Re: Ripping LPs
From: John Stumbles (dev.null_at_localhost)
Date: 09/24/04
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Date: Fri, 24 Sep 2004 20:33:16 GMT
AC wrote:
> In article <cj045e$94d$1@nntp.itservices.ubc.ca>, Bill Unruh
> <unruh@string.physics.ubc.ca> writes
>
>>the input impedance of the sound card is about 10Kohm, which means it will
>>absorb no power worth talking about from the amplifier but the preamp is
>>better impedence matched and nominal voltages are more similar (the amp
>>could put out up to 30V and your soundcard is designed for about 1V max
>>signal-- which is closer to what your preamp puts out).
>
>
> Input impedance for sound cards I have used previously was (I believe)
> about 200 or 300 ohm. I know this because I used the general technique
> described in this thread very recently with a new main board (integral
> sound) initially without success. Until I put something like 200 or 300
> ohm resistors across each (stereo) line.
What do you mean by 'without success'? Did you get nothing, or a lot of
noise with the signal? Loading the inputs with 200-300 ohm resistors
might well reduce the noise pickup relative to the wanted signal (i.e.
increase the signal-to-noise ratio) if the effective output impedance[1]
of the noise source was much higher than the output impedance of the
signal source. E.g. suppose the signal source impedance is c. 1K, the
noise impedance is c. 100K and the sound card input impedance is 10K:
then the noise voltage seen at the sound card input is 10K/(10K + 100K)
=~ 10% of the noise source voltage; the signal voltage at the input is
10K/(10K+1K) =~ 90% of the signal source voltage. Running the same
figures with an impedance of say 200 ohm at the sound card input we get
.2/(0.2+100) =~ 0.2% of the noise and 0.2/(0.2+1) =~ 17% signal.
Comparing the ratios of attenuation of noise / attenuation of signal
before we had 10%/90% =~ .1
after we have 0.2%/17% =~ .01
so the extra resistance[2] at the input has improved our s/n ratio by a
factor of 10 (which is 20dB IIRC)
There's a limit to how much we can improve the s/n ration this way: as
we decrease the resistance at the soundcard input much below the output
impedance of the signal source the attenuation of the wanted signal
increases almost as much as the attenuation of the noise. Also if we
attenuate the signal too much the electrical noise generated and picked
up in the soundcard's input electronics contributes proportionately more
to the overall noise. It is also possible that the signal source may
generate distortion if connected to too low a load resistance.
> (a shop like RadioShack or Maplin will help and advise)
ROFL ;-)
[1] the noise has to be coming from somewhere, right? It may be being
picked up in the connecting cable from the preamp to the soundcard by
the cable acting as an aerial for alternating electrical and magnetic
fields produced by e.g. transformers and switched mode power supplies in
computer PSUs and CRTs. The amount of power picked up is quite small so
if you connect the cable to a low resistance you'll get less noise
voltage than if you connect it to a high resistance. You can model this
by imagining a generator of noise voltage connected via a certain
resistance to the input of the soundcard, along with the connecting
cable. Because this is AC you may have to modify the model by including
some capacitance or inductance with the resistance: the combined effect
is know as impedance ... but we're getting a bit deeper into electronics
here :-)
[2] pedantically the extra resistor has actually given us extra
conductance (which is the inverse of resistance) i.e. we've lowered the
input resistance, not increased it.
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