Page 250 - 30105-2003 การวิเคราะห์วงจรอิเล็กทรอนิกส์ความถี่สูง

242
6.

"# $ ก
R m 5 > R
1
m
Ω
Ω
ก
6
R m 5 = 75 , R m 1 = 3.18 ;
F
ก !ก Q = R
m
B W
(4' F = 150 MHz, B = 2 MHz;
R W
×
F 150 10 6
Q = R = = 75
m
×
B 2 10 6
W
1
ก !ก (6.23a) C m 3 =
ω
Q R
m m 1
6
×
×
×
6
(4' ω = 2 F = 2 3.14 150 10 = 942 10 Rad s
π
×
R
R m 1 = 3.18 Ω
1
C m 3 = = 4.451 pF
942 10 × 75 3.18
×
6
×
1
ก !ก (6.23b) C m 4 =
ω R m 5   R m 1  
 R m 5 − R m 1 
Ω
Ω
(4' R m 1 = 3.18 , R m 5 = 75 ;
1
C m 4 = = 67.401 pF
( 942 10 × 75 )   3.18 
×
6

−
 75 3.18 
1
ก !ก (6.23c) L m 2 = + R R C
4
5 m
1 m
m
ω 2 C
m 3
1 − 12
×
×
×
L m 2 = + ( 3.18 75 67.401 10 )
4.451 10 − 12 ( 942 10 6 ) 2
×
×
L m 2 = 269.261 nH
67.401 pF,
Ω
3 R m 1 = 3.18 , L m 2 = 269.261 nH,C m 3 = 4.451 pF, C m 4 =
Ω
−
R = 75 ; $ ก !1'
2 !'" (3.18 J 3.10 ) Ω 5 ก ( J− 3.10 ) Ω 4 %
m 5
(ก
X C x = 3.10 Ω
1 1
C = = = 342.442 pF
x
ω X 942 10 ×
6
×
C x 3.10
ω 2 L C + 1
m x
ก !ก (6.25d) L :ก ! C 3 L m (new ) =
m
x
ω 2 C x
"$ " " !
' ( 6
! L !$ ก " C
(4' L m (new ) m 2 X
L = L m 2 = 269.261 nH, C = 342.442 pF;
m
X
ก

ก
ก