From: Robert Baer
X-Mailer: Mozilla 4.75 [en] (Win98; U)
Subject: Re: wattage meter exists??
References: <firstname.lastname@example.org> <3D79ACA8.CE5E286@earthlink.net>
Date: Sun, 08 Sep 2002 06:54:03 GMT
NNTP-Posting-Date: Sat, 07 Sep 2002 23:54:03 PDT
Organization: EarthLink Inc. -- http://www.EarthLink.net
"R. Lewis" wrote:
> "Robert Baer" wrote in message
> > Well, if all one wants is an estimate of power out of a sound system
> > ........then use a resistive load of the proper impedance
> > ...
> > ...
> > There is a lot of stupidity and false implications involved with most
> > "specifications" of audio amplifiers made to drive speakers and/or
> > headphones.
> > There have been some exceptions to this generality, but the makers of
> > that equipment had a lot of integrity and really cared about their
> > products and customers.
> > Virtually all audio stuff made for computers is cheap crap, some that
> > seems to work well, but junk nonetheless.
> Are you saying that my '60W full dynamic range' PC speakers, able to be
> powered either by a 6VA wall cube (supplied) or -for the man on the move
> with a laptop - 2 off D sized cells, may not be giving the full 100W across
> all the audio spectrum?
> The scoundrels.
> I have a mind to ask for my $4 back
Assume 6V supply and the speaker drivers go rail to rail (a number of
modern analog ICs now do this).
Also assume a full H-bridge driver so the speaker can be driven from
+6V to -6V (a number of modern analog ICs now do this also).
That is 12V peak to peak. Now P=(V*V)/R, so assuming 60 watts, R=2.4
ohms (unlikely as the lowest Z i have seen is 4 ohms).
Furthermore, power level is nominally specified as RMS, which is 0.707
of peak if sine wave, which is the other standard for specification.
This gives P=(0.707*V*0.707*V/R or R=4.8 ohms, which is more like a
You sure as heck will *not* get more power (100W) with less voltage (2
D cells = 3V supply).