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Measuring output watts with an oscilloscope

Jon Reck W8REA
 

I also have a measurement question - please help out an electronics novice...

I would like to accurately measure the wattage my QCX radio is putting out, with and without my brand new 50W amplifier but I do not have a suitable watt meter.
However, I do have a brand new digital siglant oscilloscope that I don't really know how to use yet. The scope does not have a built in 50 ohm impedance input, just the 1 meg ohm.

Best I can figure from the internet, I should place my 10x probe across my dummy load and read the RMS voltage, square it, and divide by 50 ohms. I've done that with my QCX 20 with varying results that I don't trust so I bet I'm doing something wrong. Also, I am concerned that I might damage the scope if I do that with the output of the 50 watt amplifier.

Any pointers?
Thanks,
Jon Reck W8REA

ajparent1/KB1GMX
 

Applying the scope across the dummy load is safe as its measuring voltage across
the resistors and that is what scopes do.

Make note that you are then measuring Peak to Peak voltage (very top of
waveform to very bottom).  IF you divide it by 1.414 you get RMS voltage
and that is used to calculate power.

so for 50W into 50 ohms that should be 50V (RMS) or 70.7V P-P.

Power is than V(rms) squared then divided by the resistance of your dummy load (usually 50 ohms).

P=V^2/R    where    R=50 ohms 

As an auxiliary project look into making a Dummy load with a power meter.  THat adds
a diode to render DC and then a meter to display the power.  Look around the net for a
design you like. 

Another project is a SWR (tadnam match ) that reads directly in watts.  Again consult 
the net for a design or kit that you like.

Allison
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Kārlis Goba
 

50W into 50 ohms is indeed 50V rms, but that would give you 141 V peak-to-peak.

Just mind that the peak-to-peak voltage (Vpp) is not the same as peak voltage (aka amplitude or maximum swing from zero) (Vp), and for a sine wave is double the peak voltage.

Vpp = 2*Vpp
Vrms = Vp / 1.414 = Vpp / 2.828

--
Karlis YL3JG

ajparent1/KB1GMX
 

Vpp is not equal to Vpp*2.

I presume you meant Vpp= 2*Vp.

I'd forgotten that as well, as I use a diode detector (rather than scope) for
power measurement so it is Vp= DC+diode drop [Vp+.06 for silicon). 
Then /1.414 for RMS  (if waveform is sine.).

The problem with scopes is often the attenuator or probes are not
compensated correctly and the result is higher or lower voltage than actual.

Allison
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jjpurdum
 

I've always seen the asterisk used to indicate multiplication and the carat (^) used for power. Did that change somewhere along the line and I missed it??

Jack, W8TEE

On Monday, January 20, 2020, 4:10:48 PM EST, ajparent1/KB1GMX <kb1gmx@...> wrote:


Vpp is not equal to Vpp*2.

I presume you meant Vpp= 2*Vp.

I'd forgotten that as well, as I use a diode detector (rather than scope) for
power measurement so it is Vp= DC+diode drop [Vp+.06 for silicon). 
Then /1.414 for RMS  (if waveform is sine.).

The problem with scopes is often the attenuator or probes are not
compensated correctly and the result is higher or lower voltage than actual.

Allison
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Jon Reck W8REA
 

Thank you, this is very helpful. And, yes, I think I will make a project of adding a power meter to my dummy load. It seems very manageable in scope and I bet I would learn from the process.

Jon W8REA

geoffrey pike
 

Typo, 0.6V for silicon
cheers
GeoffGI0GDP

On Monday, 20 January 2020, 21:10:50 GMT, ajparent1/KB1GMX <kb1gmx@...> wrote:


Vpp is not equal to Vpp*2.

I presume you meant Vpp= 2*Vp.

I'd forgotten that as well, as I use a diode detector (rather than scope) for
power measurement so it is Vp= DC+diode drop [Vp+.06 for silicon). 
Then /1.414 for RMS  (if waveform is sine.).

The problem with scopes is often the attenuator or probes are not
compensated correctly and the result is higher or lower voltage than actual.

Allison
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Jon Reck W8REA
 

So, following your suggestions, using my scope, I get an output of 3.36w from my QCX20 at 13.8 volts.
With my 50w amp connected, also at 13.8 volts, I get 14.05w, RMS voltage measured across the resistor in my Palstar dummy load. That's with silver mica capacitors in the QCX low pass filter.  (measured with a cheap MSJ QRP watt meter, I gained about 1 watt by replacing the kit capacitors).

I felt good about these numbers. I think maybe the key to consistent results was to re-set the scope to default before testing. The probe was properly compensated.

Jon W8REA

ajparent1/KB1GMX
 

Jack,

That's how its being used, * multiply, /divide, ^ raise to power.

Just trying to keep it real, anything else is irrational or imaginary.

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jjpurdum
 

I though that maybe I missed something an the position of the operator changed its interpretation. As you know, I'm into software mostly, with a Make White Smoke EE degree.

Jack, W8TEE

On Monday, January 20, 2020, 9:47:24 PM EST, ajparent1/KB1GMX <kb1gmx@...> wrote:


Jack,

That's how its being used, * multiply, /divide, ^ raise to power.

Just trying to keep it real, anything else is irrational or imaginary.

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w9ran
 

To simplify the watts calculation, just measure peak-to-peak voltage across a 50 ohm load with your (properly calibrated) oscilloscope, square that value, and divide the result by 400.

73, Bob W9RAN


On Mon, Jan 20, 2020 at 12:45 PM, ajparent1/KB1GMX wrote:
Make note that you are then measuring Peak to Peak voltage (very top of
waveform to very bottom).  IF you divide it by 1.414 you get RMS voltage
and that is used to calculate power.

so for 50W into 50 ohms that should be 50V (RMS) or 70.7V P-P.

Power is than V(rms) squared then divided by the resistance of your dummy load (usually 50 ohms).

P=V^2/R    where    R=50 ohms 

Andy Brilleaux <punkbiscuit@...>
 

Might be of use .. https://sites.google.com/site/andyg0ftd/qrp-wattmeter

Or if you can find one  find the Yaege funky little power meters come freq counters are GREAT !

https://www.ebay.co.uk/itm/YAEGE-FC-1-Portable-Frequency-Counter-10Hz-2-6GHz-/300558579779

These things are the dogz danglerz, work to 10Hz, have a TXCO and are accurate enough to measure
properly for WSPR / QRSS needs.

And damn cheap too.

73 de Andy

Andy Brilleaux <punkbiscuit@...>
 

In power mode.

Kārlis Goba
 

Hi Andy,

that's a good find, but do you know what power range the YAEGE FC-1 can measure (or safely handle)? I cannot seem to find a spec.

--
Karlis YL3JG

Andy Brilleaux <punkbiscuit@...>
 

Will measure 10w but NOT recommended for more than about 10 seconds.

You press the mode button, and it read S for dbm field strength, C = counter mode and mv, P = power.

FC1 version has a built in Lion cell.
FC1+ has three AAA cells.

The one in the link shows 3 x AAA but appears to be labelled as a plain FC1.

As always, listings on ebay can sometimes be inaccurate so take care.

There is a software menu offset for the freq counter, so when the xtal eventually ages you can recalibrate it too.

On HF you can measure 0.1Hz resolution by changing the time base.

Perfect for when you want to see wspr tones changing etc.


Jon Reck W8REA
 

Thank you, Andy

I don't understand one of the symbols on the schematic- looks like a capacitor labeled 0 micro 1-  does that mean 0.1 microfared?

Jon W8REA

Michael Clarke
 

A simple handy method, using Oscilloscope to get Volts peak to peak, and across 50 Ohm dummy load
Power in watts across a 50 ohm dummy load
approximately equals Vp-p squared divided by 100
Examples (50 ohm load)
Vp-p = 45v indicates about 5 watts
Vp-p = 64v indicates about 10 watts
Vp-p = 90v indicates about 20 watts.
73 de Mick Mi5MTC

Bill Cromwell
 

Hi,

I did the calculations using your formula and it did *NOT* work out that way. It was a usable approximation if I plugged in Volts peak instead of volts peak to peak.

73,

Bill KU8H

On 1/21/20 11:27 AM, Michael Clarke wrote:
A simple handy method, using Oscilloscope to get Volts peak to peak, and across 50 Ohm dummy load
Power in watts across a 50 ohm dummy load
approximately equals Vp-p squared divided by 100
Examples (50 ohm load)
Vp-p = 45v indicates about 5 watts
Vp-p = 64v indicates about 10 watts
Vp-p = 90v indicates about 20 watts.
73 de Mick Mi5MTC
--
bark less - wag more

Andy Brilleaux <punkbiscuit@...>
 

On Tue, Jan 21, 2020 at 04:06 PM, Jon Reck W8REA wrote:
does that mean 0.1 microfared
Yes, 0.1uf.
Not mega critical, it just damps the meter a bit.
For the resistor,I use one of those planar things that can handle 5w easily. They look like the picture below (generic photo).
The RF choke is not normally required if you stick to using it as a 0-50Mhz device.
I've attached a rough spreadsheet calculator for voltage vs power to this message if it's of any use.

The BAS50 diode is the best choice, low capacity and tests certainly show it to be more favourable than the generic
1N34A Germanium or 1N4148 types often used in basic probes.

Germanium diodes are crap really, far too temperature sensitive.
A hot summers day and you can kiss goodbye to any meaningful operation.
(Any wonder why the are not used anymore) ;-)

Have fun.



ajparent1/KB1GMX
 

Mick,

Using Vpp^2 divided by 100 is not correct as your 5W example is..
45^2=2025  divide that by 100 and you get 20.25, decidedly disagrees.

45Vpp is 5W no question there.  

Long way around is 45Vpp is 22.5V peak.  RMS is .707 times that
or 15.907V RMS.

If we take that and divide by 50(ohms) we get .318 amps
knowing voltage  already worked out as 15.907 and multiplying
by current into that 50 ohm resistor we get 5.608 watts.

That both the long way and proof using ohms law and arithmetic.

So if we want 50W power we need 50V(RMS) across 50 ohms
as that nets 1A current.  50V RMS is 70.7V peak or 141V P-P.

Or we can take 45/2.828- 15.912V rms if we square that we get 253.2
and divide by 50 we get 5.064.

Any differences are due to precision used and rounding.

The key is if one uses Volts peak we are talking peak watts
and depending on waveform the average power can be less
to very much less.

The test is apply X power measure dummy loads temp after it hits stability (stops heating).
then remove RF power and replace with DC and increase voltage till the resistor achieves the
same temperature (allow time to stabilize).  THat DC voltage corrorsponds to RMS Votlage
and across a resistor the same power (volts times amps).

This is a worth wile exercise for any electronics experimenter to do.  I've use light bulbs
and a photo meter (in a closed box) to prove this as well.  Every case the same RF power
and Same power lights (heats the filament) the light bulb the same.

Allison



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