Lect. 24: High-Frequency Response of MOSFET CS

Lect. 24: High-Frequency Response of MOSFET CS
How fast can this operate?
R L , C L due to
Q 2 and external load
Electronic Circuits 1 (09/2)
Prof. Woo-Young Choi

Lect. 24: High-Frequency Response of MOSFET CS
It can be shown from an exact (but complicated ) analysis that
'
V
−( g ) ( )
mR ⎡
L
1 −s Cgd / g ⎤
m
o
=
⎣
⎦
Vsig
1+ s{ ⎡C C ( 1 g R ) ⎤R ( C C ) R } s ⎡( C C ) C C C ⎤
⎣
+ +
⎦
+ + +
⎣
+ +
⎦
R R
' ' 2 '
gs gd m L sig L gd L L gd gs L gd sig L
Too complex. A simpler way of estimating f H , high-frequency 3-dB frequency?
Electronic Circuits 1 (09/2)
Prof. Woo-Young Choi

Lect. 24: High-Frequency Response of MOSFET CS
Open-Circuit Time Constant Method for approximating f H
1. Select one capacitor, C i , and set others to open.
2. Determine R i , the resistance seen by C i .
3. Repeat above for all capacitors.
Then,
ω
H
~
∑
i
1
CR
i
i
Electronic Circuits 1 (09/2)
Prof. Woo-Young Choi

Lect. 24: High-Frequency Response of MOSFET CS
For C gs ,
R gs = ?
R gs =R sig
Electronic Circuits 1 (09/2)
Prof. Woo-Young Choi

Lect. 24: High-Frequency Response of MOSFET CS
For C gd ,
R gd = ?
I
V
I
I
Vgs
+ V
= g V +
R '
x m gs
= −I R
gs x sig
− I R + V
=− g I R +
R
L
'
Rsig
Vx
(1 + g R + ) =
R ' R '
x m x sig
x m sig
L
L
x
x sig x
L
V
I
x
x
Rsig
= (1 + g R + ) R '= R ' + g R R ' + R
R '
m sig L L m sig L sig
L
∴ R = R (1 + g R ') +
R '
gd sig m L L
Electronic Circuits 1 (09/2)
Prof. Woo-Young Choi

Lect. 24: High-Frequency Response of MOSFET CS
For C L , R L = ?
R L =R L ’
Electronic Circuits 1 (09/2)
Prof. Woo-Young Choi

Lect. 24: High-Frequency Response of MOSFET CS
ω
Open-Circuit Time Constant Method for approximating f H
1. Select one capacitor, C i , and set others to open.
2. Determine R i , the resistance seen by C i .
3. Repeat above for all capacitors.
1 1
~ = , where τ = C R + C [ R (1 + g R ') + R '] + C R '
τ
H
CR
H H gs sig gd sig m L L L L
∑
i
i
i
Miller Effect!
Electronic Circuits 1 (09/2)
Prof. Woo-Young Choi

Lect. 24: High-Frequency Response of MOSFET CS
Determine f H for the CS shown left.
I REF =100μA, W/L=7.2μm/0.36μm
/0 k n ’=387μA/V 2 , k p ’=86μA/V 2 ,
r o1 =18kΩ, r o2 =22kΩ
C gs =20fF, C gd =5fF, C L =25fF, R sig =10kΩ.
It can be shown
R
τ
gs = C gsR
gs
= 200ps
τ = C R = 714ps
τ
gd gd gd
C L C
L
= C R =
L
= R = 10kΩ
gs sig
'
gd sig
1 m L L
142.8k
'
( g
)
R = R + g R + R = Ω
R
C
246ps
τ = τ + τ + τ = 1160ps
f
H gs gd C L
H
L
= R = 9.82kΩ
1 1
= = = 137 MHz
−12
2 πτ 2 π × 1160 ×
10
H
'
L
Electronic Circuits 1 (09/2)
Prof. Woo-Young Choi

Lect. 24: High-Frequency Response of MOSFET CS
How accurate is f H = 137MHz estimated
by the open-circuit time constant method?
From the exact analysis,
'
−
( g mR ) ⎡
L ⎣
1 − s ( C
gd / gm
)
( ) ⎤ ( ) ⎡
V
⎤
o
=
⎦
Vsig
1+ s{ ⎡C C 1 g R R C C R } s ( C C ) C C C ⎤
⎣
+ +
⎦
+ + +
⎣
+ +
⎦
R R
f
H
= 145.3MHz
' ' 2 '
gs gd m L sig L gd L L gd gs L gd sig L
What is the most influential capacitor
for f H ?
Electronic Circuits 1 (09/2)
Prof. Woo-Young Choi

Lect. 24: High-Frequency Response of MOSFET CS
Miller’s Theorem:
Two circuits below are identical assuming the rest of circuit does not change
I
V ⎛V − KV
1 1 1
1
= = I =
Z ⎜
1
Z ⎟
⎝ ⎠
VZ Z
V KV −K
1
∴ Z1
= =
1−
1
1
⎞
I
2
0 V 0 KV V KV
= − = − = I =
−
Z Z Z
Z
2 1 1 1
2 2
−KV Z
Z
1
∴
2
= =
V
1
1−
KV1 (1 − )
K
Electronic Circuits 1 (09/2)
Prof. Woo-Young Choi

Lect. 24: High-Frequency Response of MOSFET CS
CS amplifier
With Miller Theorem,
Z
Z
Z1 = , Z2
=
1 − K
1
(1 − )
K
1 1 1
=
sC sC − K
eq1 gd
1
∴ Ceq1 = Cgd (1 − K) = Cgd (1 + gm RL
')
1 1 1
=
sC sC 1−
1/ K
eq2
gd
1
∴ Ceq2
= Cgd (1 − 1 / K) = Cgd
(1 + )
g R '
C eq1 C eq2 For f H , the influence of C gd becomes larger
by factor of (1+g m R L ’) : Miller Effect
CS amplifier is slow!
m
L
Electronic Circuits 1 (09/2)
Prof. Woo-Young Choi