Page 195 - 30105-2003 การวิเคราะห์วงจรอิเล็กทรอนิกส์ความถี่สูง

187
4.

+
$% C ! 0 ก3- (
ก
%
0 %% 0 #
B
F ( L C ) = F = 30 kHz
L
B
×
I C (dc ) 5.003 10 − 3
g = = = 194.669 mS
m
k T q × − 3
B 25.7 10
β 88
r b e ′ = o = − 3 = 452.049 Ω
×
g m 194.669 10
3
×
×
R R 2 27 10 × 15 10 3
1 B
B
R BB = = = 9.642 kΩ
3
×
×
R B 1 + R B 2 27 10 + 15 10 3
) }
r
R BB { bb′ + r b e ′ + (β + 1 R E
o
R FLCB = R + + + + ( β + ) }
g
R
{ BB r bb′ r b e ′ o 1 R E
×
9.642 10 3 {2 452.049+ + (89 620× )}
R = 75 + = 8.292 kΩ
FLCB 3
2 452.049 +
×
×
9.642 10 + + (89 620 )
1 1
C = =
2 F ( L C B ) FLCB 2 3.14 30 10 × 8.292 10 3
B
π
3
×
×
×
×
R
C = 640.117 pF
B
1
ก
ก (4.26) C = C =
E
C
π
2 F R
( L C E ) FLCE
Ω
$% R FLCE = 0.5 , F ( L C E ) = F = 30 kHz;
L
1 1
C = C = = = 10.615 µF
E
C
π
3
×
×
×
2 F R 2 3.14 30 10 × 0.5
L C ) FLCE
( E
* C = == = 640.117 pF,C E = = = = C = == = 10.615 µF;
B
C
4.2.7.6 ก
(% 3 *& ก; (
: (83
*
9 3
C = 640.117 pF
0 ' C = 680 pF /
ก
B
B
C = C = 10.615 µF
0 ' C = C = 15 µF . 5
E C E C
4.2.7.7 ก
*
$ F '
H
1
ก
ก (4.21) F =
H
2 Rπ C
FH T
*& 4.23 ก 0 - F #
! ,
0 ' .5(
1 +% 2
H
1
+ , ก
ก (4.21) F =
H
π
2 R C
FH T
ก

ก
ก