Page 93 - 30105-2003 การวิเคราะห์วงจรอิเล็กทรอนิกส์ความถี่สูง

85
2.

+
×
×
R BB ( bb′ + r b e ′ ) 10.183 10 3 ( 2 1.188 10 3 )
r
R FLCB = R + = 75 + = 1.14 kΩ
g
3
×
×
+
( R BB + r bb′ + r b e ′ ) ( 10.183 10 + 2 1.188 10 3 )
1 = 1
=
F =
L F ( L C B ) 2 R FLCB B 2 3.14 1.14 10 × 150 10 − 12
π
3
×
×
×
×
C
F = 931.202 kHz
L
F - 931.202 kHz # ก 1 MHz
- '" "
L
2.1.5.8 ก

F
H
C R ! )
"
R (,5
"
%
% (

FH
T
FH
C ก ! ก % ( - '" 4
T
1
ก ก (2.17) F =
H
2 Rπ C
FH T
,
2.17 ก ! F
H
1
- .
ก ก (2.17) F =
H
π
C
2 R FH T
×
2.595 10 − 3
=
Ω
* % R BB 10.183 k , g = − = 100.972 mS;
m
×
25.7 10 3
β 120
r b e ′ = o = = 1.188 kΩ
g × − 3
m 100.972 10
 R R  ( 75 10.183 10 3 )
×
×


R FH 1 = r  bb′ + g BB  = 2 + = 76.451 Ω
×
+
 ( R + R BB )  ( 75 10.183 10 3 )


g
3
×
r R 1 1.188 10 × 76.451
′
b e FH
R FH = = = 71.828 Ω
×
3
r
( b e ′ + R FH 1 ) ( 1.188 10 + 76.451 )

×
3
R R 1.6 10 × 75
C L
R = = = 71.641 Ω
out )
( R + R L ( 1.6 10 + 75 )
×
3
C
C b c ′ = C ob = 0.65 pF
×
100.972 10 − 3 − 12
×
C b e ′ = 9 − ( 0.65 10 ) = 1.823 pF
×
×
×
2 3.14 6.5 10
C = 1.823 10 − 12 + 0.65 10 − 12 { 1+ ( 100.972 10 − 3 × 71.641 )} = 7.174 pF
×
×
×
T
1 1
F = =
H
2 Rπ FH T 2 3.14 71.828 7.174 10 − 12
×
×
×
×
C
F = 309.019 MHz
H
=
F == = 309.019 MHz # ก 250 MHz
- '" "
H
ก

ก
ก