Measuring wattage

[lowell]

I have 60v p-p unclipped on my speaker output at 8ohms with my scope. This seems insane. (60x60)/8=450watts. I have calibrated the scope. When I use an AC voltmeter it's 20v p-p. Anyone know what's up w/ my scope measurement? It's set at 1v division and the probe is at x10. On the screen it's showing 6v p-p.

[Enzo]

60v p-p is 30v peak. 30v peak is about 21v RMS. 21v across 8 ohms is about 55 watts.

21v RMS is darn close to the 20v on your meter. Your meter is not measuring peak to peak voltage.

[lowell]

damn you're good.

[Robert M. Martinelli]

Hi,
as a foot-note to Enzo's post, the math behind it:

To correctly state the RMS power, you first need to find V(RMS) - to do this, take your Vp-p reading and divide it by 2.8284.

V(RMS) = Vp-p/2,8284

Then divide the value you found by the speaker's impedance ( Z ) to get Irms.

I(RMS) = V(RMS)/Z

to find the RMS power, square I(RMS) first, then multiply it by the speaker's Z.

RMS Power = ((I(RMS)^2)*Z

Et voilà!

Hope this helps

Best regards

Bob

[pdf64]

The RMS value of an ac waveform means it has the same heating potential as dc at that value. Heat being generally what kills speakers.
For a sine wave, the rms value is 0.7071 of the peak value (the peak value being half the p-p value).
For a square wave, the rms value is the same as the peak value.
The above doesn't mean that it's ok to put dc through a speaker!

[Robert M. Martinelli]

Hi Peter,
yeah, I used 2.8284 as a factor just because I wanted to make things easier for him. ( 2.8284 being two times the square root of two, 1.4142...), this way one can simply start with the p-p value he gets on the scope.
Cheers
Bob

[Phostenix]

Hi Peter,
yeah, I used 2.8284 as a factor just because I wanted to make things easier for him. ...

You call that simple? I just use 1.4 & .7, 'cause I'm simpler than you.

[Old Tele man]

...brevity is brief!

[Enzo]

AH yes, complexity.

I once managed a branch operation for a company, and the first time I called in time cards for my employees as a new manager, I did what I had done elsewhere, I calculated hours in tenths of an hour. My staff worked like 21.6 hours or 33.9 hours. I immediately got a call from the book keeper at the home office. "What is all this .6 and .9 stuff? That is far too complicated. We calculate hours here by the quarter hour."

I simply agreed to run by their rules. I never did figure out how quarter hours was simpler then tenths. Too complex to punch in 21.6, but 21.75 is so much easier.

[Robert M. Martinelli]

You call that simple? I just use 1.4 & .7, 'cause I'm simpler than you.

Hi ST,
if you use 1.4142.... as a divider ( or 0.7071... as a multiplier ) you have to start with "peak" value, but he stated he was using the "peak-to-peak" reading, so I included the "divide by 2" factor to get the peak value ( 2.8284.... is simply two times the square root of 2 ), and "spare" him one division....as Enzo says, there are many ways of explaining the same thing, and our goal was to help Lowell, which I tried to the best of my possibilities ( not much, I know.... ). I hope the "picture" is clearer to him now.

Cheers

Bob

[lowell]

Bob I appreciate your help. even if your way seems more complex, it's still fun and nice to know other ways of getting at the same result. so everybody wins!