Page 189 - 30105-2003 การวิเคราะห์วงจรอิเล็กทรอนิกส์ความถี่สูง

181
4.

r bb′ C = C b e ′ + C b c ′ (1 g R+ m out ) R out
T
r b e ′ V b e ′ g V
m b e ′
I D 1
1 d r
1 d
C T D 2
E g
π
π
π
( -2 ) R E V ( o π − 2 )

1 4.12 . % 4==D ก (. 5 -
0 %% 1 # 1 4.5

r bb′ R FH 1 R FH
B
R FH 2
r b e ′
R g
R BB C
E T
g
β +
( F ) 1 R E

R ≈ R EP ≈ R EP 1
E

1 4.13 . % ก (
R
FH

*& 4.18 (
F # .5(
1 +%

2
H
1
+ , ก
ก (4.21) F =
H
2 Rπ FH C T
( 75 8.432 10 3 ) ( + × 3 )
+
×
×
2 75 8.432 10
$% R FH 1 = ( { R R BB ) r+ bb′ } = = 76.338 Ω
( 75 8.432 10 3 )
g
+
×
R = r + ( β + ) 1 R = 461.981+ (101 560× ) 57.021 k= Ω
FH 2 b e ′ F E
×
×
R FH 1 FH 2 76.338 57.021 10 3 76.235 Ω
R
R = = =
FH )
×
+
( R FH 1 + R FH 2 ( 76.338 57.021 10 3 )
(
×
×
3
r +
R C ( d R L ) 560 4.762 10 + 56 10 3 )
R = 1 = = 554.886 Ω
out ) ( 3 3 )
R +
×
×
+
d
( C r + R L 560 4.762 10 + 56 10
1
I C dc 5.563 10 − 3
×
( )
C b c ′ = C = C ob = 1.3 pF, g = = = 216.459 mS;
m
re
×
(k T q ) 25.7 10 − 3
B
×
g m 216.459 10 − 3 − 12
×
C b e ′ = − C b c ′ = − 1.3 10 = 113.593 pF
π
2 F × × × 6
T 2 3.14 300 10
C = C + C (1 g R+ )
T b e ′ b c ′ m out
C = 113.593 10 − 12 + 1.3 10 − 12 { 1+ ( 216.459 10 − 3 × 554.886 )} = 271.036 pF
×
×
×
T
ก

ก
ก