Page 193 - 30105-2003 การวิเคราะห์วงจรอิเล็กทรอนิกส์ความถี่สูง

185
4.

×
×
V CC TH 12 9.3 10 3
R
R B 2 = = = 14.205 kΩ
(V CC − V TH ) (12 4.144− )
ก$ * R ≥ 100R
C
L
ก
% R = 100R = 100 620 = 62 kΩ
×
L C
V − V FD )
( CC
ก
ก (4.10) I D 1 = ( R + 2R L 1 ) , I D 1 = I D 2 ;
1
−
V CC − V FD 1 12 0.6 ( 3 )
×
×
R = I D 1 − 2R = 5 10 − 6 − 2 62 10
L
1
×
R = R = 2.156 ΜΩ
1
2
Ω
Ω
Μ
Ω
* R = == = 620 , R = =Ω Ω E = = 620 , R B 1 = = = = 26.930 k , R B 2 = = = = 14.205 k , R = =Ω Ω 1 = = R = == = 2.156 ΜΩ
Μ
ΜΩΩ
Ω Ω
Ω Ω
Ω Ω Ω
Ω Ω Ω
;
C
2
R = == = 62 kΩ ΩΩ Ω
L
4.2.7.3 ก
(% 3
* 3 3 (83
*
9 3
R = 620 Ω
0 ' R = 620 Ω
C
C
R = 620 Ω
0 ' R = 620 Ω
E E
R B 1 = 26.930 kΩ
0 ' R B 1 = 27 kΩ
R B 2 = 14.205 kΩ
0 ' R B 2 = 15 kΩ
R = R = 2.156 MΩ
0 ' R = R = 2.2 MΩ
1 2 1 2
R = 62 kΩ
0 ' R = 62 kΩ
L
L
4.2.7.4 ก
*
$ : , 3'
34 $ *
ก 0 - +% # / 0 (ก (
5 ก ,
*
0
0 '
V − V )
( TH
ก
ก (2.3) I = BE
B
R TH + (β + ) 1 R E
F
I
ก
ก (2.4) I = β F B
C
}
I R +
ก
ก (2.5) V CE = V CC − { C C (β + ) 1 I R
B E
F
V − V FD )
( CC
ก
ก (4.10) I D 1 = ( R + 2R L 1 ) , I D 1 = I D 2 ;
1
!
∗

0+ V 4
; % 5 % ก ± 0.5 V
ก ก
5 . 5 # R
R
B
2
B
1
CE
ก

ก
ก