power output formula
 

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?

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

On Sun, 3 Oct 2004 09:16:53 +0100, "Ian White, G3SEK"
wrote:

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.

Okay, gentlemen, I can see where you're coming from now.
Incidentally, the (UK) definition above could be construed to allow
for some really serious QRO if one takes "normal operating conditions"
to refer to *atmospheric* conditions rather than those of the station
set-up. When's the next sunspot minima? :-}
--

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

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
=======================================

It's also very handy on 160 meters.

For example, if the antenna is just a 15-feet length of wire and 400 watts
PEP are fed into it over 115 feet of 600-ohm single-wire transmission line,
who needs concessions?

A 15-feet length of wire, all by itself, is quite efficient on 160 meters.
Nearly all of the 400 watts fed into it will be radiated and the licensing
regulations are not violated.

Furthermore, because a transmission line of that particular length
accurately does the impedance matching, a tuner becomes redundent.
----
Reg.

On Sun, 3 Oct 2004 09:16:53 +0100, "Ian White, G3SEK"
wrote:

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"

Hang on a minute, Ian!
I've just looked in your book and in section 6-5 you say in a passage
on Peak Envelope Power: "PEP is the RMS RF power level at the peak of
the modulating waveform." "RMS"? Which is it: RMS or AVERAGE??
Not getting confused, are you? ;-}
--

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

Paul Burridge wrote:
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"

Hang on a minute, Ian!
I've just looked in your book and in section 6-5 you say in a passage
on Peak Envelope Power: "PEP is the RMS RF power level at the peak of
the modulating waveform." "RMS"? Which is it: RMS or AVERAGE??

It's not a term I would use any more; but if the first sentence hadn't
tired you out, the rest of that sidebar would have told you exactly what
I meant by it.

Not getting confused, are you? ;-}

No, just getting more careful about writing things that can be
selectively misquoted :-(


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

On Sun, 3 Oct 2004 09:16:53 +0100, "Ian White, G3SEK"
wrote:


|
|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.


When our FCC changed the power limit from DC input to rf output I
assumed that the feedline was part of a "distributed" amplifier output
matching network and moved my Bird sensor to the antenna feedpoint.

Feedline loss? What feedline loss? [g]

"Reg Edwards" wrote in message
...

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
=======================================

It's also very handy on 160 meters.

For example, if the antenna is just a 15-feet length of wire and 400 watts
PEP are fed into it over 115 feet of 600-ohm single-wire transmission
line,
who needs concessions?

A 15-feet length of wire, all by itself, is quite efficient on 160 meters.
Nearly all of the 400 watts fed into it will be radiated and the licensing
regulations are not violated.

Furthermore, because a transmission line of that particular length
accurately does the impedance matching, a tuner becomes redundent.

=================================

But to keep things in proportion -

To radiate 400 watts from a 15-feet antenna wire, fed via a 115-feet,
600-ohm, single-wire, overhead transmission line, would require a
transmitter power of the order of 1.4 Megawatts.

So before ordering the materials to construct such an antenna and feedline
it would be better to forget all about it!

sorry smiley
----
Reg

On Sun, 03 Oct 2004 06:54:50 -0700, Bill Turner
wrote:

On Sun, 03 Oct 2004 03:29:17 GMT, 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.

_________________________________________________ ________

Right you are, but I'd like to know where the definition of average
power comes from. Is it the IEEE?

If the formal definition says VRMS x IRMS = average power, I suppose I
could live with that, but until then, I think it equals RMS power.

Sources, please.


Rms voltage and current are also called "the effective values".

If you have heard that average power is 1/2 of peak power we could
investigate why.

The rms voltage and rms current of a sine wave are found by
multiplying peak voltage by .707. Same for current.
If you multiply .707 by .707 that gives you .5 or 1/2.

1 volt peak Ac voltage times 1 volt peak / 1 ohm = 1 watt peak power.

1 volt peak times .707 = .707 rms volts. .707 rms volts times .707 rms
volts = .5 / 1 ohm = .5 watts average power.


"It takes twice the Ac peak power to provide the same amount of heat
as it does average DC power." Therefore 1/2 the peak Ac power is equal
to it's average power.

Note that rms voltage is defined as the amount of Ac voltage that will
cause the same amount of heating in a resistor as an amount of DC
voltage.


Also note that when describing rms voltage and its heating effects,
that it does not say amount of power required to do the same amount of
heating. It says the amount of rms voltage to do the same amount of
heating in a resistor. (thus, effective voltage) This is where many
get confused.

If you want to know the amount of power you must square the voltage
and divide by the resistance.
There is that .707 x .707 = .5 again for Ac power. (E squared /R)

2000 ARRL handbook 6.6 chapter 6, RMS VOLTAGES AND CURRENTS.

73
Gary K4FMX

On Mon, 04 Oct 2004 07:54:11 -0700, Bill Turner
wrote:

On Sun, 03 Oct 2004 16:33:46 GMT, Gary Schafer
wrote:

Rms voltage and current are also called "the effective values".

large snip

_________________________________________________ ________

You missed my question. Who defines these terms? I know that RMS,
average and effective are often used interchangeably, but who or what
organization says there is "no such thing as RMS power", and why?

No I didn't miss your question. I thought I gave you enough
information so that you could figure out the "why" yourself.

I don't know of any organization that says "there is no such thing as
rms power". What would be the point. That would be like formally
declaring "water doesn't run uphill", since there is no proof or
evidence that water runs uphill.

I also don't know of any organization that says there is such a thing
as rms power although you will often find it referred to in many
articles and advertisements. Even some of the older ARRL handbooks may
call it rms power but I don't think you will find that in the newer
versions.

Most any AC circuit theory will explain how values of a sine waves are
found and the relation of those values. The reference I quoted from
the ARRL handbook does very nicely at explaining it. It is worth a few
minutes to read. Only a couple of paragraphs will get you there. (you
will see the absence of rms power mentioned)

The thing to consider is that once you multiply an rms value by
another rms value the result is no longer an rms value.
If it were then you could use any of the rms factors to convert it to
peak etc.

Consider 1 volt peak x .707 = .707 volts rms. .707 volts x .707 volts
= .5. Divide that by 1 ohm and you have 1/2 watt.

Using rms factors, convert the rms values to peak. .707 volts x 1.414
= 1 volt peak. OK.

Now try and find peak power like you would find peak voltage from a
known rms value.
Do that with the power value that you want to call rms power and you
have .5 x 1.414 = .707 watts. Not the 1 watt peak you were looking
for.

73
Gary K4FMX

Yadda, yadda, yadda


The term "RMS POWER" while not technically correct for anything practical is
tossed about and I suspect it is 'meant' to mean true or average power as
generally understood by those schooled in the field. One of the reasons for
formal training (or understanding of that training) is so we have
terminology which we have in common. One word or phrase relates to the same
concept for everyone in the discussion.


"Peak-to-peak power" is quite meaningless. P-P voltage and current can be
measured, but power is a second order quantity requiring voltage and current
and an in-phase component as well...and multiplication of these quantities
(that's what makes it second order). This takes care of the phase
relationships and when the voltage goes negative, the current does (the
in-phase component does) and two negatives make a positive and you again
have positive power. There is NO amount of voltage or current which occurs
which is the P-P value. This is only in the eye of the beholder who chooses
to relate two parts of a waveform which occur at different times.


Question:

In a pulsed situation like a common bridge rectifier, capacitor input filter
DC supply, can the true power be determined by measuring the Irms and Vrms
(on the AC side) and doing P = Irms X Vrms ???

Anybody sure enough to say???


--
Steve N, K,9;d, c. i My email has no u's.

Ask a sensible question.
What is "true power"
Where in the circuit is it located.

Steve Nosko wrote:

The term "RMS POWER" while not technically correct for anything
practical is tossed about and I suspect it is 'meant' to mean true or
average power as generally understood by those schooled in the field.

It was somewhat in that sense that I was trying to use the term about 15
years ago. But I was shot down for a number of reasons... with all of
which, I now agree.

1. When used without any qualifier, "power" always means the value
averaged over one or more complete cycles, so it is redundant and
confusing to add "RMS" where in fact no qualifier is necessary. This is
the strongest technical reason for not using it.

2. I was using "RMS power" to mean exactly what Steve says: "average
power as generally understood by those schooled in the field", or "power
averaged by the correct use of the RMS method". However, that attempt to
condense a much bigger concept into two words just does not work.

It's even worse than a simple failure to communicate. The mere mention
of "RMS power" is guaranteed to hijack the whole discussion, and the
point you actually wanted to make will be lost forever.

3. Any term that is used in the "hi-fi" industry - even by the so-called
"good guys" - is automatically tainted.

In the end, I decided that using "RMS power" was causing *many* more bad
effects than good... and after 15 years, I'm still being punished for it
:-)

Those are the reasons why I now come down firmly on the side of "just
call it power".


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

On Tue, 05 Oct 2004 08:53:31 -0700, Bill Turner
wrote:

On Mon, 04 Oct 2004 17:45:56 GMT, Gary Schafer
wrote:

Consider 1 volt peak x .707 = .707 volts rms. .707 volts x .707 volts
= .5. Divide that by 1 ohm and you have 1/2 watt.

Ok so far.


Using rms factors, convert the rms values to peak. .707 volts x 1.414
= 1 volt peak. OK.

OK so far.


Now try and find peak power like you would find peak voltage from a
known rms value.
Do that with the power value that you want to call rms power and you
have .5 x 1.414 = .707 watts. Not the 1 watt peak you were looking
for.

You used the wrong factor. To convert RMS power to peak power (sine
wave only) the factor is 2, not 1.414. The peak power in your example
is actually one watt, just as expected.


Your answer is what my point is all about. 2 is not an rms factor.
Nowhere do you use 2 to find an rms value. But 2 is used when
associated with average values.

As Steve said in another post, power is often referred to as rms power
as it is derived from rms voltage. But it is important to realize that
there is a difference even though it may not be stated.

If it were really rms power then you should be able to find peak power
by multiplying by 1.414 as you do with rms voltage. But that doesn't
work because power is not an rms value.

73
Gary K4FMX

People who use the term "rms power" are braggarts, attempting to show-off
how psuedo scientifically and engineering-ly educated, politically-correct,
and clever they are.

Whereas all they accomplish is to display to the mostly sensible world their
abysmal ignorance.

Articles in Audio Hi-Fi magazines fall below even lower standards of silly
journalism. There ought to be a law against it !

There's only one sort of "power" and that is simply "power". I have serious
doubts about such vague and indeterminate notions as "reflected" power.

In all my experience I have never found any use for such terms.
----
Reg.

In article , "Reg Edwards"
writes:

People who use the term "rms power" are braggarts, attempting to show-off
how psuedo scientifically and engineering-ly educated, politically-correct,
and clever they are.

Whereas all they accomplish is to display to the mostly sensible world their
abysmal ignorance.

Articles in Audio Hi-Fi magazines fall below even lower standards of silly
journalism. There ought to be a law against it !


While it's a misnomer, there's a reason the audio folks use the term "RMS
power".

Back when stereo was hi-fi, amplifiers were simply rated in watts output. The
test method was simple: you fed a sine-wave audio tone into the amp and
measured the continuous output power into a matched resistor load. A "50 watt"
amplifer could deliver 50 watts of audio power continuously.

Then a bunch of things happened.....

First off, there was stereo. Audio amplifiers became 2 channel, usually with a
common power supply. Manufacturers found they could get more power if only one
channel was operating - particularly if the power supply was skimpy to begin
with. So they tested that way, and doubled the result.

Then came solid state. Some early designs could deliver amazing levels of power
into very low impedance loads - for a short time, anyway. So the idea was born
to test the amps by feeding a pulsed signal rather than a continuous tone, and
using a very low impedance dummy load.Plus one channel at a time. Plus
measuring peak power, not average power. With big capacitors in the power
supply, and all the tricks optimized, some amazing power levels could be
measured.

Of course you'd never realize such power in actual operation, but the
manufacturers rationalized that music wasn't sine waves anyway.

End result was that an amp which would have been rated as maybe a "20 watt
stereo amplifier" under the old scheme (two 20 watt channels operating
simultaneously) could be rated as high as "200 watts IHF music power" or some
such. All sorts of terms were invented, mostly to favor a particular test
method.

The serious audio folks coined the term "rms power" to mean the old way of
actually feeding a continous tone to all channels and measuring the continuous
audio power developed.

73 de Jim, N2EY

Then came solid state. Some early designs could deliver amazing levels of
power
into very low impedance loads - for a short time, anyway. So the idea was
born
to test the amps by feeding a pulsed signal rather than a continuous tone,
and
using a very low impedance dummy load.Plus one channel at a time. Plus
measuring peak power, not average power. With big capacitors in the power
supply, and all the tricks optimized, some amazing power levels could be
measured.

Of course you'd never realize such power in actual operation, but the
manufacturers rationalized that music wasn't sine waves anyway.

End result was that an amp which would have been rated as maybe a "20 watt
stereo amplifier" under the old scheme (two 20 watt channels operating
simultaneously) could be rated as high as "200 watts IHF music power" or
some
such. All sorts of terms were invented, mostly to favor a particular test
method.

That is similar to how computer speakers can get 50 to 100 watts rating out
of a wall cube and a look inside will see the speakers rated around 2 watts.

Ralph Mowery wrote:

[snip discussion of "IHF power" and other specious ratings]

That is similar to how computer speakers can get 50 to 100 watts rating out
of a wall cube and a look inside will see the speakers rated around 2 watts.

I have frequently had a good laugh at 50 to 100 watt computer speakers
fed by a single 12 VDC 750 mA wall-wart, not to mention stereo amps
with ratings plates showing 120 VAC 300 mA for an amp supposedly rated
at 100 W per channel. WHat color is the sky in _their_ universe?

--
I think anything that comes from M$ can safely be called a "coprogram".
Once it's fossilized, it's just a coprolite.
-- me, in the Monastery

On Wed, 06 Oct 2004 08:24:08 -0700, Bill Turner
wrote:

On Tue, 05 Oct 2004 17:13:54 GMT, Gary Schafer
wrote:

If it were really rms power then you should be able to find peak power
by multiplying by 1.414 as you do with rms voltage. But that doesn't
work because power is not an rms value.

_________________________________________________ ________

Here's where you are going wrong. You are only multiplying the VOLTAGE
by 1.414 and forgetting to multiply the CURRENT by 1.414. Look at the
entire formula:

P(pk) = VRMS x 1.414 x IRMS x 1.414

You can multiply the two 1.414 factors together and get - voila - 2.

So the formula becomes

P(pk) = VRMS x IRMX x 2.

And that's where the 2 comes from. What you were doing is multiplying
the peak voltage by the RMS current, which is incorrect.

Make sense?


oooooow. I think I will give up here. I made too many posts on this.

Once it becomes power it is no longer an rms value.

73
Gary K4FMX

On Wed, 06 Oct 2004 10:22:52 -0700, Bill Turner
wrote:

On Wed, 06 Oct 2004 16:17:56 GMT, Gary Schafer
wrote:

Once it becomes power it is no longer an rms value.

_________________________________________________ ________

To be unambiguous, the word "power" needs to have an adjective attached,
like "DC power", "RMS power", "Peak power", "Real power", etc, etc.

Lots of times, people get lazy and say only "power", but that leaves it
open to mis-interpretation. I'm guilty of it too.

How about "average power" the correct term.

73
Gary K4FMX

1- The term "RMS power" has no mathematical or electronically correct
meaning. It is technically meaningless. On the other hand, we often use
terms to "imply" things within a certain circle. This is very common and
accepted by many.

RAM is random access memory, but so is ROM (random access, that is) We all
know what is meant.

Though Gary used the term "RMS Power" in his math description, It is a
strange usage. Perhaps the "audio folks" coined it for some reason, but it
is not a technical term in circuits, electrical Engineering or what ever you
want to call serious electronics fields. HOWEVER, he did the correct math
and he did enough explaining that it was clear what he meant by it. I
maintain it is the meaning that is important and it is important to look
past the differences in tech-speak some may have. Discuss the concept --
work to arrive at the meaning and don't quibble over terminology if at all
possible. Too many time is see flames which jump between concept and
terminology choice. Pick a common term and move on.

Not to long ago, it was common to call average power "true power". If
circuit conditions are kept as they are for long enough, this is the power
which goes wherever. I'll leave the philosophical (sp) discussion of just
what power this is to Reg.

B- There is "Peak Power". Typically, this is the power at the peak of the
waveform (often sinewave, but not necessarily). This power DOES occur at
that time. The peak power is there. Voltage is there and current is there
at the same time and power happens. For "Peak-to-peak" it ain't so.

There is never any peak-to-peak voltage which is present at an instant
causing a peak-to-peak current to flow at that instant resulting in a
peak-to-peak power.

B2 - We usually talk about the power over one cycle because that is enough
time to know what it really is over the long term (assuming the wave sticks
around for the long term)...

3- I disagree. An adjective or modifier isn't "needed" but it sure can help
if there may be confusion as to just what the subject is.

D- What Gary said is true. "Once it becomes power it is no longer an rms
value". This is because RMS values of voltage and current are simply tools
we use to arrive at the true, average power...call it what you like We have
to because there ain't no gizmo which measures the power directly (I'll
ignore all the discussion about thermal means [or whatever] actually
measuring the power).

5- What Gary said is false - "If it were really rms power then you should be
able to find peak power by multiplying by 1.414 as you do with rms voltage.
But that doesn't
work because power is not an rms value."
--- NOPE ---
RMS has a specific meaning (in this context) and it ain't in regard to
power. It is applied to the voltage and / or current when that is what you
want to use to find power. The part about "multiplying by 1.414" can not be
applied to power. This is an invalid extension of the math behind RMS...not
applicable...apples and oranges. Applying the wrong formula to the right
(or wrong) situation.


73
--
Steve N, K,9;d, c. i My email has no u's.


"Bill Turner" wrote in message
...
On Tue, 5 Oct 2004 19:03:48 +0000 (UTC), "Reg Edwards"
wrote:

There's only one sort of "power" and that is simply "power". I have
serious
doubts about such vague and indeterminate notions as "reflected" power.

__________________________________________________ _______

Is that so? Then what do you call the instantaneous power which occurs
at the top of a cycle of AC. I call it "peak".

--
Bill W6WRT

How about:
average power =

where V and I are understood to be the effective or rms values of the
voltage and current.

http://hyperphysics.phy-astr.gsu.edu...c/powerac.html
--

"Bill Turner" wrote in message
...
On Wed, 06 Oct 2004 18:15:11 GMT, Gary Schafer
wrote:

How about "average power" the correct term.

__________________________________________________ _______

This will be the third time I've asked for an official source for this
"correct" term. If there is no reply, I shan't be asking again.

--
Bill W6WRT