Page 249 - 30105-2003 การวิเคราะห์วงจรอิเล็กทรอนิกส์ความถี่สูง

241
6.

R S
100 Ω
Z S Z R
P P C P
C S
10 pF
F = 100 MHz
R
# 4' 6.15 # R :ก ! C 5 R C
S S P P

×
6
×
6
×
π
×
"# $ ω = 2 F = 2 3.14 100 10 = 628 10 Rad s
R
×
×
6
1+ ( C Rω ) 2 1+ ( 628 10 × 10 10 − 12 × 100 ) 2
ก !ก (6.26b) R = S S 2 = 2
P
(
ω
×
6
R ( C ) 100 628 10 × × − 12 )
S S 10 10
R = 353.559 Ω
P
×
C 10 10 − 12
C = S = = 7.171 pF
P
1+ ( C Rω S S ) 2 1+ ( 628 10 × 10 10 − 12 × 100 ) 2
×
6
×
R 353.559
Z = P =
P
+
1 J ω P P ) 1 J ( 628 10 × 7.171 10 − 12 × 353.559 )
( C R
6
×
+
×

Z = 188.043 − 57.868 Ω
P
1 1
Z = R + = 100 +
S
S
ω
×
×
6
J C S J ( 628 10 × 10 10 − 12 )

Z = 188..031 − 57.871 Ω
S

Ω
Ω Ω Ω
Ω Ω

R = == = 353.559 , C = =Ω Ω = = 7.171 pF, Z = == = 188.043 − − − − 57.868 ,
P P P

Ω Ω Ω
Ω
Z = == = 188.031 − − − − 57.871 ;
S

! 6.6 ! ;2
ก 4' !' !1'
2 ก 75 Ω ก
−
4
2
4' !'" !1'
2(3.18 J 3.10 ) Ω 4' " !7' 150 MHz !'

2
42 2 MHz (6; !ก : ! 4' 3

L
C m 2
m 3
R m 5
75 Ω
C m 4 R m 1
E 3.18 Ω
S

# 4' 6.16 ก 6; !ก : ! 4' 3
ก

ก
ก