Page 149 - 30105-2003 การวิเคราะห์วงจรอิเล็กทรอนิกส์ความถี่สูง

141

3.

"#
3.16 & )

.( & '

×
×
V CC B 2 12 22 10 3
R
- ,
ก !ก (2.1) V TH = = = 7.135 V
3
×
×
( R B 1 + R B 2 ) ( 15 10 + 22 10 3 )
×
×
3
R R 2 15 10 × 22 10 3 8.918 kΩ
B
1 B
ก !ก (2.2) R TH = = =
3
×
×
( R B 1 + R B 2 ) ( 15 10 + 22 10 3 )
(V − V ) (7.135 0.6− )
ก !ก (2.3) I = TH BE = = 85.226 µA
B
×
×
R TH + (β + ) 1 R E { 8.918 10 + (121 560 )}
3
F
×
ก !ก (2.4) I = β I = ( 120 85.226 10 − 6 ) = 10.227 mA
×
C F B
ก !ก (3.23) V = 0.5V , V = 0.5V ;
CE CC RE CC
}
V CE = V CC − ( { β + ) 1 I R
F
B E
×
×
V CE = 12 − ( 121 85.226 10 − 6 × 560 )
V = 6.225 V
CE

'
" )

.( & I = == = 85.226 µA, I = == = 10.227 mA %V CE = == = 6.225 V
C
B
8
ก V <( / ก(
± 0.5 V"
+
+ 9 R
R "
!9
B
CE
1
B
2

3.6.5 ก
ก,
.
C BB ,C C BP 1
E

ก
"
X = X = X + %! 0.5 Ω !8 ,.

.' T
C E C BB C BP 1 1

T ก !ก (2.30)
2
1
C =
E
π
2 R FLCE F ( L C E )
C = C = C
E BB BP 1
, C % C
"#
3.17 + %! 9 C BB E BP 1
1
- ,
ก !ก (2.30) C =
E
π
2 R FLCE F ( L C E )
Ω
, R = X = 0.5 , F = 500 kHz;
FLCE C E ( L C E )
1 1
C = = = 0.636 µF
E
π
2 R F × × × × 3
FLCE ( L C E ) 2 3.14 0.5 500 10
" C == = C = = = = C = = = = 0.636 µF
=
E BB BP 1
3.6.6 ก
$
"# ก1 $
&@.* $I

"
J
C = 0.636 µF +;
9 ! & D
C = 0.68 µF
!
E E
ก

ก
ก