-
Smoothly to increase the d.c voltage up to the level,
when the bulb has the same glow that this one has had at the RF
current.
-
Fix the d.c voltage on PSU's meter.
-
In this case we make a decision,
the d.c voltage has the same level like that RF voltage across
the bulb.
-
Hence it is possible to calculate going from the transmitter
into the RF transformer power. The power is:
P
= V2/R,
Where:
P- a power, going to the transformer,
V-
an RF voltage across input winding,
R-
a resistance of input winding.
We
know both the dissipated power at the load and the power going
into the input winding. So, it is possible to find the efficiency.
Certainly, the indirect method gives us a big error. However it
is a very simple method, and even a beginner
radio amateur can do it.
|
Note: When the incandescent bulb at the input winding is
glowing to its full brightness, the RF power, going to the RF
transformer, is close to 10-15 watts. When the incandescent bulb
at transformer's load is glowing to a full brightness, the dissipated
RF power (in the load) is of 4-9 watts (see Tab.
6). Hence, using the indirect method, and at a full
luminescence of load's bulb of 26V/0.12A, it is possible to find
values of the efficiency in limits of 40-90 percents. To increase
the limits, it need connect a resistor (either the same bulb)
in serial together with the bulb at the input winding.
Reference:
1.
Grigorov
I.N.: Antenna. Matching and tuning.-
Moscow, RadioSoft, 2002.
ISBN: 5- 93037-
087- 7
|
|