Chapter 13
Frequency Response
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 transfer function analysis and dominant-pole
approximations of amplifier transfer functions.
• Learn partition of ac circuits into low and high-frequency
equivalents.
• Learn short-circuit and open-circuit time constant methods to
estimate upper and lower cutoff frequencies.
• Develop bipolar and MOS small-signal models with device
capacitances.
• Study unity-gain bandwidth product limitations of BJTs and
MOSFETs.
• Develop expressions for upper cutoff frequency of inverting,
non-inverting and follower configurations.
• Explore gain-bandwidth product limitations of single and
multiple transistor circuits.
2 Microelettronica – Circuiti integrati analogici 2/ed
Richard C. Jaeger, Travis N. Blalock
Copyright © 2005 – The McGraw-Hill Companies srl
Chapter Goals (contd.)
• Understand Miller effect and design of op amp frequency
compensation.
• Develop relationship between op amp unity-gain
frequency and slew rate.
• Understand use of tuned circuits to design high-Q bandpass amplifiers.
• Understand concept of mixing and explore basic mixer
circuits.
• Study application of Gilbert multiplier as balanced
modulator and mixer.
2 Microelettronica – Circuiti integrati analogici 2/ed
Richard C. Jaeger, Travis N. Blalock
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Transfer Function Analysis
F ( s) 
L


 

 s   L  s   L ... s   L 
Z1
Z 2 
Zk 



 

 s   L  s   L ... s   L 
P1
P2  
Pk 

F ( s) 
H
a  a s  a s2  ...  amsm
N
(
s
)
2
Av (s) 
 0 1
D(s) b  b s  b s2  ...  b sn
n
0 1
2
 A F ( s) F ( s)
H
mid L
Amid is midband gain between upper
and lower cutoff frequencies.
2 Microelettronica – Circuiti integrati analogici 2/ed
Richard C. Jaeger, Travis N. Blalock



 



 
1  s 1  s ...1  s 

H 
H  
H





Z 1 
Z2 
Zl 




 



 
s
s
s


1 
1 
... 1 





H 
H 
H





P1
P2  
Pl 

F ( j )  1 for    H , H ,i =1…l
H
Zi Pi
 A ( s)  A
F ( s)
L
mid L
F ( j )  1 for    L , L ,j =1…k
L
Zj Pj
 A ( s)  A
F ( s)
H
mid H
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Low-Frequency Response
s
F ( s) 
L
s 
 
L
1

2
P2
 2   2   2 2 

Z 2  Z1 Z 2
1  Z1
P2
Pole P2 is called the dominant
low-frequency pole (> all other
poles) and zeros are at frequencies
low enough to not affect L.
If there is no dominant pole at low
frequencies, poles and zeros
interact to determine L.
A ( s)  A
F ( s)  A
L
mid L
mid
1
 
2



 s 
 s  
Z 1 
Z 2 




 s 
 s  
P1
P2 

A
mid
For s=j, at L,A( j L )  2
2 Microelettronica – Circuiti integrati analogici 2/ed
Richard C. Jaeger, Travis N. Blalock
 L2  Z12  L2  Z 22 
 L2  P12  L2  P22 
 2
 4
L
L
 2   2  2 2
P2  P1 P2
1  P1
 2
 4
L
L

 

Pole L > all other pole and zero
frequencies
   2 
L
P1
P2
2  2
2  2 2
Z1
Z2
In general, for n poles and n zeros,
 
L

2  2 2
n Pn
n Zn
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Transfer Function Analysis and
Dominant Pole Approximation Example
• Problem: Find midband gain, FL(s) and fL for


s
 1
 100

A ( s)  2000
L
0.1s 1s 1000
s
• Analysis: Rearranging the given transfer function to get it in standard
form,
A ( s)  200 ss  100
L
s 10s 1000
Now,
A ( s)  A F ( s)
L
mid L
s(s 100)
F
(
s
)

and
A
 200
L
(s 10)(s 1000)
mid
Zeros are at s=0 and s =-100. Poles are at s= -10, s=-1000
1
f 
102 10002  2(02 1002  158Hz
L 2p
All pole and zero frequencies are low and separated by at least a decade.
Dominant pole is at =1000 and fL =1000/2p= 159 Hz. For
s
frequencies > a few rad/s:
A ( s)  200
L
2 Microelettronica – Circuiti integrati analogici 2/ed
Richard C. Jaeger, Travis N. Blalock
s 1000
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High-Frequency Response
s
F ( s) 
L
1 (s /

H

P3
)
P3






1  ( 2 /  2 ) 1  ( 2 /  2 )
1
H
H
Z1
Z2

2
2
2
2
1  (
/
) 1  (
/ 2 )
H
H
P1
P2
 2  2
 4
H
Pole P3 is called the dominant high1 H  H 
 2  2  2 2
1
frequency pole (< all other poles).
Z1
Z2
Z1 Z 2
 
2
 2  2
 4
H
H
H
1


If there is no dominant pole at low
 2  2  2 2
P1
P2
P1 P2
frequencies, poles and zeros interact
Pole H < all other pole
and zero frequencies
to determine H.
1
A ( s)  A
F ( s)
H
mid H

) 1  ( s /  ) 
1  ( s / 
Z 1 
Z2 

A



mid 1  ( s /  ) 1  ( s /  ) 
P1 
P2 

A
mid
For s=j, at H,A( j H )  2
2 Microelettronica – Circuiti integrati analogici 2/ed
Richard C. Jaeger, Travis N. Blalock

H

1


1

2

2
2  2  2  2
P1
P2
Z1
Z2
In general,  
H
1
1
1
 2

2
2
n  Pn
n  Zn
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Direct Determination of Low-Frequency
Poles and Zeros: C-S Amplifier
R
Vo(s)  Io(s) R   gmVgs (s)
3
  gm ( R R )
3 D s
s
1
D
R
R  (1/ sC )  R 3
3
2
3
Vgs (s)
C (R  R )
2 D 3
s C R
1 G
Vg (s) 
V (s)
s  C ( R  R ) 1 i
1 I
G
s  (1/ C R )
3 S
Vgs (s)  Vg - Vs 
Vg (s)
1
s


C (1/ gm ) R 
3
S
V (s)
Av(s)  o  A
F (s)
mid L
V (s)
i
R
G
A
  gm ( R R )
3
R

mid
D G RI
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Richard C. Jaeger, Travis N. Blalock
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Direct Determination of Low-Frequency
Poles and Zeros:
C-S
Amplifier
(contd.)
2
F (s) 
L

s  s  (1/ C R ) 
3 S 






1
1
s 
 s 

C ( R  R )  C (1/ g )

1 I
G 

m
3



1
 s 
 
R   C2 ( RD
S



 R ) 
3 
The three zero locations are: s = 0, 0, -1/(RS C3).
1
The three pole locations are: s  
,

1
1
,
C ( R  R ) C (1/ g ) R  C ( R  R )
1 I G
2 D 3
m S 
3
Each independent capacitor in the circuit contributes one pole and
one zero. Series capacitors C1 and C2 contribute the two zeros at
s=0 (dc), blocking propagation of dc signals through the amplifier.
Third zero due to parallel combination of C3 and RS occurs at
frequency where signal current propagation through MOSFET is
blocked (output voltage is zero).
2 Microelettronica – Circuiti integrati analogici 2/ed
Richard C. Jaeger, Travis N. Blalock
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Short-Circuit Time Constant Method to
Determine L
• Lower cutoff frequency for
a network with n coupling
and bypass capacitors is
given by: n 1
 
L
Midband gain and upper and lower
cutoff frequencies that define
bandwidth of amplifier are of more
interest than complete transfer function.
2 Microelettronica – Circuiti integrati analogici 2/ed
Richard C. Jaeger, Travis N. Blalock

i  1 RiS Ci
where RiS is resistance at
terminals of ith capacitor Ci
with all other capacitors
replaced by short circuits.
Product RiS Ci is shortcircuit time constant
associated with Ci.
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Estimate of L for C-E Amplifier
Using SCTC method, for C1,
R  R  ( R RCE )  R  ( R rp )
I
B in
B
1S
2
For C2,
R  R  ( R RCE
)  R  ( R ro )
2S
3
C out
3
C
R R
3 C
For C3,
rp  R
th
CC
R  R Rout  R
E
E  1
3S
o
rp  ( R R )
I B
R
E
o  1
3
1
L
i  1 RiS Ci
 
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Richard C. Jaeger, Travis N. Blalock

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Estimate of L for C-S Amplifier
Using SCTC method,
For C1,
R  R  ( R RCS )  R  R
I
1S
G in
S G
For C2,
R  R  ( R RCS
)  R  ( R ro )
D out
D
2S
3
3
R R
3 D
For C3,
1
R  R RCG

R
3S
S out
S g
m
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Richard C. Jaeger, Travis N. Blalock
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Estimate of L for C-B Amplifier
Using SCTC method,
For C1,
1
R  R  ( R RCB)  R  ( R
)
I
E in
I
E g
1S
m
For C2,
R  R  ( R RCB
)R R
2S
3
C out
3 C
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Richard C. Jaeger, Travis N. Blalock
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Estimate of L for C-G Amplifier
Using SCTC method,
For C1,
1
R  R  ( R RCG)  R  ( R
)
I
I
1S
S in
S g
m
For C2,
R  R  ( R RCG
)R R
D out
2S
3
3 D
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Richard C. Jaeger, Travis N. Blalock
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Estimate of L for C-C Amplifier
Using SCTC method,
For C1,
R  R  ( R RCC )
I
B in
1S



 R   R rp   o 1 R R  

 E 3 
I  B

For C2,
R  R  (R
E
2S 3
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Richard C. Jaeger, Travis N. Blalock


CC
R
)

R

Rout
3  E

rp  R 
th 
o 1 
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
Estimate of L for C-D Amplifier
Using SCTC method,
For C1,
R  R  ( R RCD)  R  R
I
I
1S
G in
G
For C2,
1
R  R  R RCD

R

R
2S
3 S out
3 S g
m
2 Microelettronica – Circuiti integrati analogici 2/ed
Richard C. Jaeger, Travis N. Blalock
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Frequency-dependent Hybrid-Pi Model
for BJT
Capacitance between base and
emitter terminals is:
Cp  gmt
Capacitance between base and
collector terminals is:
Cmo
Cm 
1 (V / )
CB jc
Cmo is total collector-base
junction capacitance at zero bias,
Fjc is its built-in potential.
2 Microelettronica – Circuiti integrati analogici 2/ed
Richard C. Jaeger, Travis N. Blalock
F
tF is forward transit-time of the
BJT. Cp appears in parallel with
rp. As frequency increases, for a
given input signal current,
impedance of Cp reduces vbe and
thus the current in the controlled
source at transistor output.
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Unity-gain Frequency of BJT
Ic (s)  ( gm  sCm )V (s)
be
The right-half plane transmission zero Z =
+ gm/Cm occurring at high frequency can be
neglected.
o
o
 ( s) 

s(Cp  Cm )rp 1 (s / ) 1

 = 1/ rp(Cm + Cp ) is the beta-cutoff
frequency
 

 (s) 
rp
 ( gm  sCm )I (s)
b s(C  C )r  1
p
m p

sC m 


o 1

g m 

I c (s)


 ( s) 

I (s) s(Cp  Cm )rp  1
b
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Richard C. Jaeger, Travis N. Blalock
o 
 T
s 
s 

o

gm
where     

o 
T
(Cp  Cm )rp Cp  Cm
and fT = T /2p is the unity gain bandwidth
product. Above BJT has no current gain.
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Unity-gain Frequency of BJT (contd.)
Current gain is o = gmrp at low
frequencies and has single pole rolloff at frequencies > f, crossing
through unity gain at T. Magnitude
of current gain is 3 dB below its
low-frequency value at f.
Cp 
gm

T
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Richard C. Jaeger, Travis N. Blalock
 Cm 
40I
C C
m

T
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High-frequency Model of MOSFET
o
gm
  o 

T
 (C  C )r C  C
p
m p
p
m
I (s)  ( gm  sC )Vgs (s)
GD
d
( g m  sC
)
GD
 I (s)
b s(C  C
)
GS
GD
I (s) 
 ( s)  d  T
I g (s) s



1 



T

s

1  (C
GS

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Richard C. Jaeger, Travis N. Blalock





/C
)
GD  
f 
T
mnCox "
V V 

L GS TN
W
(2/3)Cox "WL

3 mn VGS VTN

2
L2
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
Limitations of High-frequency Models
•
•
•
•
Above 0.3 fT, behavior of simple pi-models begins to deviate
significantly from the actual device.
Also, T depends on operating current as shown and is not constant as
assumed.
For given BJT, a collector current ICM exists that yield maximum fTmax.
For FET in saturation, CGS and CGD are independent of Q-point
current, so
  gm  I
T
D
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Effect of Base Resistance on Midband
Amplifiers
Base current enters the BJT through
external base contact and traverses a
high resistance region before
entering active area. rx models
voltage drop between base contact
and active area of the BJT.
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Richard C. Jaeger, Travis N. Blalock
To account for base resistance rx is
absorbed into equivalent pi model and
can be used to transform expressions for
C-E, C-C and C-B amplifiers.
i  gmv  gm
rp
v  g 'v
rp  rx be m be
r
o
gm '  gm p 
rp  rx rp  rx
rp '  rp  rx
o'  o
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Direct High-Frequency Analysis: C-E
Amplifier
R  R R  100kΩ 4.3kΩ
L 3 C
R  R R  30kΩ 10kΩ
B 1 2
The small-signal model can be simplified
by using Norton source transformation.
R
B
v v
th i R  R
I
B
v
th
is 
R  rx
th
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R R
R  I B
th R  R
I
B
rpo  rp ( R  rx )
th
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Direct High-Frequency Analysis: C-E
Amplifier
(Pole
Determination)
From nodal equations for the circuit
C  Cp
T
in frequency domain,
V (s)  Is(s)
2

( sC m - g m )

  s2  Cp  Cm  C




  C C 
L  p L 


 s Cp g  Cm  g  g m  gp   C gpo   g gpo
L
L
L
 L



High-frequency response is given by
2 poles, one finite zero and one zero
at infinity. Finite right-half plane
zero, Z = + gm/Cm > T can easily be
neglected.
For a polynomial s2+sA1+A0 with
roots a and b, a =A1 and b=A0/A1.
2 Microelettronica – Circuiti integrati analogici 2/ed
Richard C. Jaeger, Travis N. Blalock


 Cm 1  g m R

L

R
 L
rpo

R

L
C
L

rpo

A
1
  0
P1 A r C
1 po T
gm
gm
 

P2 C 1  (C / C )   C
Cp  C
L
p 
L m  L
Smallest root that gives first pole limits
frequency response and determines H.
Second pole is important in frequency
compensation as it can degrade phase
margin of feedback amplifiers.
Copyright © 2005 – The McGraw-Hill Companies srl
Direct High-Frequency Analysis: C-E
Amplifier
(Overall
Transfer Function)
( sC - g )
V (s)
m m
th
R  rx g g 1  ( s /  ) 1 ( s /  ) 
th
L po 
P1 
P2 


1  ( s /  ) 
V (s)
Z 

Vo(s)  th
( gm R rpo )
L
R  rx
g gpo 1 (s /  ) 1  ( s /  ) 
th
L
P1 
P2 

V (s) g R r
m L po
Vo(s)  - th
R  rx 1 ( s /  ) 
th
P1 

o R
L
A
A

Vo (s)
mid
mid
A ( s) 

R  rx  rp
vth
V (s) 1 (s /  ) 
th
th
P1 

Vo(s) 



1
P1 r C
po T

Dominant pole model at high
frequencies for C-E amplifier is as
shown.
2 Microelettronica – Circuiti integrati analogici 2/ed
Richard C. Jaeger, Travis N. Blalock
Copyright © 2005 – The McGraw-Hill Companies srl
Direct High-Frequency Analysis: C-E
Amplifier (Example)
•
•
•
Problem: Find midband gain, poles, zeros and fL.
Given data: Q-point= ( 1.60 mA, 3.00V), fT =500 MHz, o =100, Cm
=0.5 pF, rx =250W, CL 0
Analysis: gm =40IC =40(0.0016) =64 mS, rp = o/gm =1.56 kW.
gm
 Cm  19.9pF
f

1
 1.56MHz
P1 2pr C
2pf
po T 
T


g
1
1
1 
R  R R  100kΩ 4.3kΩ  4.12kΩ
m

 
1

L 3 C

P2 R C C  g r
g
R

m p
m po
m L 
R  R R  7.5kΩ 1kΩ  882Ω
L
B I
th
gm

P
2
f

 20.4GHz
f 
 603MHz
rpo  rp ( R  rx )  656W
Z
P
2
th
2pCm
2p

R 
R
o R

L
L  156pF
L
 C
C  Cp  Cm 1 gm R 
A


 153
T
L

L r 
vth
rpo
R  rx  rp
po 

th
Cp 
Overall gain is reduced to -135 as vth =0.882vs.
2 Microelettronica – Circuiti integrati analogici 2/ed
Richard C. Jaeger, Travis N. Blalock
Copyright © 2005 – The McGraw-Hill Companies srl
Gain-Bandwidth Product Limitations of
C-E Amplifier
•
If Rth is reduced to zero in order to increase bandwidth, then rpo would
not be zero but would be limited to approximately rx.
GBW  Av
H
•

o R

L


 R  rx  rp
 th


1


 rpoC
T







If Rth = 0, rx <<rp so that rx = rpo and CT  Cm ( gmRL )
GBW 
2 Microelettronica – Circuiti integrati analogici 2/ed
Richard C. Jaeger, Travis N. Blalock
1
rxCm
Copyright © 2005 – The McGraw-Hill Companies srl
High-Frequency Analysis: C-S
Amplifier
R R R
I G
th
R R R
L D 3
R
G
v v
th i R  R
I
G
C C
T
GS


 C 1 gm R 
GD
L


R 
L
R 
th 
1
P1 R C
th T



g
1
1
1 
m

 
1


P2 R C
C  gm R
gm R 
L GD GS 
L
th
g
  m
Z C
GD

2 Microelettronica – Circuiti integrati analogici 2/ed
Richard C. Jaeger, Travis N. Blalock

Copyright © 2005 – The McGraw-Hill Companies srl
Miller Multiplication
Vo(s)   AV (s) Is(s)  sC V (s)  Vo (s)
1
 1

I ( s)
Y(s)  s  sC (1 A)
V ( s)
1
For the C-E amplifier,
Total input capacitance = C(1+A) because C  Cp  Cm (1 A)  Cp  Cm (1 gmR )
T
L
total voltage across C is vc = vi(1+A) due to
inverting voltage gain of amplifier.
2 Microelettronica – Circuiti integrati analogici 2/ed
Richard C. Jaeger, Travis N. Blalock
Copyright © 2005 – The McGraw-Hill Companies srl
Miller Integrator
 1  A




 RC  1  A
V
Ao
Av (s)  o  

1
V
s  o
s

1
RC(1 A)
where
Assuming zero current in input
terminal of amplifier,
V V
1 in  sC (V  V )
o
in
R
Vo   AV
in
2 Microelettronica – Circuiti integrati analogici 2/ed
Richard C. Jaeger, Travis N. Blalock
o 
1
RC(1 A)
For frequencies >> o, assuming A>>1,
A o
1
Av ( s)  

s
sRC
which is the transfer function of an
integrator.
Copyright © 2005 – The McGraw-Hill Companies srl
Open-Circuit Time Constant Method to
Determine H
At high frequencies, impedances of
coupling and bypass capacitors are small
enough to be considered short circuits.
Open-circuit time constants associated
with impedances of device capacitances
are considered instead.

1
 m
H
 RioCi
i 1
where Rio is resistance at terminals of
ith capacitor Ci with all other
capacitors open-circuited.
For a C-E amplifier, assuming CL =0
Rpo  rpo
2 Microelettronica – Circuiti integrati analogici 2/ed
Richard C. Jaeger, Travis N. Blalock
R
Rmo 
 rpo  (1 gmR  L )
L r
ix
po
1
1
 

H R C R C
po p
mo m rpoCT
vx
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Gain-Bandwidth Trade-off Using
Emitter Resistor

R
R
o
L
A

 L
mid
R  rx  rp  ( o  1) R
R
E
E
th
for rp  Rth  rx and gmRE  1
gain decreases as emitter resistance
increases and bandwidth of stage will
correspondingly increase.
To find bandwidth using OCTC method:
R  rx  R
E
Req 
 th
ix
1 gm R
E
R  rx  R
E
Rpo  rp Req  th
1 gm R
E
vx
2 Microelettronica – Circuiti integrati analogici 2/ed
Richard C. Jaeger, Travis N. Blalock
Copyright © 2005 – The McGraw-Hill Companies srl
Gain-Bandwidth Trade-off Using
Emitter Resistor (contd.)
Test source ix is first split into two
equivalent sources and then
superposition is used to find vx =(vb - vc).
Assuming that o >>1 and



 R  rx    rp

th


 ( o 1) R 
E

gmRL
RL 
v

x
Rmo 
 ( R  rx )1


R

r
th
ix
 1 gm R
x
E
th


 
H
2 Microelettronica – Circuiti integrati analogici 2/ed
Richard C. Jaeger, Travis N. Blalock
1


Cp
( R  rx )
th
 1 gm R
E



R 
g R


E   C 1 m L
1

R  rx  m  1 gm R

E
th




R  
 L  
R  rx  
th

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Dominant Pole for C-B Amplifier
R R R
E I
th
R R R
L C 3
R  rx
R  rx
vx
th
Rpo  rp
 rp
 th
ix
1 gm R
1 gm R
th
th
Using split-source transformation
Assuming that o >>1 and rx << rp

v b  vc
gmRL 

Rmo 
 rx 1
  RL
ix
 1 gm R 
th 

 
H
1





R 
gm R  

C

p 1 th   C 1
L   C R
rx 

m

rx 
1 gm R   m L
 1 gm R 


th 
th  


Neglecting first term of order of 1/ T and
since last term is dominant.
1
 
H C R
m L
2 Microelettronica – Circuiti integrati analogici 2/ed
Richard C. Jaeger, Travis N. Blalock
Copyright © 2005 – The McGraw-Hill Companies srl
Dominant Pole for C-G Amplifier
R R R
th 4 I
R R R
L D 3
R
1
th 
R

GSo 1 gmR
th Gth  gm
R
R
L
GDo
1
1
 

H  C

C R

GD L
GS

 C
R


GD L
 G  gm 
 th

2 Microelettronica – Circuiti integrati analogici 2/ed
Richard C. Jaeger, Travis N. Blalock
Copyright © 2005 – The McGraw-Hill Companies srl
Dominant Pole for C-C Amplifier
R R R
B I
th
R R R
L
E 3
R  rx  R
R  rx  R
vx
L
L
th
Rpo  rp
 rp
 th
ix
1 gm R
1 gm R
L
L

Rmo  ( R  rx ) RCC  ( R  rx ) rp  ( o 1)R
in
th
th
L
 ( R  rx )
th
1
 
H
Cp
( R  rx  R )
 ( R  rx )Cm
L 1 g R
th
th

m L
A better estimate is obtained if we set RL =0
in expression for Rpo.
 
H
2 Microelettronica – Circuiti integrati analogici 2/ed
Richard C. Jaeger, Travis N. Blalock
1


Cp
( R  rx )
th
 1 gm R
L



 Cm 


1
GBW  (1) 
H C r
mx
Copyright © 2005 – The McGraw-Hill Companies srl
Dominant Pole for C-D Amplifier
R R R
G I
th
R R R
L
S 3
Substituting rp as infinite and rx as
zero in expression for emitter
follower,
 
H
2 Microelettronica – Circuiti integrati analogici 2/ed
Richard C. Jaeger, Travis N. Blalock
1

1


R
 C

th C  C R
GS  C 

R
1 gm R GS GD th th  1 g R
GD 

L
m L


Copyright © 2005 – The McGraw-Hill Companies srl
Frequency Response: Differential
Amplifier
CEE is total capacitance at emitter node
of the differential pair.
Differential mode half-circuit is similar
to a C-E stage. Bandwidth is determined
by the rpoCT product. As emitter is a
virtual ground, CEE has no effect on
differential-mode signals.
For common-mode signals, at very low
R
frequencies,
C
Acc (0) 
2R
 1
EE
Transmission zero due to CEE is
1
s    
Z
C
2 Microelettronica – Circuiti integrati analogici 2/ed
Richard C. Jaeger, Travis N. Blalock
R
EE EE
Copyright © 2005 – The McGraw-Hill Companies srl
Frequency Response: Differential
Amplifier (contd.)
 
P
rp  rx
1
R
 2R

EE  1 g
EEO
o
m
1


Cp
rx 
 1 2 gm R
EE

 2R

EE
1 

rx



gm R


C

  C 1
m


 1 2 gm R

EE

R   C
 C    EE
rx   2gm

As REE is usually designed to be large,
Common-mode half-circuit is similar to a
C-E stage with emitter resistor 2REE.
OCTC for Cp and Cm is similar to the C-E
stage. OCTC for CEE/2 is:
2 Microelettronica – Circuiti integrati analogici 2/ed
Richard C. Jaeger, Travis N. Blalock
1
1
 

P C C
p EE  C ( R  r ) Cm ( RC  rx )
m C x
2 gm
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Frequency Response: CommonCollector/ Common-Base Cascade
REE is assumed to be large and
neglected.
r r
1
CC
1
Rout  p1 x11 
g
o1
m1
r r
1
CB
2
Rin  p2 x12 
g
o2
m2
Sum of the OCTC of Q1 is:



C

p1
r 
x1
1
 1 g
m1 g

m2

2 Microelettronica – Circuiti integrati analogici 2/ed
Richard C. Jaeger, Travis N. Blalock



C


 p1

C   r 
C 
m1  x1 2
m1 





Copyright © 2005 – The McGraw-Hill Companies srl
Frequency Response: Common-Collector/
Common-Base Cascade (contd.)
Sum of the OCTC of Q2 is:



C

p2

r
x2 
1
 1 g
m2 g


m1




gm R

C
 C 1
1
m 2  1 g

m2 g

m1





  C
R
m2 C


 



C
 r  p 2
x2  2



 C 1
m2

gm R
R  
C  C 
2
r  
x2  
Combining the OCTC for Q1 and Q2, and assuming that transistors are matched,
 
H
1


rx  Cp





 Cm  2 


2 Microelettronica – Circuiti integrati analogici 2/ed
Richard C. Jaeger, Travis N. Blalock
gm R
R  
C  C 
2
rx  

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Frequency Response: Cascode Amplifier
OCTC of Q1 with load
resistor 1/ gm2 :


R C  R C  r C  r C  C 1
po1 p 1 mo1 m1 po1 T1 po1 p 1 m1


g
As IC2 = IC1, gm2 = gm1, gain of first
stage is unity. Assuming gm2 rpo1 >>1,


C  r  C  2C 
po1 T1 po1 p 1
m1 
r
OCTC of Q1, a C-B stage for ro1 >> RL
and mf>>1:
C
C  R C  p 2  (r  R )C
po2 p 2 mo2 m 2 g
x2 L m 2
m2
R
Assuming matched devices,
1
 
H r C  2C  r  R C
m  x L  m
po1 p
2 Microelettronica – Circuiti integrati analogici 2/ed
Richard C. Jaeger, Travis N. Blalock


1
m1 

g
g r 
m2 m2 po1 
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Frequency Response: MOS Current
Mirror
rpo 
1
g
R r
L
o2
m1
Cm  C
GD 2
Cp  C
C
GS1 GS 2
For matched transistors,

2 Microelettronica – Circuiti integrati analogici 2/ed
Richard C. Jaeger, Travis N. Blalock
P1

1

 2C
GS


 g
m1



 2C
r 
GD2 o2 


1
2C
r
GD2 o2
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Frequency Response: Multistage
Amplifier
• Problem:Use open-circuit and short-circuit time constant methods to
estimate upper and lower cutoff frequencies and bandwidth.
• Approach: Coupling and bypass capacitors determine low-frequency
response, device capacitances affect high-frequency response.
At high frequencies, ac model for multi-stage
amplifier is as shown.
2 Microelettronica – Circuiti integrati analogici 2/ed
Richard C. Jaeger, Travis N. Blalock
Copyright © 2005 – The McGraw-Hill Companies srl
Frequency Response: Multistage
Amplifier (Estimate of L)
SCTC for each of the six independent coupling and bypass capacitors has

 

to be determined.
R   R
R    R R 
R  R  ( R R )  10kΩ 1MΩ 
I
1S
G in1
 1.01MΩ
1
1
R R
 200W
 66.7W
2S
S1 g
0.01S
m1




R   R
R R
R 
3S  D1 O1   B2 in2 

  R




r R
r   2.69kΩ
D1 o1   B2 p 2 
R  R R r  571W
th2 B2 D1 o1
R r
th2 p 2  19.4W
R R
4S
E 2  1
o2
2 Microelettronica – Circuiti integrati analogici 2/ed
Richard C. Jaeger, Travis N. Blalock
5S


C 2 O2 


B3 in3 


  R
r    R  r  (  1)( R
R )  
o3
E3 L  
 C 2 o2   B3  p 3
 18.4kΩ
R  R R r  3.99kW
th3 B3 C2 o2
R r
th3 p 3  311W
R R R
L E 3  1
6S
o3
n
1
  
 3300rad/s
L
i  1 RiS Ci

f  L  530Hz
L 2p
Copyright © 2005 – The McGraw-Hill Companies srl
Frequency Response: Multistage
Amplifier
(Estimate
of

)
H
OCTC for each of the two capacitors associated with each transistor has to
be determined.
For M1, R  R
L1
r  478W
I12 p 2





1  g
R C  R C
C
R 
th T1 th  GS1 GD1
m1 L1



 1.07 107s
R
r  570W
I12 o1
th2
r
 r ( R  r )  610W
po2 p 2 th2 x2


R   R
R 
L2  I 23 in3 


r
 p 3  ( o3  1)( RE 3

I 23
 3.54kΩ

2 Microelettronica – Circuiti integrati analogici 2/ed
Richard C. Jaeger, Travis N. Blalock
 1.74 107s
For Q3, R
R
r  3.99kW
I 23 o2
(R  r )
R C  R C  th3 x3 C  ( R  r )C
p 3O p 3 m 3O m 3 1 g R
p 3 th3 x3 m 3
m3 EE
 1.51108s
1
  m
 3.38 106 rad/s
H
 RioCi
i 1

f  H  538kHz
H 2p
th3
For Q2, R
  R


L1  

R  
th  
R



R  

r C  r  C  C 1 g R  L2  
po2 T 2 po2  p 2 m 2
m2 L2 r  

po2  



R )  
L 
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Single-pole Op Amp Compensation
• Frequency compensation forces overall amplifier to have a
single-pole frequency response by connecting
compensation capacitor around second gain stage of the
Ao

basic op amp.
B
A ( s) 
 T
v
2 Microelettronica – Circuiti integrati analogici 2/ed
Richard C. Jaeger, Travis N. Blalock
s 
B
s 
B
Copyright © 2005 – The McGraw-Hill Companies srl
Three-stage MOS Op Amp Analysis
Input stage is modeled by its Norton
equivalent- current source Gmvdm and
output resistance Ro. Second stage
has gain of gm5ro5= mf5 and follower
output stage is a unity-gain buffer.
Vo(s) = Vb(s) = - Av2Va(s)
Gm Ro A
Vo (s) - Av2Va (s)
v2
Av (s) 


V ( s)
V ( s)
1  sRoC (1 A )
C
v2
dm
dm

Ao
T
B


s 
s 
B
B
Gm A
1
v2
 


B R C (1 A )
T C (1 A )
o C
C
v2
v2
Gm
For large Av2
 
T C
C
2 Microelettronica – Circuiti integrati analogici 2/ed
Richard C. Jaeger, Travis N. Blalock
Copyright © 2005 – The McGraw-Hill Companies srl
Transmission Zeros in FET Op Amps
Incorporating the zero determined
by gm5 in the analysis,
1  ( s /  ) 

Z 
A ( s)  (  g r ) 
m5 o5 1  ( s /  ) 
vth
P1 

g
g
m
5
 
  m5
Z C C
Tg
C
GD5
m2
1
P1 R C
o T

r 

C C
 (C  C
)1 m  o5 
T
GS 5
C GD5 
f 5 R 
o



This zero can’t be neglected due to
low ratio of transconductances of M2
and M5. Zero can be canceled by
addition of RZ =1/ gm5.
 
Z
2 Microelettronica – Circuiti integrati analogici 2/ed
Richard C. Jaeger, Travis N. Blalock
(1/ g
) R
m5
Z
C
C
Copyright © 2005 – The McGraw-Hill Companies srl
Bipolar Amplifier Compensation
• Bipolar op amp can be
compensated in the same manner as
a MOS amplifier
• Transmission zero occurs at too
high a frequency to affect the
response due to higher
transconductance of BJT that FET
for given operating current.
I
m
5
 
  C5
Z C
TI
C
C2
g
•
2 Microelettronica – Circuiti integrati analogici 2/ed
Richard C. Jaeger, Travis N. Blalock
Unity gain frequency is given by:
g
40I
20I
m
2
C
2
C1
 


T C
C
C
C
C
C
Copyright © 2005 – The McGraw-Hill Companies srl
Slew rate of Op Amp
• Slew-rate limiting is caused by
limited current available to
charge/discharge internal capacitors.
For very large Av2, amplifier behaves
dv (t )
likeI an C
integrator:
B  C dvo (t )
C1
C
dt
C
dt


 

I
dvo (t)
T
1


SR 



C
dt max
C  Gm / I1 


 amplifier,

• For CMOS
I
T 
1

SR  

 Gm / I 
1


T K
n2
SR
• For bipolar amplifier,
2 Microelettronica – Circuiti integrati analogici 2/ed
Richard C. Jaeger, Travis N. Blalock


 
 
T

 T


20
 Gm / I 
1

Copyright © 2005 – The McGraw-Hill Companies srl
Tuned Amplifiers
• Amplifiers with narrow bandwidth are often required in
RF applications to be able to select one signal from a large
number of signals.
• Frequencies of interest > unity gain frequency of op amps,
so active RC filters can’t be used.
• These amplifiers have high Q (fH and fL close together
relative to center frequency)
• These applications use resonant RLC circuits to form
frequency selective tuned amplifiers.
2 Microelettronica – Circuiti integrati analogici 2/ed
Richard C. Jaeger, Travis N. Blalock
Copyright © 2005 – The McGraw-Hill Companies srl
Single-Tuned Amplifiers
•
RLC network selects the
frequency, parallel
combination of RD, R3 and ro
set the
sC
g
VQ
( s)and bandwidth.
m
GD
Av ( s)  o 
V ( s) G  s(C  C
)  (1/ sL)
P
GD
i
G  go  G  G
P
D
3
•
o plane
Neglecting right-half
s
Q
zero,
Av (s)  A

mid 2
s  s o  o 2
Q
o 
2 Microelettronica – Circuiti integrati analogici 2/ed
Richard C. Jaeger, Travis N. Blalock
1
L(C  C )
GD
R
Q  o R (C  C )  P
P
GD  L
o
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Single-Tuned Amplifiers (contd.)
• At center frequency, s =
jo, Av = Amid.
A
  gmR   gm(ro R R )
P
mid
D 3
o
o 2 L
1
BW 


Q R (C  C
) R
P
P
GD
2 Microelettronica – Circuiti integrati analogici 2/ed
Richard C. Jaeger, Travis N. Blalock
Copyright © 2005 – The McGraw-Hill Companies srl
Use of tapped Inductor- Auto
Transformer
C and r can often be small enough to
GD
o
degrade characteristics of the tuned
amplifier. Inductor can be made to work
as an auto transformer to solve this
problem.
V ( s)
Vo(s) nV1(s)

 n2 1
I2(s) Is(s) / n
Is(s)
Z s ( s)  n2 Z p ( s)
These results can be used to transform the
resonant circuit and higher Q can be
obtained and center frequency doesn’t
shift significantly due to changes in CGD.
Similar solution can be used if tuned
circuit is placed at amplifier input instead
of output
2 Microelettronica – Circuiti integrati analogici 2/ed
Richard C. Jaeger, Travis N. Blalock
Copyright © 2005 – The McGraw-Hill Companies srl
Multiple Tuned Circuits
• Tuned circuits can be placed at both input
and output to tailor frequency response.
• Radio-frequency choke(an open circuit at
operating frequency) is used for biasing.
• Synchronous tuning uses two circuits tuned
to same center frequency for high Q.
•
BWn  BW 21/ n 1
Stagger tuning uses1 two circuits tuned to
slightly different center frequencies to
realize broader band amplifiers.
• Cascode stage is used to provide isolation
between the two tuned circuits and
eliminate feedback path between them due
to Miller multiplication.
2 Microelettronica – Circuiti integrati analogici 2/ed
Richard C. Jaeger, Travis N. Blalock
Copyright © 2005 – The McGraw-Hill Companies srl
Mixers: Conversion Gain
• Amplifiers discussed so far have always been assumed to
be linear and gain expressions involve input and output
signals at same frequency.
• Mixers are nonlinear devices, output signal frequency is
different from input signal frequency.
• A mixer’s conversion gain is the ratio of phasor
representation of output signal to that of input signals, the
fact that the two signals are at different frequencies is
ignored.
2 Microelettronica – Circuiti integrati analogici 2/ed
Richard C. Jaeger, Travis N. Blalock
Copyright © 2005 – The McGraw-Hill Companies srl
Single-Balanced Mixer
• Eliminates one of the two input
signals from the output.
i I
 I sin  t
EE EE 1
1
4
v (t )  
sin n t
2
2
n odd np
I

R sin n t


EE
C
2

4
Vo (t)  
 I R

IR
  1 C cos(n   )t  1 C cos(n   )t 
n
p
n odd 
2 1
2 1 
2
2

• No signal energy appears at1 ,
but 2 appears in output spectrum,
so circuit is single-balanced.
• Up-conversion uses component
(2-1) and down-conversion uses
(2+1) component.
2 Microelettronica – Circuiti integrati analogici 2/ed
Richard C. Jaeger, Travis N. Blalock
Copyright © 2005 – The McGraw-Hill Companies srl
Double-Balanced Mixer/ Modulator:
Gilbert Multiplier
• Double-balanced mixers don’t
contain spectral components at
either of the two input frequencies.
Vm
Vm
i

I

sin mt
i I

sin mt
C 2 BB 2R
C1 BB 2R
1
1
R
4

Vo(t) Vm C 
 cos(n   )t  cos(n   )t 
c
m
c
m


R
1 n odd np
• Modulator applications give double
sideband suppressed carrier output
signal. Amplitude-modulated signal
can also be obtained if v Vm(1 M sin mt)
1
R

4 
M
M

vo(t) Vm C 
sin
n

t

cos(
n



)
t

cos(
n



)
t


c
c
m
c
m
R
2
2

1 n odd np 
2 Microelettronica – Circuiti integrati analogici 2/ed
Richard C. Jaeger, Travis N. Blalock
Copyright © 2005 – The McGraw-Hill Companies srl