Page 128 - 30105-2003 การวิเคราะห์วงจรอิเล็กทรอนิกส์ความถี่สูง

120

3.

. 9 L % ,.

.' 10 MHz

1
ก !ก (3.1b) F =
R
2π LC
1 1 1
LC = , LC = , L = ;
π
π
π
2 F R (2 F R ) 2 C (2 F R ) 2

ก C = 10.695 nF, F = 10 MHz;
R
1 1
L = = = 23.708 nH
π
C ( 2 F R ) 2 10.695 10 − 9 ( 2 3.14 10 10 6 ) 2
×
×
×
×
" L = == = 23.708 nH
, L % C ก& 9 3.1 ! + %! 9
"#
3.2 "

9 R
L
ก. (!/
.' !8 ,.

.' (Z F R )

. (!/
.' !8 &
(Z F H ) . (!/
.' !8 &
& (Z F L )
)
- ,
ก. 9 (!/
.' !8 ,.

.' (Z F R
1
ก !ก (3.1) Z =
G + 1 + J C
ω
ω
J L
, L = 23.708 nH, C = 10.695 nF, R = 10 k ;
Ω
L
1 1 −
6
G = = = 100 10 S
×
×
R L 10 10 3
1 = 1 = 1 S
ω
6
×
×
×
J L J ( 2 3.14 10 10 × 23.708 10 − 9 ) J (1.48886 )
×
×
6
=
×
ω
×
×
J C = J ( 2 3.14 10 10 × 10.695 10 − 9 ) ( 671.646 10 − 3 ) S
×
J
1
Z =
×
×
100 10 − 6 + 1 + J ( 671.646 10 − 3 )
J (1.48886 )
1
Z =
×
×
+
×
J
100 10 − 6 − J ( 671.65 10 − 3 ) ( 671.65 10 − 3 )
1
Z F = Z = R = = 10 kΩ
L
R × − 6
100 10
" Z F R = = = = 10 kΩ ΩΩ Ω
. 9 (!/
.' !8 &
(Z F H )

3
×
×
ก !ก (3.1c) Z F H = 0.707R L − 45 = 0.707 10 10 = 7.07 − 45 kΩ
" Z = == = 7.07 − −− − 45 kΩ ΩΩ Ω
F H
ก

ก
ก