NetNews Usenet Archive 1992 #31/NN_1992_31.iso/spool/rec/audio/car/5486

Internet Message Format | 1992-12-27 | 3.4 KB

Path: sparky!uunet!spool.mu.edu!olivea!charnel!rat!ucselx!crash!cmkrnl!jeh
From: [e-mail]
Newsgroups: rec.audio.car
Subject: dB, volts vs. watts (was Re: Why so much power?
Message-ID: <[e-mail]>
Date: 27 Dec 92 02:58:43 GMT
References: <[e-mail]> <[e-mail]>
Organization: Kernel Mode Consulting, San Diego, CA
Lines: 69

In article <[e-mail]>, [e-mail] (Kirk Lindstrom) writes:
> OOPS!
> I did it again and forgot power is 10log(P1/P2) while voltage gain and
> spectrum analyzer displays are 20Log(V1/V2).  Reason is
>   Power=Voltage*current=V*I=V^2/R

Right so far.

> Thus if talking power gain where the formula is dbW=10log(P1/P2) and you
> want to convert to voltage gain, then db=10log[V1^2/V2^2]=20log[V1/V2].

I'm not sure what you're trying to say here.  Maybe you're saying correct
things but I'm misreading you.  In which case, please take this as a correction
for the benefit of the other people who misread you the same way:

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".

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.

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.

I suppose you could measure the [micro]watts drawn by a line-level input stage,
convert this to voltage, measure the corresponding output power into a speaker,
convert THAT to voltage, and find the dB gain that way, but you might as well
measure the two voltages in the first place -- it's much easier.

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.  For any gain g expressed in
decibels:

                 E2
voltage ratio = ---- = 10 ** (g / 20)
                 E1

                 P2
power ratio =   ---- = 10 ** (g / 10)
                 P1

Note that it doesn't matter whether the original dB figure was obtained from
power or voltage measurements.

For the example problem, if you look at a spectrum analyzer display and see
that the 180 Hz output is 6 dB below the 60 Hz, that indeed means that the
analyzer is measuring voltages with a 2:1 difference... or that, when this
signal is passed through a power amp and applied to a speaker, the power of
the 60 Hz component will be 4x that of the 180 Hz component (assuming that
the speaker impedance is the same at both frequencies, which isn't usually
likely for this particular pair of freqs!  Oh well...).

    --- Jamie Hanrahan, Kernel Mode Consulting, San Diego CA
Internet:  [e-mail], [e-mail], or [e-mail]
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