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'[PIC]: Minimal Vref+ value on pic16f877'
2001\09\14@055911 by Also-Antal Csaba

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Hi, all.


 I try  1.1.94V on vref+, and vref- is set to gnd and there are on an0
two resistor connected. One is 15k to vcc and second is a silicon temp
sensor (900-2000 ohm)which is connected to gnd. The voltage on an0 is
.877V and the adc measure 1023 on this pin. No matter what is the input
voltage (gnd or other). If I set the vref+ to vcc then the measured
value is good, but the resolution is unusable.

I would like to know what is the minimal useable voltage on vref+? Or
has someone met this problem?

udv
Csaba

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2001\09\14@061427 by Vasile Surducan

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see http://www.geocities.com/vsurducan/pic/f877.html

however you can't go under +2.5V on vref+

Vasile



On Fri, 14 Sep 2001, Also-Antal Csaba wrote:

{Quote hidden}

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2001\09\14@073753 by Alan B. Pearce

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You have sent this question in while I was pondering on the lack of
information on what one can do with Vref- :)

What I am trying to do is use an AD590 temperature sensor as a temperature
monitor. This gives 1uA/K and has a usable range from -55C to 150C (or
thereabouts). My usage is for the range 0C to somewhere in the region of
100C (as I am looking for over temperature monitoring on resistive loads).
Using a 10Kohm resistor as a load I will end up with a voltage output of
2.73V to around 3.73V, so I am looking to set Vref- to around 2.7V. I take
it that this will then represent the 000 reading of the ADC, and that if I
use 3.7V for Vref+, that will represent 7FF for the ADC.

I am trying to simplify the conversion from binary to an ASCII string that
represents the temperature by suitably scaling the conversion range (I can
truncate the string being transmitted to stop unnecessary "accuracy").

However the questions that spring to mind come from what has been said
previously about Vref+ needing to be greater than about 3V. Is this the
minimum voltage Vref+ to Vref- that needs to be 3V? If this is the case I
will have Vref- internally set to Vss and free up a pin, and have to
subtract the 2.73V value from the ADC before doing the binary to ASCII
conversion. What is does do is divide the accuracy by nearly 4 if I need to
do it this way.

I do have another option which is to connect the current sink resistor to a
suitable negative voltage, and by using a larger value of resistor make use
of the ADC voltage range, but would like to use this method only if I
absolutely have to.

Does anyone have any real experience with trying to do things like this and
offer advice before I commit too deeply to unnecessary circuitry.

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2001\09\14@082153 by Olin Lathrop

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> I would like to know what is the minimal useable voltage on vref+?

I don't know, but isn't this in the data sheet?  And if it's not, how
meaningful would an answer from someone on the list be unless it was an
official Microchip statement?


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

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2001\09\14@083230 by Also-Antal Csaba

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Vasile Surducan wrote:
>
> see http://www.geocities.com/vsurducan/pic/f877.html
>
> however you can't go under +2.5V on vref+

Hmm, and what about the Vref-? Is the max voltage is +2.5V there?

udv
Csaba

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2001\09\14@100753 by Dave Johnson

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Alll this stuff is in the data sheet, of course:

First, there has to be at least a 2V difference between Vref+ and
Vref-. Vref+ can range from Vdd - 2.5V to Vdd + 0.3V, Vref- can range
from Vss - 0.3V to Vref+ - 2.0V.

Dave Johnson

{Quote hidden}

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2001\09\14@102249 by Alan B. Pearce

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>Alll this stuff is in the data sheet, of course:

>First, there has to be at least a 2V difference between Vref+ and
>Vref-. Vref+ can range from Vdd - 2.5V to Vdd + 0.3V, Vref- can range
>from Vss - 0.3V to Vref+ - 2.0V.

Perhaps you could point me at which page of the 16F87x data sheet it is on
then, because the only reference to the value of Vref I could find was on
page 175 of DS30292c where the clock freq is given for Vref >= 3V. The
reference manual (DS33023a) does not seem to contain any more than the data
sheet that I could see which is why I asked the question. The mode I want to
use does not seem to be mentioned anywhere at all.

Appreciate your figures anyway.

thanks

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2001\09\14@114508 by Also-Antal Csaba

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Olin Lathrop wrote:
>
> > I would like to know what is the minimal useable voltage on vref+?
>
> I don't know, but isn't this in the data sheet?  And if it's not, how
> meaningful would an answer from someone on the list be unless it was an
> official Microchip statement?

I cant read pdf doc's on my mobil phone and I think not I the first man
who want to use it.

Thanks for your golden information.

udv
Csaba

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2001\09\14@121332 by Dave Johnson

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At 3:20 PM +0100 9/14/01, Alan B. Pearce wrote:
>Perhaps you could point me at which page of the 16F87x data sheet it is on
>then, because the only reference to the value of Vref I could find was on
>page 175 of DS30292c where the clock freq is given for Vref >= 3V.

The info is in the table on page 174 of DS30292C.

Dave Johnson

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2001\09\16@193707 by Thomas McGahee

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Personally, I would use the AD590 to provide current to an op-amp
"current to voltage converter". Scaling is a function of the feedback
resistor. For example, you are interested in a range from 0 to 100 degrees
centigrade. That is a *change* of 100 degrees Kelvin, which would be
a change of 100 ua. Let's say you want to scale this to produce
an output that varies from 0-2.50 volts in 1000 steps: you would
get .1 degree for each .025 volts. (recall that you get 2048 counts for
a 0-5.12 volt input assuming your reference is 5.12 v)

Note that using 1000 steps keeps the math simple, since only the
position of the decimal point has to be changed to accomplish the required
divide-by-ten function. 984 counts would display as 98.4 degreees celsius.

Note that the AD590 operates based on the Kelvin scale where 273
degrees K is [0 degrees centigrade]. To "bias" our "current to
voltage converter" circuit to operate in the proper way we must
provide a source of -273 ua to the input of the converter. Thus
when the AD590 is producing +273 ua, the -273 ua will cancel it
and a total of 0 ua will flow into the converter.

Then an ADDITIONAL 100 ua must get converted to 2.50 volts by using a
50K feedback resistor. Actually you would want the 50K feedback
resistance to be variable so you can tweak the gain since most
people will be using a 5 v reference instead of the 5.12 reference
assumed in this design example. A 47K resistor in series with a
6.8K pot would allow variation from 47k to 53.8k which should be
adequate.
                  zero adj                           gain adj
          8.2k      2.5k                    47k      6.8k
-2.5v ->-/\/\/\----/\/\/\------*-----------/\/\/\---/\/\/\---+
                              |    |\                        |
                              | (-)| \                       |
AD590 current source->---------*----|  \----------------------*--->- to
inverter
                                   |  /
                                (+)| /  op-amp
                           GND-----|/


Note that another inverting stage is required to flip polarity to positive.
Output of invert goes to 1k resistor to A/D i/o pin.

Requires a +/- 5v supply or greater.

Fr. Tom McGahee

{Original Message removed}

2001\09\16@201619 by Robert Rolf

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KISS.
Why not use a single supply Op-Amp that goes rail to rail (LMC6462
comes to mind)? Use the + input to bias the output to + full scale
for temp of 0 in (zero adj), and just invert the A/D conversion in
software.
Then you don't need a -5V supply.
R

Thomas McGahee wrote:
{Quote hidden}

> {Original Message removed}

2001\09\17@101047 by Alan B. Pearce

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>The info is in the table on page 174 of DS30292C.

Doh how did I miss that? Thanks for the pointer.

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2001\09\17@103221 by Alan B. Pearce

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>KISS
>Why not use a single supply Op-Amp that goes rail to rail (LMC6462
>comes to mind)? Use the + input to bias the output to + full scale
>for temp of 0 in (zero adj), and just invert the A/D conversion in
>software.
>Then you don't need a -5V supply.

>Thomas McGahee wrote:
>>
>> Personally, I would use the AD590 to provide current to an op-amp
>> "current to voltage converter". Scaling is a function of the feedback
>> resistor. For example, you are interested in a range from 0 to 100
degrees
>> centigrade. That is a *change* of 100 degrees Kelvin, which would be
>> a change of 100 ua. Let's say you want to scale this to produce
>> an output that varies from 0-2.50 volts in 1000 steps: you would
>> get .1 degree for each .025 volts. (recall that you get 2048 counts for
>> a 0-5.12 volt input assuming your reference is 5.12 v)

The kiss idea is what I am trying to do. By using a 10K resistor in the
bottom leg of the AD590 you get the current to voltage conversion done nice
and easily in a manner that should feed the A/D in the pic OK. Worst case I
will need a single voltage follower op-amp after a MUX (I will have between
8 and 16 sensors). It converts the 1uA/K to 10mV/K and I am then hoping to
introduce the offsets by setting the Vref+ and Vref- to appropriate values.
As Vref+ to Vref- needs to be 2V min, it looks like I will need to be
setting my values from -50C to +150C, and possibly doing a single constant
subtraction to get it all within an easily convertible range.

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