Page 204 - 30105-2003 การวิเคราะห์วงจรอิเล็กทรอนิกส์ความถี่สูง

196
5.

!
%BB< ก !
"0 3 !""
%2" -%,

'
ก #$ 5.1 &. !
. !

!"" 3. ( 3 0 3 B !
%BB< ก !
"0 3 !""
%2" -%,

5.1.1 ก

$%

& '

ก #$ 5.1 *. ! ก %

#$ I 0 = I r + I Z − I Z
b b e ′
b
b
z
1
1
0 = I b ( b e ′ + Z 1 ) I Z− z 1 (5.1)
r

#$ I I = g V
m b e ′
c
c

#$ I 0 = I Z + I Z + I Z − I Z + I Z
z z 1 z 2 z 3 b 1 c 2
0 = I z ( Z + Z + Z 3 ) I Z− b 1 + I Z
1
2
2
c
. ก I = g V !
V = I r
c m b e ′ b e ′ b b e ′
0 = I ( Z + Z + Z ) I Z− + g I r Z
z 1 2 3 b 1 m b b e ′ 2

0 = I b ( m b e ′ 2 Z 1 ) I+ z ( 1 Z + Z 3 ) (5.2)

Z +
g r Z −
2
∆
b z
ก ก (5.1)!
(5.2)
!
* $ D E ( )* I !
I
( r b e ′ + Z 1 ) ( Z− 1 )
∆ =
( g r Z − Z 1 ) ( 1 Z + Z 3 )
Z +
m b e ′
2
2
∆ = (r b e ′ + Z 1 )( Z + Z + Z 3 ) ( g r Z− m b e ′ 2 − Z 1 )( Z− 1 )
2
1
∆ = (r b e ′ + Z 1 )( Z + Z + Z 3 ) ( g r Z+ m b e ′ 2 − Z 1 )( )
Z

1
1
2

. ก ก

∆ ( ก " 0
∆ = ( r b e ′ + Z 1 )( Z + Z + Z 3 ) ( g r Z+ m b e ′ 2 − Z 1 )( ) 0Z 1 =
2
1
+
g r Z −
Z +
0 = ( b e ′ + Z 1 )( 1 Z + Z 3 ) ( m b e ′ 2 Z 1 )( )
r
Z
2
1
=
0 r b e ′ ( Z + Z + Z 3 ) Z Z+ 1 1 + Z Z + Z Z + g r Z Z − Z Z
m b e ′
1 1
1 3
1 2
1 2
2
1
=
0 r b e ′ ( Z + Z + Z 3 ) Z Z+ 1 2 + Z Z + g r Z Z
m b e ′
1 2
1
1 3
2

0 = r b e ′ ( 1 Z + Z 3 ) Z+ 1 ( Z + Z + g r Z 2 ) (5.3)
Z +
m b e ′
3
2
2
ก

ก
ก