti integrati analogici 2/ed
Richard C. Jaeger, Travis N. Blalock
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 -1800 is
phase margin.
Small phase margin causes excessive
peaking in closed-loop frequency
response and ringing in step response.
Copyright © 2005 – The McGraw-Hill Companies srl
Third-Order Systems
In given example, T (s) 
14
s3  s 2  3s  2
T(0) = 7, but, for high frequencies
14
14
T ( j )  ( j)3
 j
3
3
T s 







1
s

1







To
1
s

2







1
2 Microelettronica – Circuiti integrati analogici 2/ed
Richard C. Jaeger, Travis N. Blalock
s

3







As  increases, polar plot
asymptotically approaches zero along
positive imaginary axis and plot can
enclose the -1 point under any
circumstances and system is unstable.
Copyright © 2005 – The McGraw-Hill Companies srl
Gain Margin
GM 
1
T ( j
)
180
Where T ( j180)  180
GM
Gain Margin is the reciprocal of
magnitude of T(j) evaluated at
frequency for which phase shift is
1800.
2 Microelettronica – Circuiti integrati analogici 2/ed
Richard C. Jaeger, Travis N. Blalock
dB
 20log(GM)
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.
Copyright © 2005 – The McGraw-Hill Companies srl
Bode Plots
2 1019
Ab 




 s 105  s 106  s 107 








At 1.2e+6 rad/s, magnitude of loop gain is unity
and corresponding phase shift is 1450, and phase
margin is given by 1800 - 1450 = 350.
Amplifier can tolerate additional phase shift of
350 before it becomes unstable.
At 3.2e+6 rad/s, phase shift is exactly 1800 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.
2 Microelettronica – Circuiti integrati analogici 2/ed
Richard C. Jaeger, Travis N. Blalock
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.
2 1024
Ab 




 s 105  s  3106  s 108 








Assuming feedback is independent of frequency,
For 1/b =80 dB, m=850, amplifier is stable.
For 1/b =50 dB, m=150, amplifier is stable, but with
significant overshoot and ringing in its step response.
1 For 1/b =0 dB,  = -450, amplifier is unstable
20log A  20log A  20log
m
b
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  =700.
m
•
Given data: RC1=3.3 kW , RC2 =12 kW ,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 CC and pnp C-E stage. RZ is
included to remove zero associated
with CC, pnp and npn transistors are
identical, quiescent value of Vo =0,
VJC=0.75 V, MJC=0.33. Q4 and Q5
are in parallel, small signal resistances
of diode-connected Q7 and Q8 can be
neglected.
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.)
g
• Analysis: IC1 = IC2 =250 mA.
For Vo =0,voltage across RC2
=12-0.75=11.3 V and IC3
=11.3V/12 kW=938 mA. Q4
and Q5 mirror currents in Q7
and Q8 , so, IC4 = IC5 =938
mA. For Vo =0, VCE4 =12 V,
VCE5=12 V, VCE3 =11.3 V. For
VI =0, VCE2 =12.8 V, VCE1
=12-3300(0.25 mA)+0.75=
11.9 V
Small signal parameters are
found using their respective
formulae.
2 Microelettronica – Circuiti integrati analogici 2/ed
Richard C. Jaeger, Travis N. Blalock
A  m1 (2r R r )
v1 2
o1 C1  3
0.01

(696kΩ 3.3kΩ 3.07kΩ)  7.93
2
r

 4

A  g (r R 
 ( b 1)R 
v2 m2 o3 C 2  2
L 
o4

 0.0375(92kΩ 12kΩ
3.09kΩ
 (117)500W)
2
 338
( bo 1)R
(117)500
L
A r

v3
3090
 4  ( b 1)R
 (117)500
L
o4
2
2
 0.974
Av  A A A  2610
v1 v2 v3
Copyright © 2005 – The McGraw-Hill Companies srl
Operational Amplifier Compensation
Example
(contd.)
Input stage pole:
At f ,dominant pole due to C
T
f
H

1
2
1






C
0
90 . For
contributes phase shift of
m=700, other 2 poles can contribute
more phase margin of 200.
gm R
R  
C  C 
r C  Cm 2 

2
rx  




f
f





T
T
  tan1

20  tan1



 59.2MHz
 59.2MHz 
 82.5MHz 




Emitter Follower pole: Q4 and Q5 are  f  12.2MHz
T
in parallel, composite parameters areG
g  1 


  65pF
gm =0.02 S, rx =125 W, C =56.2 pF,  CC  Cm 3   m1  m1 
 2f

2


T
T


Cm =1.60 pF, Rth =1/ gm3 =267 W.
1
1
RZ =1/ gm3 =27.5 W is included to
f 


H 2
remove zero associated with CC.


C


   C  R  r 
f
( R  rx  R )
L  1 g R  m  th x  f  T  12.2MHz  4.67kHz
th

m L 
B A

2610
o
 82.5MHz






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  jo  0 T  jo  1
•
•
•
Barkhausen’s criteria state-
For sinusoidal oscillator, poles of T  j  0 Or even multiples of 3600
o
closed-loop amplifier should be at
T  jo  1
frequency 0 on j axis.
Use positive feedback through
• Phase shift around feedback loop
frequency-selective feedback
should be zero degrees and
network to ensure sustained
magnitude of loop gain must be
oscillation at 0 .
unity.
As
As
Av s 

• Loop gain greater than unity causes
1 Asb s 1 T s
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
Z ( s)
2
Vo(s)  V (s)
1 Z ( s)  Z ( s)
1
2
Vo ( s)
sRCG
T(s) 

V ( s) (1  2 R 2C 2 )  3sRC 1
I
Phase shift will be zero if (1 2R2C2) = 0,
At 0 =1/RC T(jo )   G
3
G
T(jo )  0
T(jo ) 
3
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
Vo(s)
 sCR
1
V ( s)
2
3C 3R 2 R
s
V ( s)
1
T (s)  o 
Vo'(s) 3s 2 R 2C 2  4sRC 1
Phase shift will be zero if (1 3o2R2C 2)= 0,
1
 
o
3RC
At 0
o2C 2 RR
1 R1
1
T(jo ) 

4
12 R
  V ( s) 
 sCV ' ( s) (2sC  G)

sC

 1
o 





 
V
(
s
)


(2sC

G)
0
 sC


 
 2

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
R1 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 R R
R R
2
3 4 2
2 3 2
R
R
1
1
Thus, when diodes are off, op amp gain is
slightly >3 ensuring oscillation, but, when
one diode is on, gain is reduced to

slightly<3.

R 
v
vo  v vo  v V v  o 1 2 
1
1 D 1 3

i

R 
R
R

1
3
4

3V
R
D
vo  
2 2
For positive signal at vo, D1 turns on as

R  R  R
R

2 1 4   4
voltage across R3 exceeds diode turn2 
1
R



R
R

on voltage. R4 is in parallel with R3,
1 
3  1

Same method can also be used in phase
loop gain is reduced. D2 functions
shift oscillators.
similarly at negative signal peak.
2 Microelettronica – Circuiti integrati analogici 2/ed
Richard C. Jaeger, Travis N. Blalock
Copyright © 2005 – The McGraw-Hill Companies srl
LC Oscillators: Colpitts Oscillator

 sC

s(C  C ) 1/ sL
 Vg ( s)
3
3
GD




 


s
(
C

g
)


s
(
C

C
)

g

G
V
(
s
)


0


m

3 m
1 3

  s




0





  s2 C C  C (C  C )   s (C
 C )G  GC 
GD 1 3   GD 3
3
 1 3
gm  G (C1  C3)


sL
L
=0, collect real and imaginary parts and set
them to zero.
1
CC
o 
C C
 1 3
LC
TC GD C  C
1 3
TC
C
At 0
gm R  3
C
G  1/( R ro ) C  C  C
1
3 2 GS
S
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
LC Oscillators: Hartley Oscillator

sC 1/ sL

1
/
sL
 V ( s)
2
2

 g








0   (1/ sL )  gm   (1/ sL1)  (1/ sL2)  gm  go   Vs(s) 
2







0






  sC  gm  go  
gm

1
sL s 2 L L
2
1 2
C







1 1

L L
1 2







=0, collect real and imaginary parts and set
them to zero.
1
o 
C( L  L )
1 2
L
G-S and G-D capacitances
At 0
m  1
f L
are neglected, assume no
2
mutual coupling between
Generally more gain is used to ensure
inductors.
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 RG 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











R
1
s2  s 
Z Z
L LC
1
P
S
S
Z 

C Z Z
sC s 2  s R  1
P S
P
L LC
T
Crystal: A piezoelectric device that vibrates
C C
is response to electrical stimulus, can be
C  P S
modeled electrically by a very high Q
T C C
P S
(>10,000) resonant circuit.
L, CS, R represent intrinsic series resonance
path through crystal. CP is package
capacitance. Equivalent impedance has series
resonance where CS resonates with L and
parallel resonance where L resonates with
series combination of CS and CP.
2 Microelettronica – Circuiti integrati analogici 2/ed
Richard C. Jaeger, Travis N. Blalock
Below S and above P,
crystal appears capacitive,
between S and P it exhibits
inductive reactance.
Copyright © 2005 – The McGraw-Hill Companies srl











Crystal Oscillators: Example
•
•
Problem: Find equivalent circuit elements for crystal with given parameters.
Given data: fS=5 MHz, Q=20,000 R =50 W, CP =5 pF
Analysis:
RQ 50(20,000)
L

 31.8mH
6

S 2  (510 )
1
1
C 

 31.8fF
S  2L 
2

7


S
10   (0.0318)



1
1

C C
2 (31.8mH)(31.6fF)
2 L P S
C C
P
S
 5.02MHz
f 
P
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 JFET
2 Microelettronica – Circuiti integrati analogici 2/ed
Richard C. Jaeger, Travis N. Blalock
Crystal Oscillator using BJT
Crystal Oscillator using CMOS
inverter as gain element.
Copyright © 2005 – The McGraw-Hill Companies srl