Page 159 - 30105-2003 การวิเคราะห์วงจรอิเล็กทรอนิกส์ความถี่สูง

151

3.

×
R R 840 75 68.852 Ω
C L
ก !ก (2.9) Z = R = ( R R ) = = =
out out C L R + ) +
( C R L 840 75
r
R BB ( bb′ + r b e ′ )
ก !ก (3.30) Z = R +
incf CF 2 )
( R BB + r bb′ + r b e ′
3
×
×
R R 2 24 10 × 6.9 10 3 5.359 kΩ
1 B
B
R = = =
BB )
( R B 1 + R B 2 ( 24 10 + 6.9 10 3 )
3
×
×
)
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
, C b c ′ = C ob = 0.5 pF
×
g m 193.872 10 − 3 − 12 3.910 pF
×
C b e ′ = − C ob = − 0.5 10 =
π
×
×
×
2 F T 2 3.14 7 10 9
×
×
C = C b e ′ + C b c ′ (1 g R+ m out ) 3.910 10= × − 12 + 0.5 10 − 12 { 1+ ( 193.872 10 − 3 × 68.852 )}
T
C = 11.084 pF
T
R BB = 5.359 kΩ
)
V o (0.5 g Z r R
′
m out b e BB
A V (F R ) = − = − )
E
i 1 R CF 2 ( R BB + r bb′ + r b e ′ ) R+ BB (r bb′ + r b e ′
×
×
×
×
×
0.5 193.872 10 − 3 × 68.852 515.804 5.359 10 3
A = −
(
V (F R ) 3 3
)
+
75 5.359 10 + + ) 5.359 10 (2 515.804
×
2 515.804 +
×
A = − 5.737
V (F R )
1
ก !ก (3.32) F =
H
π
2 R C
FH T
r b e ′ { CF 2 BB + r bb′ ( CF 2 + R BB )}
R
R
R
, R =
FH + + + )
R
R
R
R
{ CF 2 BB r bb′ ( CF 2 R BB )} r+ b e ′ ( CF 2 R BB
( { 3 ) ( 3 )}
×
+
×
×
515.804 75 5.359 10 + 2 75 5.359 10
R FH = ( { 3 3 )} 3
(
×
+
×
×
+
2 75 5.359 10
75 5.359 10 ) ( + × + 515.804 75 5.359 10 )
×
212.920 10 6
R FH = = 66.231 Ω
( 412.793 10 + 2.802 10 6 )
3
×
×
C = 11.084 pF
T
1
F = = 216.911 MHz
H
×
×
×
×
2 3.14 66.231 11.084 10 − 12
±
−
F R = 5.5 MHz, B W = 60 kHz ( 3 dB);
" I = == = 49.833 µA, I = == = 4.983 mA,V CE = = = = 6 V, F R = 5.5 MHz, B W = 60 kHz ,
± ± ±
±
B
C
=
= − −− −
A == = 5.737, F == = 216.911 MHz;
V (F R ) H
ก

ก
ก