From: jeh@cmkrnl.com
Newsgroups: rec.audio.car
Subject: Re: dB, volts vs. watts (was Re: Why so much power?
Date: 30 Dec 92 21:20:26 GMT
Organization: Kernel Mode Consulting, San Diego, CA

In article <26130135@hpcc01.corp.hp.com>, kirk@hpcc01.corp.hp.com (Kirk Lindstrom) writes:
> [jeh@cmknrl.com wrote:]
>>In general, gain is gain, and dB are dB.  There is no such concept as
>>"converting from dB[power] to dB[voltage]", and no need to express decibels
>>as "dBW" or "dBV" as opposed to just "dB". 
>>
> I don't agree.  Read on and see if you agree with what I write.
>
>>Consider an amplifier driving a load R with a 1-volt RMS waveform.  
>>Now suppose we increase the output to 10 volts RMS.  How many dB is that?
>>(note, all "log(x)" = log{base 10}(x))
>>
>>	From volts:  gain = 20*log(10/1) 
>>                          = 20*1 = 20 dB
>>
>>	From watts:  gain = 10*log((10*10/R)/(1*1/R))
>>                          = 10*log(100/1)
>>                          = 10*2 = 20 dB
>>
>>In other words, db[power] = db[voltage], provided that the load impedance is
>>the same for the two measurements.  
>>
> For clarity, we (myself and many I work with) usally specify what units
> we are working in since db is a relative measurement.  I believe many use
> db and dbW (and dbV) interchangealby in/with the MKS system since W=1. 
> Correct?

I'm not sure what you mean by "W=1".  

But in any case, I don't see what the "W" or "V" tells you.  What do you 
do differently based on "dbV" vs. "dbW"?

> In other words, since db is relative we must state or assume with what
> it is relative to.  

I don't see this at all.  dB is an expression of a ratio.  The original units
which were measured to get the ratio disappear completely, because they appear
on both the top and bottom of the ratio "fraction". 

Let's translate to another domain and see if it makes more (or less!) sense.
Suppose I drive to LA one week, and my odometer shows 130 miles for the trip.
Next week I drive to Las Vegas and the odometer shows that the trip was 520
miles.  If I tell you "LV is four times as far from here as LA", you don't need
to know whether I measured it in miles, kilometers, light-years, or furlongs to
apply this ratio to other measurements.  If, for example, you're more familiar
with km, and you know that LA is 200 km from here, then you can use my "4x"
ratio to learn that LV is 800 km. 

> For example, page 64 of Dec. 92 Stereophyle shows
> a McIntosh 1000 Watt power amp.  The meter on the amp has two 
> calibrations for a single scale labeled:
> 
>          WATTS        .01   .1    1   10   100  1000
>                   |----|----|-----|----|----|-----|-|=|
>       DECIBELS   -60  -50  -40   -30  -20  -10    0  +6
> 
> Here McIntosh is saying that 0db for their amp is full power and has
> the potential for +6db peaks.  

Fine.  But what you're looking at here is really a voltmeter; the "watts"
figures are for a presumed 8-ohm load.   From E=sqrt(P*R) we can find the
voltages and add another scale:

          VOLTS        .28  .89   2.8  8.9   28   89
          WATTS        .01   .1    1   10   100  1000
                   |----|----|-----|----|----|-----|-|=|
       DECIBELS   -60  -50  -40   -30  -20  -10    0  +6
 
and the dB figures remain the same.  

> (I wonder how much this amp costs?)

Don't know, except that about half the cost will be for the McIntosh name. :-)

>>In audio work, one usually does not measure amplifier gain by measuring watts 
>>in vs. watts out, because the load impedance *won't* be the same for the two 
>>measurements.  Audio amps are typically voltage amplifiers, with only the
>>final stage providing current to a low-impedance load.  
>>
>>Now, if you have dB and want to convert back to power ratio or voltage ratio,
>>then, yes, you need to apply the proper divisor.  

I should have said "...the proper divisor for the 'target' units, but this does
not depend on the original units."  ie if you have a dB figure and you want to
get a voltage ratio from it, it's the same calculation whether volts or watts
(or current, for that matter) was originally measured to come up with the dB. 

Again -- I don't see what information you're, er, gaining by being told "dBV"
vs "dBW".  

I can see that the two figures would be different if, for example, you measured
voltage gain with the amp driving an open circuit, but measured power gain
under a typical  load... but in that case you need a lot more information than
just "W" vs. "V", you need to know ALL of the test conditions for the two
measurements.  

	--- Jamie Hanrahan, Kernel Mode Consulting, San Diego CA
Internet:  jeh@cmkrnl.com, hanrahan@eisner.decus.org, or jeh@crash.cts.com
Uucp:  ...{crash,eisner,uunet}!cmkrnl!jeh
Newsgroups: rec.audio.car