Page 198 - 30105-2003 การวิเคราะห์วงจรอิเล็กทรอนิกส์ความถี่สูง

190
4.

( )
V
2 V 2 2 V − ( FD + 2I R )} 2
D
{ CC
P = 1 R = 1 1 L (4.30a)
R
1
R
1 R 1
4.2.7.9.6 ก 5 4==D # R
! P (
4% ก
2 R 2
( )
V
2 V R 2 2 V − ( FD + 2I D R L )} 2
{ CC
P R 2 = 2 = 2 2 (4.30b)
R
2 R 2
*& 4.25 ก 0 - + 5 กK ( 4==D # 0 0
×
( 2 0.25V CC ) 2 ( 2 0.25 12 ) 2
+ , ก
ก (4.27) P R = P R = = = 29.032 mW
E
C
R C 620
1
R = 620 Ω ± 5% 5! ก , # % W (125 mW)
C
8
×
( 2 0.25V CC ) 2 ( 2 0.25 12 ) 2
ก
ก (4.28) P R E = R E = 620 = 29.032 mW
1
R = 620 Ω ± 5% 5! ก , # % W (125 mW)
E
8
V
+
( 2 V R ) 2 2 V − ( BE + V R )} 2 2 {12 − (0.6 3 )} 2
{ CC
ก
ก (4.29) P R B 1 = B 1 = E = 3
×
R
B 1 R B 1 27 10
P R = 5.226 mW
B 1
1
R B 1 = 27 kΩ ± 5% 5! ก , # % W (125 mW)
8
2 V ) 2 ( 2 V BE + V R ) 2 ( 2 0.6 3+ ) 2
( R
ก
ก (4.30) P R B 2 = B 2 = E = 3 = 1.728 mW
×
R
B 2 R B 2 15 10
1
R B 2 = 15 kΩ ± 5% 5! ก , # % W (125 mW)
8
( )
V
2 V 2 2 V − ( FD + 2I R )} 2
D
{ CC
ก
ก (4.30a) P = 1 R = 1 1 L
R 1
R
1 R 1
)}

× ×
2 12 − { 0.6 + ( 2 5 10 − 6 × 62 10 3  2
×
P =     = 0.105 mW
1 R 6
×
2.2 10
P = P
R
R
1
2
1
R = 2.2 ΜΩ ± 5% 5! ก , # % W (125 mW)
1
8
1
R = 2.2 ΜΩ ± 5% 5! ก , # % W (125 mW)
2
8
* (4%
0 % &
ก

ก
ก