Page 158 - 30105-2003 การวิเคราะห์วงจรอิเล็กทรอนิกส์ความถี่สูง

150

3.

1
ก !ก (2.17) F =
H
π
C
2 R FH T
r bb′ b′ C = C b e ′ + C b c ′ (1 g R+ m out )
T
Q 1
r b e ′ C T
E
2 i

R FH
R
R
r
R
R FH = { CF 2 BB + r bb′ ( CF 2 + R BB )} b e ′
+
R
R
r
{ CF 2 BB + r bb′ ( R CF 2 + R BB )} ( R CF 2 + R BB ) b e ′
+ 3.20 ก + %! 9 R
FH
)

ก !ก (2.18) A = (0.707 A − 45
V (F H ) V (F Mid )

ก A = A
V (F Mid ) V (F R )
V o
)
−

A V (F H ) = − = (0.707 A V (F R ) (3.32)
45
E i 1

"#
3.22 ก + 3.17 + %! 9 I B , I C ,V CE , F B W , A V (F R ) , F %T = 25 C
,
H
R
A
×
×
V CC B 2 12 6.9 10 3
R
- ,
ก !ก (2.1) V TH = = = 2.679 V
3
×
×
( R B 1 + R B 2 ) ( 24 10 + 6.9 10 3 )
3
×
×
R R 2 24 10 × 6.9 10 3 5.359 kΩ
B
1 B
ก !ก (2.2) R = = =
TH )
×
3
( R B 1 + R B 2 ( 24 10 + 6.9 10 3 )
×
V − V ) ( 2.679 0.6− )
ก !ก (2.3) I = ( TH BE = = 49.833 µA
B 3
×
R TH + (β + ) 1 R E 5.359 10 + (101 360× )
F
ก !ก (2.4) I = β I = 100 49.833 10 − 6 = 4.983 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 − ( { 4.983 10 − 3 × 840 + 101 49.833 10 − 6 × 360 )}
V CE = 6 V
×
×
I C (dc ) I C (dc ) 4.983 10 − 3 × 1.60 10 − 19
ก !ก (1.9) g = = = = 193.872 mS
m
×
V T (k T q ) 1.38 10 − 23 (273 25+ )
B
β 100
ก !ก (1.10) r b e ′ = o = = 515.804 Ω
×
g m 193.872 10 − 3
ก

ก
ก