Page 228 - 30105-2003 การวิเคราะห์วงจรอิเล็กทรอนิกส์ความถี่สูง

220
5.

R
V CC TH V CC TH
R
ก ก (2.28) R = , R = ;
B 1 B 2 )
V TH (V CC − V TH
) 5.14 $ &( R , R !
R . ก I = 5 mA
B 1 B 2 E C
* + ก ก (3.23) V CE = 0.5V CC ,V RE = 0.5V CC ;
V CE = (0.5 12× ) 6 V=

V RE = (0.5 12× ) 6 V=

×
I 5 10 − 3
I = C = = 50 µA
B
β F 100
0.5V CC 6
ก ก (3.24) R E = = = 1.188 kΩ
( β F + ) 1 I B 101 50 10 − 6
×
×
3
×
×
ก ก (2.26) R = 15R = 15 1.188 10 = 17.82 kΩ
TH E
ก ก (2.27) V = I R +V + ( β + ) 1 I R
TH B TH BE F B E
×
×
×
×
×
V TH = ( 50 10 − 6 × 17.82 10 3 ) + 0.6 + ( 101 50 10 − 6 × 1.188 10 3 )
V TH = 7.49 V
×
×
R
V CC TH 12 17.82 10 3
ก ก (2.28) R B 1 = = = 28.551 kΩ
V TH 7.49
×
×
V CC TH 12 17.82 10 3
R
R B 2 = = = 47.417 kΩ
(V CC −V TH ) (12 7.49− )
$
$- &( /
R E = 1.188 kΩ $- &( / R E = 1.1 kΩ

R = 28.55 kΩ $- &( / R = 27 kΩ
B 1 B 1

R B 2 = 47.417 kΩ $- &( / R B 2 = 47 kΩ
Ω Ω Ω
Ω
Ω
Ω Ω Ω
Ω Ω Ω
R = == = 1.1 k , R = = = = 27 k , R = = = = 47 k ;
Ω
E B 1 B 2
5.5.3 ก
5
$ 6 C
3
C 3. 500 kHz !ก(0
!
C &( $
(
3 3

C 2 ,. !"( ! * 500 kHz ก& ( 0
,
7ก
10

C
C = 2 (5.31)
3
10

) 5.15 $ &( C
3
C
* + ก ก (5.31) C = 2
3
10
ก

ก
ก