Page 194 - 30105-2003 การวิเคราะห์วงจรอิเล็กทรอนิกส์ความถี่สูง

186
4.

*
*& 4.21 0 - +% # / 0 .5(4%$ %
×
×
R
V CC B 2 12 15 10 3
+ , ก
ก (2.1) V TH = = = 4.285 V
3
×
( R B 1 + R B 2 ) ( 27 10 + 15 10 3 )
×
×
×
3
R R 2 27 10 × 15 10 3
1 B
B
ก
ก (2.2) R TH = = = 9.642 kΩ
3
( R B 1 + R B 2 ) ( 27 10 + 15 10 3 )
×
×
V − V ) (4.285 0.6− )
( TH
ก
ก (2.3) I = BE = = 56.847 µA
B
+
×
R TH + ( β + ) 1 R E ( 9.642 10 3 ) (89 620 )
×
F
ก
ก (2.4) I = β F B 88 56.847 10 − 6 = 5.003 mA
×
×
I =
C
}
ก
ก (2.5) V = V − I R + ( β + ) 1 I R
CE CC { C C F B E
) (
×
×
×
V CE = 12 − ( { 5.002 10 − 3 × 620 + 89 56.847 10 − 6 × 620 )}
V CE = 5.761 V
V − V )
−
( CC FD 1 12 0.6
ก
ก (4.10) I D = I D = = = 4.905 µA
1 2 R + ) 6 ( 3 )
×
×
×
( 1 2R L 2.2 10 + 2 62 10
* I = == = 56.847 µA, I = == = 5.003 mA, I D 1 = = = = I D 2 = = = = 4.905 µA .5(V CE = == = 5.761 V/ V
CE
C
B
ก5 6 V ! , 4%
4.2.7.5 ก
(

) C B , C #% C
E
C

$% ก
%
C - +
F #
B
L
1
ก
ก (4.18) F ( L C B ) = 2 Rπ FLCB B
C
1
% & C = (4.25)
B
2 Fπ R
L C ) FLCB
( B
1
ก
ก (4.19) F =
( L C E )
π
2 R FLCE C E
Ω
ก
%
R FLCE = 0.5 , F ( L C ) = F L ;
E
1
% & C = C = (4.26)
C
E
2 Fπ R
( L C
E ) FLCE
, C .5( C
*& 4.22 (
C B E C
1
+ , ก
ก (4.25) C =
B
π
R
2 F ( L C B ) FLCB
ก

ก
ก