Hi there Barry,

B> I have been running the 16 bit Burr Brown A/D at an oversampling rate of=20
30 readings per sample. Someone mentioned on a previous listing that=20
oversampling was the equivalent of a higher resolution A/D.=20
    If you had a constant signal with no noise and a perfect A/D giving 16=20
bits, calling the converter would produce the same integer number n every=20
time. Averaging would produce the same integer.

    If you have a constant signal, but up to two bits of noise on the=20
converter, you will get a series of 16 bit readings of  n-2, n-1, n, n+1, n+=
2=20
in random order. There will be more n readings than either n+1 or n-1 etc.=20
--- you get a statistical distribution and you are down to something like 14=
=20
bit confidence accuracy on one reading. This is not good news. To halve the=20
average error on one single reading, you have to average four readings=20
together. For each factor of two, you need to average four times the previou=
s=20
number --- to get 1/4 the average error, takes 16 readings.
=20
    If you add 16 off 16 bit readings together you get a 20 bit integer=20
number. You can a) use this number b) convert to 'real' numbers and divide b=
y=20
16 c) stay with integers, div 16 and loose any extra accuracy of a size less=
=20
than 1 bit and accept the 'rounding error' on the last bit d) stay with=20
integers, div 4 and use the 18 bit integer --- the last two or three bits of=
=20
the 20 bit number will be noise that you don't want to know about anyway.
=20
B> I agree with the consept that multiple readings will give one a more=20
accurate sample value but I wonder how I would get more than a count of =20=
=B1=20
32768 (higher than 16 bit)?=20

    The only way to represent an accuracy of less than 1 bit with a 16 bit=20
integer is to use a larger total number, 17, 18 bits etc.=20
    If you have a couple of bits of noise on a 16 bit A/D, you can get more=20
than '16 bit accuracy' by taking and averaging enough readings 128, 512 ??=20
This can take a long time and you also have to ask yourself if your signal i=
s=20
going to stay steady for this length of time. You need to have a filter on=20
your analogue input line which rejects any signal or noise of a frequency=20
greater than your sample time.=20

    In the real world, it can be an advantage to average enough readings to=20
be reasonably certain of getting an accuracy within the last bit of the=20
converter, but beyond this, it starts to take a lot of time and effort and=20
you are usually better off using a higher accuracy A/D converter.

    When you have a lot of mains hum on the signal, which maybe changes from=
=20
time to time, another technique which may be worth considering is to take an=
d=20
average enough samples to exactly cover one mains cycle, but this does limit=
=20
the sample rate to less than the mains frequency per second. If you have 60=20
Hz mains and a 20 mu Sec sample rate, you could set the coverter to sample=20
say eight channels in succession, 84 times. The average contribution to all=20
channels from mains hum over this period will be zero. In calculating the=20
sample rate, you have to take into account set-up times and settling times,=20
or you may also loose accuracy that way. Sampling eight channels with=20
suitable pauses 20 times equally spread out over a single cycle would=20
probably be better.=20
=20
    Hope that this makes things clearer.

    Regards Chris

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