Page 187 - 30105-2003 การวิเคราะห์วงจรอิเล็กทรอนิกส์ความถี่สูง

179
4.

×
×
75 8.432 10 3
$% R FLCE 1 = = 0.736 Ω
( 75 8.432 10 3 ) (100 1+ )
×
+
+
×
 2 461.981 560 5.329
Ω
Ω
R FLCE 2 = 0.736 +   = 5.329 , R FLCE 3 = = 5.278 ;
+
+
 (100 1 )  (560 5.329 )
3
+
×
×
R FLCE = ( 5.278 4.762 10 + 56 10 3 ) = 60.767 k , CΩ E = 0.22 µF;
1
=
F ( L C E ) 2 3.14 60.767 10 × 0.22 10
−
6
×
×
3
×
×
F ( L C E ) = 11.911 kHz
* F ( L C ) = == = 11.911 kHz
E
4.2.4.3 ก
(

) F
( L C C )

X
$% C - +

0 %% 0 $% ,
5 กก F
C ( L C C ) C C

ก -
0 0 C /
0 % ก5 ! R ก 1 4.11 (4%
C FLCC

ก
C C
R FLCC
D 1
r
1 d
R L
R FLCC = R + r + R L
1 d
C

1 4.11 . % ก (
R FLCC
R
R FLCC = R + r +
C
L
d
1
X C C = R FLCC
1
! ก X C C = 2 F ( L C C ) C
π
C
1
R FLCC =
2 Fπ C
L C ) C
( C
1 (4.20)
% & F ( L C C ) = 2 R FLCC C
π
C
*& 4.16 ก 1 4.5 .5( 1 4.11 (
F ( L C C )
1
+ , ก
ก (4.20) F =
( L C C ) 2 Rπ
C
FLCC C
ก

ก
ก