Page 142 - 30105-2003 การวิเคราะห์วงจรอิเล็กทรอนิกส์ความถี่สูง

134

3.

, F = F ( R L 21 ) , L = L 21 , C = (C BT C b e ′ ) C= BT + C b e ′ ;
R

1
F ( R L ) = (3.13)
21
2π L 21 (C BT + C b e ′ )

"#
3.9 ก + 3.14 + %! 9 g % r
b e ′
m
I I 5.012 10 − 3
×
( )
( )
- ,
ก !ก (1.9) g = C dc = C dc = = 195.019 mS
m
×
V T (k T q ) 25.7 10 − 3
B
β 120
ก !ก (1.10) r b e ′ = o = = 615.324 Ω
×
g m 195.019 10 − 3
Ω
Ω Ω Ω
" g = == = 195.019 mS, r b e ′ ′′ ′ = = = = 615.324 ;
m
3.5.2.2 ก
$

'
F R (L )
12
F ( R L 12 ) 9 !8 ,.

.' &
9 L ก( ก L C CT (ab) )
( 12
12
1
ก !ก (3.1b) F =
R
2π LC
, F = F ( R L 12 ) , L = L 12 ,C = C CT (ab) ;
R
1

$
F ( R L 12 ) = (3.14)
2π
L C
12 CT (ab)
3.5.2.3 ก
$

'
C CT (ab)
)
C CT (ab) & ก? + % 8 ก ก C ! +D!3 !(&
9 (ab)
CT
"
+ 3.11 % 3.14

2
2 C N
ก !ก (3.11) C N S = P P
S
2
C N
C = S S
P 2
N P
, C = C ,C = C , N = N , N = N ;
P CT (ab) S CT P 12 S 22
2
C CT N 22 (3.15)

$
C CT (ab) = 2
N 12

3.5.2.4 ก
$

'
R L (ab )
9 !&

8 ก ก R ! +D!3 !( &
(ab)
9
R L (ab) L
"
+ 3.11 % 3.14

ก

ก
ก