Miller effect

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 C
1
decreases and the voltage divider consists
of C
1
and R
1
attenuates V
in
to a greater extent at the output.
4

5. 5
Gain Roll-off: Common Source

The capacitive load, C
L
, 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
æ ö
=- ç ¸
è ø

6. 6
BODE PLOT
When we hit a zero, ω
zj
, the Bode magnitude rises with a
slope of +20dB/dec.
When we hit a pole, ω
pj
, the Bode magnitude falls with a
slope of -20dB/dec


÷
÷
ø
ö
ç
ç
è
æ
+
÷
÷
ø
ö
ç
ç
è
æ
+
÷
÷
ø
ö
ç
ç
è
æ
+
÷
÷
ø
ö
ç
ç
è
æ
+
=
21
21
0
11
11
)(
pp
zz
ss
ss
AsH
ww
ww

7. 7
Example: Bode Plot
The circuit only has one pole (no zero) at 1/(R
D
C
L
), so the
slope drops from 0 to -20dB/dec as we pass ω
p1
.
LD
p
CR
1
1=w

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

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 C
F
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 A
v
is the gain from node 1 to 2, then a floating impedance
Z
F
can be converted to two grounded impedances Z
1
and Z
2
.
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
Z
2
=Z
F
/(1­(V
1
/V
2
))

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
w
F
Dm
D
out
C
Rg
R
÷
÷
ø
ö
ç
ç
è
æ
+
=
1
1
1
w

14. THANK YOU!
14