EECS140 ANALOG CIRCUIT DESIGN
LECTURES ON FREQUENCY RESPONSE
University of California
Berkeley
College of Engineering
Department of Electrical Engineering
and Computer Science
Robert W. Brodersen
EECS140
Analog Circuit Design
Lectures
on
FREQUENCY RESPONSE
ROBERT W. BRODERSEN
EECS140 ANALOG CIRCUIT DESIGN
LECTURES ON FREQUENCY RESPONSE
FR-1
Bode Plots
Solve impedance network transfer function.
V OUT( ω )
H ( ω ) = -----------------V IN ( ω )
V IN ( ω )
( H(ω) , Vout(ω) & Vin(ω) are phasors )
R
VOUT ( ω )
C
1 ⁄ jωC
V OUT ( ω )
1 ⁄ jωC
H ( ω ) = -----------------= --------------------------V IN ( ω )
R + 1 ⁄ jωC
1 ⁄ jωC
----------------------H( ω) =
1 + jωRC
Convert H(ω) to polar coordinates, |H(ω)| < θ
H ( ω ) = [ H ( ω ) ( H∗ ( ω ) ) ]
 Im{H(ω) } 
θ = atan  ------------------------
R
{
H
(
ω
)
}
 e

ROBERT W. BRODERSEN
12
N(ω)
IF H ( ω ) = ------------- then
D(ω)
R e ⋅ { H ( ω ) } = R e ⋅ { N ( ω ) ⋅ D∗ ( ω ) }
I M ⋅ { H ( ω ) } = I M ⋅ { N ( ω ) ⋅ D∗ ( ω ) }
EECS140 ANALOG CIRCUIT DESIGN
LECTURES ON FREQUENCY RESPONSE
FR-2
Bode Plots (Cont.)
( H ( ω ) ⋅ H∗ ( ω ) )
1
1
=  ----------------------- ⋅  ----------------------- 
1 + jωRC
1 – jωRC
1 /2
1
=  ----------------------------2
 1 + ( ωRC) 
Bode Plot Magnitude
1/2

 1/2


1
=  -------------------------2
ω - 
 1 +  -------
 ω 3dB 
H ( ω ) d B = 20 ⋅ log H ( ω )
ω » ω3dB
-3dB
1/2
20dB/Decade
6dB/Octave
ωp = 1/RC = ω3dB
ROBERT W. BRODERSEN
ω
1
H ( ω ) ≈ ---------------2
ω 
 ------- ω 3dB
ω 3d B
= -------ω
6dB/Octave - drops by 2 every
time frequency doubles
EECS140 ANALOG CIRCUIT DESIGN
LECTURES ON FREQUENCY RESPONSE
Bode Plots (Cont.)
1
H ( ω ) = -------------------- = H ( ω ) ⋅ exp ( jθ ( ω ) )
ω
1 + j --------ω 3dB
ω 
 1 – j -------

ω
3dB
1
H ( ω ) = -------------------- ⋅ ------------------------ω - 1 – j -------ω1 + j -------ω 3dB
ω 3dB
ω
1 – j --------ω3dB
= ------------------------2
ω
1 +  ---------
 ω 3 d B
 Im H ( ω ) 
θ ( ω ) = atan  ------------------
 R eH ( ω ) 
ROBERT W. BRODERSEN
FR-3
EECS140 ANALOG CIRCUIT DESIGN
LECTURES ON FREQUENCY RESPONSE
FR-4
Bode Plots (Cont.)
1
Re { H ( ω ) } = -------------------------2
ω
1 +  ---------
 ω 3dB
ω
–  ---------
ω 3dB
------------------------Im { H ( ω ) } =
2
ω
1 +  ---------
 ω 3dB
 ω 
θ ( ω ) = arc tan  – --------- 
 ω 3d B 
 ω 
= – arc tan  --------- 
 ω 3d B 
ROBERT W. BRODERSEN
0.1ω 3dB
log ( ω )
ω 3dB
10ω 3dB
0
θ
-π/4
-π/2
Linear Approximation
EECS140 ANALOG CIRCUIT DESIGN
LECTURES ON FREQUENCY RESPONSE
FR-5
Bode Plots (Cont.)
0 Phase
Magnitude
90 Phase difference
Magnitude Change
t = 0
t = 0
R
1
----------------------H(ω) =
1 + jωRC
C
ROBERT W. BRODERSEN
H ( ω ) = ( H ( ω ) ⋅ H∗ ( ω ) )
1/2
EECS140 ANALOG CIRCUIT DESIGN
LECTURES ON FREQUENCY RESPONSE
FR-6
Bode Plots (Cont.)
1 Pole Summary R
1
H ( ω ) = -------------------ω
1 + j --------ω 3d B
1
ω 3dB = -------RC
C
log ( ω )
H ( ω ) dB
0.1ω 3dB
0dB
0
20dB/Decade
1∝ --ω
-10dB
-20dB
θ
-π/4
-π/2
ω 3dB
ROBERT W. BRODERSEN
ω 3dB
10ω 3dB
EECS140 ANALOG CIRCUIT DESIGN
LECTURES ON FREQUENCY RESPONSE
FR-7
Bode Plots (Cont.)
2 Poles
R1
Ideal Unity gain Buffer
R2
ν IN
C1
ROBERT W. BRODERSEN
νOUT
1
C2
EECS140 ANALOG CIRCUIT DESIGN
LECTURES ON FREQUENCY RESPONSE
FR-8
Bode Plots (Cont.)
Two Poles:


1
H ( ω ) =  ------------------

ω
1
+
j
------

ω p1


1
⋅  ------------------  = H ω ( ω ) ⋅ H ω ( ω )

ω
1
+
j
------

ω p2
p1
p2
20 ⋅ log H ( ω ) = 20 ⋅ log ( H ω ( ω ) ⋅ H ω ( ω ) )
p2
p1
= 20 ⋅ log H ω ( ω ) + 20 ⋅ log H ω ( ω )
ω p1
ω p2
p2
p1
1--ω
1
-----2
ω
ROBERT W. BRODERSEN
EECS140 ANALOG CIRCUIT DESIGN
LECTURES ON FREQUENCY RESPONSE
FR-9
Bode Plots (Cont.)
H ( ω ) = [ H ω ⋅ exp ( jθω ) ] ⋅ [ H ω ⋅ exp ( jθω ) ]
p2
p2
p1
p1
= H ω ⋅ H ω ⋅ exp ( j [ θ ω + θω ] )
p1
p2
p1
p2
= H ( ω ) ⋅ exp ( jθ ( ω ) )
θ ( ω ) = θ ω + θω
p1
p2
0.1ω p1
0
ω p1
10ω p1
0.1ω p2
ω p2
10ω p2
log ( ω )
θ
-π/4
-π/2
-3π/4
-π
ROBERT W. BRODERSEN
EECS140 ANALOG CIRCUIT DESIGN
LECTURES ON FREQUENCY RESPONSE
FR-10
Bode Plots (Cont.)
R1
R2
ν IN
ν OUT
1
C1
C2
If we have 180 degree phase shift we have a problem.
(+)
ν IN
(-)
ROBERT W. BRODERSEN
+
The negative feedback will turn into positive feedback.
ν OUT
Can’t have positive feedback in a loop with gain > 1
EECS140 ANALOG CIRCUIT DESIGN
LECTURES ON FREQUENCY RESPONSE
FR-11
Bode Plots (Cont.)
ω p1 = ω p2 = ω p3 = ω p
10ω p
0.1ω p
ωp
0
log ( ω )
θ
3 poles on top
of each other
-π/4
-π/2
Likely unstable circuit
-3π/4
Can’t kill gain here
without adding phase shift.
-π
-5π/4
-3π/2
ROBERT W. BRODERSEN
EECS140 ANALOG CIRCUIT DESIGN
LECTURES ON FREQUENCY RESPONSE
Bode Plots (Cont.)
Zero at zero frequency, pole at 1/RC
C
ν IN
ν OUT
R
H ( ω ) dB
1
R
jωRC
jωRC
-------------------= ----------------------- = ------------------1
1 + jωRC
ω
1 + j -------R + ---------jωC
1
-------RC
1
ω p = -------RC
log ( ω )
ROBERT W. BRODERSEN
FR-12
EECS140 ANALOG CIRCUIT DESIGN
LECTURES ON FREQUENCY RESPONSE
Single Zero at ωz
Left half plane zero
ω
H ( ω ) =  1 + j -----
ωZ
π/2
θ
π/4
FR-13
ω
θ = arc tan  ----- 
 ω Z
0.1ω Z
ωZ
0
θ
0
Right half plane zero
ω
H ( ω ) =  1 – j -----
ωZ
H(ω)
log ( ω )
-π/4
-π/2
For both:
dB
ω
H ( ω ) =  1 +  ----- 
ωZ
2
∝ω
1
0.1ω Z
ROBERT W. BRODERSEN
10ω Z
ωZ
10ω Z
log ( ω )
1/2
EECS140 ANALOG CIRCUIT DESIGN
LECTURES ON FREQUENCY RESPONSE
FR-14
Bode Plots (Cont.)
R1
ν IN
ν OUT
C
R2
C
ν IN
ROBERT W. BRODERSEN
R2  ⋅ ---------------------------------------1
H ( ω ) =  --------------- R1 + R 2 1 + jω ( R 1 || R 2 )C
1 Pole, 1 Zero
ν OUT
R1
1 Pole
R2
1 + jωR1 C R2  ⋅ ---------------------------------------H ( ω ) =  --------------- R1 + R 2 1 + jω ( R 1 || R 2 )C
EECS140 ANALOG CIRCUIT DESIGN
LECTURES ON FREQUENCY RESPONSE
FR-15
Bode Plots (Cont.)
1 Pole, 1 Zero
H ( ω ) dB
ωp > ωz
ωp = ωz
ωp < ωz
ω
1 + j ----ωZ
H ( ω ) = ----------------ω
1 + j ----ωp
ROBERT W. BRODERSEN
EECS140 ANALOG CIRCUIT DESIGN
LECTURES ON FREQUENCY RESPONSE
FR-16
Capacitances
εSiO
F
-----2 = 10 –4 ----C OX = -------- = 0.1 fF
tOX
µ
m2
G
2
CGS
C GD
S
C SB
D
CGB
C DB
– 10 F
CGSO = 5 ×10 ---m
– 10 F
CGDO = 5 ×10 ---m
– 10 F
CGBO = 4 ×10 ---m
F
CJ = 10 –4 ----m2
PB = φ B = 0.8V
B
ROBERT W. BRODERSEN
EECS140 ANALOG CIRCUIT DESIGN
LECTURES ON FREQUENCY RESPONSE
Capacitances (Cont.)
Linear :
Sat :
CGS = 2
--- ⋅ COX ⋅ L ⋅ W + CGSO ⋅ W
3
CGD = CGDO ⋅ W
CJSW ⋅ PS
CJ ⋅ AS
-----------------------------+
CS B = -------------------------MJSW
MJ
V
V
BS
BS
 1 + ------- 
 1 + -------


PB
PB
CGB = CGBO ⋅ L
COX ⋅ L ⋅ W
C GS = ------------------------+ CGSO ⋅ W
2
C OX ⋅ L ⋅ W
C GD = ------------------------+ CGDO ⋅ W
2
(similar for C DB )
MJ = 1--- (default)
2
MJSW = 3 (default)
PS = Perimeter of Source
AS =
Area of Source
CGBO ≡
ROBERT W. BRODERSEN
Capacitance of gate to bulk overlap
FR-17
EECS140 ANALOG CIRCUIT DESIGN
LECTURES ON FREQUENCY RESPONSE
FR-18
Capacitances (Cont.)
Layout
Source
Drain
W
2λ
λ
L
2λ
4λ
2λ
(Minimum size device, W/L = 2)
Area of Source
Area of Drain
= AS = 4λ ⋅ W
= AD = AS
Perimeter of Source
ROBERT W. BRODERSEN
= PS = 8λ + W
EECS140 ANALOG CIRCUIT DESIGN
LECTURES ON FREQUENCY RESPONSE
Capacitances (Cont.)
M1 1 2 3 4 NMOS L=2u W=2u
+ AS=4p AD=4p PS=6u PD=6u
G
S
D
CGSO
CGDO
Capacitor (in linear)
ROBERT W. BRODERSEN
FR-19
EECS140 ANALOG CIRCUIT DESIGN
LECTURES ON FREQUENCY RESPONSE
Miller Approximation i
FR-20
C
C
it
ν in
–A
ν OUT
νt
ν OUT = – A ⋅ ν in
i C = C ⋅ d ( ν in – ν OUT) = C ⋅ d ( ν in + A ⋅ ν OUT ) = C ⋅ ( 1 + A ) ⋅ dν in
dt
dt
dt
ν in
–A
νt
C( 1 + A)
ROBERT W. BRODERSEN
ν OUT
EECS140 ANALOG CIRCUIT DESIGN
LECTURES ON FREQUENCY RESPONSE
Miller Approximation (Cont.)
C
FR-21
C
RL
ν OUT
ν in
ν OUT
νin
R IN
g m ⋅ νgs
RL
R IN
RIN
ν in
C ( 1 + g m ⋅ ROUT )
ROBERT W. BRODERSEN
ν OUT









RL
ν OUT
g m ⋅ R OUT
--------- = ---------------------------------------------------------------------------------------------ν in
1 + jω [ RIN C ( 1 + g m ⋅ ROUT ) + R OUT C ]
CMILLER
1
ω p = --------------------------------------------R IN C ( 1 + g m ⋅ ROUT )
EECS140 ANALOG CIRCUIT DESIGN
LECTURES ON FREQUENCY RESPONSE
Inverter
FR-22
RL
CGD
R IN
CGD
ν OUT ν in
R IN
νOUT
ν in
CD
CGB
C GS
CG
g m ⋅ νin
R OUT = R L || r o
CG = C GB + C GS
CD ignored (not usually possible)
ROBERT W. BRODERSEN
RL
EECS140 ANALOG CIRCUIT DESIGN
LECTURES ON FREQUENCY RESPONSE
Inverter (Cont.)
R IN
ν1
ν in
ZG
FR-23
ZGD
ν OUT
g m ⋅ ν in
RL
ν1 – ν in ν 1 ν1 – ν OUT
---------------- + ----- + ------------------- = 0
R IN
ZG
Z GD
GD
1 – C
-------
gm 
ν OUT
--------- = – g m ⋅ R OUT ⋅ --------------------------------------------------------------------------------------------------------------------------------------------------------νin
1 + jω { [ CGD ( 1 + g m ⋅ R L ) + CG ] ⋅ RIN + RL CGD } – ω 2 R OUT RIN C G C GD
 1 – jω C
 1 – jω C
GD
GD
--------------

gm 
gm 
= ----------------------------------------------------- = ---------------------------------------------------------------------ω-  ⋅  1 + j -----ω-
1 -----1 -
1
 1 + j -----2 1
 ------+
–
ω
----------1
jω
+

 ω p1 ω p2
ω p1 
ω p2
ω p1 ω p2
ROBERT W. BRODERSEN
EECS140 ANALOG CIRCUIT DESIGN
LECTURES ON FREQUENCY RESPONSE
Inverter (Cont.)







1
ω p1 = – ------------------------------------------------------------------------------------------R IN ⋅ [ C G + C GD ( 1 + g m ROUT ) ] + R L C GD
CMILLER
ω p2
1
1
= – ------------------ – -------------------------------------------R OUT CGD 
1- C
R OUT || RIN || ---G

g m
g
ω Z = --------m
C GD
ω
 1 + j ----
ω Z
H ( ω ) = ----------------------------------------------------ω- ⋅  1 + j ------ω
 1 + j -----
ω p1 
ω p2
ROBERT W. BRODERSEN
FR-24
EECS140 ANALOG CIRCUIT DESIGN
LECTURES ON FREQUENCY RESPONSE
FR-25
Inverter (Cont.)
ω
 1 + j ----
ω Z
H ( ω ) = ----------------------------------------------------ω- ⋅  1 + j -----ω-
 1 + j -----
ω p1 
ω p2
s Z = – jω Z
s p1 = – jω p1
s
 1 – --
s Z
H ( s ) = ------------------------------------------- 1 – --s---  ⋅  1 – ---s-- 

s p1 
s p2
X
g m----ωp2 ≈ –
CG
ROBERT W. BRODERSEN
s p2 = – jω p2
X
1
--------------------ω p1 ≈ –
R IN CMILLER
O
gm
-------ωZ =
C GD
EECS140 ANALOG CIRCUIT DESIGN
LECTURES ON FREQUENCY RESPONSE
FR-26
Inverter (Cont.)
RL
RL
RIN
ν OUT
ν in
CG
R IN
1
--------------------ω p1 ≈ –
R IN C MILLER
1
----gm
ω p1
ω p2
1
----------1
----- CG
gm
as Cgd increases
ROBERT W. BRODERSEN
EECS140 ANALOG CIRCUIT DESIGN
LECTURES ON FREQUENCY RESPONSE
FR-27
Inverter (Cont.)
RL
Cµ
ν OUT
R IN
ν in
ROUT = R L || r o
CD
Cπ
Case 1 (Miller Capacitance not important) :







R IN C π, ROUT C D » RIN ⋅ ( 1 + g m ROUT )C µ
C MILLER
1
ω p1 = -----------RIN C π
ROBERT W. BRODERSEN
1
ω p2 = ---------------ROUT C D
EECS140 ANALOG CIRCUIT DESIGN
LECTURES ON FREQUENCY RESPONSE
Inverter (Cont.)
1
Z OUT = R L || r o || -----------jωC D
Z OUT
ro
CD
A ν = – g m ⋅ ZOUT
1
ω p = ---------------R OUT CD
ROBERT W. BRODERSEN
1
= R OUT || -----------jωC D
R OUT
-----------jωCD
R OUT
= – g m ⋅ ---------------------------= – g m ⋅ ------------------------------1
1
+
jωR
C
OUT
D
ROUT + -----------jωC D
FR-28
EECS140 ANALOG CIRCUIT DESIGN
LECTURES ON FREQUENCY RESPONSE
Inverter (Cont.)
FR-29
R IN
ν in
ν' in
Cπ
1
-----------ν in'
jωC π
1
------ = ------------------------= ---------------------------ν in
1
1 + jωRIN Cπ
R IN + -----------jωCπ
1
ω p = -----------R IN Cπ
ROBERT W. BRODERSEN
EECS140 ANALOG CIRCUIT DESIGN
LECTURES ON FREQUENCY RESPONSE
FR-30
Inverter (Cont.)
Case 2 (Large Cd) :
R OUTC D » R IN ( 1 + g m ROUT )C µ, R IN Cπ
1
ω p1 = ---------------ROUT C D
1
ω p2 = ----------------------------RIN ( Cπ + Cµ )
Case 3 (Large Cµ) :
R IN ( 1 + g m ROUT )C µ » R OUTC D, R IN Cπ
1
ω p1 = -----------------------------------------RIN ( 1 + g m R OUT )C µ
1
ω p2 = ---------------------------1
----- ( Cπ + CD )
gm
For case 2 and 3,
g
ω ZERO = -----m
Cµ
ROBERT W. BRODERSEN
X
C µ large
gm
ω p2 ≈ ---------------------( Cπ + CD )
X
X
C µ = 0 Cµ = 0
1
1
--------------------------R OUTC D RIN Cπ
1
ω p1 ≈ --------------------R IN CMILLER
X
C µ large
O
g
ω Z = -----m
Cµ
EECS140 ANALOG CIRCUIT DESIGN
LECTURES ON FREQUENCY RESPONSE
FR-31
Source Follower
R IN
νg
ν in
g mb ( – νOUT )
Cπ
g m ( νg – ν OUT )
ν OUt
RS
1
ν g =  ----------------------------  ⋅ ( νin – ν OUT ) + ν OUT
 1 + jωR IN Cπ
νOUT
ν g – νOUT
--------- = ------------------+ g m ⋅ ν g – ( 1 + χ ) ⋅ g m ⋅ νOUT
RS
1
-----------jωCπ
ROBERT W. BRODERSEN
ν OUT
EECS140 ANALOG CIRCUIT DESIGN
LECTURES ON FREQUENCY RESPONSE
Source Follower (Cont.)
1
ν OUT ⋅  ----- + ( 1 + χ ) ⋅ g m + jωCπ = ( jωC π + gm ) ⋅ ν g
 RS








 1 + jω C
-----π

gm
νOUT
gmR S
--------- = ----------------------------------------- ⋅ ----------------------------------------------------------------ν in
1 + ( 1 + χ ) ⋅ g m RS 1 + jωR -------------------------------------Cπ ( 1 + χ ⋅ g m RS )
IN
1 + ( 1 + χ )g m R S
1–A
g
ω Z = -----m
Cπ
1
ω p = -----------------------------R IN Cπ ( 1 – A )
g m RS
A = ------------------------------------1 + ( 1 + χ )gm R S
ROBERT W. BRODERSEN
g m RS
------------------------------------Aν =
1 + ( 1 + χ )g m R S
FR-32
EECS140 ANALOG CIRCUIT DESIGN
LECTURES ON FREQUENCY RESPONSE
R IN
FR-33
CGD
ν in
Source Follower
again with CSB
ν OUT
CGB
C GS
RS
Small Signal :
RIN
CS B
νg
νin
g m ( ν g – νOUT )
CGS
C G = CGD + CGB
CSB
ROBERT W. BRODERSEN
ν OUT
RS
EECS140 ANALOG CIRCUIT DESIGN
LECTURES ON FREQUENCY RESPONSE
Source Follower (Cont.)
FR-34
νin – ν G
---------------- = ν G ⋅ jωCG + ( ν G – νOUT ) ⋅ jωCGS
RIN
νOUT
( ν G – ν OUT ) ⋅ jωC GS – g m ( ν G – ν OUT ) = --------- + ν OUT ⋅ jωC SB
RS
gm RS
CGS
------------------- ⋅ 1 + jω ------1 + gm RS
gm
ν
OUT
-------- = ---------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------ν in
RIN C GS
R S ( CGS + C S B )
C GS CG + C SB ( CG + CGS )
------------------1 + jω RIN C G +
+ ------------------------------- – ω 2 RS RIN ----------------------------------------------------1 + g m RS
1 + g m RS
1 + gm R S
let denominator
ω
ω
=  1 + j ----  ⋅  1 + j ---- 

p 1 
p 2
2
1
1
ω
= 1 + jω  ---- + ---- – -------- p 1 p2 p 1 p 2
ROBERT W. BRODERSEN
EECS140 ANALOG CIRCUIT DESIGN
LECTURES ON FREQUENCY RESPONSE
Source Follower (Cont.)
FR-35
for p 1 and p 2 widely separated, if we assume that p 1 is the dominant pole,
1 1
---- » ---p1 p2
1
1
p 1 = ------------------------------------------------------------------------------ = ------------------------------------------------------------------------------RIN C GS
RS ( C GS + C SB )
RIN C GS
- + ------------------------------- + RO ( CGS + CS B )
R IN C G + ------------------RIN C G + ------------------1 + g m RS
1 + gm RS
1 + gmR S
where,
1- || R
R O = ---S
gm
thus,
R IN CGS
R IN CG + -------------------- + R O ( C GS + C SB )
1 + gm RS
p 2 = -----------------------------------------------------------------------------RO R IN [ C GS CG + CS B CG + C SB C GS ]
ROBERT W. BRODERSEN
EECS140 ANALOG CIRCUIT DESIGN
LECTURES ON FREQUENCY RESPONSE
Source Follower (Cont.)
FR-36
2 limiting cases,
Case 1 :
C GS
RIN  C G + -------------------- » RO ( CGS + CSB )

1 + g m RS
1
ω p 1 = --------------------------------------------CGS
R IN  CG + --------------------

1 + g m RS
Case 2 :
C GS 
RO ( CGS + CSB ) » RIN  C G + ------------------
1 + g m R S
ω p1
1
-------------------------------=
R O ( C GS + C S B )
ROBERT W. BRODERSEN
Miller cap = CGS ⋅ ( 1 – A )
gmR S
A = -------------------1 + gmR S
1
gm RS
1 – A = 1 – -------------------- = -------------------1 + gmR S
1 + gm RS
EECS140 ANALOG CIRCUIT DESIGN
LECTURES ON FREQUENCY RESPONSE
FR-37
Common Gate
RS
ν OUT
ν in
+-
RL
C GS
C SB
C GD
C DB
assume r o → ∞
RS
ν in
νS
ν OUT
+-
RL
CS
ROBERT W. BRODERSEN
CD
EECS140 ANALOG CIRCUIT DESIGN
LECTURES ON FREQUENCY RESPONSE
Common Gate (Cont.)
KCL@ν S
ν in – ν S
---------------= ν S ⋅ jωC S + g m νS
RS
ν OUT
gm ν S = ν OUT ⋅ jωCD + -------RL
g m RL ------------------ν
1 + gmR S
OUT
-------- = -----------------------------------------------------------------------------ν in
RS CS 
( 1 + jωR L CD ) ⋅  1 + jω ------------------
1 + g m R S
no zeros, poles @
KCL@νOUT
1
p 1 = ----------RLCD
1
1
p 2 = -------------------------- = --------------------------RS - C
1- C
 R || ---------------------S
S
S

1 + g m RS
g m
ROBERT W. BRODERSEN
FR-38
EECS140 ANALOG CIRCUIT DESIGN
LECTURES ON FREQUENCY RESPONSE
Common Gate (Cont.)
@ ν OUT → Req = RL
C eq = CD
1
@ ν S → R eq = RS || ----gm
Ceq = CS
1
1
p 1 = ------------- = ----------R eq Ceq
RLCD
1
1
p 2 = ------------- = --------------------------R eq Ceq
1
 R S || ---- CS

g m
Since all caps go to ground, finding poles reduces to finding
Req’s and Ceq’s at the nodes.
ROBERT W. BRODERSEN
FR-39
EECS140 ANALOG CIRCUIT DESIGN
LECTURES ON FREQUENCY RESPONSE
Cascode - reduces the Miller probelm
RL
CGD2
νOUT
(2)
VBIAS
CD2
C GS2
RIN
νin
CGD1
(1 )
CGS1
ROBERT W. BRODERSEN
C D1
FR-40
EECS140 ANALOG CIRCUIT DESIGN
LECTURES ON FREQUENCY RESPONSE
Cascode (Cont.)
If we assume that RS is very large, then,
At (1) :
Req = R S
C eq = CGS1 + C GD1 ⋅ ( 1 – A )
1- = – 1
A = – g m ⋅ ---gm
1
p 1 = ------------R eq Ceq
C eq = CGS1 + 2 ⋅ CGD1
At (2) :
R eq = RL || r o ← large
C eq = CGD2 + CD2
1
p 2 = ------------R eq Ceq
No large Miller multiplication of a capacitance!
ROBERT W. BRODERSEN
FR-41
EECS140 ANALOG CIRCUIT DESIGN
LECTURES ON FREQUENCY RESPONSE
Zero Value Time Constant Analysis
FR-42
General case of dominant pole approximation. Use this technique
for complex circuits where we can’t identify node with large Req
and Ceq.
Strategy :
1) Set all caps Cj=0 except for Ci
2) Find resistance seen by Ci
3) Calculate RiCi for all caps
4) ω
3dB
1
= -------------∑ Ri C i
ROBERT W. BRODERSEN
EECS140 ANALOG CIRCUIT DESIGN
LECTURES ON FREQUENCY RESPONSE
FR-43
ZVTC with Source Follower
RIN
CGD
νin
ν OUT
C GB
CGS
RS
RIN
CSB
νg
νin
g m ( ν g – νOUT )
CGS
C G = CGD + C GB
CSB
ROBERT W. BRODERSEN
νOUT
RS
EECS140 ANALOG CIRCUIT DESIGN
LECTURES ON FREQUENCY RESPONSE
ZVTC with Source Follower
C1 :
R1 = RIN
R IN
CG
C1 = CG
C2 :
1
R2 = RS || ----gm
gm ( – νOUT )
ν OUT
C SB
ROBERT W. BRODERSEN
RS
C2 = CSB
FR-44
EECS140 ANALOG CIRCUIT DESIGN
LECTURES ON FREQUENCY RESPONSE
ZVTC with Source Follower (Cont.)
C3 :
FR-45
R IN
+
νt
it
g m ( νg – ν OUT )
-
C GS
do small signal analysis to find
ν OUT
RS
νt
---- = R3 seen by C GS
it
R IN - where R = R || ---1R 3 = RO + ------------------O
S
1 + gm RS
gm
C3 = CGS
hence,
1
1
ω 3 d B = ---------------------------------------------- = ------------------------------------------------------------------------------------R 1 C1 + R 2 C 2 + R 3 C3
R IN -
R IN CG + RO CS B +  RO + ------------------C

1 + g m R S GS
ROBERT W. BRODERSEN