[AMPS] Audio amp power

[Dave D'Epagnier]

	----------
	From:  John Nelson [SMTP:John_Nelson@compuserve.com]
	Sent:  Monday, January 17, 2000 3:38 PM
	To:  Dave D'Epagnier; QRO list
	Subject:  RE: [AMPS] Audio amp power

	Message text written by "Dave D'Epagnier"

	>John, mind telling us how rms power is calculated for a single
tone?<

	I can try, but I'm hoping that cleverer people than me will tell us
where
	I've got this wrong...

	>The average power is calculated as Vrms squared divided by R. This
average
	power is also how much heat is delivered to the load. So what is rms
power,
	and how is it calculated...<

	As I understand it, an alternating current of (say) 1A is that
which, if it
	flows in a non-reactive resistance, produces heat energy at an
average rate
	equal to that produced by 1A DC flowing in the same resistance.
Since the
	rate of heat production is I^2R, the equivalent alternating current
is such
	that the square root of the average I^2 gives the value required,
which is
	therefore called the root-mean-square or RMS value. 

	This is the RMS value of the current, not power. Power (average
power that is) is calculated by I^2R, where I is the rms value of the
current.

	Sorry, but I still don't buy "RMS Watts". 


	73,

Dave K0QE


	From memory, if i is the instantaneous value of alternating current
flowing
	in a non-reactive resistance R, the heat developed in time dt is
dw=i^2Rdt.
	The mean rate of heating (i.e. the mean power) over a complete
period of t
	is given by:

	P = 1/t {T  dw = 1/t {t i^2R.dt = I^2R
	          } 0              }o 

	(the curly brackets are the best I can do for an integral!)

	>what does it physically mean?<

	In the case of a sine wave, I suspect it's where the factor 0.707
comes
	from (i.e. Vmax or Imax/root2). 

	Anyone have a copy of Langford-Smith on the bookshelf? My copy
appears to
	have warped into hyperspace, and IIRC there's a good treatment of
the
	subject in there somewhere.   

	73 John
	GW4FRX

	 












	 





	

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