Page 109 - 30105-2003 การวิเคราะห์วงจรอิเล็กทรอนิกส์ความถี่สูง

101
2.

1 1
ω
C = =
G
R FLCG ( R + R GG )
g
1
%
Cω G =
( R + R GG )
g
V o
A = −
V (F L )
E
g
g Z JR GG 1
m out
×
A V (F L ) = −  ( R + ) ( R + R )


+

 1 J g R GG  g GG
 ( R + R GG ) 


g
g Z JR GG g Z R 90
m out GG
m out
A = − = −
V (F L )
( R + R GG )(1 J+ ) 1 ( R + R ) ( ) 1 2 + ( ) 1 2 tan − 1 1
g
g
GG
1
)

90 −
g Z R (0.707 g Z R
45
90
m out GG
m out GG
A V (F L ) = − 45 = − )
( R + R GG ) 2 ( R + R GG
g
g
V o 45 R GG
)
A V (F L ) = − = − (0.707 g Z × )
m out
E g ( R + R GG
g
V o R GG
(
- ก A V (F Mid ) = − E g = − g Z × ( R + R GG )
m out
g
V o
45
A V (F L ) = − = 0.707A V (F Mid ) (2.57)
E g
V o
,
2.28 ก ,% 2.21 , A V (F L ) = −
E
g
V o
- .
ก ก (2.57) A V (F L ) = − = 0.707A V (F Mid ) 45
E
g
V o
)
−
A V (F L ) = − E g = (0.707 )( 2.095 = − 1.481 45
V o
= − −− −
= − −− −
A V (F L ) == = E g == = 1.481 45
2.2.4 ก

#$!
%&&'
ก
#( +,
,
2.2.4.1 ก
)
#
*
F
H
F -
"
# C
"

ก ! C
H rss iss
ก (,5
C (% ก ! (C iss + C g R ) ก

,% 2.31 2.32
rss m out
T
ก

ก
ก