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Hints, Tips and Solutions (Do NOT post requests for help here) If you have any useful general hints and tips for vintage technology repair and restoration, please share them here. PLEASE DO NOT POST REQUESTS FOR HELP HERE!

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Old 14th Sep 2021, 8:57 pm   #1
Skywave
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Arrow Measuring power in a known load with a 'scope.

Measuring power in a known load with a 'scope - the easy way. *

So you've got your repaired audio amp. on the bench ** and you need to check its output watts when it is connected to its appropriate load - typically a 8 ohm resistor of adequate power rating - and your audio sig. gen. is connected to the amplifier's input.

The common method is to measure the peak-peak voltage across that load R with the 'scope and then do the necessary arithmetic:
1. Measure the Vp-p.
2. Calculate Vpeak: Vp-p ÷ 2
3. Calculate Vrms: Vpeak ÷ 1.414
4. Calculate the watts in R: (Vrms)² ÷ R.
Unless you're outstanding at mental arithmetic, that involves a fair bit of work with a calculator or W-H-Y.

Here's the easy way:
Power in R = (Vp-p)² ÷ 8*R
which for a common 8 ohm load reduces to (Vp-p)² ÷ 64.

A numerical example; method 1 (assumes R = 8 ohms):
1. Vp-p = 10 v.
2. Vpeak = 5v.
3. Vrms = 5 ÷ 1.414 = 3.536
4. Watts in R = (3.536)² ÷ 8 = 1.56.

A numerical example; method 2 (assumes R = 8 ohms) :
1. Vp-p = 10 v.
2. (Vp-p)² = 100
3. Watts in R = 100 ÷ 64 = 1.56.

Now I could have reduced all of the above by only simply stating:
Power in load = (Vp-p)² ÷ 8*R without the details, but I went into detail to justify and illustrate my opening claim and to try to avoid the need for any detailed explanation. (But I am happy to explain if required).
Finally, as to why (Vrms)² ÷ R = (Vp-p)² ÷ 8*R, I leave that up the reader to justify that for himself: it's not difficult.

* All of this post assumes that you do not have an instrument to measure the output power directly - such as the Marconi TF893A power meter.
** All that follows from here is obviously not limited only to audio power amps.: my choice of such is just a typical example.

Al. / Sept. 14th.
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Old 10th Dec 2021, 11:14 pm   #2
Radio1950
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Default Re: Measuring power in a known load with a 'scope.

Hi Skywave

I use this type of measurement a lot in RF, but within the specs of probe and CRO.


Always have to have a calculator ready.

Your math removes one step, which is good.

The reason for the "8" is that Vpk/Vrms = SQR2.
And SQR2^6=8.
This is like saying 1+1=2.


And for 50 ohm circuits, Power directly in milliwatts = 2.5 x (Epk-pk)^2.

For example 2.5 x (0.63 x 0.63)^2 = 1 mW in 50 ohms.



Back to your scenario, if you take freq down towards DC, Epk/Erms approaches "1" and in the limit, power = (E^2)/R, which is familiar.

I wonder who was the first person in history to use that equation?
.
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Old 11th Dec 2021, 1:33 am   #3
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Default Re: Measuring power in a known load with a 'scope.

Quote:
Originally Posted by Skywave View Post
* All of this post assumes that you do not have an instrument to measure the output power directly - such as the Marconi TF893A power meter.
It also assumes that you are using quite an old scope . The modern ones will show an RMS value while displaying the waveform. The only modern scopes I have are USB ones with very limited bandwidth, but OK for audio frequencies.

B
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Old 11th Dec 2021, 4:06 am   #4
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Default Re: Measuring power in a known load with a 'scope.

I thought that we were only allowed to use a "vintage" CRO.
It's prescribed in the Forum Agreement, part iiia, sect b, sub-section 14, dated 1/4/2005.
R1950.
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Old 11th Dec 2021, 1:21 pm   #5
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Default Re: Measuring power in a known load with a 'scope.

I designed this little spreadsheet calculator to quickly take peak to peak scope measurements and show all the results. In the end it is useful for all ohms law and power calculations.

Yes I know there is no such thing as Peak Watts but the term still gets quoted. Please do not turn this thread into another discussion on Peak Power.

Al
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Old 11th Dec 2021, 2:27 pm   #6
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Default Re: Measuring power in a known load with a 'scope.

Quote:
Originally Posted by Alistair D View Post
Yes I know there is no such thing as Peak Watts but the term still gets quoted
There is, though - in a resistive load (which is what's been considered all along in this thread) peak watts is peak volts x peak amps.

For sine waveforms, it's exactly double the average power.

Why could it be important? Well, suppose you are making an efficient 10W audio amp using Class D (switching) techniques, you'd have to be sure the power supply is up to the job. It might need only to supply 10W average to the amplifier stage, but it would have to be capable of supplying 20W without going into current limit or anything nasty like that.
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Old 11th Dec 2021, 4:13 pm   #7
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Default Re: Measuring power in a known load with a 'scope.

Thanks Al, that calculator will come in handy.
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Old 11th Dec 2021, 4:31 pm   #8
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Default Re: Measuring power in a known load with a 'scope.

Quote:
Originally Posted by Radio1950 View Post
I thought that we were only allowed to use a "vintage" CRO.
It's prescribed in the Forum Agreement, part iiia, sect b, sub-section 14, dated 1/4/2005.
R1950.
There would be some very unhappy people on the forum if they had to use vintage test gear on their vintage radios . Much of my own test gear are other people's hand-me-downs .

In my periodic maintenance activity on my AVO VCM, I do find the USB scopes showing RMS voltages damn useful!

B
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Old 12th Dec 2021, 12:57 am   #9
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Question Re: Measuring power in a known load with a 'scope.

Quote:
Originally Posted by Radio1950 View Post
I thought that we were only allowed to use a "vintage" CRO.
It's prescribed in the Forum Agreement, part iiia, sect b, sub-section 14, dated 1/4/2005.
Where do I find this 'Forum Agreement' document, please? I'd simply like to read it.
Al. / Dec. 11th.
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Old 12th Dec 2021, 1:05 am   #10
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Default Re: Measuring power in a known load with a 'scope.

Don't bother, it only gives you the power in foot-poundals per fortnight, or in RAC horsepower. (The latter having no unambiguous conversion factor to anything of any use. Not for discussion here, but look it up)

David
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Old 12th Dec 2021, 1:09 am   #11
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Default Re: Measuring power in a known load with a 'scope.

Quote:
Originally Posted by Skywave View Post
Where do I find this 'Forum Agreement' document, please? I'd simply like to read it.
Al. / Dec. 11th.
It can only be found by Australians, who we know to be standing upside down .

B
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Old 12th Dec 2021, 9:54 am   #12
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Default Re: Measuring power in a known load with a 'scope.

Back on topic please.
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Old 12th Dec 2021, 10:47 am   #13
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Default Re: Measuring power in a known load with a 'scope.

Scopes measure voltage, and that can only be related to power if you know the load impedance (real part) that the voltage is expressed across. If that's an RF mismatch, or a loudspeaker, then the whole issue is a lot more complicated.

From a point of waveshapes, if the waveform is one of the regular shapes, then you can calculate a conversion factor from peak-peak to RMS volts and use that to get the RMS value. For complicated waveforms or ones which keep changing, then a digitising scope with an RMS arithmetic function is the best option.


Scopes as a whole, are not terribly accurate, being in the few percent class at best, but they do tell you a lot in your first eyeful. I got brought up with the habt of first reaching for the scope. Only rarely, if I needed more accuracy did a meter get involved. Multimeters were on their ohms range >95% of the time.

So it all comes down to what sort of power do you want to measure, what waveforms, what frequency ranges etc.

Quite often voltage measurement is all you need. Loudspeakers are made to be as flat as possible with a constant voltage drive, not a constant power input. Their impedance and efficiency goes all over the shop. We try to design audio amplifiers with large damping factors, and that means low output impedances, and that means they try to act like voltage sources.

We don't care a lot about the output impedance of RF amplifiers, but we do care about getting the load applied to them accurately to what the amplifier is intended to drive. People chirp on about the maximum power transfer theorem, but the max power theorem doesn't include irritating little details like whether the amplifier will blow up before you get to that condition, or whether the efficiency is lousy. So RF amps are routinely NOT matched, care is taken to load them correctly into the impedance they were designed to drive. What Z source the amplifier presents ceases to be important, the correct load reflects nothing back down the line, so there is nothing to behave badly if the amplifier Zout isn't the same as the line to the load.

Oscilloscopes are the easiest thing to prod aroound with, but then you have to interpret what you see.

David
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Old 15th Dec 2021, 8:08 am   #14
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Default Re: Measuring power in a known load with a 'scope.

I do a lot of measuring OP power, with an 8r load 3v RMS is roughly 1w, 5v = 3w, 9v 10w, 12v 20w, 20v 50w & 30v 100w, I've got to a point where I can say that's X watts..... on a good day. Other days I'm scratching parts of my anatomy, looking puzzled and trying to figure out where I left the calculator.

Andy.
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Old 15th Dec 2021, 8:27 am   #15
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Default Re: Measuring power in a known load with a 'scope.

I love it.
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Old 15th Dec 2021, 11:05 pm   #16
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Arrow Re: Measuring power in a known load with a 'scope.

Quote:
Originally Posted by Diabolical Artificer View Post
I do a lot of measuring OP power, with an 8r load 3v RMS is roughly 1w, 5v = 3w, 9v 10w, 12v 20w, 20v 50w & 30v 100w, etc.
Andy.
If you are regularly using only 8 ohms as a load and have a 'scope, it's a simple matter to produce a graph that shows the relationship between peak-peak volts and watts in the load.
Al.
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Old 17th Dec 2021, 11:38 pm   #17
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Default Re: Measuring power in a known load with a 'scope.

I just measure Voltage ( 1KHz ) across an 8 Ohm dummy load with my Avo 8.

Then - V/R = I. I x V= Watts!

Seems to work out pretty close .

Or am I missing the intricacies?
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Old 18th Dec 2021, 12:36 am   #18
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Default Re: Measuring power in a known load with a 'scope.

I tend to use 400 CPS , just like my text books said.
I use 1 kHz when playing in the sand pit.
I find doing these measurements very easy as I always have a couple of meters set up to go, right off the bat. One digital just to check, but my preference is an analogue meter for the measurements. The CRO is only to "try" tell me when the waveshape is clipping/squashing/distorting. In any case a few percent is perfectly usable for anybody here ( I would think ), especially in audio. Does it really matter if its 50.25 watts or 49.75 watts ?. After all, to make it twice as loud requires 525 watts !!, although the voicecoils in my speakers, and even the 100 watt dummy load I use would vaporise long before that!!

Joe
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Old 18th Dec 2021, 4:18 pm   #19
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Arrow Re: Measuring power in a known load with a 'scope.

From post #17:
Quote:
Originally Posted by thermionic View Post
Or am I missing the intricacies?
This thread is about measuring power in a known load whilst using a 'scope: it is not about various methods of determining that power.
The big advantage of the 'scope method is that one can see the waveform across the load and unless that waveform is reasonably distortion free, other methods - such as yours - could easily give incorrect results.

Al.
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Old 18th Dec 2021, 9:20 pm   #20
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Default Re: Measuring power in a known load with a 'scope.

Al, I totally agree that using the scope to observe the waveform is essential, and I also do this in addition to the Avo 8. The difference is that I’m not using the ‘scope for voltage measurement as I find my method described above easier for my simple brain to work out!

I take your comment on the title of the thread, and apologise for skewing it somewhat!

Cheers.

SimonT.
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