Page 184 - 30105-2003 การวิเคราะห์วงจรอิเล็กทรอนิกส์ความถี่สูง

176
4.

π
π
V o ( -2 ) ! . % 0ก
R , ( -2 ) (
4% ก
ก (4.16)
π
π
L
1 ก - R
L
% & V o ( -2 ) = I R (4.17)
π
π
d
L
1
*& 4.13 (
I .5(V
! ก
% E = 25 mV
π
π
d 1 o ( -2 ) i peak
E g r R
′
i m b e C
+ , ก
ก (4.16) I =
d 1 )
R +
R i ( -2 ) ( C r + R L
π
π
d
1
×
(k T q ) 1.38 10 − 23 × (273 25+ )
$% r = B = = 4.762 kΩ
d
×
×
1 I D 1 1.60 10 − 19 × 5.397 10 − 6
Ω
Ω
r b e ′ = 461.981 , R = 560 ;
C
Ω
R i ( -2 ) = 57.008 k , E = 25 mV peak ;
π
π
i
( 25 10 − 3 × 216.459 10 − 3 × 461.981 560 )
×
×
×
)
I d 1 = 57.023 10 3 ( 560 4.762 10 + 56 10
×
3
×
×
+
3
I d 1 = 400.369 nA peak
ก
ก (4.17) V o ( -2 ) = I R
π
π
d
L
1
$% I d 1 = 400.369 nA peak
×
×
V o ( -2 ) = I R = 400.369 10 − 9 × 56 10 3
π
π
d
L
1
V o π π = 22.420 mV peak
( -2 )
* I d 1 = = = = 400.369 nA peak ,V o ππ π = = = = 22.420 mV peak ;
π
π ππ
π
( -2 )
4.2.4 ก

!!" ก
#$$%&
*& 3* ,
4.2.4.1 ก
(

) F
( L C B )
$% C ( - +

0 %% 0 $% ,
5 กก F (
X
B ( L C B ) C B

ก -
0 0 C /
0 % ก5 ! R ก 1 4.9 (4%
B FLCB

ก
R E ( d + R L )
r
R EP = 2
( R + r d 2 + R L )
E
R R 2
E
R EP 1 =
R + R 2
E
ก

ก
ก