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1 Lecture14-Frequency Response
EE105 – Fall 2014 Microelectronic Devices and Circuits
Prof. Ming C. Wu
511 Sutardja Dai Hall (SDH)
2 Lecture14-Frequency Response
Amplifier Frequency Response Transfer Function Analysis
Av s( ) =N s( )D s( )
=a0 + a1s+ a2s
2 +...+ amsm
b0 + b1s+ b2s2 +...+ bns
n
Av s( ) = AmidFL s( )FH s( )Amid is the midband gain between the lower and upper cutoff frequencies ωL and ωH .
FL s( ) =s+ωZ1
L( ) s+ωZ 2L( ) ⋅ ⋅ ⋅ s+ωZk
L( )s+ωP1
L( ) s+ωP2L( ) ⋅ ⋅ ⋅ s+ωPk
L( )
FH s( ) =1+ s
ωZ1H
"
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&' 1+ s
ωZ 2H
"
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&'⋅ ⋅ ⋅ 1+ s
ωZlH
"
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%
&'
1+ sωP1
H
"
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%
&' 1+ s
ωP2H
"
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&'⋅ ⋅ ⋅ 1+ s
ωPlH
"
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&'
FH jω( ) →1 for ω <<ωZiH ,ωPi
H , j = 1... l
∴AL s( ) ≅ AmidFL s( )
FL jω( ) →1 for ω >>ωZjL ,ωPj
L , j = 1... k
∴AH s( ) ≅ AmidFH s( )
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3 Lecture14-Frequency Response
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.
FL s( ) ≅ ss+1000
Amid = 200ωL ≅1000 rad / sec
4 Lecture14-Frequency Response
Lower Cutoff Frequency Calculations If there is no dominant pole at low frequencies,the poles and zeros interact to determine the lower cutoff frequency ωL. For example, suppose:
AL s( ) = AmidFL s( ) = Amids+ωZ1( ) s+ωZ 2( )s+ωP1( ) s+ωP2( )
For s = jωL, AL jωL( ) = Amid2
12=
ωL2 +ωZ1
2( ) ωL2 +ωZ 2
2( )ωL
2 +ωP12( ) ωL
2 +ωP22( )
12=ωL
4 + ωZ12 +ωZ 2
2( )ωL2 +ωZ1
2 ωZ 22
ωL4 + ωP1
2 +ωP22( )ωL
2 +ωP12 ωP2
2
Lower cutoff frequency ωL will begreater than all the individual polezero frequencies.
∴ωL ≅ ωP12 +ωP2
2 − 2ωZ12 − 2ωZ 2
2 In general, for n poles and n zeros,
ωL ≅ ωPn2
n∑ − 2 ωZ1
2
n∑
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5 Lecture14-Frequency Response
Dominant Pole Example
• Problem: Find midband gain, FL(s) and fL for
• Analysis: Rearranging the given transfer function into standard form,
AL s( ) = 2000s s100
+1!
"#
$
%&
0.1s+1( ) s+1000( )
AL s( ) = 200s s+100( )
s+10( ) s+1000( )= AmidFL s( )→ Amid = 200 FL s( ) =
s s+100( )s+10( ) s+1000( )
Zeros: s = 0 and s = -100 Poles: s = -100 and s = -1000
The poles and zeros are all widely separated. ∴AL s( ) ≅ 200 ss+1000
The dominant pole is at ω =1000 and fL ≅10002π
=159 Hz.
The more exact calculation is fL =1
2π102 +10002 − 2×02 − 2×1002 =158 Hz
6 Lecture14-Frequency Response
High-Frequency Dominant Pole
• The lowest of all high frequency poles is called the dominant high-frequency pole.
• If there is no dominant pole at high frequencies, the poles and zeros interact to determine ωH.
AH s( ) ≅ AmidFH s( )
FH s( ) ≅ 11+ s ωP3( )
ωH ≅ωP3
AH s( ) = AmidFH s( ) = Amid1+ s
ωZ1
!
"#
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%& 1+ s
ωZ 2
!
"#
$
%&
1+ sωP1
!
"#
$
%& 1+ s
ωP2
!
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$
%&
For s = jωH , Amid jωH( ) = Amid2
12=
1+ ωH
ωZ1
!
"#
$
%&
2'
())
*
+,,
1+ ωH
ωZ 2
!
"#
$
%&
2'
())
*
+,,
1+ ωH
ωP1
!
"#
$
%&
2'
())
*
+,,
1+ ωH
ωP2
!
"#
$
%&
2'
())
*
+,,
1ωH
≅1ωP1
2 +1ωP2
2 −2ωZ1
2 −2ωZ 2
2
For the general case of n poles and n zeros,
1ωH
≅1ωPn
2n∑ − 2 1
ωZn2
n∑
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7 Lecture14-Frequency Response
High Frequency Dominant Pole Approximation
AH s( ) = AmidFH s( )Amid = 200
FH s( ) ≅ 11+ s 106 =
106
s+106
ωH ≅106 rad / sec
fH =ωH
2π≅159 kHz
8 Lecture14-Frequency Response
Low-Frequency Poles and Zeros Direct Calculation: C-S Amplifier
C2
C3
C2
Vo s( ) = Io s( )R3 = −gmVgs s( )RD
RD + 1 sC3( )+ R3
R3
Vo s( ) = −gm RD R3( ) s
s+ 1C3 RD + R3( )
Vgs s( )
Vg s( ) =sC1RG
sC1 RI + RG( )+1Vi s( )
Node Eq. at Source: gm (Vg −Vs ) =VsRs+ sC2Vs
Vgs s( ) =Vg −Vs =s+ 1
RSC2
s+ 11gm
RS"
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%
&'C2
Vg s( )
Av s( ) =Vo s( )Vi s( )
= AmidFL s( )
Amid = −gm RD R3( ) RGRI + RG
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9 Lecture14-Frequency Response
Low-Frequency Poles and Zeros Direct Calculation: C-S Amplifier (cont.)
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).
FL s( ) =s2 s+ 1
RSC2
!
"#
$
%&
s+ 1RI + RG( )C1
'
())
*
+,,s+ 1
1 gm RS( )C2'
())
*
+,,s+ 1
RD + R3( )C3
'
())
*
+,,
The three zero locations are: s = 0, 0, −1/ RSC2
The three pole locations are: s = − 1RI + RG( )C1
, − 11 gm RS( )C2
, − 1RD + R3( )C3