EE105 – Fall 2014
Microelectronic Devices and Circuits
Prof. Ming C. Wu
wu@eecs.berkeley.edu
511 Sutardja Dai Hall (SDH)
Lecture14-Frequency Response
1
Amplifier Frequency Response
Transfer Function Analysis
FL ( s ) =
(s + ω ) (s + ω ) ⋅⋅⋅ (s + ω )
(s + ω ) (s + ω ) ⋅⋅⋅ (s + ω )
L
Z1
L
Z2
L
Zk
L
P1
L
P2
L
Pk
"
s %"
s % "
s %
$1+ H '$1+ H ' ⋅⋅⋅ $1+ H '
# ω Z1 &# ω Z 2 & # ω Zl &
FH ( s ) =
"
s %"
s % "
s %
$1+ H '$1+ H ' ⋅⋅⋅ $1+ H '
# ω P1 &# ω P 2 & # ω Pl &
Av ( s ) =
N ( s ) a0 + a1s + a2 s 2 +... + am s m
=
D ( s ) b0 + b1s + b2 s 2 +... + bn s n
FH ( jω ) →1 for ω << ω ZiH , ω PiH , j = 1... l
∴ AL ( s ) ≅ Amid FL ( s )
Av ( s ) = Amid FL ( s ) FH ( s )
Amid is the midband gain between the lower
and upper cutoff frequencies ω L and ω H .
Lecture14-Frequency Response
FL ( jω ) →1 for ω >> ω ZjL , ω PjL , j = 1... k
∴ AH ( s ) ≅ Amid FH ( s )
2
1
Low Frequency Dominant Pole Approximation
•  A Dominant Pole exists if one of the low frequency poles is much
larger than the others.
–  In the graph below case ω = 1000 rad/sec is a dominant pole.
All other poles and zeros are at low enough frequencies that
they do not affect the lower cutoff frequency ωL.
s
s +1000
= 200
FL ( s ) ≅
Amid
ω L ≅ 1000 rad / sec
Lecture14-Frequency Response
3
Lower Cutoff Frequency Calculations
If there is no dominant pole at low frequencies,
the poles and zeros interact to determine the
lower cutoff frequency ω L .
For example, suppose:
AL ( s ) = Amid FL ( s ) = Amid
( s + ω Z1 ) ( s + ω Z 2 )
( s + ω P1 ) ( s + ω P 2 )
For s = jω L , AL ( jω L ) =
1
=
2
(ω
(ω
Amid
2
2
L
2
+ ω Z1
) (ω L2 + ω Z2 2 )
2
L
2
+ ω P1
) (ω L2 + ω P2 2 )
4
2
2
2
2
2
1 ω L + (ω Z1 + ω Z 2 )ω L + ω Z1ω Z 2
= 4
2
2
2 ω L + (ω P1
+ ω P2 2 )ω L2 + ω P1
ω P2 2
Lower cutoff frequency ω L will be
greater than all the individual pole
zero frequencies.
2
2
∴ω L ≅ ω P1
+ ω P2 2 − 2ω Z1
− 2ω Z2 2
In general, for n poles and n zeros,
ωL ≅
∑ω
n
Lecture14-Frequency Response
2
Pn
2
− 2∑ω Z1
n
4
2
Dominant Pole Example
•  Problem: Find midband gain, FL(s) and fL for
! s
$
s#
+1&
" 100 %
AL ( s ) = 2000
(0.1s +1) ( s +1000)
•  Analysis: Rearranging the given transfer function into standard
form,
s ( s +100 )
s ( s +100 )
AL ( s ) = 200
= Amid FL ( s ) → Amid = 200
FL ( s ) =
( s +10) ( s +1000)
( s +10) ( s +1000)
Zeros: s = 0 and s = -100
Poles: s = -100 and s = -1000
The poles and zeros are all widely separated.
The dominant pole is at ω = 1000 and fL ≅
The more exact calculation is fL =
∴ AL ( s ) ≅ 200
s
s +1000
1000
= 159 Hz.
2π
1
10 2 +1000 2 − 2 × 0 2 − 2 ×100 2 = 158 Hz
2π
Lecture14-Frequency Response
5
High-Frequency Dominant Pole
•  The lowest of all high frequency poles is
called the dominant high-frequency pole.
AH ( s ) ≅ Amid FH ( s )
FH ( s ) ≅
1
1+ ( s ω P3 )
ω H ≅ ω P3
•  If there is no dominant pole at high
frequencies, the poles and zeros interact
to determine ωH.
!
s $!
s $
#1+
&#1+
&
" ω Z1 %" ω Z 2 %
AH ( s ) = Amid FH ( s ) = Amid
!
s $!
s $
#1+
&#1+
&
" ω P1 %" ω P 2 %
A
For s = jω H , Amid ( jω H ) = mid
2
Lecture14-Frequency Response
2
2
' !
$ *' !
$*
)1+ # ω H & ,)1+ # ω H & ,
)( " ω Z1 % ,+)( " ω Z 2 % ,+
1
=
2
2
' !
2
$ *' !
$*
)1+ # ω H & ,)1+ # ω H & ,
)( " ω P1 % ,+)( " ω P 2 % ,+
1
1
1
2
2
≅
+ 2 − 2 − 2
2
ωH
ω P1
ω P 2 ω Z1 ω Z 2
For the general case of n poles and n zeros,
1
≅
ωH
1
∑ω
n
2
Pn
− 2∑
n
1
2
ω Zn
6
3
High Frequency Dominant Pole
Approximation
AH ( s ) = Amid FH ( s )
Amid = 200
FH ( s ) ≅
1
10 6
=
6
1+ s 10
s +10 6
ω H ≅ 10 6 rad / sec
ω
fH = H ≅ 159 kHz
2π
Lecture14-Frequency Response
7
Low-Frequency Poles and Zeros
Direct Calculation: C-S Amplifier
Vo ( s ) = I o ( s ) R3 = −gmVgs ( s )
Vo ( s ) = −gm ( RD R3 )
RD
R3
RD + (1 sC3 ) + R3
s
s+
Vgs ( s )
1
C3 ( RD + R3 )
sC1RG
Vg ( s ) =
Vi ( s )
sC1 ( RI + RG ) +1
Node Eq. at Source: gm (Vg −Vs ) =
1
RS C2
Vgs ( s ) = Vg −Vs =
Vg ( s )
1
s+
"1
%
RS 'C2
$
# gm
&
s+
C3
C2
C2
Av ( s ) =
Vo ( s )
= Amid FL ( s )
Vi ( s )
Amid = −gm ( RD R3 )
Lecture14-Frequency Response
Vs
+ sC2Vs
Rs
RG
RI + RG
8
4
Low-Frequency Poles and Zeros
Direct Calculation: C-S Amplifier (cont.)
FL ( s ) =
!
1 $
s2 # s +
&
" RS C2 %
*'
'
*'
*
1
1
1
,)s +
)s +
,)s +
,
)( ( RI + RG ) C1 ,+)( (1 gm RS ) C2 ,+)( ( RD + R3 ) C3 ,+
The three zero locations are: s = 0, 0, −1 / RS C2
The three pole locations are: s = −
1
1
1
, −
, −
R
+
R
C
1
g
R
C
R
+
( I G ) 1 ( m S ) 2 ( D R3 ) C3
Each independent capacitor in the circuit contributes one pole and one zero.
Series capacitors C1 and C3 contribute the two zeros at s = 0 (dc), blocking
propagation of dc signals through the amplifier. The third zero due to the
parallel combination of C2 and RS occurs at frequency where signal current
propagation through the MOSFET is blocked (output voltage is zero).
Lecture14-Frequency Response
9
5