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...