Page 115 - 30105-2003 การวิเคราะห์วงจรอิเล็กทรอนิกส์ความถี่สูง

107
2.

V R V R V
-
I = G 2 . - I = G 1 . I = DD (
- ก R R
)ก ก
1
G
2
G
R G 2 R G 1 (R G 1 + R G 2 )
V R G 1 = V DD − V R G 2
V − V
R = DD R G 2
G 1
I
V R
* ( - ก I = G 2
R G 2
V
V DD − V R ( DD − V R ) R G 2
R = G 2 = G 2
G 1
I V R
G 2
V − V ) R
( DD R G 2 G 2 (2.66)

$
R =
G 1
V R G 2

,
2.31 ก ,% 2.34 ก , R , R R R
,
D S G 1 G 2
−
- .
ก ก (2.61) V R S = 0.586 V P = 0.586 4 = 2.344 V
=
×
ก ก (2.62) V DS = 0.5V DD = 0.5 20 10 V
ก ก (2.63) V = V − V + V ) = 20 − (10 2.344+ ) 7.656 V=
R D DD ( DS R S
"
ก .
. I = 0.5I = 0.5 8 10 − 3 = 4 mA
×
×
D DSS
V R 7.656
ก ก (2.64) R = D = = 1.914 kΩ
D − 3
×
I D 4 10
V R 2.344
ก ก (2.65) R = S = = 586 Ω
S
×
I D 4 10 − 3
ก .
." R G 2 = 47 kΩ
ก ก (2.65a) V = V = 0.293V + 0.586 V ((H3 (/ ' ( +
)
R G 2 G P P
V = {0.293 ( )} ( 0.586 4− 4 + − ) 1.172 V=
R G 2

V − V ) R ) × 3
( DD R G 2 G 2 (20 1.172 47 10−
ก ก (2.66) R G 1 = = = 755.047 kΩ
V R 1.172
G 2
R = == = 1.914 k , R = =Ω Ω S = = 586 , R G 2 = = = = 47 k , R G 1 = = = = 755.047 k ;
Ω
Ω Ω Ω
Ω Ω Ω
Ω
Ω
Ω Ω Ω
Ω Ω
D
2.2.5.3 ก
)+

$ )D

E
'" % A-1
; <
ก ก.
R = 1.914 kΩ (,5
4
R = 2 kΩ
D
D
ก

ก
ก