Page 221 - 30105-2003 การวิเคราะห์วงจรอิเล็กทรอนิกส์ความถี่สูง

213
5.

5.3.3 ก
5
$ 6 C B , C $ C
E
C
0 C !
C 3. ( 500 kHz ก &( C ( ก " 10 (

B
E
B
* C !
C ( ก " C
3 E 3
C = 10C 3 , C = C 3 ;
B
E
C " ,
500 kHz
ก
ก &( X &( $

C
C
C
0.5-10 Ω
1
C = (5.24)
C
π
2 F X C C
o
) 5.8 $ &( C B , C !
C
E
C
1
* + ก ก (5.24) C =
C
π
2 F X
o C C
0 F = 500 kHz, X C C = 10 ;
Ω
o
1
C = = 0.0318 µF
C
2 3.14 500 10 × 10
×
3
×
×
×
C = 10C = 10 280.952 10 − 12 = 0.0028 µF
×
3
B
C = C = 280.952 pF
3
E
!
$
$- &( /
C = 0.0028 µF $- &( / C = 0.0033 µF ก
B
B
C = 0.0318 µF $- &( / C = 0.033 µF ก
C C
C = 280.952 pF $- &( / C = 270 pF ก
E
E
C = == = 0.0033 µF, C = == = 0.033 µF !
C = == = 270 pF
E
C
B

5.3.4 ก
5
$ 6 β ββ β 1
F
X
ก ก (5.7) β = 1
o
X 2
0 X = L = 256 µH, X = L = 9 µH;
1 1 2 2
×
L 256 10 − 6
β o = 1 = = 28.444
×
L 2 9 10 − 6

&( β % ก * ) *

! (ก( ก ก

o
β o > 28.444 0
. ก$ 80 100
) Q & β = $ 80 100

1 F β o
ก

ก
ก