Page 191 - 30105-2003 การวิเคราะห์วงจรอิเล็กทรอนิกส์ความถี่สูง

183
4.

F T ( B ) q
k T
ก
ก (2.20) I C (dc ) =
π
F
R FH H (2 F C R + ) 1
′
b c out
T
$% ก
% F = 500 kHz, R FH = R g , C b c ′ = C ob = C .5( R out = R
H
C
re
ก 5! ก I (0 5! ก ก I 4% ก
ก (2.20) .5( I (
C C (dc ) C (dc )
*
(0 4
ก 0.5I C (MAX) !
5 %9 # / 0
$% ก
% V CE = 0.5V CC ,V RC = 0.25V CC ,V RE = 0.25V CC ; (4.22)
I
I = C
B
β F
0.25V CC
R = (4.23)
C
I
C
0.25V 0.25V
R = CC ≈ CC (4.24)
E β +
( F ) 1 I B I C
*& 4.19 ก
% I
C
F T ( B ) q
k T
+ , ก
ก (2.20) I C (dc ) =
R F ( 2 F C Rπ + ) 1
′
FH H T b c out
Ω
$% F = 500 kHz, R = 75 , C = C = C = 1.3 pF, R = R = 620 Ω;
H FH b c ′ ob re out C
F T ( B ) q
k T
I C (dc ) =
R FH H ( 2 F C Rπ T b c out + ) 1
F
′
300 10 × 25.7 10 − 3
×
6
×
I C (dc ) =
) }
6
×
×
×
×
×
×
75 500 10 3 ( { 2 3.14 300 10 × 1.3 10 − 12 × 620 + 1
I C (dc ) = 81.635 mA
)
(

−
3
×
×
! ก I C (MAX)
30 mA I 5! ก (0 0 ก 0.5 30 10 ก - 15 mA ก$ *
C
ก
%
R
620 Ω

C
0.25V
ก
ก (4.23) R = CC
C
I C
×
0.25V 0.25 12
I = CC = = 4.838 mA
C
R C 620
* 5! ก I = == = 4.838 mA
C
4.2.7.2 ก
(

) R C , R E , R B 1 , R B 2 , R R #% R
,
2
L
1
/ 0 (0 4% -ก 0 & % , &
ก (. 5 53ก 0 *
+
*
ก

ก
ก