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?

Your formula is off by a factor of 4. 24 vpp across 8 ohms = 9 watts.
Sorry 'bout that.

But who cares -- is it loud enough or isn't it?

Roy Lewallen, W7EL

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?

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?

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?

How do you know?

73
Gary K4FMX

On Wed, 29 Sep 2004 08:07:07 -0700, "Micro MegaWatt"
wrote:

Just a thought -- the 8 ohms is Z impedance -- not R resistance

http://www.epanorama.net/documents/a...impedance.html

Should expain it

--
One Watt

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


"Gary Schafer" wrote in message
...
How do you know?

73
Gary K4FMX

On Wed, 29 Sep 2004 08:07:07 -0700, "Micro MegaWatt"
wrote:

Just a thought -- the 8 ohms is Z impedance -- not R resistance

The article explains how to determine speaker impedance. Not how to
measure amplifier power.

My point was "how do you know" the OP did not use an 8 ohm resistor
for his power measurements. He did not say.

The common notation for power is E squared /R.

E squared /Z will not give you the correct answer unless Z is purely
resistive.

You can not just measure voltage across a complex impedance and
determine power. It is more complicated. You must then also know the
phase angle that the reactance presents along with the resistance.

73
Gary K4FMX


On Wed, 29 Sep 2004 09:19:22 -0700, "Micro MegaWatt"
wrote:

http://www.epanorama.net/documents/a...impedance.html

Should expain it

Gary Schafer wrote: You can not just measure voltage across a complex
impedance and
determine power. It is more complicated. You must then also know the
phase angle that the reactance present along with the resistance.


Exactly my point - if he used the speaker and not an 8-ohm non inductive
resistor as the load

--
One Watt

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


"Gary Schafer" wrote in message
...
The article explains how to determine speaker impedance. Not how to
measure amplifier power.

My point was "how do you know" the OP did not use an 8 ohm resistor
for his power measurements. He did not say.

The common notation for power is E squared /R.

E squared /Z will not give you the correct answer unless Z is purely
resistive.

You can not just measure voltage across a complex impedance and
determine power. It is more complicated. You must then also know the
phase angle that the reactance presents along with the resistance.

73
Gary K4FMX


On Wed, 29 Sep 2004 09:19:22 -0700, "Micro MegaWatt"
wrote:

http://www.epanorama.net/documents/a...impedance.html

Should expain it

On Thu, 30 Sep 2004 09:41:43 -0700, Bill Turner
wrote:

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.

_________________________________________________ ________

Peak-to-peak voltage has no meaning when computing power. There is no
such thing as peak-to-peak power.

You can use either the peak voltage or RMS voltage, and in either case
the formula is E(squared)/R. Your answer will be either peak power or
RMS power.

Since everyone is being a little picky, there is also no such thing as
RMS power. When you use RMS voltage you get average power. :)

73
Gary K4FMX

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?