Page 188 - 30105-2003 การวิเคราะห์วงจรอิเล็กทรอนิกส์ความถี่สูง

180
4.

+
=
R +
×
3
×
$% R FLCC = ( C r + R L ) ( 560 4.762 10 + 56 10 3 ) = 61.322 kΩ
d
1
C = 0.22 µF
C
1 = 1
=
F ( L C C ) 2 R FLCC C 2 3.14 61.322 10 × 0.22 10 − 6
π
3
×
×
×
×
C
F ( L C C ) = 11.803 kHz
* F = == = 11.803 kHz
( L C C )
4.2.4.4 ก
(

) F '
L
$% F # !
0 %% 0 # ก
L
0 %% 0 # F , F .5( F

1 +%
F #
( L C B ) ( L C E ) ( L C C ) L

*& 4.17 (
F #
L
+ , 5! ก
1 +% F ( L C B ) = 45.660 kHz, F ( L C C ) = 11.803 kHz, F ( L C E ) = 11.911 kHz;
* F # ! F = == = 45.660 kHz
L ( L C B )

4.2.5 ก

!!" ก
#$$%&
*& 3$7
F !
0 %% 1
X
ก - R $% R !
0 -

FH
H
C
FH
T
0 X .5(;5% ก5 (
. % V
5%5
5! 0.707 # V
ก5
C
o
o
T
( , , ( -2 )π π # ก (
* % . % 1 4.12 .5( 4.13
ก 1 4.13
. %
ก 4% % &
g BB
g
R = ( { R R ) r+ } = R R + r bb′ ( R + R BB )
FH 1 g BB bb′ )
( R + R BB
g
R FH 2 = r b e ′ + ( β + ) 1 R
E
F
R FH 1 FH 2
R
R FH = ( R FH 1 R FH 2 ) =
( R FH 1 + R FH 2 )
! ก R

0 1
ก
! 0 # ก -0 0
0
5( & 4% (4%

1
r +
R = R r + R )} = R C ( d 1 R L )
out { C ( d 1 L )
R +
( C r + R L
d
1
1
ก
ก (2.17) F = (4.21)
H
π
2 R C
FH T
ก

ก
ก