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?

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?

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.

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

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

Why not? p-p power is essentially "peak envelope power" isn't it? Are
you saying there's one rule for RF and another for audio??
--

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

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

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

Why not? p-p power is essentially "peak envelope power" isn't it?

No.

--
73 from Ian G3SEK 'In Practice' columnist for RadCom (RSGB)
http://www.ifwtech.co.uk/g3sek

Bill Turner wrote:

On Thu, 30 Sep 2004 23:32:54 GMT, Gary Schafer
wrote:


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. :)


__________________________________________________ _______

Then what do you get when you use average voltage?

IIRC for a sine wave,

VRMS = .707 x peak
VAVG = .615 x peak

--
Bill W6WRT



Misleading answers.

If you _really_ want to get the actual average power delivered to the
load you need to measure the voltage and current, multiply them and
average (or multiply them, average and measure...). Using voltages
assumes that your load is resistive. Using Vrms = 0.707 * Vpeak assumes
sine waves, etc.

--

Tim Wescott
Wescott Design Services
http://www.wescottdesign.com

Paul Burridge wrote:

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


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


Why not? p-p power is essentially "peak envelope power" isn't it? Are
you saying there's one rule for RF and another for audio??

A: There is such a thing as peak-to-peak power, it's just not generally
very useful -- an example would be a 10Vrms sinewave drive to a 10 ohm
capacitor. The instantaneous power going into the capacitor would peak
at 5W and the instantaneous power going out of the capacitor would peak
at 5W. If you defined delivered power as positive and returned power as
negative then the peak-to-peak power would be 10W -- but you would
almost never care.

B: P-P audio power is _not_ the same thing as "peak envelope power", at
least not as defined in the US. Peak envelope power is average power of
the RF at the peak of the envelope -- so if you are running sine-wave
modulated AM with a PEP of 1500 watts into 50 ohms the peak _envelope_
power happens at an _rms_ RF voltage of 274V -- but the _peak_ power
happens at the RF voltage peak of 387V, or 3kW.

--

Tim Wescott
Wescott Design Services
http://www.wescottdesign.com

On Sat, 02 Oct 2004 21:48:00 +0100, Paul Burridge
wrote:

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

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

Why not? p-p power is essentially "peak envelope power" isn't it? Are
you saying there's one rule for RF and another for audio??

The definition of peak envelope power (PEP) is: "The average power
contained in one RF cycle at the crest of the modulation envelope".
(note that the definition says "AVERAGE power" not RMS power) This is
from the FCC definition.

If your SSB transmitter puts out 1500 watts pep with voice it also
will put out 1500 watts with full carrier inserted and no modulation.
(assuming power supplies do not sag)

Same maximum rf power with the carrier only difference is that with
the carrier all the rf cycles are at full power all the time. With
voice the rf reaches the same power levels but it is turned on and off
by the voice. The power levels also vary with the amplitude of the
voice. At full modulation the power levels of the rf cycles reach the
same levels as they do with the carrier. They just don't stay there as
long.

Thus from the definition "average power contained in one rf cycle at
the crest of the modulation envelope". It tells you that the average
rf power is measured at the maximum part of the modulation peak or in
other words the maximum average power that the transmitter reaches on
voice peaks. (if only one rf cycle reaches that power level that
qualifies)

This is one reason why I say that it is important to realize that
there is no such thing as rms power. If you read the FCC definition of
pep it does not say rms power, it says average power. Lots of people
refer to power as rms power.
Lots of people get confused over this thinking pep is something else.

73
Gary K4FMX

On Sat, 02 Oct 2004 21:48:00 +0100, Paul Burridge
wrote:

Why not? p-p power is essentially "peak envelope power" isn't it? Are
you saying there's one rule for RF and another for audio??

PEP is the average power of one RF cycle at maximum modulation.

It is defined as pX in ITU-R radio regulations (RR 151).

Paul OH3LWR

Gary Schafer wrote:

The definition of peak envelope power (PEP) is: "The average power
contained in one RF cycle at the crest of the modulation envelope".
(note that the definition says "AVERAGE power" not RMS power) This is
from the FCC definition.


The definition used by OFCOM, the UK licensing authority, contains the
same words: "The average power ... in one RF cycle at the crest of the
modulation envelope" - but it also contains two useful loopholes.

The full wording is: "The average power supplied to the antenna by a
transmitter during one RF cycle at the crest of the modulation envelope
taken under normal operating conditions."

That means UK amateurs are explicitly permitted to allow for feedline
loss (very handy at UHF and higher) and abnormal transients aren't
counted. Given our 400W PEP output limit, we need all the concessions we
can get.


--
73 from Ian G3SEK 'In Practice' columnist for RadCom (RSGB)
http://www.ifwtech.co.uk/g3sek