miller effect- razavi book

1. 1
ANALOG CMOS IC DESIGN
MILLER EFFECT
BY
KANCHERLA SAIKIRAN

2. 2
CONTENTS
High Frequency Roll-off of Amplifier
BODE PLOT
Pole Identification
Miller’s Theorem
Examples

3. 3
High Frequency Roll-off of Amplifier
 As frequency of operation increases, the gain of amplifier
decreases. This chapter analyzes this problem.

4. 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.
4

5. 5
Gain Roll-off: Common Source
 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.
1
||out m in D
L
V g V R
C s
 
= −  ÷
 

6. 6
BODE PLOT
 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










+







+






+





+
=
21
21
0
11
11
)(
pp
zz
ss
ss
AsH
ωω
ωω

7. 7
Example: Bode Plot
 The circuit only has one pole (no zero) at 1/(RDCL), so the
slope drops from 0 to -20dB/dec as we pass ωp1.
LD
p
CR
1
1 =ω

8. 8
Pole Identification Example I
inS
p
CR
1
1 =ω
LD
p
CR
1
2 =ω

9. 9
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.

10. 10
Miller’s Theorem
The Miller’s theorem establishes that in a linear circuit, if there exists a branch
with impedance Z, connecting two nodes with nodal voltages ­V
1
and V
2
, we can
replace this branch by two branches connecting the corresponding nodes to ground
by impedances respectively Z / (1­K) and KZ / (K­1).

11. 11
 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.
PROOF
The current flowing from 1 to 2 is given by
V
1
­V
2
/Z
F
The current flowing from 1 to ground is
V
1
/Z
1
By equating the both currents
(V
1
­V
2
)/Z
F
= V
1
/Z
1
Z
1
= Z
F
/(1­(V
2
/V
1
))
Similarly

12. 12
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.
v
F
A
Z
Z
−
=
1
1
v
F
A
Z
Z
/11
2
−
=

13. 13
Example: Miller Theorem
( ) FDmS
in
CRgR +
=
1
1
ω
F
Dm
D
out
C
Rg
R 





+
=
1
1
1
ω

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