Frequency Response of Amplifiers
•General Considerations
• Miller Effect
• Association of Poles with Nodes
•Common Source Stage
•Source Follower
•Differential Pair
Hassan Aboushady
University of Paris VI
References
• B. Razavi, “Design of Analog CMOS Integrated Circuits”,
McGraw-Hill, 2001.
H. Aboushady
University of Paris VI
1
Miller Effect
• Miller’s Theorem
with
• Proof
Av =
VY
VX
we have
Z1 =
Z2 =
Z
1 − Av−1
Z
V
1− Y
VX
V X − VY V X
=
Z
Z1
Z1 =
VY − V X VY
=
Z
Z2
Z2 =
H. Aboushady
Z
1 − Av
Z
V
1− X
VY
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Example 1
• Calculate the input capacitance Cin:
Z=
1
sC F
1
sC F
Z1 =
1+ A
Cin = C F (1 + A)
Av =
VY
should be calculated at the frequency of interest.
VX
To simplify calculations we usually use low frequency value of Av.
Miller’s theorem cannot be used simultaneously to calculate
input-output transfer function and the output impedance.
H. Aboushady
University of Paris VI
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Association of Poles with Nodes
Vout2
Vout1
VM ( s ) =
H. Aboushady
Vin ( s )
1
Vin ( s )
=
1 sCin 1 + sRS Cin
RS +
sCin
VN ( s ) =
Vout1 ( s )
1 + sR1C N
VP ( s ) =
Vout 2 ( s )
1 + sR2C P
Vout
A1
A2
1
( s) =
Vin
1 + sRS Cin 1 + sR1C N 1 + sR2C P
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Association of Poles with Nodes
Vout2
Vout1
Vout
A1
A2
1
( s) =
Vin
1 + sRS Cin 1 + sR1C N 1 + sR2C P
ω1 =
1
RS Cin
ω2 =
1
R1C N
ω3 =
1
R2C P
3 poles:
each determined by the total capacitance seen from each node to
ground multiplied by the total resistance seen at the node to ground
H. Aboushady
University of Paris VI
3
Example 2
• Calculate the pole associated with node X:
The total equivalent capacitance seen
from X to ground: C = C (1 + A)
X
The pole frequency:
F
ωX =
1
1
=
RS C X RS C F (1 + A)
H. Aboushady
University of Paris VI
Example 3
• Neglecting channel length modulation,
compute the transfer function of
the common gate stage with parasitic capacitances
Parasitic capacitances at node X:
Input resistance of a
common gate amplifier:
Pole frequency at node X:
Rin =
C S = CGS 1 + CSB1
1
g m1 + g mb1
ωX =
1
⎛
(CGS1 + CSB1 )⎜⎜ RS //
⎝
Parasitic capacitances at node Y:
Pole frequency at node X:
H. Aboushady
ωY =
⎞
1
⎟
g m1 + g mb1 ⎟⎠
C D = C DG1 + C DB1
1
(C DG1 + CDB1 )RD
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Example 3 (cont.)
Low-frequency gain of a common gate stage
neglecting channel length modulation:
Av 0 =
(g m + g mb )RD
1 + (g m1 + g mb1 )RS
The overall transfer function is given by:
Vout ( s )
Av 0
=
Vin ( s ) ⎛
s ⎞⎛
s
⎟⎟⎜⎜1 +
⎜⎜1 +
⎝ ωin ⎠⎝ ωout
⎞
⎟⎟
⎠
Note that if we do not neglect r01 , the input and output nodes
interact, making it difficult to calculate the poles.
H. Aboushady
University of Paris VI
Common Source Stage
Neglecting channel length modulation and
applying the Miller’s theorem on CGD , we have:
The total capacitance at node X:
C X = CGS + (1 − Av )CGD
where,
Av = − g m RD
The 1st pole frequency:
ω p1 =
RS (CGS
1
+ (1 + g m RD )CGD )
The total capacitance at the output node:
(
)
Cout = C DB + 1 − Av−1 CGD ≈ C DB + CGD
The 2nd pole frequency:
H. Aboushady
ω p2 =
1
RD (C DB + CGD )
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Common Source Stage
The transfer function:
Vout ( s )
− g m RD
=
Vin ( s ) ⎛
⎞
⎞⎛
⎜1 + s ⎟⎜1 + s ⎟
⎜ ω ⎟⎜ ω ⎟
p1 ⎠⎝
p2 ⎠
⎝
r0 and any load capacitance can be easily included.
Sources of error (approximation):
• we have not considered the existence of zeros in the circuit
• the amplifier gain varies with frequency
H. Aboushady
University of Paris VI
Common Source : “exact “ Transfer Function
To obtain the exact transfer function:
Applying Kirchoff Current Law (KCL):
VX − Vin
+ sCGSVX + sCGD (VX − Vout ) = 0
RS
⎛
1 ⎞
⎟⎟Vout = 0
sCGD (Vout − V X ) + g mV X + ⎜⎜ sC DB +
R
D ⎠
⎝
H. Aboushady
University of Paris VI
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Common Source : “exact” 1st pole
After some manipulations, we get:
(sCGD − g m )RD
Vout
=
2
Vin RS RDξ s + [RS (1 + g m RD )CGD + RS CGD + RS CGS + RD (CGD + C DB )]s + 1
with
ξ = CGS CGD + CGS C DB + CGD C DB
Writing the
denominator as:
Assuming
⎛ 1
⎛
1 ⎞⎟
s ⎞⎟⎛⎜
s ⎞⎟
s2
1+
=
+⎜
+
D = ⎜1 +
s +1
⎟
⎜ω
⎜ ω ⎟⎜ ω ⎟ ω ω
p1 ⎠⎝
p1 ⎠
p1 p 2
⎝ p1 ω p 2 ⎠
⎝
ω p1 << ω p 2
ω p1 ≈
1
RS (1 + g m RD )CGD + RS CGD + RS CGS + RD (CGD + C DB )
Compare this result with ωin calculated using Miller’s Theorem
H. Aboushady
University of Paris VI
Common Source : “exact” 2nd pole
(sCGD − g m )RD
Vout
=
2
Vin RS RDξ s + [RS (1 + g m RD )CGD + RS CGD + RS CGS + RD (CGD + C DB )]s + 1
with
having
and
then
ξ = CGS CGD + CGS C DB + CGD C DB
⎛ 1
1 ⎞⎟
+⎜
+
s +1
ω p1ω p 2 ⎜⎝ ω p1 ω p 2 ⎟⎠
1
ω p1 ≈
RS (1 + g m RD )CGD + RS CGD + RS CGS + RD (CGD + C DB )
D=
s2
ω p2 =
1
1
RS RDξ ω p1
ω p2 =
RS (1 + g m RD )CGD + RS CGS + RD (CGD + C DB )
RS RD (CGS CGD + CGS C DB + CGD C DB )
H. Aboushady
University of Paris VI
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Comparison between “exact” and Miller’s theorem
1st pole:
exact
If
RD (CGD + C DB )
is negligible
2nd pole:
Miller
exact
ω p1 =
ω p1 =
ω p2 =
RS (CGS
1
+ (1 + g m RD )CGD ) + RD (CGD + C DB )
RS (CGS
1
+ (1 + g m RD )CGD )
RS (1 + g m RD )CGD + RS CGS + RD (CGD + C DB )
RS RD (CGS CGD + CGS C DB + CGD C DB )
if CGS >> (1 + g m RD )CGD +
ω p2 ≈
H. Aboushady
Miller
ω p2 =
RD
(CGD + CDB )
RS
CGS
RD (CGS CGD + CGS C DB + CGD C DB )
1
RD (C DB + CGD )
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Common Source : transfer function zero
After some manipulations, we get:
(sCGD − g m )RD
Vout
=
2
Vin RS RDξ s + [RS (1 + g m RD )CGD + RS CGD + RS CGS + RD (CGD + C DB )]s + 1
ωz =
H. Aboushady
gm
CGD
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Source Follower
High frequency equivalent circuit
X
Applying Kirchoff Current Law (KCL)
at the output node:
sCGSV1 + g mV1 = sC LVout
Vout + V1 − Vin
+ sCGD (Vout + V1 ) + sCGSV1 = 0
RS
Vout ( s )
g m + sCGS
=
Vin ( s ) RS (CGS C L + CGS CGD + CGD C L )s 2 + ( g m RS CGD + C L + CGS )s + g m
at node X:
H. Aboushady
University of Paris VI
Source Follower Input Impedance
VX =
⎛
IX
g I
+ ⎜⎜ I X + m X
sCGS ⎝
sCGS
⎞ 1
⎟⎟
⎠ sC L
Input Impedance:
Z in =
g
1
1
+
+ 2 m
sCGS sC L s CGS C L
Note the negative resistance:
H. Aboushady
− gm
ω CGS C L
2
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Source Follower Output Impedance
Neglecting CGD
Applying KCL
sCGSV1 + g mV1 + I X = 0
RS
V1
+ sCGS (V1 − V X ) = 0
RS
1/gm
Output Impedance:
Z out =
sCGS RS + 1
sCGS + g m
Zout increases with frequency
It contains an inductive component
H. Aboushady
University of Paris VI
Source Follower Output Impedance Equivalent C t
Equivalent circuit of source follower output impedance:
at ω = ∞ ⇒ Z1 = R1 + R2
at ω = 0 ⇒ Z1 = R2
Z1 =
sLR1
+ R2
sL + R1
⎛
1 ⎞
⎟⎟
R1 = ⎜⎜ RS −
g
m ⎠
⎝
H. Aboushady
⎛1
1 ⎞
sL⎜⎜ + ⎟⎟ + 1
R R2 ⎠
Z1 = ⎝ 1
sL
1
+
R1 R2 R2
R2 =
1
gm
L=
Z out =
CGS
gm
sCGS RS + 1
sCGS + g m
⎛
1 ⎞
⎜⎜ RS −
⎟⎟
g
m ⎠
⎝
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Differential Pair
Differential inputs:
same as common source stage
H. Aboushady
University of Paris VI
Differential Pair
Common-Mode inputs:
Low frequency ACM with M1-M2 mismatch:
ACM =
∆g m R D
(g m1 + g m 2 )rO 3 + 1
High frequency ACM with
M1-M2 mismatch:
ACM
where
H. Aboushady
⎛
1 ⎞
⎟
∆g m ⎜⎜ RD //
sC L ⎟⎠
⎝
=
⎛
⎞
(g m1 + g m 2 )⎜⎜ rO 3 // 1 ⎟⎟ + 1
sC P ⎠
⎝
C P = C DG 3 + C DB 3 + C SB1 + CSB 2
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