EECS140 ANALOG CIRCUIT DESIGN
LECTURES ON FREQUENCY RESPONSE
EECS140 ANALOG CIRCUIT DESIGN
LECTURES ON FREQUENCY RESPONSE
FR-1
Bode Plots
University of California
Berkeley
Solve impedance network transfer function.
V OUT ( ω )
H ( ω ) = -----------------VI N ( ω )
College of Engineering
Department of Electrical Engineering
and Computer Science
VI N ( ω )
Robert W. Brodersen
EECS140
( H(ω) , V out(ω) & V in(ω) are phasors )
R
Analog Circuit Design
V O U T( ω )
C
Lectures
on
FREQUENCY RESPONSE
1 ⁄ jωC
H ( ω ) = [ H ( ω ) ( H∗ ( ω ) ) ]
ROBERT W. BRODERSEN
1
-2
N(ω)
IF H ( ω ) = ------------- t h e n
D( ω )
R e ⋅ { H ( ω ) } = R e ⋅ { N ( ω ) ⋅ D∗ ( ω ) }
I M ⋅ { H ( ω )} = IM ⋅ { N ( ω ) ⋅ D ∗ ( ω ) }
ROBERT W. BRODERSEN
EECS140 ANALOG CIRCUIT DESIGN
LECTURES ON FREQUENCY RESPONSE
1
1
=  -----------------------  ⋅  ----------------------- 
 1 + jωR C   1 – jωR C 
1/ 2
1
=  ----------------------------2
 1 + ( ωR C ) 
Bode Plot Magnitude
H (ω)
dB
1/ 2




1
=  -------------------------2

 ω  
 1 +  --------ω 3 dB 
1 /2
ω » ω 3dB
1 /2
20dB/Decade
6dB/Octave
ωp = 1/RC = ω3dB
ω
1
H ( ω ) ≈ ----------------2
ω 
 -------- ω 3 d B
LECTURES ON FREQUENCY RESPONSE
Bode Plots (Cont.)
1
H ( ω ) = -------------------- = H ( ω ) ⋅ exp ( jθ ( ω ) )
ω
1 + j --------ω 3dB
ω 
 1 – j --------
ω 3 d B
1
H ( ω ) = -------------------- ⋅ ------------------------ω
ω
1 – j --------1 + j --------ω 3dB
ω 3d B
= 20 ⋅ log H ( ω )
-3dB
EECS140 ANALOG CIRCUIT DESIGN
FR-2
Bode Plots (Cont.)
ROBERT W. BRODERSEN
1 ⁄ jωC
H ( ω ) = ----------------------1 + jωR C
Convert H(ω) to polar coordinates, |H(ω)| < θ
 Im { H ( ω ) } 
θ = atan  ------------------------
 Re { H ( ω ) } 
( H( ω ) ⋅ H∗( ω ) )
V O U T( ω )
1 ⁄ jωC
H ( ω ) = -----------------= --------------------------V IN ( ω )
R + 1 ⁄ jωC
ω 3dB
= --------ω
6dB/Octave - drops by 2 every
time frequency doubles
ω
1 – j --------ω3 d B
= -------------------------2
ω 
1 +  -------- ω 3 d B
 Im H ( ω ) 
θ ( ω ) = atan  ------------------
 Re H ( ω ) 
ROBERT W. BRODERSEN
FR-3
EECS140 ANALOG CIRCUIT DESIGN
LECTURES ON FREQUENCY RESPONSE
FR-5
Bode Plots (Cont.)
0 Phase
Magnitude
log ( ω )
0.1 ω 3d B
ω
–  -------- ω 3 d B
Im { H ( ω ) } = -------------------------2
ω 
1 +  -------- ω 3 d B
LECTURES ON FREQUENCY RESPONSE
FR-4
Bode Plots (Cont.)
1
R e { H ( ω ) } = -------------------------2
ω

1 +  --------- 
ω 3dB
EECS140 ANALOG CIRCUIT DESIGN
ω 3dB
90 Phase difference
Magnitude Change
10ω3 dB
0
θ
 ω 
θ ( ω ) = a r c tan  – --------- 
 ω 3dB
t = 0
t = 0
-π/4
-π/2
R
Linear Approximation
1
H ( ω ) = ----------------------1 + jωR C
 ω 
= – a r c tan  --------- 
 ω 3dB
C
ROBERT W. BRODERSEN
LECTURES ON FREQUENCY RESPONSE
EECS140 ANALOG CIRCUIT DESIGN
LECTURES ON FREQUENCY RESPONSE
FR-6
Bode Plots (Cont.)
1
H ( ω ) = -------------------ω
1 + j --------ω 3dB
1
ω3 d B = -------RC
2 Poles
C
log ( ω )
H (ω ) dB
0.1ω3 dB
0dB
0
20dB/Decade
1
∝ ---ω
θ
ω 3dB
10ω 3dB
R1
Ideal Unity gain Buffer
R2
νI N
-π/4
ω 3dB
ROBERT W. BRODERSEN
νOUT
1
C1
-π/2
ROBERT W. BRODERSEN
FR-7
Bode Plots (Cont.)
1 Pole Summary R
-20dB
1/ 2
ROBERT W. BRODERSEN
EECS140 ANALOG CIRCUIT DESIGN
-10dB
H ( ω ) = ( H ( ω ) ⋅ H∗( ω ) )
C2
EECS140 ANALOG CIRCUIT DESIGN
LECTURES ON FREQUENCY RESPONSE


1
H ( ω ) =  ------------------ 

ω
 1 + j ------- 
ωp 1
H ( ω ) = [ H ω ⋅ exp ( jθ ω ) ] ⋅ [ H ω ⋅ exp ( jθ ω ) ]
p1
p1
p2
p2
p2
p1
p2
= H ( ω ) ⋅ exp ( jθ ( ω ) )
θ (ω ) = θω + θ ω
p2
p1
= 20 ⋅ log H ω ( ω ) + 20 ⋅ log H ω ( ω )
ω p1
ω p2
p1
p1
= H ω ⋅ H ω ⋅ exp ( j [ θ ω + θ ω ] )
p2
20 ⋅ log H ( ω ) = 20 ⋅ log ( H ω ( ω ) ⋅ H ω ( ω ) )
p1
FR-9
Bode Plots (Cont.)


1
⋅  ------------------ = H ω ( ω ) ⋅ H ω ( ω )
ω
 1 + j ------- 
ω p2
p1
LECTURES ON FREQUENCY RESPONSE
FR-8
Bode Plots (Cont.)
Two Poles:
EECS140 ANALOG CIRCUIT DESIGN
p2
0.1 ω p1
p2
ω p1
10 ω p1
0.1 ω p2
ω p2
10ω p 2
0
log ( ω )
θ
-π/4
-1ω
-π/2
1
----ω2
-3π/4
-π
ROBERT W. BRODERSEN
ROBERT W. BRODERSEN
EECS140 ANALOG CIRCUIT DESIGN
LECTURES ON FREQUENCY RESPONSE
FR-10
Bode Plots (Cont.)
R1
EECS140 ANALOG CIRCUIT DESIGN
0
νOUT
1
FR-11
Bode Plots (Cont.)
ω p 1 = ω p 2 = ωp 3 = ω p
10 ωp
ωp
0.1 ωp
R2
ν IN
LECTURES ON FREQUENCY RESPONSE
log ( ω )
θ
3 poles on top
of each other
-π/4
C1
C2
-π/2
Likely unstable circuit
-3π/4
Can’t kill gain here
without adding phase shift.
If we have 180 degree phase shift we have a problem.
ν IN
(+)
-
(-)
+
-π
The negative feedback will turn into positive feedback.
νO U T
Can’t have positive feedback in a loop with gain > 1
-5π/4
-3π/2
ROBERT W. BRODERSEN
ROBERT W. BRODERSEN
EECS140 ANALOG CIRCUIT DESIGN
LECTURES ON FREQUENCY RESPONSE
Zero at zero frequency, pole at 1/RC
ν IN
νO U T
dB
1
Single Zero at ω z
Left half plane zero
ω
H ( ω ) =  1 + j ----- 

ω Z
π/2
θ
0
θ
0
Right half plane zero
ω
H ( ω ) =  1 – j ----
ω Z
1
ω p = -------RC
H (ω)
log ( ω )
-π/2
For both:
ω
H ( ω ) =  1 +  ----- 
ωZ
2
∝ω
1
0.1ω Z
ROBERT W. BRODERSEN
10ω Z
-π/4
dB
log ( ω )
ωZ
10 ω Z
1 /2
log ( ω )
ROBERT W. BRODERSEN
EECS140 ANALOG CIRCUIT DESIGN
LECTURES ON FREQUENCY RESPONSE
EECS140 ANALOG CIRCUIT DESIGN
FR-14
R1
LECTURES ON FREQUENCY RESPONSE
FR-15
Bode Plots (Cont.)
Bode Plots (Cont.)
1 Pole, 1 Zero
νI N
νO U T
C
R2
C
νI N
1 Pole
R2 
1
⋅ ----------------------------------------H ( ω ) =  ---------------R 1 + R 2 1 + jω ( R 1 || R 2 )C
H (ω)
ω p > ωz
dB
ω p = ωz
1 Pole, 1 Zero
νO U T
R1
ROBERT W. BRODERSEN
FR-13
ω
θ = a r c tan  ----- 
ωZ
ωZ
0.1 ω Z
π/4
jωR C
jω R C
R
-------------------= ----------------------- = ------------------1 + jωR C
ω
1
1 + j -------R + ---------1
jωC
-------RC
R
H (ω )
LECTURES ON FREQUENCY RESPONSE
FR-12
Bode Plots (Cont.)
C
EECS140 ANALOG CIRCUIT DESIGN
R2
R 2  ⋅ ----------------------------------------1 + jωR 1 C
H ( ω ) =  --------------- R 1 + R 2  1 + jω ( R 1 || R 2 )C
ω p < ωz
ω
1 + j ----ωZ
H ( ω ) = ----------------ω
1 + j ----ωp
ROBERT W. BRODERSEN
EECS140 ANALOG CIRCUIT DESIGN
LECTURES ON FREQUENCY RESPONSE
EECS140 ANALOG CIRCUIT DESIGN
LECTURES ON FREQUENCY RESPONSE
FR-16
Capacitances
Sat :
εSiO
fF
F
C O X = -------- = 0.1 -----2 = 10– 4 -----2
tO X
µ
m
G
CGS
CG D
S
C SB
C G S = 2- ⋅ C O X ⋅ L ⋅ W + C G S O ⋅ W
3
2
D
CGB
CD B
C GD = C G D O ⋅ W
C G S O = 5 ×10– 10 -F
--m
C J S W ⋅ PS
C J ⋅ AS
-----------------------------C SB = -------------------------MJSW
MJ +
V
BS
B
S
1 + V
 1 + ------- 
------

P B
P B
–10 F
C G D O = 5 ×10 ---m
C G B O = 4 ×10 –10 -F
--m
C GB = C G B O ⋅ L
F
C J = 10 – 4 -----2
m
PS = Perimeter of Source
AS =
P B = φ B = 0.8V
COX ⋅ L ⋅ W
+ CGSO ⋅ W
C G S = ------------------------2
C OX ⋅ L ⋅ W
C G D = ------------------------- + C G D O ⋅ W
2
(similar for C DB )
1
M J = -- (default)
2
MJSW = 3 (default)
Area of Source
CGBO ≡
B
FR-17
Capacitances (Cont.)
Linear :
Capacitance of gate to bulk overlap
ROBERT W. BRODERSEN
ROBERT W. BRODERSEN
EECS140 ANALOG CIRCUIT DESIGN
LECTURES ON FREQUENCY RESPONSE
EECS140 ANALOG CIRCUIT DESIGN
LECTURES ON FREQUENCY RESPONSE
FR-18
Capacitances (Cont.)
Layout
Source
Drain
M1 1 2 3 4 NMOS L=2u W=2u
+ AS=4p AD=4p PS=6u PD=6u
W
G
2λ
2λ
S
D
λ
L
4λ
FR-19
Capacitances (Cont.)
CGSO
2λ
CGDO
(Minimum size device, W/L = 2)
Area of Source = AS = 4 λ ⋅ W
Area of Drain
= A D = AS
Perimeter of Source
ROBERT W. BRODERSEN
Capacitor (in linear)
= PS = 8 λ + W
ROBERT W. BRODERSEN
EECS140 ANALOG CIRCUIT DESIGN
LECTURES ON FREQUENCY RESPONSE
Miller Approximation i
EECS140 ANALOG CIRCUIT DESIGN
Miller Approximation (Cont.)
FR-20
C
C
it
ν OUT
νOUT
FR-21
C
RL
C
–A
ν in
LECTURES ON FREQUENCY RESPONSE
ν OUT
ν in
RI N
νt
ν in
RL
g m ⋅ ν gs
R IN
i C = C ⋅ d ( ν i n – ν O U T) = C ⋅ d ( ν in + A ⋅ ν O U T ) = C ⋅ ( 1 + A ) ⋅ dν i n
dt
dt
dt
ν in
RL
ν OUT
R IN
ν OUT
–A
ν OUT
gm ⋅ R O U T
--------- = ----------------------------------------------------------------------------------------------ν in
1 + jω [ R I N C ( 1 + g m ⋅ R O U T ) + R O U T C ]









ν O U T = – A ⋅ ν in
C MILLER
ν in
νt
C(1 + A)
1
ω p = --------------------------------------------R I N C( 1 + g m ⋅ R O U T )
C ( 1 + g m ⋅ RO U T )
ROBERT W. BRODERSEN
ROBERT W. BRODERSEN
EECS140 ANALOG CIRCUIT DESIGN
LECTURES ON FREQUENCY RESPONSE
Inverter
EECS140 ANALOG CIRCUIT DESIGN
LECTURES ON FREQUENCY RESPONSE
FR-22
Inverter (Cont.)
R IN
ν1
ν in
νOUT
RL
R IN
C GD
ν O U T ν in
R IN
CGD
νO U T
g m ⋅ ν in
RL
ZG
FR-23
ZG D
g m ⋅ ν in
RL
ν1 – ν OUT
ν 1 – ν i n ----ν
---------------+ 1 + ------------------= 0
R IN
ZG
Z GD
ν in
CD
C GB
C GS
CG
R O U T = R L || r o
C G = C GB + C G S
C D ignored (not usually possible)
ROBERT W. BRODERSEN
C GD
 1 – -------
gm 
ν
OUT
--------- = – g m ⋅ R O U T ⋅ --------------------------------------------------------------------------------------------------------------------------------------------------------ν in
1 + jω { [ C G D ( 1 + g m ⋅ R L ) + C G ] ⋅ R I N + R L C G D } – ω2 R O U T R I N C G C G D
C GD
 1 – jω C
G D
 1 – jω --------------

gm 
gm 
= ----------------------------------------------------- = ---------------------------------------------------------------------ω
1
1
 1 + j ---ω
1
----  ⋅  1 + j ------1 + jω  ------- + -------  – ω 2 ---1---- ------
 ωp 1 ω p 2 
ω p 1 
ω p2
ωp 1 ω p 2
ROBERT W. BRODERSEN
EECS140 ANALOG CIRCUIT DESIGN
LECTURES ON FREQUENCY RESPONSE
FR-25
Inverter (Cont.)
ω
 1 + j ----
ω Z
H ( ω ) = ----------------------------------------------------ω
ω 
 1 + j ------⋅ 1 + j ------- 

ω p2
ωp 1 







1
= – ------------------------------------------------------------------------------------------R I N ⋅ [ C G + C G D ( 1 + g m R O U T) ] + R L C G D
C MILLER
ω p2
LECTURES ON FREQUENCY RESPONSE
FR-24
Inverter (Cont.)
ω p1
EECS140 ANALOG CIRCUIT DESIGN
1
1
= – ------------------ – -------------------------------------------R OUT CGD 
R || R I N || --1---  C G
 OUT
g m
s Z = – jω Z
s p 1 = – jω p1
 1 – -s- 

sZ
H ( s ) = -------------------------------------------s
s 
 1 – ----⋅ 1 – ----- 

sp 2
s p 1 
gm
ω Z = -------CGD
ω
 1 + j ----
ω Z
H ( ω ) = ----------------------------------------------------ω
 1 + j ---ω
----  ⋅  1 + j ------
ω p1 
ωp 2
X
gm
ω p 2 ≈ – -----CG
s p 2 = – jω p2
O
X
1
ω p1 ≈ – --------------------RI N C M I L L E R
gm
ωZ = -------CGD
ROBERT W. BRODERSEN
ROBERT W. BRODERSEN
EECS140 ANALOG CIRCUIT DESIGN
LECTURES ON FREQUENCY RESPONSE
FR-26
Inverter (Cont.)
RL
FR-27
RL
Cµ
ν OUT
RI N
νi n
--1--gm
CG
R IN
LECTURES ON FREQUENCY RESPONSE
Inverter (Cont.)
RL
R IN
ν OUT
EECS140 ANALOG CIRCUIT DESIGN
νi n
R O U T = R L || r o
CD
Cπ
1
ω p 1 ≈ – --------------------R I N CMILLER
ω p1
ω p2
1
----------1
----- C G
gm
Case 1 (Miller Capacitance not important) :







R IN C π , R O U T C D » R I N ⋅ ( 1 + g m R O U T )C µ
as Cgd increases
C MILLER
1
ω p 1 = -----------RI N C π
ROBERT W. BRODERSEN
ROBERT W. BRODERSEN
1
ωp 2 = ---------------R OUT CD
EECS140 ANALOG CIRCUIT DESIGN
LECTURES ON FREQUENCY RESPONSE
EECS140 ANALOG CIRCUIT DESIGN
LECTURES ON FREQUENCY RESPONSE
Inverter (Cont.)
FR-28
Inverter (Cont.)
FR-29
RI N
ν in
1
Z O U T = R L || r o || -----------jωC D
ZO U T
ν' in
Cπ
ro
1
= R O U T || -----------jωC D
CD
A ν = –gm ⋅ ZO U T
R OUT
-----------R OUT
jωC D
= – g m ⋅ ---------------------------- = – g m ⋅ -------------------------------1 + jω R O U T C D
1
-----------R OUT +
jωC D
1
-----------1
jωC
ν
π
in '
------ = ------------------------= ---------------------------1
1
+
jω
RI N C π
νi n
R I N + -----------jω C π
1
ω p = ---------------RO U T C D
1
ω p = -----------RI N C π
ROBERT W. BRODERSEN
ROBERT W. BRODERSEN
EECS140 ANALOG CIRCUIT DESIGN
LECTURES ON FREQUENCY RESPONSE
EECS140 ANALOG CIRCUIT DESIGN
FR-30
Inverter (Cont.)
RI N
1
ω p 2 = ----------------------------R I N ( Cπ + Cµ )
g m ( νg – νOUT )
νO U t
R IN ( 1 + g m R O U T )C µ » R O U T C D , R I N C π
1
= -----------------------------------------R I N ( 1 + g m R O U T)C µ
ω p2
RS
1
= ---------------------------1
----- ( C π + C D )
gm
For case 2 and 3,
g
ω Z E R O = -----m
Cµ
ROBERT W. BRODERSEN
gmb( –ν OUT)
Cπ
Case 3 (Large Cµ) :
ω p1
νg
ν in
R O U T C D » R IN ( 1 + g m R O U T )C µ , R I N C π
X
C µ large
gm
ωp 2 ≈ ----------------------( C π + CD)
X
X
C µ = 0 Cµ = 0
1
1
--------------------------R O U TC D
R I N Cπ
1
ω p 1 ≈ --------------------R I N C MILLER
X
C µ large
O
gm
ω Z = ----Cµ
FR-31
Source Follower
Case 2 (Large Cd) :
1
ω p 1 = ---------------R O U TC D
LECTURES ON FREQUENCY RESPONSE
1
ν g =  ---------------------------- ⋅ ( ν in – ν O U T) + ν O U T
1 + jωR I N C π
ν
νg – ν OUT
OUT
--------= ------------------+ g m ⋅ ν g – ( 1 + χ ) ⋅ g m ⋅ νO U T
RS
1
-----------jωC π
ROBERT W. BRODERSEN
ν OUT
EECS140 ANALOG CIRCUIT DESIGN
LECTURES ON FREQUENCY RESPONSE
EECS140 ANALOG CIRCUIT DESIGN
FR-32
Source Follower (Cont.)
RI N
1
ν O U T ⋅  ----- + ( 1 + χ ) ⋅ g m + jωC π = ( jωC π + g m ) ⋅ ν g
RS

Source Follower
again with C SB







Small Signal :
1–A
RI N
C GS
gm R S
A ν = -------------------------------------1 + (1 + χ )g m R S
CS B
νg
ν in
g m ( νg – νOUT )
C GS
CG = CGD + C G B
gm R S
A = -------------------------------------1 + (1 + χ )g m R S
C SB
νO U T
RS
ROBERT W. BRODERSEN
ROBERT W. BRODERSEN
EECS140 ANALOG CIRCUIT DESIGN
LECTURES ON FREQUENCY RESPONSE
Source Follower (Cont.)
νOUT
( ν G – ν O U T ) ⋅ j ωC G S – g m ( ν G – ν O U T ) = --------- + ν O U T ⋅ jωC SB
RS
g m R S ⋅ 1 + jω C
GS
-------------------------1 + g m RS
gm
ν
OUT
--------- = ---------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------ν in
R I N CGS
R S ( CGS + C S B )
G SC G + C SB ( C G + C G S )
1 + jω R I N C G + -------------------+ ------------------------------– ω 2 R S R IN C
----------------------------------------------------1 + g mR S
1 + g m RS
1 + gm R S
ω
ω
=  1 + j ---- ⋅  1 + j ----

p 1 
p 2
1
1
ω2
= 1 + jω  ---- + ----  – --------p1 p2
p1p2
EECS140 ANALOG CIRCUIT DESIGN
FR-34
ν in – ν G
---------------= ν G ⋅ jωC G + ( ν G – ν O U T ) ⋅ jω C G S
R IN
ROBERT W. BRODERSEN
νO U T
C GB
RS
g
ω Z = -----m
Cπ
let denominator
FR-33
CGD
ν in
 1 + jω C
-----π 

gm
νO U T
g
mRS
--------- = ------------------------------------------ ⋅ ----------------------------------------------------------------ν in
1 + (1 + χ ) ⋅ g m R S 1 + jωR C
π ( 1 + χ ⋅ g m R S)
IN -------------------------------------1 + (1 + χ )g m R S
1
ω p = -----------------------------R I N C π( 1 – A )
LECTURES ON FREQUENCY RESPONSE
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
---- » ---p 1 p2
1
1
p 1 = ------------------------------------------------------------------------------ = ------------------------------------------------------------------------------R IN C G S
R S ( C G S + C SB )
RI N C G S
R IN C G + -------------------- + ------------------------------R I N C G + -------------------- + R O ( C G S + C S B )
1 + g mR S
1 + gm R S
1 + g m RS
where,
R O = --1--- || R S
gm
thus,
R IN C G S
R IN C G + -------------------+ R O ( C G S + C SB )
1
+ g mR S
p 2 = ------------------------------------------------------------------------------R O R IN [ C G S C G + C S B C G + C S B C G S ]
ROBERT W. BRODERSEN
EECS140 ANALOG CIRCUIT DESIGN
LECTURES ON FREQUENCY RESPONSE
Source Follower (Cont.)
EECS140 ANALOG CIRCUIT DESIGN
LECTURES ON FREQUENCY RESPONSE
FR-36
FR-37
Common Gate
2 limiting cases,
RS
ν OUT
Case 1 :
CG S
R I N  C G + --------------------  » R O ( C G S + C SB )

1 + g m R S
ω p1
1
= --------------------------------------------C GS
R I N  C G + -------------------- 
1 + gm R S
ν in
Miller cap = C G S ⋅ ( 1 – A )
+
-
RL
CGS
CS B
gm R S
A = -------------------1 + g m RS
CG D
C DB
assume r o → ∞
g mR S
1
1 – A = 1 – -------------------= -------------------1 + gm R S
1 + gm R S
Case 2 :
RS
ν in
C GS
R O ( C G S + C SB ) » R I N  C G + -------------------- 

1 + g m RS 
νS
ν OUT
+
-
RL
CS
CD
1
ω p1 = -------------------------------R O ( CG S + CS B )
ROBERT W. BRODERSEN
ROBERT W. BRODERSEN
EECS140 ANALOG CIRCUIT DESIGN
LECTURES ON FREQUENCY RESPONSE
Common Gate (Cont.)
KCL@ ν S
FR-38
LECTURES ON FREQUENCY RESPONSE
Common Gate (Cont.)
ν in – ν S
---------------- = ν S ⋅ jω C S + g m ν S
RS
νOUT
g m ν S = ν O U T ⋅ jω C D + --------RL
g
m RL
-------------------1 + g m RS
νO U T
--------= -----------------------------------------------------------------------------ν in
RS C S 
( 1 + jωR L C D ) ⋅  1 + jω -------------------1 + g m RS 
no zeros, poles @
KCL@ ν O U T
1
p 1 = ----------RL C D
1
1
p 2 = -------------------------- = --------------------------R
S
 R S || --1---  C S
-------------------- C S
1 + gm R S

g m
ROBERT W. BRODERSEN
EECS140 ANALOG CIRCUIT DESIGN
@ ν O U T → R eq = R L
C eq = C D
1
@ ν S → R eq = R S || ----gm
Ce q = C S
1
1
p 1 = ------------- = ----------R e q C eq
R L CD
1
1
p 2 = ------------- = --------------------------R eq C eq
 R || --1---  C S
 S 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
EECS140 ANALOG CIRCUIT DESIGN
LECTURES ON FREQUENCY RESPONSE
FR-40
FR-41
Cascode (Cont.)
If we assume that R S is very large, then,
RL
At (1) :
CGD 2
R eq = R S
νOUT
( 2)
Ce q = CGS1 + C GD1 ⋅ ( 1 – A )
V BIAS
CD2
C G S2
C eq = C G S 1 + 2 ⋅ C G D 1
R IN
C GD1
νi n
1
p 1 = ------------R eq C eq
A = – g m ⋅ --1--- = – 1
gm
( 1)
At (2) :
R eq = R L || r o ← large
C D1
CGS1
1
p 2 = ------------R eq C eq
C eq = C G D 2 + C D 2
No large Miller multiplication of a capacitance!
ROBERT W. BRODERSEN
ROBERT W. BRODERSEN
EECS140 ANALOG CIRCUIT DESIGN
LECTURES ON FREQUENCY RESPONSE
Zero Value Time Constant Analysis
EECS140 ANALOG CIRCUIT DESIGN
LECTURES ON FREQUENCY RESPONSE
FR-42
R IN
General case of dominant pole approximation. Use this technique
for complex circuits where we can’t identify node with large R eq
and C eq.
CG D
νi n
νOUT
C GB
C GS
Strategy :
RI N
νg
ν in
2) Find resistance seen by Ci
g m ( νg – νOUT )
3) Calculate RiC i for all caps
3dB
C GS
C G = C G D + C GB
1
= -------------∑ R iC i
C SB
ROBERT W. BRODERSEN
CS B
RS
1) Set all caps Cj=0 except for Ci
4) ω
FR-43
ZVTC with Source Follower
ROBERT W. BRODERSEN
νO U T
RS
EECS140 ANALOG CIRCUIT DESIGN
LECTURES ON FREQUENCY RESPONSE
ZVTC with Source Follower
EECS140 ANALOG CIRCUIT DESIGN
ZVTC with Source Follower (Cont.)
FR-44
C3 :
C1 :
LECTURES ON FREQUENCY RESPONSE
+
R 1 = R IN
CG
C1 = C G
C GS
do small signal analysis to find
1
R 2 = R S || ----gm
g m ( – νOUT )
νOUT
ROBERT W. BRODERSEN
g m ( νg – νOUT )
-
C2 :
C SB
it
νt
RI N
C2 = CS B
FR-45
RI N
νO U T
RS
ν
-- t = R 3 seen by C G S
it
1
R IN
where R O = R S || ----R 3 = R O + -------------------gm
1 + gm R S
C 3 = C GS
hence,
RS
1
1
ω 3 d B = ---------------------------------------------- = ------------------------------------------------------------------------------------R 1 C 1 + R2 C 2 + R3 C 3
R IN

R I N C G + R O C S B + R O + --------------------  C G S

1 + g m RS 
ROBERT W. BRODERSEN