Page 107 - 30105-2003 การวิเคราะห์วงจรอิเล็กทรอนิกส์ความถี่สูง

99
2.

R
R = R +
FLCD D L
g V R = R + R )
m
D C D FLCD ( D L

R D R L R D C D

R L
S
R FLCD
(/
,% 2.29 , R
FLCD

,
2.26 ก ,% 2.21 , F ( L C G ) , F ( L C S ) F ( L C C )
1
- .
ก ก (2.54) F ( L C G ) =
π
2 R
C
FLCG G
=
Ω
+
Ω
×
* % R GG = 43.532 k , R FLCB = ( R + R GG ) ( 75 43.532 10 3 ) = 43.607 k ;
g
C = 100 pF
G
1 1
F ( L C ) = =
G 2 R FLCG G 2 3.14 43.607 10 × 100 10 − 12
π
×
3
×
×
×
C
F ( L C G ) = 36.516 kHz
1
ก ก (2.55) F ( L C S ) = 2 R FLCS C S
π
R S 750 281.663 Ω
* % R = = =
FLCS
(1 g R+ m S ) 1+ ( 2.217 10 − 3 × 750 )
×
1 1 5.653 kHz
F ( L C S ) = = − 6 =
π
×
×
×
×
2 R
FLCS C S 2 3.14 281.663 0.1 10
1
ก ก (2.56) F ( L C ) =
π
D 2 R FLCD C D
×
3
×
=
* % R FLCD = ( R + R L ) ( 1.8 10 + 2 10 3 ) = 3.8 kΩ
D
1 1
F = = = 891.577 Hz
( L C D ) 3 − 6
π
×
×
×
×
2 R FLCD C D 2 3.14 3.8 10 × 0.047 10
F ( L C ) = = = = 36.516 kHz, F ( L C ) = = = = 5.653 kHz, F ( L C ) = = = = 891.577 Hz;
G S D
2.2.3.4 ก
)
#
*
F
L

* 3& ก
F ( L C G ) , F ( L C S ) F ( L C D ) %
) (,5

F
L

ก

ก
ก