[sdiy] A/D converter with good accuracy for a reasonable price?
Karl Ekdahl
elektrodwarf at yahoo.se
Fri May 6 14:02:51 CEST 2011
All; thank you so much for your input!
Here's a thought tho; i already have a good D/A in my system, and i used that in conjunction with a S&H to make a SAR A/D - it seemed to me like it didn't work at first, hence why i asked about the "real" A/D alternatives. However, now it does indeed seem to work quite well.
But to avoid future troubles i wonder if there's any hidden dark sides to do it this way? I realize that i'm loosing processing power but that doesn't bother me.
Thanks!
Karl
--- Den fre 2011-05-06 skrev Scott Gravenhorst <music.maker at gte.net>:
> Från: Scott Gravenhorst <music.maker at gte.net>
> Ämne: Re: [sdiy] A/D converter with good accuracy for a reasonable price?
> Till: synth-diy at dropmix.xs4all.nl
> Datum: fredag 6 maj 2011 01:43
> jon schatz <jon at divisionbyzero.com>
> wrote:
> >On 5/5/2011 3:02 PM, Scott Gravenhorst wrote:
> >
> >> If you are averaging (for example) four ADC
> readings, you collect them,
> >> add them together and divide by four (or right
> shift by 2 bits).
> >
> >you can also do some exponential filtering:
> >
> >newResult = ( prevResult * (n-1) + curAdcReading) / n;
> >
> >which has the advantage of giving you a new result
> every cycle instead
> >of every n cycles (which the above averaging would do).
> it also means
> >that you'll never get a change greater than 1/n *
> resolution which can
> >be a good thing for reducing noise but a bad thing if
> you want to track
> >sharp changes.
>
> Yeah, I was thinking that too after I hit send. The
> filter is a single pole lowpass
> IIR filter and is analogous to a simple RC lowpass
> filter. The higher the value of n,
> the lower the "bandwidth" of the filter, that is more
> suppression of noise at higher
> frequencies. So it really depends on what you need
> from the sample stream. I think
> that for reading the value off of pots, the exponential
> filter method can be nicely
> advantageous, and as Jon points out, there is a loss of
> slew rate, but for pots, how
> fast can you turn them anyway?
>
>
>
> -- ScottG
> ________________________________________________________________________
> -- Scott Gravenhorst
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