Thanks for the explaination. Now it makes sense how a bridge
amplifier works, it combines the positive and negative
peaks to actually get twice the voltage across the load
from the same power supply voltage.

Hi Ken,

It is fine to measure peak to peak voltage but you can only use peak
voltage to find power. Looking at a scope it is easier to find peak to
peak than it is to find just peak as there is no line or mark where
zero voltage is. Once you find peak to peak then just divide by 2 and
you have peak.

Consider that your positive peak goes from zero to 12 volts and back
to zero. The negative peak goes from zero to - 12 volts and back to
zero. Although there is 24 volts peak to peak they do not happen at
the same time. One is half a cycle later than the other. So you can
only use one at a time. When the positive peak is at +12 volts there
is no negative peak at that time. No power there.

Remember that the horizontal movement of the trace on the scope
represents time. Draw a vertical line down the screen from a positive
peak and you will see that there is nothing on the negative side at
that time. Do the same with a negative peak. Draw a vertical line up
and you will see that there is nothing on the positive side at that
time.

At any given time all you have to work with is either a positive or a
negative voltage. Not both at once.
You can use either the positive peak or the negative peak voltage (1/2
of peak to peak).

By the way, if the amplifier is not perfectly linear the positive and
negative peaks may not be exactly the same. You could have some
clipping on the positive peaks and not the negative peaks as an
example.
Then you don't have a perfect sine wave and you will get an error
trying to find rms voltage too as the .707 only works for a pure sine
wave.

You first take 1/2 of peak to peak to find peak voltage. Then multiply
the peak voltage by .707 to find rms voltage. Then square the rms
voltage and divide that by the resistance. That gives power. 9 watts
in your case.

73
Gary K4FMX

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.

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?