Most of us are building for ourselves and not marketing products so many
common amp specs are not that important. However, I'm interested in how
power is calculated and would like to see some RAT feedback on this
subject.

There are two power calculations:
P=V^2/R (call this 'voltage power')
P=I^2*R (call this 'current power')

Let's take an example of a constant-current Class A output stage:

Load = 4 ohms
Peak-to-Peak (P-P) output voltage before clipping is 8 volts.
Using the voltage power formula:
(8*8)/4=16 watts

However, it's customary to represent voltage with the root mean square
(RMS) value. So 8 P-P becomes 5.66v rms when divided by 1.414 (sq. root
of 2). Now, (5.66*5.66)/4=8 watts

The constant current of the primary is 100ma. The output transformer
has a winding ratio of 35:1 so this calculates to 3.5a at the secondary.

Using the current power calculation:
(3.5*3.5)*4= 49 watts

Q1:Would it make sense to calculate the current power using a reduced RMS
value? So 3.5/1.414=2.48a. Using this value:
(2.48*2.48)*4=25 watts

Q2: In the end, I imagine that the amp power would be described using the
_lower_ of the two calculations. I would say that this is an 8 watt
amp.

Comments?

"Paul D. Spiegel" wrote

...Load = 4 ohms
Peak-to-Peak (P-P) output voltage before clipping is 8 volts.
Using the voltage power formula:
(8*8)/4=16 watts

However, it's customary to represent voltage with the root mean
square
(RMS) value. So 8 P-P becomes 5.66v rms when divided by 1.414
(sq. root
of 2). Now, (5.66*5.66)/4=8 watts...

RMS values of voltage *and* current *must* be used in the general
equation P=V.I, from which both of your variants can be derived
using ohms law.

That is because rms * rms = ms, the mean square. Since power is the
square, the result is mean power. Note that "rms power" is a common
misnomer. It's not a matter of custom, but of mathematics and
coherence.

...The constant current of the primary is 100ma. The output
transformer
has a winding ratio of 35:1 so this calculates to 3.5a at the
secondary.

Using the current power calculation:
(3.5*3.5)*4= 49 watts...

In your current calculation, you have calculated the dissipation due
to the bias, mostly that of the valves. For the dissipation by the
load, which is what you want, you need the current variation due to
the signal. You will find this will give the same answer as your
voltage calculation as long as you use rms current.

cheers, Ian

Ian Iveson wrote:

"Paul D. Spiegel" wrote


...Load = 4 ohms
Peak-to-Peak (P-P) output voltage before clipping is 8 volts.
Using the voltage power formula:
(8*8)/4=16 watts

However, it's customary to represent voltage with the root mean
square
(RMS) value. So 8 P-P becomes 5.66v rms when divided by 1.414
(sq. root
of 2). Now, (5.66*5.66)/4=8 watts...


RMS values of voltage *and* current *must* be used in the general
equation P=V.I, from which both of your variants can be derived
using ohms law.

That is because rms * rms = ms, the mean square. Since power is the
square, the result is mean power. Note that "rms power" is a common
misnomer. It's not a matter of custom, but of mathematics and
coherence.


...The constant current of the primary is 100ma. The output
transformer
has a winding ratio of 35:1 so this calculates to 3.5a at the
secondary.

Using the current power calculation:
(3.5*3.5)*4= 49 watts...


In your current calculation, you have calculated the dissipation due
to the bias, mostly that of the valves. For the dissipation by the
load, which is what you want, you need the current variation due to
the signal. You will find this will give the same answer as your
voltage calculation as long as you use rms current.

cheers, Ian



In addition to Ivan's information, in order to calculate true rms voltage to use
in your power calculation, the rms value of a SINE wave is .707 times the PEAK
voltage. The peak voltage is simply half of the peak-to-peak voltage.
(.707 is 1/1.414. this factor is only valid for sine waves.)

Paul

--
return email paulbabiak at rogers dot com

PS, I wrote:

"...you need the current variation due to
the signal. You will find this will give the same answer as your
voltage calculation as long as you use rms current..."

which is ambiguous. You can't use the supply current variation
because this is not directly related to the signal. Use the ac
current through each half of the primary and the transformed load
resistance or, more simply, the secondary current and load
resistance, to derive power output.

cheers, Ian

On Sun, 17 Jul 2005 18:00:29 +0000 (UTC), "Paul D. Spiegel"
wrote:

Let's take an example of a constant-current Class A output stage:

Load = 4 ohms
Peak-to-Peak (P-P) output voltage before clipping is 8 volts.
Using the voltage power formula:
(8*8)/4=16 watts

Error number one.

However, it's customary to represent voltage with the root mean square
(RMS) value. So 8 P-P becomes 5.66v rms

Error number two.

The constant current of the primary is 100ma. The output transformer
has a winding ratio of 35:1 so this calculates to 3.5a at the secondary.

Using the current power calculation:
(3.5*3.5)*4= 49 watts

Error number three.

Comments?

Really, I think that a primary text would be the most useful place
to start. In the USA, the ARRL handbook, refined over the decades
and available at in public library, is tough to beat.

Good fortune,

Chris Hornbeck

On Mon, 18 Jul 2005 00:06:11 GMT, Chris Hornbeck
wrote:

In the USA, the ARRL handbook, refined over the decades
and available at in public library, is tough to beat.

"at ANY public library." Sorry.

Chris Hornbeck

On Sun, 17 Jul 2005 18:00:29 +0000 (UTC), "Paul D. Spiegel" wrote:

Most of us are building for ourselves and not marketing products so many
common amp specs are not that important. However, I'm interested in how
power is calculated and would like to see some RAT feedback on this
subject.

There are two power calculations:
P=V^2/R (call this 'voltage power')
P=I^2*R (call this 'current power')

Let's take an example of a constant-current Class A output stage:

Load = 4 ohms
Peak-to-Peak (P-P) output voltage before clipping is 8 volts.
Using the voltage power formula:
(8*8)/4=16 watts

Nope - you can only do this if you work in marketing!

However, it's customary to represent voltage with the root mean square
(RMS) value. So 8 P-P becomes 5.66v rms when divided by 1.414 (sq. root
of 2).

Nope - 8 P-P = 4 P = 2.8 vrms

Now, (5.66*5.66)/4=8 watts

2.8 ^ 2 / 4 = 2

The constant current of the primary is 100ma. The output transformer
has a winding ratio of 35:1 so this calculates to 3.5a at the secondary.

Nope - the 100ma is the DC bias and signal input, useless info for you here.

Using the current power calculation:
(3.5*3.5)*4= 49 watts

Q1:Would it make sense to calculate the current power using a reduced RMS
value? So 3.5/1.414=2.48a. Using this value:
(2.48*2.48)*4=25 watts

Q2: In the end, I imagine that the amp power would be described using the
_lower_ of the two calculations. I would say that this is an 8 watt
amp.

The above is a waste of time... it is a 2 watt amp.

Comments?

Back to the drawing board...

wrote:

The above is a waste of time... it is a 2 watt amp.

It is a waste of time at any other level as well, except as a
generalized indication - there is no such thing as any simple voltage
or current measurement & accurate formula for power output of a complex
music signal, unless you use or mock up a calorimeter for its output.

Evil idea: make output power calorimeter kits for audiophiles...

The above is a waste of time... it is a 2 watt amp.


It is a waste of time at any other level as well, except as a
generalized indication - there is no such thing as any simple voltage
or current measurement & accurate formula for power output of a complex
music signal, unless you use or mock up a calorimeter for its output.



** The general aim is to ascribe a max POWER rating to a particular
amplifier under defined load conditions.

You are describing a measurement of the total ENERGY in some unspecified
piece of music programme.

Stick those goal posts back where you found them - Mr anonymous smartarse.

Go figure out the difference between energy and power - anytime.





............ Phil

Bob I think you've nailed it properly.
For a peak ac wave, peak/2 x .707 gives you the rms.
Then take the rms value-square it and divide by load.
So yes 8 volts peak/2 =4x.707=2.82x2.82/4=1.99 watts.!!
Dan
wrote in message
...
On Sun, 17 Jul 2005 18:00:29 +0000 (UTC), "Paul D. Spiegel"
wrote:

Most of us are building for ourselves and not marketing products so many
common amp specs are not that important. However, I'm interested in how
power is calculated and would like to see some RAT feedback on this
subject.

There are two power calculations:
P=V^2/R (call this 'voltage power')
P=I^2*R (call this 'current power')

Let's take an example of a constant-current Class A output stage:

Load = 4 ohms
Peak-to-Peak (P-P) output voltage before clipping is 8 volts.
Using the voltage power formula:
(8*8)/4=16 watts

Nope - you can only do this if you work in marketing!

However, it's customary to represent voltage with the root mean square
(RMS) value. So 8 P-P becomes 5.66v rms when divided by 1.414 (sq. root
of 2).

Nope - 8 P-P = 4 P = 2.8 vrms

Now, (5.66*5.66)/4=8 watts

2.8 ^ 2 / 4 = 2

The constant current of the primary is 100ma. The output transformer
has a winding ratio of 35:1 so this calculates to 3.5a at the secondary.

Nope - the 100ma is the DC bias and signal input, useless info for you
here.

Using the current power calculation:
(3.5*3.5)*4= 49 watts

Q1:Would it make sense to calculate the current power using a reduced RMS
value? So 3.5/1.414=2.48a. Using this value:
(2.48*2.48)*4=25 watts

Q2: In the end, I imagine that the amp power would be described using the
_lower_ of the two calculations. I would say that this is an 8 watt
amp.

The above is a waste of time... it is a 2 watt amp.

Comments?

Back to the drawing board...

Phil Allison wrote:


The above is a waste of time... it is a 2 watt amp.


It is a waste of time at any other level as well, except as a
generalized indication - there is no such thing as any simple voltage
or current measurement & accurate formula for power output of a complex
music signal, unless you use or mock up a calorimeter for its output.



** The general aim is to ascribe a max POWER rating to a particular
amplifier under defined load conditions.

You are describing a measurement of the total ENERGY in some unspecified
piece of music programme.

Stick those goal posts back where you found them - Mr anonymous smartarse.

Go figure out the difference between energy and power - anytime.

Amusing.

- You've heard of goalposts but you don't know what a stopwatch is;

- You apparently have never measured electrical power by
straightforward physical means in your life;

- And a year ago you raved for days in you usual cretinous lout style
concerning the uselessness, fantasy and irrelevance of RMS power
measurement.

Time wounds all heels. :-)

Phil Allison wrote:

The above is a waste of time... it is a 2 watt amp.


It is a waste of time at any other level as well, except as a
generalized indication - there is no such thing as any simple voltage
or current measurement & accurate formula for power output of a complex
music signal, unless you use or mock up a calorimeter for its output.


** The general aim is to ascribe a max POWER rating to a particular
amplifier under defined load conditions.

You are describing a measurement of the total ENERGY in some
unspecified
piece of music programme.

Stick those goal posts back where you found them - Mr anonymous
smartarse.

Go figure out the difference between energy and power - anytime.

Amusing.


** Go **** yourself - that will be amusing to watch.


- You've heard of goalposts but you don't know what a stopwatch is;


** What drivel.


- You apparently have never measured electrical power by
straightforward physical means in your life;


** Course I have - like using a cup of water in a microwave to test the
magnetron.


- And a year ago you raved for days in you usual cretinous lout style
concerning the uselessness, fantasy and irrelevance of RMS power
measurement.


** Not me - ****head.

Better lay off those mind bending drugs you are taking.




............ Phil

Allison,

My aim is true......


(Elvis Costello, on the likelihood he could stuff an American football
goalpoast up Phil's ass on the first try....)