#### Transcript Document

```Chapter 11 Frequency Response
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11.1
11.2
11.3
11.4
11.5
11.6
11.7
11.8
11.9
Fundamental Concepts
High-Frequency Models of Transistors
Analysis Procedure
Frequency Response of CE and CS Stages
Frequency Response of CB and CG Stages
Frequency Response of Followers
Frequency Response of Cascode Stage
Frequency Response of Differential Pairs
1
Chapter Outline
CH 11 Frequency Response
2
High Frequency Roll-off of Amplifier
 As frequency of operation increases, the gain of amplifier
decreases. This chapter analyzes this problem.
CH 11 Frequency Response
3
Example: Human Voice I
Natural Voice
Telephone System
 Natural human voice spans a frequency range from 20Hz to
20KHz, however conventional telephone system passes
frequencies from 400Hz to 3.5KHz. Therefore phone
conversation differs from face-to-face conversation.
CH 11 Frequency Response
4
Example: Human Voice II
Path traveled by the human voice to the voice recorder
Mouth
Air
Recorder
Path traveled by the human voice to the human ear
Mouth
Air
Ear
Skull
 Since the paths are different, the results will also be
different.
CH 11 Frequency Response
5
Example: Video Signal
High Bandwidth
Low Bandwidth
 Video signals without sufficient bandwidth become fuzzy as
they fail to abruptly change the contrast of pictures from
complete white into complete black.
CH 11 Frequency Response
6
Gain Roll-off: Simple Low-pass Filter
 In this simple example, as frequency increases the
impedance of C1 decreases and the voltage divider consists
of C1 and R1 attenuates Vin to a greater extent at the output.
CH 11 Frequency Response
7
Gain Roll-off: Common Source
Vout

1 
  g mVin  RD ||

C
s
L 

 The capacitive load, CL, is the culprit for gain roll-off since
at high frequency, it will “steal” away some signal current
and shunt it to ground.
CH 11 Frequency Response
8
Frequency Response of the CS Stage
Vout
g m RD

Vin
RD2 C L2 2  1
 At low frequency, the capacitor is effectively open and the
gain is flat. As frequency increases, the capacitor tends to
a short and the gain starts to decrease. A special
frequency is ω=1/(RDCL), where the gain drops by 3dB.
CH 11 Frequency Response
9
Example: Figure of Merit
F .O.M . 
1
VT VCC C L
 This metric quantifies a circuit’s gain, bandwidth, and
power dissipation. In the bipolar case, low temperature,
supply, and load capacitance mark a superior figure of
merit.
CH 11 Frequency Response
10
Example: Relationship between Frequency
Response and Step Response
H  s  j  
1
R12C12 2  1

t 
Vout  t   V0 1  exp
 u t 
R1C1 

 The relationship is such that as R1C1 increases, the
bandwidth drops and the step response becomes slower.
CH 11 Frequency Response
11
Bode Plot

s 
s 
1 
1 
 
 z1   z 2 

H ( s)  A0



s
s
1 
1 

    
p1 
p2 

 When we hit a zero, ωzj, the Bode magnitude rises with a
slope of +20dB/dec.
 When we hit a pole, ωpj, the Bode magnitude falls with a
slope of -20dB/dec
CH 11 Frequency Response
12
Example: Bode Plot
 p1
1

RD C L
 The circuit only has one pole (no zero) at 1/(RDCL), so the
slope drops from 0 to -20dB/dec as we pass ωp1.
CH 11 Frequency Response
13
Pole Identification Example I
 p1
 p2 
1

RS Cin
Vout

Vin
CH 11 Frequency Response
1  
1
RD C L
g m RD
2
 p21 1   2  p2 2 
14
Pole Identification Example II
 p1
1


1 
 RS ||
Cin
gm 

CH 11 Frequency Response
 p2
1

RD C L
15
Circuit with Floating Capacitor
 The pole of a circuit is computed by finding the effective
resistance and capacitance from a node to GROUND.
 The circuit above creates a problem since neither terminal
of CF is grounded.
CH 11 Frequency Response
16
Miller’s Theorem
ZF
Z1 
1  Av
ZF
Z2 
1  1 / Av
 If Av is the gain from node 1 to 2, then a floating impedance
ZF can be converted to two grounded impedances Z1 and Z2.
CH 11 Frequency Response
17
Miller Multiplication
 With Miller’s theorem, we can separate the floating
capacitor. However, the input capacitor is larger than the
original floating capacitor. We call this Miller multiplication.
CH 11 Frequency Response
18
Example: Miller Theorem
1
in 
RS 1  g m RD C F
CH 11 Frequency Response
out 
1

1 
C F
RD 1 
 g m RD 
19
High-Pass Filter Response
Vout

Vin
R1C1
R12C1212  1
 The voltage division between a resistor and a capacitor can
be configured such that the gain at low frequency is
reduced.
CH 11 Frequency Response
20
Example: Audio Amplifier
Ci  79.6nF
CL  39.8nF
Ri  100K
g m  1 / 200
 In order to successfully pass audio band frequencies (20
Hz-20 KHz), large input and output capacitances are
needed.
CH 11 Frequency Response
21
Capacitive Coupling vs. Direct Coupling
Capacitive Coupling
Direct Coupling
 Capacitive coupling, also known as AC coupling, passes
AC signals from Y to X while blocking DC contents.
 This technique allows independent bias conditions between
stages. Direct coupling does not.
CH 11 Frequency Response
22
Typical Frequency Response
Lower Corner
CH 11 Frequency Response
Upper Corner
23
High-Frequency Bipolar Model
C  Cb  C je
 At high frequency, capacitive effects come into play. Cb
represents the base charge, whereas C and Cje are the
junction capacitances.
CH 11 Frequency Response
24
High-Frequency Model of Integrated Bipolar
Transistor
 Since an integrated bipolar circuit is fabricated on top of a
substrate, another junction capacitance exists between the
collector and substrate, namely CCS.
CH 11 Frequency Response
25
Example: Capacitance Identification
CH 11 Frequency Response
26
MOS Intrinsic Capacitances
 For a MOS, there exist oxide capacitance from gate to
channel, junction capacitances from source/drain to
substrate, and overlap capacitance from gate to
source/drain.
CH 11 Frequency Response
27
Gate Oxide Capacitance Partition and Full Model
 The gate oxide capacitance is often partitioned between
source and drain. In saturation, C2 ~ Cgate, and C1 ~ 0. They
are in parallel with the overlap capacitance to form CGS and
CGD.
CH 11 Frequency Response
28
Example: Capacitance Identification
CH 11 Frequency Response
29
Transit Frequency
gm
2f T 
CGS
gm
2f T 
C
 Transit frequency, fT, is defined as the frequency where the
current gain from input to output drops to 1.
CH 11 Frequency Response
30
Example: Transit Frequency Calculation
2fT 
3 n
VGS  VTH 
2
2L
L  65nm
VGS  VTH  100m V
 n  400cm2 /(V .s )
fT  226GHz
CH 11 Frequency Response
31
Analysis Summary
 The frequency response refers to the magnitude of the
transfer function.
 Bode’s approximation simplifies the plotting of the
frequency response if poles and zeros are known.
 In general, it is possible to associate a pole with each node
in the signal path.
 Miller’s theorem helps to decompose floating capacitors
into grounded elements.
 Bipolar and MOS devices exhibit various capacitances that
limit the speed of circuits.
CH 11 Frequency Response
32
High Frequency Circuit Analysis Procedure
 Determine which capacitor impact the low-frequency region
of the response and calculate the low-frequency pole
(neglect transistor capacitance).
 Calculate the midband gain by replacing the capacitors with
short circuits (neglect transistor capacitance).
 Include transistor capacitances.
 Merge capacitors connected to AC grounds and omit those
that play no role in the circuit.
 Determine the high-frequency poles and zeros.
 Plot the frequency response using Bode’s rules or exact
analysis.
CH 11 Frequency Response
33
Frequency Response of CS Stage
with Bypassed Degeneration
Vout
 g m RD RS Cb s  1
s  
VX
RS Cb s  g m RS  1
 In order to increase the midband gain, a capacitor Cb is
placed in parallel with Rs.
 The pole frequency must be well below the lowest signal
frequency to avoid the effect of degeneration.
CH 11 Frequency Response
34
Unified Model for CE and CS Stages
CH 11 Frequency Response
35
Unified Model Using Miller’s Theorem
CH 11 Frequency Response
36
Example: CE Stage
RS  200
I C  1m A
  100
C  100 fF
C   20 fF
CCS  30 fF
 p,in  2  516MHz
 p,out  2  1.59GHz
 The input pole is the bottleneck for speed.
CH 11 Frequency Response
37
Example: Half Width CS Stage
W  2X
 p ,in 
 p ,out
CH 11 Frequency Response
1
C
 g R C 
RS  in  1  m L  XY 
2  2 
 2 
1

 Cout 
2  C XY 


RL 
 1 

2
g
R
2
m L 



38
Direct Analysis of CE and CS Stages
gm
|  z |
C XY
1
|  p1 |
1  g m RL C XY RThev  RThevCin  RL C XY  Cout 
1  g m RL C XY RThev  RThevCin  RL C XY  Cout 
|  p 2 |
RThev RL Cin C XY  CoutC XY  Cin Cout 
 Direct analysis yields different pole locations and an extra
zero.
CH 11 Frequency Response
39
Example: CE and CS Direct Analysis
 p1
1

1  g m1 rO1 || rO 2 C XY RS  RS Cin  rO1 || rO 2 (C XY  Cout )
1  g m1 rO1 || rO 2 C XY RS  RS Cin  rO1 || rO 2 (C XY  Cout )
 p2 
RS rO1 || rO 2 Cin C XY CoutC XY  Cin Cout 
CH 11 Frequency Response
40
Example: Comparison Between Different Methods
RS  200
CGS  250 fF
CGD  80 fF
C DB  100 fF
g m  150 
1
 0
RL  2 K 
Dominant Pole
Miller’s
Exact
 p,in  2  571MHz
 p,in  2  264MHz
 p,in  2  249MHz
 p,out  2  428MHz
 p,out  2  4.53GHz
 p,out  2  4.79GHz
CH 11 Frequency Response
41
Input Impedance of CE and CS Stages
1
1
Z in 
|| r Z in 
CGS  1  g m RD CGD s
C  1  g m RC C s
CH 11 Frequency Response
42
Low Frequency Response of CB and CG Stages
Vout
g m RC Ci s
s  
1  g m RS Ci s  g m
Vin
 As with CE and CS stages, the use of capacitive coupling
leads to low-frequency roll-off in CB and CG stages
(although a CB stage is shown above, a CG stage is
similar).
CH 11 Frequency Response
43
Frequency Response of CB Stage
 p, X
1


1 
 RS ||
C X
gm 

C X  C
 p ,Y
rO  
CH 11 Frequency Response
1

RL CY
CY  C  CCS
44
Frequency Response of CG Stage
1
 p , Xr   
O 
1 
 RS ||
C X
gm 

C X  CGS  CSB
 p,Y
rO  
1

RL CY
CY  CGD  C DB
 Similar to a CB stage, the input pole is on the order of fT, so
rarely a speed bottleneck.
CH 11 Frequency Response
45
Example: CG Stage Pole Identification
 p, X 
1

1 
 RS ||
C SB1  CGD1 
g m1 

CH 11 Frequency Response
 p ,Y 
1
1
C DB1  CGD1  CGS 2  C DB2 
g m2
46
Example: Frequency Response of CG Stage
RS  200
CGS  250 fF
CGD  80 fF
C DB  100 fF
g m  150 
 p , X  2  5.31GHz
 0
 p ,Y  2  442MHz
1
Rd  2 K
CH 11 Frequency Response
47
Emitter and Source Followers
 The following will discuss the frequency response of
emitter and source followers using direct analysis.
 Emitter follower is treated first and source follower is
derived easily by allowing r to go to infinity.
CH 11 Frequency Response
48
Direct Analysis of Emitter Follower
Vout
Vin
C
1
s
gm
 2
as  bs  1
CH 11 Frequency Response
RS
C  C  C  C L  C C L 
a
gm
C  R S
b  RS C  
 1 
gm 
r
 CL

 gm
49
Direct Analysis of Source Follower Stage
Vout
Vin
CGS
1
s
gm
 2
as  bs  1
CH 11 Frequency Response
RS
CGDCGS  CGDC SB  CGS C SB 
a
gm
CGD  C SB
b  RS CGD 
gm
50
Example: Frequency Response of Source Follower
RS  200
C L  100 fF
CGS  250 fF
CGD  80 fF
C DB  100 fF
g m  150 
1
 0
CH 11 Frequency Response
 p1  2  1.79GHz  j 2.57GHz
 p 2  2  1.79GHz  j 2.57GHz
51
Example: Source Follower
Vout
Vin
CGS
1
s
gm
 2
as  bs  1
RS
CGD1CGS1  (CGD1  CGS1 )(C SB1  CGD2  C DB2 )
a
g m1
b  RS CGD1 
CGD1  C SB1 C GD 2 C DB 2
g m1
CH 11 Frequency Response
52
Input Capacitance of Emitter/Source Follower
rO  
C / CGS
Cin  C / CGD 
1  g m RL
CH 11 Frequency Response
53
Example: Source Follower Input Capacitance
1
Cin  CGD1 
CGS1
1  g m1 rO1 || rO 2 
CH 11 Frequency Response
54
Output Impedance of Emitter Follower
VX RS r C s  r  RS

IX
r C s    1
CH 11 Frequency Response
55
Output Impedance of Source Follower
V X RS CGS s  1

I X CGS s  g m
CH 11 Frequency Response
56
Active Inductor
 The plot above shows the output impedance of emitter and
source followers. Since a follower’s primary duty is to
lower the driving impedance (RS>1/gm), the “active
inductor” characteristic on the right is usually observed.
CH 11 Frequency Response
57
Example: Output Impedance
rO  
VX rO1 || rO 2 CGS 3 s  1

IX
CGS 3 s  g m3
CH 11 Frequency Response
58
Frequency Response of Cascode Stage
Av, XY
 g m1

 1
g m2
C x  2C XY
 For cascode stages, there are three poles and Miller
multiplication is smaller than in the CE/CS stage.
CH 11 Frequency Response
59
Poles of Bipolar Cascode
 p, X
1

RS || r 1 C 1  2C1 
 p ,out 
CH 11 Frequency Response
 p ,Y 
1
1
CCS1  C 2  2C1 
g m2
1
RL CCS 2  C  2 
60
Poles of MOS Cascode
 p, X 
1



g m1 
CGD1 
RS CGS1  1 
g m2 



 p ,Y 
CH 11 Frequency Response
 p ,out 
1
RL C DB 2  CGD2 
1
1
g m2



g m2 
CGD1 
C DB1  CGS 2  1 
g m1 



61
Example: Frequency Response of Cascode
RS  200
CGS  250 fF
CGD  80 fF
C DB  100 fF
g m  150 
1
 p , X  2  1.95GHz
 0
 p ,Y  2  1.73GHz
RL  2 K 
 p ,out  2  442MHz
CH 11 Frequency Response
62
MOS Cascode Example
 p, X 
1



g m1 
CGD1 
RS CGS1  1 
g m2 



 p ,Y 
1 
C DB1  CGS 2
CH 11 Frequency Response
g m2 
 p ,out 
1
RL C DB 2  CGD2 
1


g m2 
CGD1  CGD3  C DB3 
 1 
g m1 


63
I/O Impedance of Bipolar Cascode
1
Z in  r 1 ||
C 1  2C1 s
CH 11 Frequency Response
Z out
1
 RL ||
C 2  CCS 2 s
64
I/O Impedance of MOS Cascode
1
Z in 


 g m1 
CGD1  s
CGS1  1 
 g m2 


CH 11 Frequency Response
Z out
1
 RL ||
CGD2  C DB2 s
65
Bipolar Differential Pair Frequency Response
Half Circuit
 Since bipolar differential pair can be analyzed using halfcircuit, its transfer function, I/O impedances, locations of
poles/zeros are the same as that of the half circuit’s.
CH 11 Frequency Response
66
MOS Differential Pair Frequency Response
Half Circuit
 Since MOS differential pair can be analyzed using halfcircuit, its transfer function, I/O impedances, locations of
poles/zeros are the same as that of the half circuit’s.
CH 11 Frequency Response
67
Example: MOS Differential Pair
 p, X 
 p ,Y 
 p ,out
CH 11 Frequency Response
1
RS [CGS1  (1  g m1 / g m 3 )CGD1 ]
1



g m3 
CGD1 
C DB1  CGS 3  1 
g m1 



1

RL C DB3  CGD3 
1
g m3
68
Common Mode Frequency Response
Vout
g R R C  1
 m D SS SS
VCM
RSS CSS s  2 g m RSS  1
 Css will lower the total impedance between point P to
ground at high frequency, leading to higher CM gain which
degrades the CM rejection ratio.
CH 11 Frequency Response
69
Tail Node Capacitance Contribution
 Source-Body Capacitance of
M1, M2 and M3
 Gate-Drain Capacitance of M3
CH 11 Frequency Response
70
Example: Capacitive Coupling
Rin2  RB 2 || r 2    1RE 
L1 
1
 2  542Hz 
r 1 || RB1 C1
CH 11 Frequency Response
L 2
1

   22.9 Hz 
RC  Rin2 C2
71
Example: IC Amplifier – Low Frequency Design
Rin2 
CH 11 Frequency Response
RF
1  Av 2
L1 
g m1RS1  1
 2  42.4MHz
RS1C1
L 2 
1
 2  6.92MHz
RD1  Rin2 C2
72
Example: IC Amplifier – Midband Design
vX
  g m1 RD1 || Rin2   3.77
vin
CH 11 Frequency Response
73
Example: IC Amplifier – High Frequency Design
 p1  2  (308 MHz)
 p 2  2  (2.15 GHz)
 p3
1

RL 2 (1.15CGD 2  C DB 2 )
 2  (1.21 GHz)
CH 11 Frequency Response
74
```