Page 196 - 30105-2003 การวิเคราะห์วงจรอิเล็กทรอนิกส์ความถี่สูง

188
4.

$% R BB = 9.642 kΩ
( 75 9.462 10 3 ) ( + × 3 )
×
×
+
2 75 9.462 10
R = ( { R R ) r+ } = = 76.410 Ω
( 75 9.462 10 3 )
FH 1 g BB bb′
×
+
R FH 2 = r b e ′ + ( β + ) 1 R = 452.049 + (89 620× ) = 55.632 kΩ
E
F
×
×
R
R FH 1 FH 2 76.410 55.632 10 3 76.305 Ω
R = = =
FH )
( R FH 1 + R FH 2 ( 76.410 55.632 10 3 )
+
×
×
( k T q ) 25.7 10 − 3
r = r d = B = = 5.239 kΩ
d
×
1
2
I D 4.905 10 − 6
1
(
3
×
×
r +
R C ( d R L ) 620 5.239 10 + 62 10 3 )
=
R out = 1 ) ( 3 3 ) = 614.335 Ω
×
+
R +
×
( C r + R L 620 5.239 10 + 62 10
d
1
Ω
C b c ′ = C = C ob = 1.3 pF, g = 194.669 mS, r b e ′ = 452.049 ;
re
m
×
g m 194.669 10 − 3 − 12
×
C b e ′ = − C b c ′ = − 1.3 10 = 102.027 pF
2 F T 2 3.14 300 10 6
π
×
×
×
C = 102.027 10 − 12 + 1.3 10 − 12 { 1+ ( 194.669 10 − 3 × 614.335 )} = 258.796 pF
×
×
×
T
1
F = = 8.063 MHz
H − 12
×
×
×
×
2 3.14 76.305 258.796 10
6

1 +% 2 2F = 2 8.063 10 =16.126 MHz
×
=
×
H

* F = == = 8.063 MHz 1 ก 500 kHz ! , 4% .5(
1 +% 2 4% 16.126 MHz
H

4.2.7.8 ก
*
$ :)%&ก<) !!" '
34 $ *
!
5 %9 # / 0 # ( ,
ก (2.32),
*
(2.33).5( (2.34)0 -
V CC ≤ V CEO
I < 0.5I
C C (MAX)
V I < P
CE C D
*& 4.24 ก 0 - + 5 กK ( 4==D # / 0 *
+ , ก
ก (2.32), (2.33).5( (2.34)
V CC ≤ V CEO
12 V ≤ 20 V
ก

ก
ก