Page 171 - 30105-2003 การวิเคราะห์วงจรอิเล็กทรอนิกส์ความถี่สูง

163
4.

*& 4.2 ก 1 4.1 .5( 4.2 (
C
! E

500 kHz
CM
g
×
L 13 ( N 23 + N 33 ) 2 1.40 10 − 6 (79 79+ ) 2
+ , ก
ก (4.4) L = =
43 2 2
N 13 ( ) 7
L 43 = 713.257 µH

1
ก
ก (4.5a) C CM = 2
( )
π
L 43 ( 2 F R L 43 )
$% F R (L ) = 500 kHz, L 43 = 713.257 µH;
43
1
C CM = 2 = 142.198 pF
×
×
×
×
713.257 10 − 6 ( 2 3.14 500 10 3 )
* C = == = 142.198 pF
CM
4.1.1.3 ก
(

) B
W (L 43 )

ก 1 4.2 (0 - # R 0 C 4% .ก ก R 4
CM g
}
0 .
L ก5 R .5( R (r + R + R ) 4 0 .
L ก5 R
23 gL 23 ( { g L 23 ) d 1 1 L 43 BW
k T ) q
ก
ก (1.20) r = ( B
d
I
D
&
B
% r = r = (k T q ) (4.5b)
d 1 d
I D
$% r !
0 +% # 4%$ % D
d
1 1
ก
ก (3.10) Z P = Z S , Z = Z N 2 P ;
S
P
N 2 N 2 N 2
P S S
$% Z = R BW , Z = R g L ) ( d 1 R + R L ) , N = N 23 + N 33 , N = N 23 ;
r +
S
1
P
P
S
( 23
2
R N 23
g
R =
g L ) 2
( 23
N
13
r
R g L ) ( d + R + R L ) ( N + N ) 2
1
% & R = ( 23 1 × 23 33 (4.5c)
BW ) 2
R ( ( + r + R + R N 23
g L 23 ) d 1 1 L
$% R !
0 0 C 0 .
L
BW CM 43
R g L ) !
0 ก L 4 L 23
13
( 23
ก

ก
ก