Page 92 - 30105-2003 การวิเคราะห์วงจรอิเล็กทรอนิกส์ความถี่สูง

84
2.

+
×
×
r
R BB ( bb′ + r b e ′ ) 10.183 10 3 ( 2 1.188 10 3 ) 1.14 kΩ
R = R + = 75 + =
FLCB g ) 3 3
×
+
×
( R BB + r bb′ + r b e ′ ( 10.183 10 + 2 1.188 10 )
F ( L C B ) = F = 1 MHz
L
1 = 1
C =
B 2 Rπ FLCB F ( L C B ) 2 3.14 1.14 10 × 1 10
×
×
6
×
×
3
C = 139.680 pF
B
1
ก ก (2.30) C =
E
π
2 R FLCE F ( L C E )
Ω
* % R FLCE = X C E = 0.5 , F ( L C E ) = F = 1 MHz;
L
1 1
C = = = 0.3184 µF
E
π
×
×
×
×
2 R FLCE F ( L C E ) 2 3.14 0.5 1 10 6
1
ก ก (2.31) C =
C
π
2 R FLCC F ( L C C )
Ω
* % R FLCC = 0.5 , F ( L C E ) = F = 1 MHz;
L
1 1
C = = = 0.3184 µF
C
π
×
×
×
×
2 R FLCC F ( L C C ) 2 3.14 0.5 1 10 6
C = == = 139.680 pF,C E = 0.3184 µF, C = == = 0.3184 µF;
B
C
2.1.5.6 ก
)+
, กG )
# F $ )D

E
('
( ก
ก ! "
%
* '" A-4
; <
ก ก.
C = 139.680 pF (,5
4
C = 150 pF ( &ก
B
B
C = 0.3184 µF (,5
4
C = 0.33 µF %
%
E E
C = 0.3184 µF (,5
4
C = 0.33 µF %
%
C
C
2.1.5.7 ก

F
L
C % .
" % ! )
"
ก !ก % ( -
B
" '" (ก+!, ) 4
1
ก ก (2.13) F = F =
L ( L C B )
π
C
2 R FLCB B
,
2.16 ! F ( - '" 4
L
1
- .
ก ก (2.13) F = F ( L C ) =
L
π
B
C
2 R
FLCB B
Ω
* % R BB = 10.183 k , C = 150 pF;
B
ก

ก
ก