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antenna
resistance
is greater
than
70
ohms.
When
the
antenna
resistance
is less
than
70
ohms,
the
reactance
of L107
should
be equal
to the
capacitive
reactance
of
the
antenna
plus
35
ohms.
2.
Determine
the
inductance
of L IO 7 using
the
formula:
159 x,
L=--fi
3.
From
Figure
6,
determine
the
number
of
turns
to be shorted.
4.
From
Figure
4 or
5, determine
the
required
inductance
of
L106,
using
the
known
frequency
and
antenna
resistance.
5.
Using
Figure
6, determine
the
turns
on L106
to
be
shorted,
to
obtain
the
inductance
required
in Step
4.
6.
Determine
the
required
shunt
capacitance
from
the
known
frequency
and
antenna
resistance,
using
Figure
7.
7.
Using
the
capacitance
from
the previous
step,
and
referring
to Table
5, determine
the
manner
of
connecting
Cl 20,
C 12 1, and
C 122
for
the
closest
approximation
to the required
value.
8.
Determine
the
antenna
current:
I-E--
9.
Determine
the
voltage
across
the
capacitance:
I---
--__-
E,. =I,
\/
35"
+
Ra2
when
Ra is less than
70 ohms
or
E,. =
I,R,
when
R,
is greater
than
70 ohms
IO.
Calculate
the
current
in each
capacitor
used
and
make
sure
it does
not
exceed
the
rating
deter-
mined
from
Table
6.
shunt
As
an
example
of
the
preceding
calculations,
assume
that
the
power
output
from
the
transmitter
is 250
watts
and
that
the
transmitter
is operating
into
an antenna
whose
impedance
is 20 -j30
ohms.
Assume
the frequency
to be
1250
kc.
I.
The
antenna
resistance
is less than
70 ohms,
therefore,
the
reactance
of L107
is 30
+
35 =
65
ohms.
2.
The
inductance
of L 107
is
15 9x,,
159
(65)
L = ------=
------.
=
8.3
microhenries.
f
1250
3.
From
Figure
6, 26 turns
must
be shorted.
4.,
From
Figure
5,
the
inductance
of
LlO6
should
be 3.8
microhenries.
5.
From
Figure
6, 3 1 turns
must
be shorted.
6.
From
Figure
7, the required
shunt
capacitance
is 6200
mmf.
Note
that
since
the
antenna
resist-
ance,
20
ohms,
is
less
than
70
ohms,
the
curve
marked
"ANT
RES.
=
LESS
THAN
70
OHMS"
on Figure
7 is to be used.
7.
From
Table
5, 5900
mmf
is found
to be the
nearest
approximation
to
the
required
6200
mmf.
Therefore,
Cl 20 and
Cl 2 1 should
be connected
in
parallel.
8.
1, =
L/r=
\/see=
3.54
amperes
__
-. -
9.
E,. = I,
35"
+ Ra"
=
3.54
\/--.
~-
35'
+
20"
II
142 volts
IO.
(Cl
20)
I,. = 6.28fCE,
=
6.28
( 1,250,OOO)
(2000)
(IO)
lz (142)
=
2.23
amperes
(C121)
I;. = 6.28
(1,250,OOO)
(3900)
(lo)--"
(142)
=
4.35
amperes
Reference
to Table
6 indicates
that
the
currents
are not
excessive.
Check
the
output
frequency
on
the
frequency
monitor
connected
to J IO I.
Any
slight
frequency
adjustment
necessary
may
be made
by
rotation
of
capacitor
Cl
in the
oscillator
stage.
Potentiometer
Rl 12 adjusts
the
level
to the
monitor.
Modulation
may
be adjusted
by
using
the
I OOO-
cycle
input
signal
connected
to terminals
TBA-2
and
TBA-3.
One
hundred
per
cent
modulation
may
be checked
in any
of the
usual
methods.
Cathode
currents
in the
modulator
are
available
in positions
7, 8, and
9 of
the
METER
switch.
Monitor
level
may
be adjusted
by
setting
the
secondary
taps
on
T 103
as necessary.
When
all
adjustments
are
completed,
operate
all
panel
switches
to the
OFF
position.
15
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