Page 88 - 30105-2003 การวิเคราะห์วงจรอิเล็กทรอนิกส์ความถี่สูง

80
2.

F T ( B ) q
k T

$
I C (dc ) = (2.20)
π
R FH H (2 F C R + ) 1
F
′
b c out
T
2.1.5.1 ก
ก.
$
I
C
" ( - ก กก I C (dc ) % , ก ก (2.20) #

I C (dc ) " ก&
0.5I C (MAX) (3- , ; %
& (

F T ( B ) q
k T
ก ก (2.20) I C (dc ) =
π
F
R FH H (2 F C R + ) 1
′
b c out
T
ก .
." F = 250 MHz, R FH = R g , R out = R L , C b c ′ = C ob ;
H
,
2.12 , I
C
F T ( B ) q
k T
- .
ก ก (2.20) I C (dc ) =
R F (2 F C Rπ + ) 1
′
FH H T b c out

Ω
* % F = 250 MHz, R FH = R = 75 , R out = R = 75 , C b c ′ = C ob = 0.65 pF,T = 25 C;
Ω
H
g
L
A
k T
F T ( B ) q
I C (dc ) =
R F (2 F C Rπ + ) 1
′
FH H T b c out
×
6.5 10 × 25.7 10 − 3
9
×
I C (dc ) =
) }
9
×
×
×
×
×
×
75 250 10 6 ( { 2 3.14 6.5 10 × 0.65 10 − 12 × 75 + 1
I C (dc ) = 2.979 mA
* ( - ก I = == = 2.5 mA
C
2.1.5.2 ก
ก.
$
R C , R E , R (+# R
B
B
2
1
"

$ %
.
' $
( ก
)
"
"
( +ก( , 2.5 mA ก .

//0 ก ,
$

ก .
V CE = 0.5V CC (2.21)
V RC = 0.35V CC (2.22)
V RE = 0.15V (2.23)
CC
0.35V CC
R = (2.24)
C
I
C
0.15V CC
R = (2.25)
E
( β + ) 1 I B
F
ก

ก
ก