Page 135 - 30105-2003 การวิเคราะห์วงจรอิเล็กทรอนิกส์ความถี่สูง

127

3.

L L
E 1 2
g
N 1 N 2

+ 3.7
"-
! +

"#
3.6 ก + 3.7 C = 2.15 nF, C = 70 pF, R = R = 75 , L = 112.58 nH, N = 10 T,
Ω
L
2
1
g
1
1
N = 2 T; + %! 9
2
ก. !8 ,.

.' %
!!

& E
9
g
(
'
.
'
- ,
ก. !8 ,.

.'
1
ก !ก (3.1b) F =
R
2π LC
, L = L = 112.58 nH
1
C = C + C = ( 2.15 10 − 9 + 2.8 10 − 12 ) = 2,152.8 pF
×
×
1 2P
1
F =
R
×
( 2 3.14× ) ( 112.58 10 − 9 × 2,152.8 10 − 12 )
×
F = 10.228 MHz
R
" F = == = 10.228 MHz
R
C
&9 E ก C +
+;
C ก !ก (3.11)
2
2
2P
g
2
C N
2 2 S S ;
C N S = C N P , C =
P
S
P
N 2 P
, C = C 2P , C = C = 70 pF, N = N = 10 T, N = N = 2 T;
1
P
S
2
P
S
2
2 × − 12 2
C N 2 70 10 ( ) 2
2
C 2P = = = 2.8 pF
N 2 1 ( ) 2
10
R
&9 E ก R +
+;
R ก !ก (3.10)
L g L LP
Z P = Z S
N 2 N 2
P S
2
Z N P
Z =
S
P
N 2 S
, Z = R ,Z = R = 75 , N = N = 10 T, N = N = 2 T;
Ω
P LP S L P 1 S 2
ก

ก
ก