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Old 14th Sep 2021, 8:57 pm   #1
Skywave
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Join Date: Jun 2006
Location: Chard, South Somerset, UK.
Posts: 7,457
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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