p33_015.pdf
(
67 KB
)
Pobierz
Chapter 33 - 33.15
1
/
2
π
√
LC
, th
e small
er value of
C
gives the larger value of
f
.Consequently,
f
max
=1
/
2
π
√
LC
min
,
f
min
=1
/
2
π
√
LC
max
,and
f
min
=
√
C
max
√
C
min
=
√
365 pF
f
max
√
10 pF
=6
.
0
.
(b) An additional capacitance
C
is chosen so the ratio of the frequencies is
r
=
1
.
60 MHz
0
.
54 MHz
=2
.
96
.
Since the additional capacitor is in parallel with the tuning capacitor, its capacitance adds to that
of the tuning capacitor. If
C
is in picofarads, then
√
C
+365pF
√
C
+10pF
=2
.
96
.
The solution for
C
is
(2
.
96)
2
(10 pF)
(2
.
96)
2
C
=
(365 pF)
−
1
=36pF
.
We solve
f
=1
/
2
π
√
LC
for
L
. For the minimum frequency
C
= 365 pF + 36 pF = 401 pF and
f
=0
.
54 MHz. Thus
−
L
=
1
(2
π
)
2
Cf
2
=
1
10
6
Hz)
2
=2
.
2
×
10
−
4
H
.
(2
π
)
2
(401
×
10
−
12
F)(0
.
54
×
15. (a) Since t
he
frequency of oscillation
f
is related to the inductance
L
and capacitance
C
b
y
f
=
Plik z chomika:
kf.mtsw
Inne pliki z tego folderu:
p33_002.pdf
(59 KB)
p33_011.pdf
(64 KB)
p33_005.pdf
(66 KB)
p33_018.pdf
(73 KB)
p33_017.pdf
(71 KB)
Inne foldery tego chomika:
chap01
chap02
chap03
chap04
chap05
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