Page 118 - 30105-2003 การวิเคราะห์วงจรอิเล็กทรอนิกส์ความถี่สูง

110
2.

1 1
C = = = 3.594 pF
G 3 6
π
×
×
×
×
2 R FLCG F ( L C G ) 2 3.14 44.303 10 × 1 10
1
ก ก (2.68) C =
S
π
2 R FLCS F ( L C )
S
Ω
* % R FLCS = 0.5 , F ( L C S ) = F = 1 MHz;
L
1
C = = 0.3184 µF
S
×
×
× ×
2 3.14 0.5 1 10 6
1
ก ก (2.69) C =
D
π
2 R FLCD F ( L C D )
3
3
Ω
×
×
* % R FLCD = R + R = 2 10 + 1.5 10 = 3.5 k , F ( L C D ) = F = 1 MHz;
L
D
L
1
C = = 45.495 pF
D 3 6
2 3.14 3.5 10 × 1 10
×
×
×
×
C = == = 3.594 pF, C = == = 0.3184 µF, C = == = 45.495 pF;
S
G
D
2.2.5.6 ก
)+
, กG )
# F $ )D

E
C = 3.594 pF (,5
4
C = 3.9 pF ( &ก
G
G
C = 0.3184 µF (,5
4
C = 0.33 µF %
%
S S
C = 45.495 pF (,5
4
C = 47 pF ( &ก
D
D
2.2.5.7 ก

F (+# F
H
L
1
ก ก (2.54) F =
( L C G ) 2 Rπ FLCG G
C
1
F = F =
L ( L C G ) 2 Rπ FLCG G
C
1
ก ก (2.58) F =
H
π
2 R FH T
C
∗. (. ) F % ! " - ก . (% ก ! % ก .
-
"
L
"
F % ! " - ก . (% ก ! % ก .
-
H
,
2.34 ก ,% 2.34 ! F F
H
L
1
- .
ก ก (2.54) F ( L C G ) =
π
2 R
C
FLCG G
1
F = F =
L ( L C G ) π
2 R
C
FLCG G
ก

ก
ก