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The original query was to "test power level (amps) at antenna?".
Assuming he means watts instead of amps, we still have to figure out what he means
by "at antenna".
If he wants to know the power available at the antenna,
your 50 ohm dummy load and scope will work fine.
If he wants to know how many watts are actually going into the antenna, not so much.
Your scope might work for this if you add a 1 ohm resistor in series to sense the current.
Then measure voltage and current at the antenna, and the phase difference, assuming
you have a two channel scope.
Multiply rms voltage, rms current, and the sine of the phase angle between the two to get real power.
##################################
Allow me a short rant here, as I find the subject interesting,
and key to understanding what a nanovna is.
Composing this note helped me to think about it.
The key passages of the Bruene article
are on the first page:
#######################################
Textbooks tell us that the voltage on a line can be considered to have two components, a forward component ... and a reflected component.
#######################################
An important thing to note is that at any point along the line the reflected components of voltage and current are exaclty 180 degrees out of phase.
#######################################
How the Directional Coupler Works
The directional coupler can sense either the forward or reflected component by taking advantage of the fact that the reflected components of voltage and current are 180 degrees out of phase while the forward components are in phase. A small voltage derived from the current in the line is added to a sample of the voltage across the line. If these two samples have the right amplitude relationship, the two reflected components cancel. The sum then represents only the forward component. By reversing the phase of the current sample 180 degrees, the forward components cancel and the result is the sum of only the reflected components.
#######################################
Those first two items are stated as known textbook facts,
Bruene does not attempt to explain why they are true.
Here is one way to go about it:
Think about 5 volts DC into a 50 ohm resistive load, we would measure 5v/50 = 0.1 amps into the load.
And the power burned by that 50 ohms would be 5*0.1 = 0.5 Watts.
Assume we apply 5 volts DC to a black box, and we measure -0.1 amps, which is to say that the
current is flowing the opposite direction. This means there is something unexpected about the box,
it is somehow sending 0.5 watts in the opposite direction down our wire toward the 5 volt battery.
Perhaps the box has a 10 volt battery inside, in series with a 50 ohm resistor.
That works for AC as well as DC, we might have 0.1 amps rms travelling from the black box to
a 10mhz 5 volt rms signal source. The black box in this case would have to contain a second
10mhz source that is exactly synchronized with our 5 volt rms source.
We can have 10 mhz signals travelling in both directions simultaneously, as is the case when
the black box reflects part of the 10mhz energy back to the signal source.
At this point we have caught up with Bruene, being able to differentiate forward power from reverse power
by noting that reverse power has voltage and current that are 180 degrees out of phase.
The other curiosity in the Bruene article is that these SWR meters add RF voltage to RF current,
which makes about as much sense to a physicist as adding bananas to miles.
For a result of power in watts we would have to multiply voltage and current,
but you can't do a multiply with coils and caps and resistors.
Let's think again about our 5 volts DC driving a 50 ohm resistor, resulting in a 5/50 = 0.1 amp current.
Assume we measure the 0.1 amp current by passing it through a 1 ohm resistor, resulting in 0.1*1 = 0.1 volts across it.
And assume we scale the 5 volts down to 0.1 volts by passing it through a resistive divider of 49 + 1 ohms in series.
Now, if we add the 0.1 volt current reading to the 0.1 volt voltage reading, we get 0.2 volts, but only if the resistor is 50 ohms.
And if the current is going in the wrong direction, we get zero volts when we add the two readings.
If we have currents going in both directions simultaneously, all we see is the forward component.
Jerry, KE7ER
On Sun, Sep 27, 2020 at 10:04 AM, Howard Fidel wrote:

I measure power into my 50 ohm dummy load by transmitting in CW and measuring
the RMS voltage on my scope. P = V*V/50.

The original question was sparked by having a radio (HF frequencies) that no
one can hear transmissions from. The transmission line includes surge
protectors, baluns, and HF coax and connectors. Of course there could be
dozens of different reasons for the problem but I wanted to troubleshoot by
removing the antenna from the surge protector (the “last” component in
line) and measuring for real power there. Yup, the question should have been
watts and not amps.

I have a DVM and can take a measurement between SO-239 inner conductor and
shield, if that will tell me anything.

If I find expected TX power at the last component before antenna, then I’ll
know that none of the transmission line components are bad and that the radio
is producing power. If not, then I’ll work backwards along the transmission
line to zero in on the problem.

On 9/27/20 10:22 AM, David Eckhardt wrote:

This method is highly likely more accurate than the Bird Watt Meters at 5%

of full scale and other wattmeters offered into the amateur market.
Excellent procedure!

Assuming the dummy load is 50 ohms (viz all that discussion on the list about
the loads for the NanoVNA)

Most oscilloscopes don't have very tight tolerances on their amplitude
accuracy.
A Rigol DS1000 (a popular sub kilobuck scope) says:
DC gain accuracy is +/- 3%
(and a bunch of other stats that are basically related to that DC spec) It is,
after all, an 8 bit ADC.

A decent analog scope like the venerable Tek 465B is also 3%

Then you have the input impedance: 1 Meg +/- 2% in parallel with 13 pF +/-
3pF.
13 pF is 122 ohms at 10 MHz - so you use a 10x probe, but then the 2% of the
1Meg against the series Z of the probe starts to be a problem.

5% is about what you'd probably wind up with using the scope + load approach

Dave - W?LEV
On Sun, Sep 27, 2020 at 5:04 PM Howard Fidel <howard@...> wrote:

I measure power into my 50 ohm dummy load by transmitting in CW and
measuring the RMS voltage on my scope. P = V*V/50.

If we have currents going in both directions simultaneously, all we see is the
forward component.

I measure power into my 50 ohm dummy load by transmitting in CW and measuring
the RMS voltage on my scope. P = V*V/50.

On 9/27/20 10:22 AM, David Eckhardt wrote:

This method is highly likely more accurate than the Bird Watt Meters at

5%

of full scale and other wattmeters offered into the amateur market.
Excellent procedure!

Assuming the dummy load is 50 ohms (viz all that discussion on the list
about the loads for the NanoVNA)

Most oscilloscopes don't have very tight tolerances on their amplitude
accuracy.
A Rigol DS1000 (a popular sub kilobuck scope) says:
DC gain accuracy is +/- 3%
(and a bunch of other stats that are basically related to that DC spec)
It is, after all, an 8 bit ADC.

A decent analog scope like the venerable Tek 465B is also 3%



Then you have the input impedance: 1 Meg +/- 2% in parallel with 13 pF
+/- 3pF.
13 pF is 122 ohms at 10 MHz - so you use a 10x probe, but then the 2% of
the 1Meg against the series Z of the probe starts to be a problem.



5% is about what you'd probably wind up with using the scope + load
approach

Dave - W?LEV

On Sun, Sep 27, 2020 at 5:04 PM Howard Fidel <howard@...> wrote:

I measure power into my 50 ohm dummy load by transmitting in CW and
measuring the RMS voltage on my scope. P = V*V/50.

I measure power into my 50 ohm dummy load by transmitting in CW and
measuring the RMS voltage on my scope. P = V*V/50.

Jeff,
You are correct,.
Power delivered to the load is Forward-Power minus Reflected-Power.

On Sat, Sep 26, 2020 at 04:42 PM, Jerry Gaffke wrote:

I'm not so sure any of the SWR bridges would accurately measure power
with an antenna of other than 50 ohms.

Please correct me if I'm wrong, but isn't the power delivered to a load simply
Pfwd - Pref?

- Jeff, k6jca