Page 174 - 30105-2003 การวิเคราะห์วงจรอิเล็กทรอนิกส์ความถี่สูง

166
4.

R  E N 
R = L  i 33 − V FD  − R
L
2
V o π ) N 13 2 
( -2π 

ก
%
R = R 2x ,V o ( -2 ) = V ox ;
π
π
2
R  E N 33 
i
L
% & R 2x =  − V FD  − R (4.9)
L
V ox  N 13 2 
$% R ! 0 0 R 1ก 5 . 5 0
ก 5 . 5 V
2x
ox
2

V ! . % 4==D 0ก
R 0
0 ก
ox L
*& 4.4 ก 1 4.1,4.2 .5(4.3
! E = 180 mV peak ,V FD 1 = 0.35 V .5(V FD 2 = 0.25 V
i
ก. (
V o (0- ) .5(V o ( -2 )
π
π
π

#.
-.0 R .5( R !
V = V
π
π
π
1 2 o (0- ) o ( -2 )

+ , ก. (
V o (0- ) .5(V o ( -2 )
π
π
π
 E N V FD 


ก
ก (4.6) V o (0- ) = I R = R L  i 23 − 1 
π
D
L
R +
R +
1
  ( 1 R L ) N 13 ( 1 R L )  
 − 3 
×
3 

V o (0- ) = 1 10  180 10 × 79 − 0.35 
×
(
π
+
+
×
×
 7 500 1 10 3 ) ( 500 1 10 3 ) 


(
3
×
×
×
V o (0- ) = 1 10 1.354 10 − 3 − 233.333 10 − 6 ) = 1.12 V peak
π
 E N V FD 


ก
ก (4.7) V o π π = I D R = R L  i 33 − 2 
L
( -2 )
2
  N 13 ( R + R L ) ( R + R L )  
2
2
 − 3 
×
3  180 10 × 79 0.25 

×
V o ( -2 ) = 1 10  −
(
π
π
×
+
+
×
 7 500 1 10 3 ) ( 500 1 10 3 ) 


(
×
3
×
×
V o ( -2 ) = 1 10 1.354 10 − 3 − 166.666 10 − 6 ) = 1.187 V peak
π
π
* V = = = = 1.12 V ,V = = = = 1.187 V ;
π π π
π
( -2 ) π π
π
o (0- ) peak o π ππ π π peak

#.
-.0 R .5( R
V ก -V $% 0 5%V
ก -V
π
π
π
π
π
π
1 2 o (0- ) o ( -2 ) o ( -2 ) o (0- )
 R  E N   

ก
ก (4.9) R 2x =  L  i 33 − V FD   − R
L
  V ox  N 13 2  
$% V ox = V o (0- ) = 1.12 V,V FD = 0.25 V;
π
2
 1 10  180 10 − 3 × 79   
×
3
×

3
×
R 2x =    − 0.25  − 1 10 = 590.561 Ω

  1.12  7  

* (0 5 R ก %
500 Ω ΩΩ Ω 5 590.561 Ω ≈ ≈Ω ≈ ≈ 600 Ω ΩΩ Ω
Ω Ω
2
ก

ก
ก