Page 223 - 30105-2003 การวิเคราะห์วงจรอิเล็กทรอนิกส์ความถี่สูง

215
5.

1
ก ก (5.10b) F =
o
2π C L
T 3
1
C L =
T
3
(2 Fπ o )
1
) C = (5.27)
T 2
π
L ( 2 F )
3 o
,
5.4.1 ก
5
$ 6 C C , L $ RFC
1 2 3 1
ก ก (5.25) C = nC
1 2
C T ( +n ) 1
ก ก (5.26) C 2 =
n
1
ก ก (5.27) C =
T 2
L (2 Fπ )
3 o
ก RFC = 4L
1 3
) 5.9 ก #$ 5.15 $ &( C C L !
RFC . ก F = 500 kHz,
,
,
1 2 3 1 o
L = 360 µH !
n = 1
3
1
* + ก ก (5.27) C =
T 2
L 3 (2 Fπ o )
0 F = 500 kHz, L = 360 µH;
o 3
1
C = = 281.733 pF
T 2
×
×
×
×
360 10 − 6 ( 2 3.14 500 10 3 )
×
C ( n + ) 1 281.733 10 − 12 (1 1+ )

ก ก (5.26) C = T n = 1 = 563.466 pF
2
ก ก (5.25) C = nC = ( ) 1 563.466 10× − 12 = 563.466 pF
1 2
RFC = 4L = 4 360 10 − 6 = 1.44 mH
×
×
1
3
$
$- &( /

C 1 = C 2 = 563.466 pF $- &( / C 1 = C 2 = 560 pF ก
C = == = C = == = 560 pF ก , L = == = 360 µH, RFC = == = 1.44 mH;

2
1
1
3
5.4.2 ก
5
$ 6 R $ R
S D
0 ก ' * B Q 3 )
2
+V
ก ก (2.42) V G =V GS RS
. V = 0 V
G
ก

ก
ก