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'[PIC]:A/D with Vref+'
2001\10\10@113738 by Keong

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Hi

I am trying to do an A/D with the input source to A/D
channel 0 of 16F873 and I have configured ADCON1 to
have 00000101 with a vref+ of 1V from volt divider and
vref- to be Vss. However I get a all '1's output from
my output ports even though the input volt is less
than 1V. Cant figure out the problem here.

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2001\10\10@114359 by Dale Botkin

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For one thing, your Vref+ must be at least 2V higher than Vref-.

Dale
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On Wed, 10 Oct 2001, Keong wrote:

> Hi
>
> I am trying to do an A/D with the input source to A/D
> channel 0 of 16F873 and I have configured ADCON1 to
> have 00000101 with a vref+ of 1V from volt divider and
> vref- to be Vss. However I get a all '1's output from
> my output ports even though the input volt is less
> than 1V. Cant figure out the problem here.

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2001\10\10@114950 by Keong

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But then, is there a method to do an A/D on an input
signal of 0V to 1V on the 16F873 using Vrefs?

thanx

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2001\10\10@115619 by Mike Mansheim

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> I am trying to do an A/D with the input source to A/D
> channel 0 of 16F873 and I have configured ADCON1 to
> have 00000101 with a vref+ of 1V from volt divider and
> vref- to be Vss. However I get a all '1's output from
> my output ports even though the input volt is less
> than 1V. Cant figure out the problem here.

tris set to input?

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2001\10\10@120507 by Keong

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yup, in fact all pins in porta are set to input


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wrote:
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2001\10\10@121301 by Byron A Jeff

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On Wed, Oct 10, 2001 at 08:49:12AM -0700, Keong wrote:
> But then, is there a method to do an A/D on an input
> signal of 0V to 1V on the 16F873 using Vrefs?

Sure. Set Vref to exactly 2 volts and lose one bit of accuracy. Instead of
10 bits you only get 9. You can shift the 9th bit into the upper byte of
the A/D and round the 10th bit in. It's an 8 bit result but pretty much has
9 bits of accuracy.

I did this with fairly good results when reading a LM34 temp sensor. It's
output voltage didn't get higher than 1V and I used a Vref of 5.12V.

BAJ
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2001\10\10@142300 by Dale Botkin

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On Wed, 10 Oct 2001, Keong wrote:

> But then, is there a method to do an A/D on an input
> signal of 0V to 1V on the 16F873 using Vrefs?

Yes, as noted in the other reply you can use Vref of 2V and you'll have a
0-512 reading, or 9-bit resolution instead of 10-bit.  Other than that you
can use an op amp to amplify the input signal; of course then you need to
divide the ADC reading by the gain of your amp - which again gets you
9-bit resolution!  I don't think there's a way around this - in other
words to get full 10-bit resolution on a 1V signal.  I think your input
signal would have to be 2V or better to get you full ADC resolution, since
that's the minimum Vref.

Dale

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2001\10\10@150749 by Olin Lathrop
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> But then, is there a method to do an A/D on an input
> signal of 0V to 1V on the 16F873 using Vrefs?

Not if you want full range of the A/D without external parts.  To get the
full A/D range from a 0 to 1V signal requires an external amplifier.


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2001\10\10@164816 by Olin Lathrop

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> Yes, as noted in the other reply you can use Vref of 2V and you'll have a
> 0-512 reading, or 9-bit resolution instead of 10-bit.  Other than that you
> can use an op amp to amplify the input signal; of course then you need to
> divide the ADC reading by the gain of your amp - which again gets you
> 9-bit resolution!

Huh!?  You end up with a 0 - 1023 value that maps to the full range of the
input voltage, which is 0 to 1V in this case.  That's what he wanted in the
first case.  There's an arbitrary scale factor between the EMF expressed in
volts and the A/D reading.  A scale factor of 512 doesn't seem more
desirable than 1024.  If you're doing calibration, you're going to apply an
offset and scale anyway.

> I don't think there's a way around this - in other
> words to get full 10-bit resolution on a 1V signal.  I think your input
> signal would have to be 2V or better to get you full ADC resolution, since
> that's the minimum Vref.


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2001\10\10@181807 by Dale Botkin

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On Wed, 10 Oct 2001, Olin Lathrop wrote:

> > Yes, as noted in the other reply you can use Vref of 2V and you'll have a
> > 0-512 reading, or 9-bit resolution instead of 10-bit.  Other than that you
> > can use an op amp to amplify the input signal; of course then you need to
> > divide the ADC reading by the gain of your amp - which again gets you
> > 9-bit resolution!
>
> Huh!?  You end up with a 0 - 1023 value that maps to the full range of the
> input voltage, which is 0 to 1V in this case.

Nope, if you amplify the input by two you'll get a 0 - 1023 value that
maps to TWICE the full range of the (original) input voltage, and
represents twice what it would normally.  In other words, with a 2V
reference and a 1V unamplified input signal, one ADC step equals (roughly)
1.95mV, and you won't see an ADC reading over 512.  Amplify the input
signal x2 and you will get ADC readings up to 1023, but now each ADC step
represents 3.9mV.  Same result, no?  Just different numbers.  That was my
point.  Admittedly I should have responded when I had more time to
clarify, I shouldn't try to multitask when responding to posts.

Dale

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2001\10\10@183515 by David Duffy

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>On Wed, 10 Oct 2001, Olin Lathrop wrote:
>
> > > Yes, as noted in the other reply you can use Vref of 2V and you'll have a
> > > 0-512 reading, or 9-bit resolution instead of 10-bit.  Other than
> that you
> > > can use an op amp to amplify the input signal; of course then you need to
> > > divide the ADC reading by the gain of your amp - which again gets you
> > > 9-bit resolution!
> >
> > Huh!?  You end up with a 0 - 1023 value that maps to the full range of the
> > input voltage, which is 0 to 1V in this case.

Dale replied
>Nope, if you amplify the input by two you'll get a 0 - 1023 value that
>maps to TWICE the full range of the (original) input voltage, and
>represents twice what it would normally.  In other words, with a 2V
>reference and a 1V unamplified input signal, one ADC step equals (roughly)
>1.95mV, and you won't see an ADC reading over 512.  Amplify the input
>signal x2 and you will get ADC readings up to 1023, but now each ADC step
>represents 3.9mV.  Same result, no?  Just different numbers.  That was my
>point.  Admittedly I should have responded when I had more time to
>clarify, I shouldn't try to multitask when responding to posts.

How did the ADC step suddenly double? Vref is still 2V. Step is still 1.95mV.
Amplifying the input signal up to make is closer to Vref gives better
resolution.
Surely measuring a 2V (max) signal with a 2V Vref uses all of the available
bits (0-1023) where a 1V (max) signal only loses the MSB (0-511). The signal
being amplified doesn't make it lose any resolution. The original analog range
doesn't come into it at all. If you can amplify it to better match Vref, do
it !
Regards...

___________________________________________
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U8, 9-11 Trade St, Cleveland 4163 Australia
Ph: +61 7 38210362   Fax: +61 7 38210281
New Web: http://www.audiovisualdevices.com.au
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2001\10\10@184549 by Bob Ammerman

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Olin,

You have misinterpreted thie comment below:

> > Yes, as noted in the other reply you can use Vref of 2V and you'll have
a
> > 0-512 reading, or 9-bit resolution instead of 10-bit.  Other than that
you
> > can use an op amp to amplify the input signal; of course then you need
to
> > divide the ADC reading by the gain of your amp - which again gets you
> > 9-bit resolution!

The user's intent was to use a legal Vref+ value (ie: 2V with Vref- = 0V),
which means that the input signal, with varies from 0 to 1V would result in
a 9 bit value (0..511 or so). In effect, trading off one bit of input
precision for the need to have an amplifier.

Bob Ammerman
RAm Systems
(contract development of high performance, high function, low-level
software)

Olin said:

> Huh!?  You end up with a 0 - 1023 value that maps to the full range of the
> input voltage, which is 0 to 1V in this case.  That's what he wanted in
the
> first case.  There's an arbitrary scale factor between the EMF expressed
in
> volts and the A/D reading.  A scale factor of 512 doesn't seem more
> desirable than 1024.  If you're doing calibration, you're going to apply
an
> offset and scale anyway.
>
> > I don't think there's a way around this - in other
> > words to get full 10-bit resolution on a 1V signal.  I think your input
> > signal would have to be 2V or better to get you full ADC resolution,
since
{Quote hidden}

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2001\10\10@191505 by Dale Botkin

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{Quote hidden}

Sorry, you're right -- the ADC step itself doesn't double, its
relationship to the original signal, which is half of Vref, does.  Now
you're taking 1023 steps to measure twice the original signal, rather than
512 steps to measure the original signal.  It's the same thing, isn't it?

Dale

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2001\10\10@191524 by Herbert Graf

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{Quote hidden}

       Huh??? In both cases your "one ADC step" equals 1.95mV, multiplying the
input signal by a gain of 2 simply brings the 1V signal to the 2V level, the
"step" size doesn't change unless you change the ref. TTYL

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2001\10\10@202213 by David Duffy

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At 06:13 PM 10/10/2001 -0500, you wrote:
> > > > Huh!?  You end up with a 0 - 1023 value that maps to the full range
> of the
> > > > input voltage, which is 0 to 1V in this case.
> >
> > Dale replied
> > >Nope, if you amplify the input by two you'll get a 0 - 1023 value that
> > >maps to TWICE the full range of the (original) input voltage, and
> > >represents twice what it would normally.  In other words, with a 2V
> > >reference and a 1V unamplified input signal, one ADC step equals (roughly)
> > >1.95mV, and you won't see an ADC reading over 512.  Amplify the input
> > >signal x2 and you will get ADC readings up to 1023, but now each ADC step
> > >represents 3.9mV.  Same result, no?  Just different numbers.  That was my
> > >point.  Admittedly I should have responded when I had more time to
> > >clarify, I shouldn't try to multitask when responding to posts.

David:
{Quote hidden}

Dale:
>Sorry, you're right -- the ADC step itself doesn't double, its
>relationship to the original signal, which is half of Vref, does.  Now
>you're taking 1023 steps to measure twice the original signal, rather than
>512 steps to measure the original signal.  It's the same thing, isn't it?

The fact that the input signal is doubled makes no difference to the
resolution.
Ignore the actual input range - it's just being scaled up to get a better
result.
It might as well be a 100V signal scaled down to 2V. Does this make sense?
Regards...

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2001\10\11@075833 by Thomas McGahee

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By the way, instead of using a 2 v reference, use a 2.048 v reference
and then each "step" will be .001 volt. This simplifies any math
that you have to do later.

Other useful reference values are: 2.56 v, 4.096 v, and 5.12 v.

Certain types of conversion steps can also be simplified by choosing the
reference voltage carefully. In effect you are doing part of the
overall conversion at the analog stage instead of the digital stage.

In general, choose a reference value that makes each "step" value
a "nice" number.

Fr. Tom McGahee

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2001\10\11@085325 by Olin Lathrop

face picon face
> Nope, if you amplify the input by two you'll get a 0 - 1023 value that
> maps to TWICE the full range of the (original) input voltage, and
> represents twice what it would normally.  In other words, with a 2V
> reference and a 1V unamplified input signal, one ADC step equals (roughly)
> 1.95mV, and you won't see an ADC reading over 512.  Amplify the input
> signal x2 and you will get ADC readings up to 1023, but now each ADC step
> represents 3.9mV.

No, this is backwards.  Each A/D step will still be 2 / 1023 = 1.95mV at the
PIC pin because nothing has changed there (still using 2V reference).
However, that A/D step size gets *divided* by the opamp gain when projected
back to the original signal.  Each step now corresponds to a 978uV change in
the input signal.  Another way to see this is that the 0 to 1V input signal
maps to the full range of the A/D, regardless of how we got there.  1V /
1023 = 978uV.

Unless I needed very high accuracy, I would multiply the input signal by a
little less than 5 and dispense with Vref altogether.  In otherwords, use
the 5V power regulator as the reference.


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Olin Lathrop, embedded systems consultant in Littleton Massachusetts
(978) 742-9014, RemoveMEolinTakeThisOuTspamembedinc.com, http://www.embedinc.com

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2001\10\11@085338 by Olin Lathrop

face picon face
> Sorry, you're right -- the ADC step itself doesn't double, its
> relationship to the original signal, which is half of Vref, does.  Now
> you're taking 1023 steps to measure twice the original signal, rather than
> 512 steps to measure the original signal.  It's the same thing, isn't it?

If you multiply the original signal by 2 and use a Vref of 2V, you are
taking 1023 steps to measure the whole original signal.


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2001\10\11@154522 by Dale Botkin

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There, you see, finally Olin made sense of my addled attempt at answering
the OP's question.  Thanks, Olin.  Yes, if you amplify the original signal
you get *higher* resolution relative to the original signal.  In a project
just recently finished we had to divide a higher (0-20V) input signal by
4, which gives you *lower* resolution relative to the original signal.
OK, did I finally get it right?  8-P

Damn, it's a good thing I'm not actually working on a PIC project right
now, I could blow the house up.  Two weeks off work to move into a new
house is counter-productive - you get no "vacation" from work, and there's
two weeks' worth of crap waiting when you return to work as well.  That
and being away from my bench and the projects for even longer and the rust
rally sets in, doesn't it?

Dale
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On Thu, 11 Oct 2001, Olin Lathrop wrote:

> > Sorry, you're right -- the ADC step itself doesn't double, its
> > relationship to the original signal, which is half of Vref, does.  Now
> > you're taking 1023 steps to measure twice the original signal, rather than
> > 512 steps to measure the original signal.  It's the same thing, isn't it?
>
> If you multiply the original signal by 2 and use a Vref of 2V, you are
> taking 1023 steps to measure the whole original signal.

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2001\10\11@185008 by David Duffy

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Dale wrote:
>There, you see, finally Olin made sense of my addled attempt at answering
>the OP's question.  Thanks, Olin.  Yes, if you amplify the original signal
>you get *higher* resolution relative to the original signal.  In a project
>just recently finished we had to divide a higher (0-20V) input signal by
>4, which gives you *lower* resolution relative to the original signal.
>OK, did I finally get it right?  8-P

The resolution does not change because of the analog scaling applied.
It's determined by how well the max input is matched to the Vref. The
smaller the input max is compared with Vref, the worse the resolution.
Forget what the original input range was - it's not relevant after scaling.

>Damn, it's a good thing I'm not actually working on a PIC project right
>now, I could blow the house up.  Two weeks off work to move into a new
>house is counter-productive - you get no "vacation" from work, and there's
>two weeks' worth of crap waiting when you return to work as well.  That
>and being away from my bench and the projects for even longer and the rust
>rally sets in, doesn't it?

Dale plugs in his new PIC controlled toaster into new house...
WOOF !  SIZZLE !  POP !   :-)
Regards...

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2001\10\12@022531 by Vasile Surducan

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On Wed, 10 Oct 2001, Olin Lathrop wrote:

> > But then, is there a method to do an A/D on an input
> > signal of 0V to 1V on the 16F873 using Vrefs?
>
> Not if you want full range of the A/D without external parts.  To get the
> full A/D range from a 0 to 1V signal requires an external amplifier.
>
>
 There is a method Olin, without any external operational amplifier.
You need only to sum the input signal with a 2V dc. This is quite simple
if you have an AC input signal. However method depends on the source
impedance of the input signal.
Another tested method is to generate a zero impedance floated point
somewhere at 2V. ( but this one required an OA ) Thus you can measure both
positive and negative signals in an unorthodoxe way. Of course diferential
amplifiers is doing the same thing.

Vasile

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2001\10\12@030447 by David Duffy

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>Olin Lathrop wrote:
>
> > > But then, is there a method to do an A/D on an input
> > > signal of 0V to 1V on the 16F873 using Vrefs?
> >
> > Not if you want full range of the A/D without external parts.  To get the
> > full A/D range from a 0 to 1V signal requires an external amplifier.

Vasile wrote:
>   There is a method Olin, without any external operational amplifier.
>You need only to sum the input signal with a 2V dc. This is quite simple
>if you have an AC input signal. However method depends on the source
>impedance of the input signal.

By this do you mean shift the original up closer to Vref? How does this
increase resolution? The shifted signal is still 1V in amplitude. Maybe
I've misunderstood what you mean. I can't see how shifting it makes a
difference. Only a signal with same range as Vref gives full resolution.
The 1V signal is half of the min Vref, so there's no other way to get the
full A/D resolution without amplifying it. Am I missing something here?
Regards...

___________________________________________
David Duffy        Audio Visual Devices P/L
U8, 9-11 Trade St, Cleveland 4163 Australia
Ph: +61 7 38210362   Fax: +61 7 38210281
New Web: http://www.audiovisualdevices.com.au
___________________________________________

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2001\10\12@033511 by Vasile Surducan

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Of course you have right David, I was thinking just to the first question,
resolution can't be improved only if the signal have the same amplitude
like reference. For 87x, unfortunately the +vref can't be +1V.
Vasile

On Fri, 12 Oct 2001, David Duffy wrote:

{Quote hidden}

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2001\10\12@041551 by Vasile Surducan

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On Fri, 12 Oct 2001, Vasile Surducan wrote:

> Of course you have right David, I was thinking just to the first question,
> resolution can't be improved only if the signal have the same amplitude
> like reference. For 87x, unfortunately the +vref can't be +1V.
 But no one stop you to set -vref = +2.0 V and +vref = +4V.
 Summing to variable input ( which is 0 to 1V ) a constant +2V will alow
 to improve the resolution better than measuring 0...1V with -vref = gnd
 and +vref = vcc

 I forgot to tell you that in my last mail. ( also small negative signals
 can be measured
 see: http://www.geocities.com/vsurducan/electro/PIC/f877.html )

 Vasile




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2001\10\12@071557 by Olin Lathrop

face picon face
>   There is a method Olin, without any external operational amplifier.
> You need only to sum the input signal with a 2V dc. This is quite simple
> if you have an AC input signal. However method depends on the source
> impedance of the input signal.
> Another tested method is to generate a zero impedance floated point
> somewhere at 2V. ( but this one required an OA ) Thus you can measure both
> positive and negative signals in an unorthodoxe way. Of course diferential
> amplifiers is doing the same thing.

You can't sum voltages with passive components, only average (or weighted
average) them.  This will always attenuate the input signals at least a
little bit.  With a resistor or two you could level shift the 0-1V signal to
a different range (and loose some range in the process), but that doesn't
solve the problem.  A 0-1V range can't be converted to the full A/D range by
a PIC, no matter where in the 0-5V range this input voltage is presented.
The restriction comes from the spec that Vref+ - Vref- must be at least 2V
to ensure the full accuracy.


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Olin Lathrop, embedded systems consultant in Littleton Massachusetts
(978) 742-9014, EraseMEolinspamspamspamBeGoneembedinc.com, http://www.embedinc.com

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2001\10\12@071615 by Olin Lathrop

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>   But no one stop you to set -vref = +2.0 V and +vref = +4V.
>   Summing to variable input ( which is 0 to 1V ) a constant +2V will alow
>   to improve the resolution better than measuring 0...1V with -vref = gnd
>   and +vref = vcc

True, but why not just Vref+ = 2V, Vref- = GND, and use the orginal 0-1V
signal unmodified.  That will give you 9 bits of accuracy on a 16F87x, same
as your method above.


********************************************************************
Olin Lathrop, embedded systems consultant in Littleton Massachusetts
(978) 742-9014, olinSTOPspamspamspam_OUTembedinc.com, http://www.embedinc.com

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2001\10\15@021913 by Vasile Surducan

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On Fri, 12 Oct 2001, Olin Lathrop wrote:

> >   But no one stop you to set -vref = +2.0 V and +vref = +4V.
> >   Summing to variable input ( which is 0 to 1V ) a constant +2V will alow
> >   to improve the resolution better than measuring 0...1V with -vref = gnd
> >   and +vref = vcc
>
> True, but why not just Vref+ = 2V, Vref- = GND, and use the orginal 0-1V
> signal unmodified.  That will give you 9 bits of accuracy on a 16F87x, same
> as your method above.
>
 Because you can't set Vref+ below +2.5 V. I haven't read the last data
sheet for 877 because I can't see the Microchip site but my tests were
done on two Microchip engineering sample and one purchased chip ( when
there was no info in datasheet about the range for Vref+ and Vref- )
I've quess nothing was change from then.
Also I found that SMP bit from SSPSTAT register is only READABLE ( tested
with simulator, and debugged with LED ) even datasheet say different.

Best regards, Vasile

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