Page 170 - 30105-2003 การวิเคราะห์วงจรอิเล็กทรอนิกส์ความถี่สูง

162
4.

V − V FD V V FD
2
−
=
I D 2 = ( R + R L 2 ) ( R + 2 R L ) ( R + 2 R L )
2
2
2
E N
! ก V = i 33
2
N 13
V
&
E N
FD
% I D 2 = ( R + i R L 33 ) N 13 − ( R + 2 R L ) (4.3)
2
2
*& 4.1 ก 1 4.2 (
E i (V peak )
! ก
% I D = 1 mA peak
1
R +
I D ( 1 R L ) N + V FD N 13
13
+ , ก
ก (4.2a) E = 1 1
i
N
23
$% E ! . % 4==D D
.ก L # T
i 13 3
Ω
Ω
R = 500 , R = 1 k , N = 7 T, N = 79 T,V = 0.35 V;
13
L
23
1
FD
1
{ 7 1 10 − 3 ( 500 1 10 3 )} (0.35 7+ × )
×
×
+
×
E = = 163.924 mV
i peak
79
* E = == = 163.924 mV peak
i
4.1.1.2 ก
(

) F
R (L 43 )
F !
$/. /* 0 .
L ก % ก (L C ).5(
R (L 43 ) 43 43 CM
0 - # L ! (
# C
43 CM
L 13 = N 2 13
L 43 ( N 23 + N 33 ) 2
L 13 ( N 23 + N 33 ) 2
L 43 = (4.4)
N 2
13
1
ก
ก (3.1b) F =
R
2π LC
$% F = F R (L 43 ) , L = L 43 , C = C CM ;
R

1 (4.5)
% & F R (L 43 ) =
2π
L C
43 CM
1
C CM = 2 (4.5a)
( )
π
L 43 ( 2 F R L 43 )

ก

ก
ก