Page 79 - 30105-2003 การวิเคราะห์วงจรอิเล็กทรอนิกส์ความถี่สูง

71
2.

F
,
2.7 ก ,% 2.1 , F ( L C C ) L

1
- .
ก ก (2.15) F ( L C ) =
π
C
C
2 R FLCC C
)
3
=
×
* % R FLCC = ( R + R L ) ( 1 10 + 75 = 1,075 Ω
C
1
F ( L C C ) =
π
C
2 R
FLCC C
1
F ( L C ) =
C × × × × − 12
2 3.14 1, 075 820 10
F ( L C C ) = 180.641 kHz
(
- ก F ( L C B ) = 1.065 MHz, F ( L C E ) = 2.453 kHz, F ( L C C ) = 180.641 kHz;

$
F = F = 1.065 MHz
L ( L C B )

F ( L C C ) = == = 180.641 kHz
F - F ( L C B ) = == = 1.065 MHz
L

2.1.3.5 ก
)
#
*
A V (F L )

A V (F L ) -

"
.
V E g
o
% E " !
"
&( . (H3 ก ! ) * C
ก ! ) * C

g B E
C
$
< ('
( ก ! C . ก (
& % '"
& "
%
"
ก 50 Ω
C B
(ก ก !
"
# '" E R
"
. - 75 Ω
$
ก , A V (F L ) g g

r bb′ b′

I I I
i b r b e ′ V b e ′ g V c
m b e ′
E g

,% 2.14 , A %
ก ! ) * C
V (F L ) B

ก ,% 2.14 R ('
( ก ! ,% 2.9 " ก
FLCB
R BB (r bb′ + r b e ′ )
R FLCB = R +
g
( R BB + r bb′ + r b e ′ )
R BB (r bb′ + r b e ′ )
ก .
." Z BB =
( R BB + r bb′ + r b e ′ )

ก

ก
ก