On Thu, 30 Sep 2004 20:50:49 -0400, Ken Scharf
wrote:

Hmm, I actually thought the power output was 1/2 of what
I had calculated, so I'm a little confused.

I measured the 24v p-p using an oscilloscope (one of
those new fangled ones that actually let you set cursor
lines on the screen to bracket the waveform and it then
shows the voltage measurement on the screen, in this case
24v pp). I used an 8 ohm non-inductive 50w resistor
as a load (actually two of them, one per channel).

Now if I take 24 volts and reduce it by .707 I get 16.96,
and if I square that 287.9, and divide that by 8
(e**2)/r the result is 36. OK, I'm still missing
something. Maybe .707 * 12v (why HALF the pp voltage?),
yeah that gives just about 9 watts out.

Half pp because that gives you "peak power" which is a defined term
and not up for redefinition. Your P-P RMS power out into 8 ohms is
about 18W (but I've no calculator handy and my mental arithmetic these
days is very, very rusty.
--

"What is now proved was once only imagin'd." - William Blake, 1793.