Ron wrote:
> Bill made a point that the current to voltage ratio is set when the load
> impedance is 50
> ohms and that is where the current and voltage pickups in the Bird(tm) meet
> equal pickup
> to null an in phase load when measuring reflected power. Even though your
> load is in
> phase , the coupling of voltage and current are not equal at 1000 ohms and so
> you
> reading are the sum and difference of the pick-up from those magnitudes.
> ---
> Ron
>
>
Hi Ron,
Yes that is exactly the point I am saying. Even though the load may be
perfectly flat, no
reflected power, the meter will show what would seem to be reflected power when
there is
none. This is because the meter is set up to read at 50 ohms.
As Mike said "Reflected power is a bit of a red herring".
I chose the 1000 ohms so that it would give a very
exaggerated reading from what we may normally see so as to highlight what is
happening.
The short line section in the watt meter is such a small fraction of a
wavelength at 80 and
even 20
meters as to be negligible to cause a mismatch. It is only but a few inches.
The whole point in this is that the reflected power that is read on the watt
meter can not
be
distinguished from normal errors that the nature of the meter coupling circuit
gives us.
As Mike mentioned, a 1000 ohm load on a 50 ohm circuit gives a 20:1 VSWR. The
100 watt
forward and 80 watt reflected readings could indicate this mismatch but we have
no way of
knowing if there is really any reactive load. the readings are completely
meaningless other
than the ratio.
I found an excellent article that explains directional coupler operation in
August 1964 of
QST
entitled "the monimatch and swr".
It explains that the voltage developed from the current pickup is matched by
the voltage
pickup by the capacitive coupling of the pickup. This is done by the resistor
in the circuit
, the size and distance of the coupling loop from the line. When the two
voltages are
exactly the same the two voltages will add in the forward position.
In the reverse position, because the voltage generated by the current pickup is
180 degrees
from the capacitive pickup voltage they will null in the reverse direction.
This balance can
only be obtained at one impedance. The line impedance that the coupler is set
up for.
As the impedance seen by the coupler changes the two voltages are no longer
equal. This will
show higher forward reading and higher reflected reading as they become
unbalanced.
The big catch is that the voltages become unbalanced no matter if there is
reactance or not.
The line may be perfectly flat, no swr, but as long as the resistance seen by
the meter is
not 50 ohms you will get reflected power indications on the meter. This was
shown to happen
with my 1000 ohm resistor and no line. There is no reflected power from the
resistor but the
meter can not distinguish from a resistance different from it's design
impedance and
reflected energy.
Here are some figures from the article:
For simplicity we will assume that the voltage and current picked up by the
loop are the
same as
those on the line.
50 OHM LOAD:
For 100 watts into 50 ohms V = 70.7 I = 1.41
The voltage generated in the loop by the current is adjusted so that it equals
70.7, the
same as the
voltage picked up by the capacitive coupling.
In the forward position 70.7 and 70.7 volts add to give 141.4 volts. This
voltage will make
the
meter indicate 100 watts.
In the reverse position there is 70.7 volts from the capacitive coupling and
-70.7 volts
from the
current pickup. 70.7 -70.7 = 0 The meter will show 0 watts reverse power.
100 OHM LOAD:
Now with a 100 ohm load and the same 100 watts into the load these are the
voltages in the
circuit:
capacitive pickup V = 100
Current = 1 amp
Current pickup V = 50
Forward position will be 100 + 50 = 150 volts. This will cause the meter to
read 112.5
watts.
In the reverse position the capacitive voltage will be 100 V
The current pickup V will be 50.
100 - 50 = 50 This will cause the meter to read 12.5 watts reverse power.
With a 25 ohm load the 2 voltages (from the capacitive coupling and current
pickup) will be
the
same as for the 100 ohm load but will be opposite from the 100 ohm load. In
other words the
voltage
from the capacitive pickup will be 50 and from the current will be 100. The
meter results
will be the
same when the two add for forward power and subtract for reverse. The meter
will read the
same
112.5 watts and 12.5 watts.
REACTIVE LOAD:
We can have a load of 40 ohms resistive in series with 30 ohms inductive and
get the same
meter readings.
SUMMARY:
The meter can not distinguish between reflected power and a load of a different
impedance
than the design impedance of the meter.
I think this whole discussion started with the question of how much power got
reflected back
to the transmitter with no termination at the watt meter. As we have seen by
some of the
measurements we can have very high reflected power readings on the watt meter
and not have
any reflected power present. We can even have very low power being generated by
the
transmitter and have very high reflected power readings on the meter. And these
loads may be
only resistive loads.
Even if there is a transmission line connected to the meter we still do not
know how much
reflected power we have. Changing the length of line between the meter and a
reactive load
will change the reflected power indication of the meter.
73
Gary K4FMX
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