Page 162 - 30105-2003 การวิเคราะห์วงจรอิเล็กทรอนิกส์ความถี่สูง

154

3.

1
ก !ก (2.30) C =
E
π
2 R FLCE F ( L C E )
1
ก !ก (2.31) C =
C
π
2 R FLCC F ( L C C )
3.7.4.5 ก
"
+
F
H
I I
ก !ก (1.9) g = C (dc ) = C (dc )
m
V T (k T q )
B
β
ก !ก (1.10) r b e ′ = o
g
m
R R
C L
ก !ก (2.9) Z out = R out = )
R +
( C R L
R R 2
B
1 B
R =
BB )
( R B 1 + R B 2
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 ′
1
ก !ก (2.17) F =
H
π
2 R FH T
C
3.7.4.6 ก
"
+ A
V (F R )
)
V o (0.5 g Z r R
′
m out b e BB
ก !ก (3.31) A = − = −
V (F R ) E i 1 R CF 2 ( R BB + r bb′ + r b e ′ ) R BB ( bb′ + r b e ′ )
+
r
"#
3.23 ก + 3.21 ก
, "- /(,. . !(ก01 & '
k T
F T ( B ) q
- ,
ก !ก (2.20) I =
C (dc )
π
R F ( 2 F C R + ) 1
′
FH H T b c out
Ω
ก
"
F = 5.5 MHz, R = R = 75 , R = R = 75 ;
Ω
H FH CF 2 out L

C b c ′ = C ob = 1 pF, F = 2 GHz,T = 25 ;
T
A
2 10 × 25.7 10 − 3
×
×
9
I C (dc ) =
) }
×
×
×
×
×
×
9
75 5.5 10 6 ( { 2 3.14 2 10 × 1 10 − 12 × 75 + 1
I C (dc ) = 64.163 mA
β o T
V
ก !ก (3.33) I C (dc ) =
Z − R + r )
( incf CF 2 bb′
Ω
Ω
Ω
ก
"
R CF 2 = 75 , Z incf = 600 , r bb′ = 2 ,V = 25.7 mV;
T
×
×
85 25.7 10 − 3 4.15 mA
I C (dc ) = =
(600 75 2− + )
ก

ก
ก