Page 205 - 30105-2003 การวิเคราะห์วงจรอิเล็กทรอนิกส์ความถี่สูง

197
5.

. ก Z = JX 1 , Z = JX !
Z = JX
3
2
2
1
3
0 = r b e ′ ( JX + JX + JX 3 ) + JX 1 ( JX + JX + g r JX 2 )
m b e ′
2
3
2
1
2
0 = Jr b e ′ ( X + X + X 3 ) + J X 1 ( X + X + g r X 2 )
m b e ′
1
2
2
3
0 = Jr b e ′ ( X + X + X 3 ) − X 1 ( X + X + g r X 2 )
m b e ′
1
2
3
2
0 = Jr b e ′ ( X + X + X 3 ) − X 1 { X 2 (1 g r+ m b e ′ ) + X 3 } (5.4)
1
2
ก ก (5.4) G , 3 ก $ &( & 0 !
* . &( &

ก ก
0 = Jr b e ′ ( X + X + X 3 )
2
1
Jr
b e ′
0 = X + X + X (5.4a)
3
1
2
ก ก (5.4) 3 ก $ &( g r *

m b e ′
0 = − X 1 { X 2 (1 g r+ m b e ′ ) + X 3 }
X
−
1
0 = X 2 (1 g r+ m b e ′ ) + X 3 (5.4b)
X = − X 3 (5.5)
2
(1 g r+ m b e ′ )
.
ก ก (5.5) X ( 3 ก " X 0 ก &
" !
X 2 < −X
3
2
3
ก ก (5.5)! &( ก (5.4a)

X
0 = X − 3 + X 3
1
+
(1 g r )
m b e ′
X
X = 3 − X 3
1
(1 g r+ m b e ′ )
X − X (1 g r+ ) X − X − X g r
X = 3 3 m b e ′ = 3 3 3 m b e ′
1
+
+
(1 g r ) (1 g r )
m b e ′
m b e ′
g r
X ( m b e ′ )
X = − 3 (5.6)
1
(1 g r+ m b e ′ )
< −X
ก ก (5.6) X ( 3 ก " X !
X
1 3 1 3
ก

ก
ก