Just want to add something. I was wrong about the voltage "wandering" a bit
over "millivolts". That is only how it looks on my DVM. It is actually very
noisy. Here is an application note, PDF, on the signal:
http://ewww.motorola.com/brdata/PDFDB/docs/AN1646.pdf

This is good news then, correct?

Thanks for all the help.

John


----- Original Message -----
From: "Russell McMahon" <apptech@PARADISE.NET.NZ>
To: <PICLIST@MITVMA.MIT.EDU>
Sent: Tuesday, February 18, 2003 2:53 AM
Subject: Re: [EE]: 12 bit ADC or amplify for 10 bit ADC?


> > > > > The voltage wanders a bit but only in the realm of  milliseconds,
> and
> > > by
> > > > > millivolts. I want to take many sample and average them.
>
> > > > You need at least a 11 bit A/D.  A 10 bit A/D alone can not do what
> you
> > > > asked, regardless of how you scale the input or what reference
voltage
> > >     you pick.
>
> > > Not really. :o) If he can add/subtract an (adjustable)
> > > % to the reference voltage it should be *possible* to
> > > interpolate another bit in accuracy. It's not easy and
> > > would be messy in software but IT IS very possible.
>
> > You can achieve this by simply oversampling.  IIRC you need to sample
4^n
> > times where n is the extra number of bits you require, so 4 times for 1
> bit,
> > 16 times for two bits etc.
>
> Alas no. Oversampling will not help IF the signal is more stable than an
> LSB. More info needed here but the above "wandering" signal is required to
> be sampled to about 12 bits AFAIR. You have to ADD RANDOM NOISE to dither
> the signal between the two adjacent bit positions. The time it spends
closer
> to one or other bit position will relate to its relative distance from
each
> position. If it just sits in about the same place (no noise added) it will
> convert to the same value each time.
>
> eg    A2D has 1 mV steps say.
> Mean Vin = 1.5432
> There is 0.2 mV variation in samples, approx equally distributed.
>
> Range is therefore 1.5431 to 1.5433
>
> ALL values will convert to 1.543 in this (contrived) example.
> Adding +/- 0.5 mV of random noise will spread the values from 1.5426 to
> 1.5438
> 1.5426 to 1.5435 will convert as 1.543
> 1.5435 to 1.5438 will convert as 1.544
> Mean value over many samples will be 1.54325 which is wrong (due to dither
> in original signal) but near the true mean value of 1.5432.
>
>
>
>         Russell McMahon
>
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