Page 89 - 30105-2003 การวิเคราะห์วงจรอิเล็กทรอนิกส์ความถี่สูง

81
2.

( - ก .
R TH = 15R (2.26)
E
ก ก (2.2a) V = I R + V + I R
TH B TH BE E E
V TH = I R + V BE + ( β + ) 1 I R (2.27)
B E
B TH
F
R
V CC TH V CC TH ; (2.28)
R
R B 1 = , R B 2 =
V V − V )
TH ( CC TH
,
2.13 ก .
R C , R E , R R
B
2
1
B
- .
ก ก (2.21), (2.22) (2.23)
×
=
V = 0.5 12 6 V
CE
×
V RC = 0.35 12 = 4.2 V
=
×
V = 0.15 12 1.8 V
RE
×
I 2.5 10 − 3
I = C = = 20.833 µA
B
β 120
F
×
0.35V 0.35 12
ก ก (2.24) R = CC = = 1,680 Ω
C − 3
×
I C 2.5 10
×
0.15V 0.15 12
ก ก (2.25) R = CC = = 714.061 Ω
E − 6
( β + ) 1 I B 20.833 10 (120 1+ )
×
F
×
=
ก ก (2.26) R TH = 15R = 15 714.061 10.710 kΩ
E
ก ก (2.27) V = I R + V + ( β + ) 1 I R
TH B TH BE F B E
×
×
V TH = ( 20.833 10 − 6 × 10.710 10 3 ) + 0.6

×
×
+ ( 121 20.833 10 − 6 × 714.061 )
V TH = 2.623 V
×
×
R
V CC TH 12 10.710 10 3 48.997 kΩ
ก ก (2.28) R = = =
B 1
V TH 2.623
×
×
V CC TH 12 10.710 10 3
R
R = = = 13.705 kΩ
B 2 − ) )
V
( CC V TH (12 2.623−
Ω Ω
Ω Ω Ω
R = == = 1.68 k , R = =Ω Ω E = = 714.061 , R B 1 = = = = 48.997 k , R B 2 = = = = 13.705 k ;
Ω
Ω
Ω
Ω Ω Ω
Ω Ω Ω
C
2.1.5.3 ก
)+

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E
(
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