Page 227 - 30105-2003 การวิเคราะห์วงจรอิเล็กทรอนิกส์ความถี่สูง

219
5.

5.5.1 ก
5
$ 6 C C $ L
1 , 2 3

0 "&( F $- &( & *

!
&( L
3
o
ก ก (5.25) C = nC
1 2
C T ( +n ) 1
ก ก (5.26) C 2 =
n
1

ก ก (5.27) C =
T
L 3 (2 Fπ o ) 2
) 5.13 ก #$ 5.16 $ &( C C !
L ก F = 500 kHz, L = 360 µH
,
3
o
2
1
3
!
n = 1
1
* + ก ก (5.27) C =
T 2
π
L (2 F )
3 o
0 F = 500 kHz, L = 360 µH, n =
1;
o 3
1
C = 2 = 281.733 pF
T
×
×
×
×
360 10 − 6 ( 2 3.14 500 10 3 )
×
C T ( +n ) 1 281.733 10 − 12 (1 1+ )
ก ก (5.26) C 2 = = = 563.466 pF
n 1
ก ก (5.25) C = nC = ( ) 1 563.466 10× − 12 = 563.466 pF
1 2
$
$- &( /
C = C = 563.466 pF $- &( / C = C = 560 pF ก
1 2 1 2
C = == = C = == = 560 pF ก !
$
L = == = 360 µH $- X -TAL = == = 500 kHz
1 2 3 3

5.5.2 ก
5
$ 6 R B 1 , R $ R
B
E
2
. ก I = 5 mA
C
ก ก (3.23) V CE = 0.5V CC ,V RE = 0.5V CC ;
I C
I B =
β
F
0.5V
ก ก (3.24) R = CC
E
(β + ) 1 I B
F
ก ก (2.26) R = 15R
TH E
ก ก (2.27) V TH = I R + V BE + ( β + ) 1 I R
B TH
F
B E
ก

ก
ก