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.

BTW, I expected by the size of the power transformer I used,
and how hot the heat sinks get, that it wasn't giving more
than 10w per channel. And for driving a set of compact
speakers to just play music from my computer, that's
good enough.



Micro MegaWatt wrote:
Just a thought -- the 8 ohms is Z impedance -- not R resistance

--
One Watt

To steal ideas from one person is plagiarism;
to steal from many is research.
-- Comedian Steven Wright


"Gary Schafer" wrote in message
...

If you are looking for average power out the formula is E squared/ R
Where E is rms volts. Rms voltage = .707 of peak voltage. Peak voltage
is .5 of PP.

So your power out is around 9 watts.

73
Gary K4FMX


On Tue, 28 Sep 2004 19:16:26 -0400, Ken Scharf
wrote:


I built a small audio power amp to drive my computer
speakers. With one channel driven into 8 ohms the PP
voltage measured at 1khz was 24v before clipping.
With both channels driven it was 20v.

I think the formula for power output was ((vpp/sqr 2)**2)/R
which would give me 36W one channel, and 25w per channel with
both driven.
(not bad for a 2n3055/mj2955 pair running at +- 15 volts)

Is my math correct?