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'[EE]:A2d and PGA? Theory?'
2000\08\09@095044 by Graham North

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Hi could someone please explain to me how this works:

I have been given a 16 bit A2D converter and a Programmable gain amplifier,
and have been told to measure an input signal (+-10 Volts) to an accuracy of
18 bits!

Someone told me that by uping the gain to 2, I would in effect get another
bit.

Could someone help me here please, I am lost!

(Sorry to the people who find this trivial, if it were Monday I would blame
it on Monday brain fade.  Oh well guess I'm having Wednesday brain fade.
Just can't get my head round this!)

Thanks

Graham North

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2000\08\09@095948 by Andrew Kunz

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Take multiple samples (LOTS of them) with a known "noise" added on to dither the
input and you will gain bits.  It's a trade-off of time vs # bits.

Andy









Graham North <.....graham.northKILLspamspam@spam@LANDINST.COM> on 08/09/2000 09:48:35 AM

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Subject: [EE]:A2d and PGA? Theory?








Hi could someone please explain to me how this works:

I have been given a 16 bit A2D converter and a Programmable gain amplifier,
and have been told to measure an input signal (+-10 Volts) to an accuracy of
18 bits!

Someone told me that by uping the gain to 2, I would in effect get another
bit.

Could someone help me here please, I am lost!

(Sorry to the people who find this trivial, if it were Monday I would blame
it on Monday brain fade.  Oh well guess I'm having Wednesday brain fade.
Just can't get my head round this!)

Thanks

Graham North

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2000\08\09@105504 by M. Adam Davis

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Use a microcontroller to interface to both of them.  Here's the basic idea:

Perform two tests:

Measure the voltage with the range of -10 to 10 volts.

That will give you 16 bits of resolution over a 20 volt range.  That will give
you a resolution of .000305 volts per count.  You want 18 bits, which gives you
.0000763 volts percount.

Now think of your 20 volt range as four 5 volt ranges:
-10 to -5
-5 to 0
0 to 5
5 to 10

You know which of these four ranges the voltage is in from your initial test.
You simply need to use the 16 bit resolution of the A/D to read the 5 volt range
the voltage lies in.  This might be done with your programmable gain amplifier,
or by switching in some resister ladders or a precision voltage
reference/source.

In this second test you are only measuring 16 bits over a 5v range, which gives
you .0000763 volts per count.

I hope this helps!

-Adam

Graham North wrote:
{Quote hidden}

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2000\08\09@121549 by Barry King

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Graham,

This makes me think of the way the successive approximation A/Ds
work.  You make an approximation, then subtract off the approximation
from the input.  If you can subtract a reference from the signal
accurately ahead of the PGA, then you might be able to do this:

1) Make a first cut conversion of the signal at a gain of 1.  The two
MSBs of the result tell you what quarter of the range you are in.
Remember that (two bit) result.

2) Depending of which quarter, subtract either 0, 0.25, 0.5, or 0.75
of the range from the input.  The result is always in the 1/4 of the
range.

3) Now reset the PGA to 4 times the gain.  Convert the the result to
16 bits.  Now your two bit quarter plus the scaled up 16 bit result
are an 18 bit result.

One can imagine variants where the PGA is used to create the voltages
at the A/D Hi and Low ref.  If the A/D low ref can be floated as high
at 3/4 the input range, that would let you get away from a seperate
subtractor amp.

All this said, I wonder if that can possibly be better than
1) sell PGA
2) sell 16 bit converter
3) Buy 18 + bit delta/sigma converter, implement that.
4) buy beer with the difference.

I'll be interested to see what other ideas folks come up with.

------------
Barry King, KA1NLH
NRG Systems "Measuring the Wind's Energy"
http://www.nrgsystems.com
Check out the accumulated (PIC) wisdom of the ages at:
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2000\08\09@122617 by Bob Ammerman

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Correct me if I am wrong...

How accurate is this Progammable Gain Amp?

Unless it is extremely accurate, you will be introducing a bias or scale
error into the signal that will be quite a bit more than 1 LSB at 18bits,
won't you?

Bob Ammerman
RAm Systems

{Original Message removed}

2000\08\09@133706 by Olin Lathrop

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> I have been given a 16 bit A2D converter and a Programmable gain
amplifier,
> and have been told to measure an input signal (+-10 Volts) to an accuracy
of
> 18 bits!

That's quite a challange!  18 bits implies you have to be accurate to within
4 parts per million, or 76uV over your 20 volt range.  At these levels many
issues have to be dealt with very carefully.  This is not something you
should attempt unless you have a good background in precision analog
electronics.  Seriously, this needs to be done by an expert.

At 4ppm there are many sources of errors that you usually don't give much
thought to at 10 bits accuracy.  For example, even special temperature
compensated resistors will drift a few ppm per degree C.  So if you end up
with any resistors controlling an offset or a gain, you will be out of spec
with just a degree or two change in temperature.  Check the amplifier offset
voltage drift spec over time and over temperature and compare that to your
76uV accuracy requirement.  You will need to put the entire analog section
into a temperature controlled enclosure.  This is usually done with a
resistive heater with thermister sensor and feedback.  The easiest would be
to get this off the shelf, but I'm not sure where to point you.  And where
are you going to get a 4ppm voltage reference for the A/D?

Systematic offset and gain errors can be corrected for in software, but you
still need to make sure the system doesn't drift more than 4ppm after
calibration.  Where is the reference signal going to come from to do the
calibration?

I just noticed that you only have a 16 bit A/D.  You can't get 18 bits out
of a 16 bit D/A.  Do you mean each of the 16 bit transitions has be accurate
to within 1/8 LSB?  Most A/Ds are only speced to within 1/2 LSB.  It sounds
like this requires a high precision slope converter.  What are your speed
requirements?  Speed can be traded off with accuracy to some extent.

Again, I don't mean to put you down or be insulting, but given the question
you asked you are in way over your head.  Go get an expert to do this and to
learn from, then maybe you can be the expert next time.


*****************************************************************
Olin Lathrop, embedded systems consultant in Devens Massachusetts
(978) 772-3129, spamBeGoneolinspamBeGonespamcognivis.com, http://www.cognivis.com

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2000\08\09@133710 by Olin Lathrop

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> Perform two tests:
>
> Measure the voltage with the range of -10 to 10 volts.
>
> That will give you 16 bits of resolution over a 20 volt range.  That will
give
> you a resolution of .000305 volts per count.  You want 18 bits, which
gives you
> .0000763 volts percount.
>
> Now think of your 20 volt range as four 5 volt ranges:
> -10 to -5
> -5 to 0
> 0 to 5
> 5 to 10
>
> You know which of these four ranges the voltage is in from your initial
test.
> You simply need to use the 16 bit resolution of the A/D to read the 5 volt
range
> the voltage lies in.  This might be done with your programmable gain
amplifier,
> or by switching in some resister ladders or a precision voltage
> reference/source.
>
> In this second test you are only measuring 16 bits over a 5v range, which
gives
> you .0000763 volts per count.

This sounds nice in theory, but I don't think this is practical at 18 bit
accuracy.  You are going to end up with lots of problems where the 5 volt
ranges meet.  The more components you throw at it, the worse it will get.
All the "little" source of error become significant at this accuracy level.


*****************************************************************
Olin Lathrop, embedded systems consultant in Devens Massachusetts
(978) 772-3129, RemoveMEolinspamTakeThisOuTcognivis.com, http://www.cognivis.com

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2000\08\09@133927 by Oliver Broad

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Simplest answer I can think of is that you're supposed to turn the gain up
when the signal is low. This doesn't really increase the accuracy as a near
10v signal could still only be measured to 16 bit accuracy, but it improves
the dynamic range, since a small signal is still measured with high
resolution. If the result is then scaled by dividing it by the PGA gain then
I suppose the effective resolution has increased.

It's important to remember the difference between accuracy and resolution
too, I think you need something special as a reference to have even 16 bit
accuracy.


Oliver.

{Original Message removed}

2000\08\09@141029 by M. Adam Davis

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Ok, I've just looked at a programmable gain amplifier
http://www-s.ti.com/sc/psheets/slos214b/slos214b.pdf

From the circuit on page 20 it appears as though one can give it an input, a
reference voltage, and a gain setting (via three pins).

If you develop a precision programmable voltage source which gives its output to
the vref of the gain amplifier, then you can choose your voltage range.  I used
the 5 volt ranges as an exmaple for the general theory.

Find the initial voltage using 16 bits over the range of -10v to 10v.  Set the
programmable voltage soure to 1 volt below your initial reading.  Set the gain
such that your A/D converter sees the range of 0 to 2.5 volts (or something
similar).  Take another reading, throw away the LSB, and put the two readings
together to come up with an 18 bit reading.

Now, I understand that there are a LOT of sources for error/noise here.  A lot
of people have pointed these issues out.  Anyone involved in electronics who has
played with an oscilliscope *knows* how much noise is generated in any piece of
wiring, which is much larger than the single bit quantity being measured here.

Rather than write a huge essay showing the details of a solution to the problem
and all the little gotchas, I prefer to write about the general theory behind
one option.  Others on the piclist are sure to chime in with info which they
feel I've missed, and the person who asked the question gets to synthesize the
tomes of information a little bit at a time, instead of my own version all at
once.

But I suppose I should have mentioned that when you are measuring differences of
76uV, you might experience difficulty with noise.  ;-)

-Adam

Olin Lathrop wrote:
> This sounds nice in theory, but I don't think this is practical at 18 bit
> accuracy.  You are going to end up with lots of problems where the 5 volt
> ranges meet.  The more components you throw at it, the worse it will get.
> All the "little" source of error become significant at this accuracy level.

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2000\08\09@210720 by Olin Lathrop

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> Ok, I've just looked at a programmable gain amplifier
> http://www-s.ti.com/sc/psheets/slos214b/slos214b.pdf
>
> From the circuit on page 20 it appears as though one can give it an input,
a
> reference voltage, and a gain setting (via three pins).
>
> If you develop a precision programmable voltage source which gives its
output to
> the vref of the gain amplifier, then you can choose your voltage range.  I
used
> the 5 volt ranges as an exmaple for the general theory.
>
> Find the initial voltage using 16 bits over the range of -10v to 10v.  Set
the
> programmable voltage soure to 1 volt below your initial reading.  Set the
gain
> such that your A/D converter sees the range of 0 to 2.5 volts (or
something
> similar).  Take another reading, throw away the LSB, and put the two
readings
> together to come up with an 18 bit reading.
>
> Now, I understand that there are a LOT of sources for error/noise here.  A
lot
> of people have pointed these issues out.  Anyone involved in electronics
who has
> played with an oscilliscope *knows* how much noise is generated in any
piece of
> wiring, which is much larger than the single bit quantity being measured
here.
>
> Rather than write a huge essay showing the details of a solution to the
problem
> and all the little gotchas, I prefer to write about the general theory
behind
> one option.  Others on the piclist are sure to chime in with info which
they
> feel I've missed, and the person who asked the question gets to synthesize
the
> tomes of information a little bit at a time, instead of my own version all
at
> once.
>
> But I suppose I should have mentioned that when you are measuring
differences of
> 76uV, you might experience difficulty with noise.  ;-)

While I agree with your general theory, I submit that it is unrealizable
with any available parts to 18 bits accuracy.  The chosen amp is
particularly ill suited to the job.  According to the specs at the link you
provided, it has an input offset voltage drift of 10uV/degC, and that was
only specified as "typical" without any maximum.  The accuracy of the
selected gains is only +-7.5% at a fixed temperature of 25C.

I think the only realizable solution short of finding an off the shelf part
(I doubt it), is a slope converter.  These are slow but can be made very
accurate with care.  These also have the advantage of essentially
calibrating to a reference every conversion, since drift is such an issue at
this accuracy level.  I noticed that the orignal poster never answered my
question about his speed requirements, so we don't know if this method will
be fast enough.

In short, my answer to how to use this 16 bit A/D and this programmable gain
amp to make an 18 bit A/D is "You can't".  It takes a whole different
topology to achieve that.


*****************************************************************
Olin Lathrop, embedded systems consultant in Devens Massachusetts
(978) 772-3129, RemoveMEolinEraseMEspamEraseMEcognivis.com, http://www.cognivis.com

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2000\08\09@215349 by Spehro Pefhany

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At 07:01 PM 8/9/00 -0400, you wrote:

>I think the only realizable solution short of finding an off the shelf part
>(I doubt it), is a slope converter.

Delta-sigma converters such as the LTC2410 will do >18 bit linearity, no
missing
codes to 24 bits, price is fairly reasonable (<$10.00).

I agree it's not reasonable to do it with the parts being discussed.

Best regards,


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