Page 68 - 30105-2003 การวิเคราะห์วงจรอิเล็กทรอนิกส์ความถี่สูง

60
2.

V = I R + V + I R (2.2a)
TH B TH BE E E

(
- ก I = I + I C , I = β F B , I = (β + ) 1 I
I
;
C
B
E
E
B
F
V TH = I R + V BE + ( β + ) 1 I R
B E
F
B TH
) }
V − V = I R + ( β + 1 R
TH BE B { TH F E
V − V )

$
I = ( TH BE (2.3)
B
R TH + (β + ) 1 R E
F
2.1.1.2 ก
)
#
*
I (+#V
CE
C
ก , ก I
V ก ,% 2.5 " ก
C CE

I = β I (2.4)
C F B
V CC = I R + V CE + (β + ) 1 I R
C C
F
B E
}
I R +

$
V CE = V CC − { C C ( β + ) 1 I R (2.5)
B E
F
, I V
,
2.1 ก ,% 2.1 , V TH , R TH , I B C CE
×
×
R
V CC B 2 12 10 10 3
- .
ก ก (2.1) V = = = 3 V
TH ) 3 3
×
×
( R B 1 + R B 2 ( 30 10 + 10 10 )
3
×
×
R R 2 30 10 × 10 10 3
B
1 B
ก ก (2.2) R = = = 7,500 Ω
TH ) 3 3
×
×
( R B 1 + R B 2 ( 30 10 + 10 10 )
(V − V ) (3 0.6− )
ก ก (2.3) I = TH BE =
B
3
×
R TH + (β + ) 1 R E 7.5 10 + (100 1 470+ )
F
I = 43.66 µA
B
I =
ก ก (2.4) I = β F B 100 43.66 10 − 6 = 4.366 mA
×
×
C
}
I R +
ก ก (2.5) V CE = V CC − { C C ( β + ) 1 I R
B E
F
×
V CE = 12 − ( { 4.366 10× − 3 × 1 10 3 ) ( 101 43.66 10+ × × − 6 × 470 )}
V = 5.561 V
CE
Ω Ω
V TH = = = = 3 V, R TH = = = = 7.5 k , I = =Ω Ω B = = 43.66 µA, I = == = 4.366 mA,V CE = = = = 5.561 V;
C
ก

ก
ก