Page 117 - 30105-2003 การวิเคราะห์วงจรอิเล็กทรอนิกส์ความถี่สูง

109
2.

V = V = − 1.108 V
GS GS 1
 V  2 3  − 1.108  2
×
ก ก (2.46) I = I DSS  1− GS  = 7.812 10 −  1−  = 4.083 mA
D
 V P   − 4 
ก ก (2.45) V DD = I R + V DS + I R
D D
D S
I R +
V DS = V DD − ( D D I R )
D S

×
×
+
V DS = 20 − ( { 4.083 10 − 3 × × 3 ) ( 4.083 10 − 3 × 560 )}
2 10
V = 9.547 V
DS
= − −− −
=
) %
(/ V GS == = 1.108 V, I == = 4.083 mA V DS = == = 9.547 V
D
2.2.5.5 ก
)
#
*
C G , C (+# C
S
D

"
* ก .
. C ! ) ก
"
G
1
ก ก (2.54) F =
( L C G )
π
2 R FLCG G
C
$

C = 1 (2.67)
G
2 Rπ FLCG F ( L C G )
1
ก ก (2.55) F ( L C S ) = 2 Rπ FLCS C S
1

$
C = (2.68)
S
π
2 R FLCS F ( L C )
S
* % R FLCS = 0.5 Ω
1
ก ก (2.56) F =
( L C D )
π
C
2 R FLCD D
1

$
C = (2.69)
D
π
2 R F
FLCD ( L C D )
,
2.33 ก ,% 2.34 , C G ,C C
D
S
1
- .
ก ก (2.67) C =
G
π
2 R F
FLCG ( L C G )
3
×
×
R R 2 750 10 × 47 10 3 44.228 kΩ
1 G
G
* % R = = =
GG
×
×
( R G 1 + R G 2 ) ( 750 10 + 47 10 3 )
3
3
+
×
R FLCG = R + R GG = 75 44.228 10 = 44.303 kΩ
g
F = F = 1 MHz
( L C G ) L
ก

ก
ก