Page 179 - 30105-2003 การวิเคราะห์วงจรอิเล็กทรอนิกส์ความถี่สูง

171
4.

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
F
B E
V CC = I R + V FD 1 + ( D 1 + I D 2 ) R
I
D
1
L
1
I + I = 2I )
D
D
! ก I (0 ก - I ! +% # 4%$ % & 2 0 ก (4% ( D 1 D 2 D 1
1
2
4
5; R
L
V CC = I R + V FD 1 + 2I R L
D
1
D
1
1
V − V = I ( R + 2R )
CC FD 1 D 1 1 L
V − V )
( CC FD 1 ; (4.10)
I = , I = I
D 1 ( R + 2R L ) D 1 D 2
1
, I .5( I
*& 4.9 ก 1 4.5, 4.6 (
I B , I C ,V CE D D
1 2
×
×
V CC B 2 12 13 10 3 4.216 V
R
+ , ก
ก (2.1) V TH = = =
3
×
×
( R B 1 + R B 2 ) ( 24 10 + 13 10 3 )
×
×
3
R R 2 13 10 × 24 10 3 8.432 kΩ
1 B
B
ก
ก (2.2) R TH = = =
3
( R B 1 + R B 2 ) ( 13 10 + 24 10 3 )
×
×
−
( 4.216 0.6 )
ก
ก (2.3) I = = 55.637 µA
B
3
×
8.432 10 + (101 560× )
×
I =
ก
ก (2.4) I = β F B 100 55.637 10 − 6 = 5.563 mA
×
C
I R +
ก
ก (2.5) V CE = V CC − { C C (β + ) 1 I R }
F
B E
) (
×
×
×
V CE = 12 − ( { 5.563 10 − 3 × 560 + 101 55.637 10 − 6 × 560 )}
V = 5.737 V
CE
−
V − V FD ) (12 0.6 )
( CC
ก
ก (4.10) I D 1 = 1 ) =
R +
6
×
×
×
( 1 2R L { 2 10 + ( 2 56 10 3 )}
I D 1 = 5.397 µA
I D 1 = I D 2 = 5.397 µA
* I = == = 55.637 µA, I = == = 5.563 mA, V CE = = = = 5.737 V .5( I D 1 = = = = I D 2 = = = = 5.397 µA
C
B
π ππ
π
4.2.2 ก

!!" ก
#$$%&
ก% (0- )
1
ก 4.7 D ก (. .0 D 4
ก (.
(
*ก 4%

1
2
%
&
!
0 (
# - ก -ก %$% 5( & R , (0- )
*
R i (0- ) BB π
π
ก

ก
ก