Page 91 - 30105-2003 การวิเคราะห์วงจรอิเล็กทรอนิกส์ความถี่สูง

83
2.

C % .
" % ! 3 ก ! ก % ." & & (
E
; 3(,5
ก % //0 ก !

C % .
" % ('- //0 ก ! , R
C
L
1
ก ก (2.13) F ( L C B ) =
π
C
2 R
FLCB B
1
C = (2.29)
B
π
2 R FLCB F ( L C B )
* % F -
"
L
R BB ( r bb′ + r b e ′ ) R R
R FLCB = R + , F ( L C ) = F R BB = B 1 B 2
g
L
( R BB + r bb′ + r b e ′ ) B ( R B 1 + R B 2 )
1
ก ก (2.14) F =
( L C E ) 2 R FLCE C E
π
1
C = (2.30)
E
2 Rπ FLCE F ( L C E )
* % F -
"
L
Ω
R FLCE = X C E = 0.5 , F ( L C E ) = F L ;
1
ก ก (2.15) F =
( L C C )
π
2 R FLCC C
C
1
C = (2.31)
C
2 Rπ FLCC F ( L C C )
Ω
* % R = X = 0.5 , F = F ;
FLCE C E ( L C E ) L
F -
"
L
,
2.15 ก .
C C C
B , E C
1
- .
ก ก (2.29) C =
B
π
2 R FLCB F ( L C B )
10.183 kΩ
* % R BB =
×
2.595 10 − 3
g = = 100.972 mS
m
×
25.7 10 − 3
β 120
r b e ′ = o = = 1.188 kΩ
×
g m 100.972 10 − 3
ก

ก
ก