Page 164 - 30105-2003 การวิเคราะห์วงจรอิเล็กทรอนิกส์ความถี่สูง

156

3.

ก
"
+ & ,

+ "

×
×
V CC B 2 12 9.1 10 3 2.792 V
R
ก !ก (2.1) V = = =
TH )
( R B 1 + R B 2 ( 30 10 + 9.1 10 3 )
3
×
×
×
3
×
R R 2 30 10 × 9.1 10 3 6.982 kΩ
B
1 B
ก !ก (2.2) R TH = = =
3
×
×
( R B 1 + R B 2 ) ( 30 10 + 9.1 10 3 )
V − V ) ( 2.792 0.6− )
( TH
ก !ก (2.3) I = BE = = 49.861 µA
B 3
×
×
R TH + ( β + ) 1 R E 6.982 10 + (86 430 )
F
×
ก !ก (2.4) I = β I = 85 49.861 10 − 6 = 4.238 mA
×
C F B
}
I R +
ก !ก (2.5) V CE = V CC − { C C ( β + ) 1 I R
B E
F
) (
×
×
×
V CE = 12 − ( { 4.238 10 − 3 × 1,000 + 86 49.861 10 − 6 × 430 )}
V = 5.918 V
CE
" & )

.( & ' I = == = 49.861 µA, I = == = 4.238 mA %V = == = 5.918 V
B C CE
ก
"
+
F
H
×
I C (dc ) I C (dc ) 4.238 10 − 3
ก !ก (1.9) g = = = = 164.902 mS
m
×
V (k ) q 25.7 10 − 3
T B
β o 85
ก !ก (1.10) r b e ′ = = − = 515.457 Ω
×
g m 164.902 10 3
×
R R 1,000 75 69.767 Ω
C L
ก !ก (2.9) Z = R = = =
out out R + ) (1,000 75+ )
( C R L
3
×
×
R R 2 30 10 × 9.1 10 3 6.982 kΩ
1 B
B
R BB = = =
×
3
×
( R B 1 + R B 2 ) ( 30 10 + 9.1 10 3 )
R
R
r
R 2 BB + r bb′ ( CF 2 + R BB )} b e ′
{ CF
ก !ก (3.29) R FH =
+
R
R
r
R
R
{ CF 2 BB + r bb′ ( CF 2 + R BB )} ( CF 2 + R BB ) b e ′
{ 75 6.982 10 + 2 75 6.982 10 3 )} 515.457
(
×
3
×
+
×
R =
{ 75 6.982 10 + 2 7.057 10 3 )} ( 7.057 10 3 ) 515.457
(
FH
3
+
×
×
×
×
3
537.764 10 × 515.457
×
R FH = = 66.397 Ω
( 537.764 10 + 3.637 10 6 )
3
×
×
1
ก !ก (2.17) F =
H
π
2 R FH T
C
×
g m 164.902 10 − 3 − 12
×
, C b c ′ = C ob = 1 pF, C b e ′ = − C ob = − 1 10 = 12.192 pF;
π
×
×
×
2 F T 2 3.14 2 10 9
×
×
C = C b e ′ + C b c ′ (1 g R+ m out ) 12.192 10= × − 12 + 1 10 − 12 { 1+ ( 164.902 10 − 3 × 69.767 )}
T
C = 24.696 pF
T
ก

ก
ก