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'[PIC]:strange analog in problem'
2002\06\11@121619
by
Micro Eng
|
Using the F877, I have PortA configured for analog inputs, and had 10K pots.
Seemed to work ok, I had a watch window set up, and a breakpoint right after
the read sequences (for 4 pots).
I've changed the pots to 50K now, and upped the aquisition time, but the
values seem to be way off now, and are affected by each other.
ie...
if I read just one, it is close...but not exact. If I put the second pot in
the loop to read, the first one goes way off scale. I've read the actual
voltages on the pins, and they are not changing.
Anyone ever seen where just reading a second port affects the value of the
first one?
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2002\06\11@123629
by
Lyle Killough
. . . Disclaimer . . . I have never even read the F877 data sheet
Are you maybe having trouble charging the input capacitance of the ADC
through 50k? I kind of remember a 5k recommended maximum on the data
sheets for some of the other PICs with analog inputs.
Lyle
> {Original Message removed}
2002\06\11@124431
by
Drew Vassallo
>I've changed the pots to 50K now, and upped the aquisition time, but the
>values seem to be way off now, and are affected by each other.
As I recall, the maximum recommended input impedence for analog sources is
10k, according to the datasheets. I'm sure higher impedances would require
longer acquisition times, but I'm NOT sure that there are no other problems.
I'd keep it around 5k to 10k just to be safe. I'm sure someone with
"internal" knowledge of the PIC's construction will chime in and debunk
Microchip's recommendations, but I tend to stick by their datasheets just to
be safe.
--Andrew
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2002\06\11@124438
by
Dale Botkin
On Tue, 11 Jun 2002, Lyle Killough wrote:
> . . . Disclaimer . . . I have never even read the F877 data sheet
I have...
> Are you maybe having trouble charging the input capacitance of the ADC
> through 50k? I kind of remember a 5k recommended maximum on the data
> sheets for some of the other PICs with analog inputs.
10K. That's the spec maximum input impedance, meaning you can use a 20K
pot at most (right??).
Dale
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2002\06\11@124659
by
Peter Betts
See note 3 under A/D Acquisition Requirements on the 16F877 datasheet.
QUOTE: "The maximum recommended impedance for analog sources is 10K
This is required to meet the pin leak-age specification."
This may have something to do with it. Pin leakage? Maybe that's what you are seeing.
Pete
> Using the F877, I have PortA configured for analog inputs,
> and had 10K pots.
> I've changed the pots to 50K now
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2002\06\11@125220
by
Thomas McGahee
You must:
1) Switch to the first analog input
2) WAIT long enough for the circuitry to settle
3) Initiate a conversion cycle
4) Wait for end of conversion
5) Read the first analog input.
1') Switch to the second analog input
2') WAIT long enough for the circuitry to settle
3') Initiate a conversion cycle
4') Wait for end of conversion
5') Read the second analog input
Step 2 is the one that most people leave out, and it
cause the readings to be way out of whack.
Fr. Thomas McGahee
{Original Message removed}
2002\06\11@125701
by
Brendan Moran
Little higher in this PIC
From page 116, section 11.1 of the 16F87x datasheet,
The source impedance (RS) and the internal sampling switch (RSS)
impedance directly affect the time required to charge
the capacitor CHOLD. The sampling switch (RSS)
impedance varies over the device voltage (VDD), see
Figure 11-2. The maximum recommended impedance
for analog sources is 10 kohm. As the impedance
is decreased, the acquisition time may be decreased.
this acquisition must be done before the conversion
can be started.
Hope that helps,
Brendan
{Original Message removed}
2002\06\11@125912
by
Dwayne Reid
|
At 10:14 AM 6/11/02 -0600, Micro Eng wrote:
>Using the F877, I have PortA configured for analog inputs, and had 10K pots.
>Seemed to work ok, I had a watch window set up, and a breakpoint right after
>the read sequences (for 4 pots).
>
>I've changed the pots to 50K now, and upped the aquisition time, but the
>values seem to be way off now, and are affected by each other.
>
>ie...
>
>if I read just one, it is close...but not exact. If I put the second pot in
>the loop to read, the first one goes way off scale. I've read the actual
>voltages on the pins, and they are not changing.
>Anyone ever seen where just reading a second port affects the value of the
>first one?
You have 2 potential problems: input impedance and acquisition
time. Microchip specifies that your source impedance should be no higher
than 10K - you have a worst case source impedance of 25K (when the pot is
in the exact middle). The cure is simple - add a 0.1 - 1 uF cap from the
analog input to gnd. The cap supplies the charge required when the input
is sampled and restores full accuracy to the a/d input.
The 2nd problem is acquisition time. You have to allow a minimum setup
time between when you select the desired analog input pin and when you
start the a/d conversion. This time is specified fully in the data sheet
but is typically 30 us or so. In other words, do this: select a/d pin, do
something useful or waste time for at least 30 us, start a/d conversion.
Note that when you are processing the result of a conversion that has
already completed, you can select the next desired a/d channel BEFORE you
actually start processing the data. Changing the a/d input does not
corrupt the a/d result register (ADRES).
Deal with both of those issues and see if the problems go away.
dwayne
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2002\06\11@143421
by
David Minkler
Dale,
Wrong. 40k max. Assume the pot is wired from a pair of stiff sources
(Vcc/Vdd and GND for example). Worst case (highest impedance) will be
with the wiper at mid-range. Thats two 20k resistors in parallel for an
equivalent wiper (input) impedance of 10k (as recommended in the data
sheets). Of course, I haven't seen a lot of 40k pots advertised lately
(25k is a standard) so in that sense, maybe you are (approximately)
right.
Regards,
Dave
Dale Botkin wrote:
> 10K. That's the spec maximum input impedance, meaning you can use a 20K
> pot at most (right??).
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2002\06\11@145504
by
Dale Botkin
|
Hrmm... not sure I'm with you on that one. With the wiper centered, your
input would still see 20K Ohms to either source. Maybe I'm just slow or
theory-challenged, but I don't see how two resistors to two different
sources would qualify as "parallel". I could be wrong though, here I am
posting from work again, which results in my being wrong about 3 out of 4
times it seems.
Dale
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On Tue, 11 Jun 2002, David Minkler wrote:
> Wrong. 40k max. Assume the pot is wired from a pair of stiff sources
> (Vcc/Vdd and GND for example). Worst case (highest impedance) will be
> with the wiper at mid-range. Thats two 20k resistors in parallel for an
> equivalent wiper (input) impedance of 10k (as recommended in the data
> sheets). Of course, I haven't seen a lot of 40k pots advertised lately
> (25k is a standard) so in that sense, maybe you are (approximately)
> right.
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2002\06\11@145918
by
Brendan Moran
Isn't that only how it works under AC cnditions, where both DC sources are
considered the same?
Brendan
{Original Message removed}
2002\06\11@152252
by
Olin Lathrop
> Using the F877, I have PortA configured for analog inputs, and had 10K
pots.
> Seemed to work ok, I had a watch window set up, and a breakpoint right
after
> the read sequences (for 4 pots).
>
> I've changed the pots to 50K now, and upped the aquisition time, but the
> values seem to be way off now, and are affected by each other.
Microchip specifies the A/D source should be 10Kohm or less. Yours could be
2.5 times higher, so you should either use 20Kohm pots max, or buffer the
voltages.
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(978) 742-9014, http://www.embedinc.com
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2002\06\11@154124
by
Harold M Hallikainen
|
I don't have an F877 datasheet handy (the 18c452 datasheet I DO have
handy says 2.5k max for its 10 bit A/D). But assuming it is indeed a 10k
max, that'd allow a 40k pot. The maximum Thevinin resistance is Rpot/4.
If you have a 40k pot, the Thevinin resistance is 20k//20k or 10k.
For more on Thevenin Equivalents, see
http://www.hallikainen.org/rw/theory/theory5.html
Harold
On Tue, 11 Jun 2002 11:42:29 -0500 Dale Botkin <.....daleKILLspam
@spam@BOTKIN.ORG> writes:
{Quote hidden}> On Tue, 11 Jun 2002, Lyle Killough wrote:
>
> > . . . Disclaimer . . . I have never even read the F877 data
> sheet
>
> I have...
>
> > Are you maybe having trouble charging the input capacitance of the
> ADC
> > through 50k? I kind of remember a 5k recommended maximum on the
> data
> > sheets for some of the other PICs with analog inputs.
>
> 10K. That's the spec maximum input impedance, meaning you can use a
> 20K
> pot at most (right??).
>
> Dale
>
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2002\06\11@154532
by
Olin Lathrop
> Microchip specifies that your source impedance should be no higher
> than 10K - you have a worst case source impedance of 25K (when the pot is
> in the exact middle). The cure is simple - add a 0.1 - 1 uF cap from the
> analog input to gnd. The cap supplies the charge required when the input
> is sampled and restores full accuracy to the a/d input.
Not so fast. Yes, a large cap will hold the pin voltage constant during the
brief acquisition interval, but accuracy may still be off and the "cross
talk" between channels may become worse. If the A/D input is constantly
cycled between channels, then a current path is effectively created from
each channel to the previous channel. The magnitude of this current is
proportional to the capacitor in the PIC and the full input channels cycle
frequency. Given such a steady charge pump pin current, the external
capacitor will average to an offset error of the current times the source
impedence. Since the large external capacitor is holding the offset error,
it will not have time to decay away during a normal acquisition time.
*****************************************************************
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(978) 742-9014, http://www.embedinc.com
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2002\06\11@154853
by
Olin Lathrop
> Wrong. 40k max. Assume the pot is wired from a pair of stiff sources
> (Vcc/Vdd and GND for example). Worst case (highest impedance) will be
> with the wiper at mid-range. Thats two 20k resistors in parallel for an
> equivalent wiper (input) impedance of 10k (as recommended in the data
> sheets). Of course, I haven't seen a lot of 40k pots advertised lately
> (25k is a standard) so in that sense, maybe you are (approximately)
> right.
Oh, yeah, duh! I made the same mistake in an earlier post.
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2002\06\11@155208
by
Olin Lathrop
> > Using the F877, I have PortA configured for analog inputs, and had 10K
> pots.
> > Seemed to work ok, I had a watch window set up, and a breakpoint right
> after
> > the read sequences (for 4 pots).
> >
> > I've changed the pots to 50K now, and upped the aquisition time, but the
> > values seem to be way off now, and are affected by each other.
>
> Microchip specifies the A/D source should be 10Kohm or less. Yours could
be
> 2.5 times higher, so you should either use 20Kohm pots max, or buffer the
> voltages.
As someone else pointed out, this is off by a factor of two. The highest
impedence from a 50Kohm pot tied to ground and supply at the fixed ends is
12.5Kohms, not 25Kohms. You should not use anything higher than a 40Kohm
pot, assuming the fixed ends are driven with 0 impedence.
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(978) 742-9014, http://www.embedinc.com
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2002\06\11@162838
by
David Minkler
|
Brendan and Dale,
The pot with the centered wiper, end terminals connected to STIFF
sources (like Vcc and GND), is a Thevenin equivalent to Rpot/4 from (in
the case of this example) a source voltage of Vcc/2.
This offers you some interesting possibilities for de-linearizing a
pot. In one motion control application we needed a pot to control a
(large) servo so that with the pot centered we got no (very little)
motion. With the pot turned half way from centered to either side (1/4
and 3/4) we got 10% velocity and at the extremes we got full velocity.
Using a 5k pot with the end terminals connected to + and -10V with a 232
ohm de-linearizing resistor from the wiper to ground we got exactly the
curve we were hoping for. Not much response near center ... lots of
response at the ends.
Regards,
Dave
Brendan Moran wrote:
>
> Isn't that only how it works under AC cnditions, where both DC sources are
> considered the same?
>
> Brendan
>
> {Original Message removed}
2002\06\11@165416
by
Dwayne Reid
|
At 01:53 PM 6/11/02 -0500, Dale Botkin wrote:
>Hrmm... not sure I'm with you on that one. With the wiper centered, your
>input would still see 20K Ohms to either source. Maybe I'm just slow or
>theory-challenged, but I don't see how two resistors to two different
>sources would qualify as "parallel".
yeah - but they are indeed in parallel in terms of the impedance seen at
the wiper. I made the same mistake 'cuz I was in too much of a hurry.
dwayne
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2002\06\11@165419
by
Dwayne Reid
|
At 03:42 PM 6/11/02 -0400, Olin Lathrop wrote:
> > Microchip specifies that your source impedance should be no higher
> > than 10K - you have a worst case source impedance of 25K (when the pot is
> > in the exact middle). The cure is simple - add a 0.1 - 1 uF cap from the
> > analog input to gnd. The cap supplies the charge required when the input
> > is sampled and restores full accuracy to the a/d input.
>
>Not so fast. Yes, a large cap will hold the pin voltage constant during the
>brief acquisition interval, but accuracy may still be off and the "cross
>talk" between channels may become worse.
Nope - I disagree here. Been there, done that, spent rather a long time
verifying performance. Cross talk is definitely NOT made worse by adding
the cap.
But I should qualify my statement: the addition of the cap reduces the
effect of series impedance so long as the sample rate is rate is slow
enough that the charge removed (or added) during the sample process is
completely restored before the next sample occurs. A quick look at the
ratio between how long the sample switch is closed on a particular channel
vs how long it is open suggests that this is not a problem if sampling only
2 channels at the fastest possible sample rate and definitely not a problem
is sampling 4 channels or more.
Its absolutely not a problem with 100K series resistors and 1 uF caps when
sampling channels at a 1 ms rate - again, I spent a lot of time verifying
accuracy in one project before deciding it was OK. But take away the
capacitors - not acceptable.
dwayne
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(780) 489-3199 voice (780) 487-6397 fax
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2002\06\11@170833
by
Doug Hemingway
|
A "stiff" power supply means one that has nearly zero impedance. That means
that both ends of the pot, for purposes of analysis, are at the same
impedance point (since a power supply can be represented as a pure voltage
source with either a parallel or series resistor equal to the output
impedance). Ahhh, Thevenin and Norton! How I miss them--NOT!!
Doug Hemingway
Dale Botkin at EraseMEdalespam_OUT
TakeThisOuTBOTKIN.ORG wrote on 6/11/02 2:53 PM:
> Hrmm... not sure I'm with you on that one. With the wiper centered, your
> input would still see 20K Ohms to either source. Maybe I'm just slow or
> theory-challenged, but I don't see how two resistors to two different
> sources would qualify as "parallel". I could be wrong though, here I am
> posting from work again, which results in my being wrong about 3 out of 4
> times it seems.
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2002\06\11@171733
by
Dave Dilatush
|
Micro Eng wrote...
>Using the F877, I have PortA configured for analog inputs, and had 10K pots.
>Seemed to work ok, I had a watch window set up, and a breakpoint right after
>the read sequences (for 4 pots).
>
>I've changed the pots to 50K now, and upped the aquisition time, but the
>values seem to be way off now, and are affected by each other.
This is a symptom of not allowing enough acquisition time.
Whatever acquisition time you're allowing, try increasing it by a
factor of ten or so, just as an experiment, and see if the
problem goes away.
If it does, you've found the problem.
Others have pointed out in this thread that Microchip recommends
the PIC A/D inputs be driven from no more than 10K ohms
impedance.
That's correct, but understand that 10K is merely the maximum
driving impedance allowable for the A/D to perform at its full
rated accuracy- it's **NOT** a maximum beyond which operation
will cease or go haywire or cause the heavens to fall or cause
the PIC to emit smoke and noxious gases.
For source impedances a lot higher than 10K ohms there can be
significant DC errors due to pin leakage, especially at high
temperatures. In some applications a small error would be
tolerable, in some it would not. In your case, you are not
exceeding the 10K maximum spec by enough that you would expect
see any significant DC errors; at midscale (i.e., worst case),
the source impedance seen at your pot wipers is only 12.5K ohms.
My best guess is that before, with your 10K pots, whatever
acquistion time you were providing was marginally adequate; and
now, with 50K pots, it isn't enough and the readings are
"bleeding over" into one another.
Hope this helps...
DD
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2002\06\11@172811
by
Dave Dilatush
Dale wrote...
>Maybe I'm just slow or theory-challenged, but I don't
>see how two resistors to two different sources would
>qualify as "parallel".
"Slow" you're definitely not; theory-challenged, perhaps.
For impedance-determining purposes the two halves of the pot do
indeed appear as if they were connected in parallel.
All this is related to a thing called Thevenin's Theorem:
http://hyperphysics.phy-astr.gsu.edu/hbase/electric/thevenin.html
DD
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2002\06\11@193921
by
Sean H. Breheny
Hi all,
Perhaps I'm not understanding something here, but how does the cap help
with the leakage current? Isn't their always leakage current on the pin,
regardless of whether you are sampling or not?
Sean
At 02:50 PM 6/11/2002 -0600, you wrote:
Nope - I disagree here. Been there, done that, spent rather a long time
{Quote hidden}>verifying performance. Cross talk is definitely NOT made worse by adding
>the cap.
>
>But I should qualify my statement: the addition of the cap reduces the
>effect of series impedance so long as the sample rate is rate is slow
>enough that the charge removed (or added) during the sample process is
>completely restored before the next sample occurs. A quick look at the
>ratio between how long the sample switch is closed on a particular channel
>vs how long it is open suggests that this is not a problem if sampling only
>2 channels at the fastest possible sample rate and definitely not a problem
>is sampling 4 channels or more.
>
>Its absolutely not a problem with 100K series resistors and 1 uF caps when
>sampling channels at a 1 ms rate - again, I spent a lot of time verifying
>accuracy in one project before deciding it was OK. But take away the
>capacitors - not acceptable.
>
>dwayne
>
>--
>Dwayne Reid <
dwayner
spam_OUTplanet.eon.net>
>Trinity Electronics Systems Ltd Edmonton, AB, CANADA
>(780) 489-3199 voice (780) 487-6397 fax
>
>Celebrating 18 years of Engineering Innovation (1984 - 2002)
> .-. .-. .-. .-. .-. .-. .-. .-. .-. .-
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>
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2002\06\11@200237
by
Bob Ammerman
|
----- Original Message -----
From: "Dwayne Reid" <@spam@dwaynerKILLspam
PLANET.EON.NET>
To: <KILLspamPICLISTKILLspam
MITVMA.MIT.EDU>
Sent: Tuesday, June 11, 2002 12:54 PM
Subject: Re: [PIC]:strange analog in problem
> At 10:14 AM 6/11/02 -0600, Micro Eng wrote:
> >Using the F877, I have PortA configured for analog inputs, and had 10K
pots.
> >Seemed to work ok, I had a watch window set up, and a breakpoint right
after
> >the read sequences (for 4 pots).
> >
> >I've changed the pots to 50K now, and upped the aquisition time, but the
> >values seem to be way off now, and are affected by each other.
> >
> >ie...
> >
> >if I read just one, it is close...but not exact. If I put the second pot
in
> >the loop to read, the first one goes way off scale. I've read the actual
> >voltages on the pins, and they are not changing.
> >Anyone ever seen where just reading a second port affects the value of
the
> >first one?
>
> You have 2 potential problems: input impedance and acquisition
> time. Microchip specifies that your source impedance should be no higher
> than 10K - you have a worst case source impedance of 25K (when the pot is
> in the exact middle). The cure is simple - add a 0.1 - 1 uF cap from the
> analog input to gnd. The cap supplies the charge required when the input
> is sampled and restores full accuracy to the a/d input.
This is _not_ a complete cure because, in addition to the CHOLD charge time,
there is leakage on the input pin that can pull the voltage away from where
the pot sets it.
Bob Ammerman
RAm Systems
{Quote hidden}> The 2nd problem is acquisition time. You have to allow a minimum setup
> time between when you select the desired analog input pin and when you
> start the a/d conversion. This time is specified fully in the data sheet
> but is typically 30 us or so. In other words, do this: select a/d pin, do
> something useful or waste time for at least 30 us, start a/d conversion.
>
> Note that when you are processing the result of a conversion that has
> already completed, you can select the next desired a/d channel BEFORE you
> actually start processing the data. Changing the a/d input does not
> corrupt the a/d result register (ADRES).
>
> Deal with both of those issues and see if the problems go away.
>
> dwayne
>
> --
> Dwayne Reid <
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> Trinity Electronics Systems Ltd Edmonton, AB, CANADA
> (780) 489-3199 voice (780) 487-6397 fax
>
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2002\06\11@200246
by
Bob Ammerman
> Hrmm... not sure I'm with you on that one. With the wiper centered, your
> input would still see 20K Ohms to either source. Maybe I'm just slow or
> theory-challenged, but I don't see how two resistors to two different
> sources would qualify as "parallel". I could be wrong though, here I am
> posting from work again, which results in my being wrong about 3 out of 4
> times it seems.
>
> Dale
Here is a rough analogy to help you out....
Think of the two halves of the pot as compression springs, with spring
tension inversly proportional to resistance.
Now the way that they resist motion of the point at which they connect is
certainly dependent on both of them.
Bob Ammerman
RAm Systems
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2002\06\11@205822
by
Dwayne Reid
|
At 07:36 PM 6/11/02 -0400, Sean H. Breheny wrote:
>Hi all,
>
>Perhaps I'm not understanding something here, but how does the cap help
>with the leakage current? Isn't their always leakage current on the pin,
>regardless of whether you are sampling or not?
Leakage current is not the primary error source here. The S/H capacitor on
the input of the a/d convertor presents a sudden 50 pF load to the input -
the droop that is caused while that cap charges is the error source. Try
it sometime - write some code that reads the voltage on an analog
pin. Apply some arbitrary voltage to an a/d input through a 10K resistor
and look at voltage on the pin with a scope. Now change the resistor to
100K and look at it again. Now add a 1 uF cap between the pin and ground
and look once more.
dwayne
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2002\06\11@205829
by
Dwayne Reid
|
At 01:34 PM 6/11/02 -0400, Bob Ammerman wrote:
> >
> > You have 2 potential problems: input impedance and acquisition
> > time. Microchip specifies that your source impedance should be no higher
> > than 10K - you have a worst case source impedance of 25K (when the pot is
> > in the exact middle). The cure is simple - add a 0.1 - 1 uF cap from the
> > analog input to gnd. The cap supplies the charge required when the input
> > is sampled and restores full accuracy to the a/d input.
>
>This is _not_ a complete cure because, in addition to the CHOLD charge time,
>there is leakage on the input pin that can pull the voltage away from where
>the pot sets it.
True. I did not mean to imply that the cap cures all the effects of too
high a source impedance and perhaps I should have stated that in my initial
post. But the predominant error source seems to be from the S/H cap and
not leakage.
The message that started this thread referred to a specific problem that
was caused by either of 2 problems: not enough acquisition time or too high
a source impedance. I goofed on the source impedance - too quick on the
keyboard. But *if* the source impedance had in fact been 25K and if the
error was caused by that too-high source impedance (not acquisition time),
adding the cap as I stated would have cured the problem.
dwayne
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2002\06\11@214132
by
Tal (Zapta)
If I recall correctly, the general idea is that (perfect) voltage sources
have 0 impedance and (perfect) current sources have infinite impedance. If
you consider the 0 impedance of the voltage source, the two sides are
actually 'connected' in parallel.
I hope I got it right. ;-)
Tal
> {Original Message removed}
2002\06\12@012156
by
Dale Botkin
2002\06\12@014616
by
Peter Betts
|
I've got a bit lost now but wasn't part of the original problem that the seperate ADC channels were being "mixed up" and I believed this to be caused by the >10K input impendance which fails to meet the required pin leak-age specification and thus the pins leak?
See note 3 under A/D Acquisition Requirements on the 16F877 datasheet.
QUOTE: "The maximum recommended impedance for analog sources is 10K
This is required to meet the pin leak-age specification."
Surely no matter how much aquisition time you allow once you've got leakage your stuck!
Isn't the solution to simply reduce the input impedance?
I understand the S/H idea but why do you think the problem has got nothing to do with leakage?
How else do you explain the interaction BETWEEN channels? Or is it because all channels use the same S/H circuitry or something?
Regards
Pete
> >Perhaps I'm not understanding something here, but how does
> the cap help
> >with the leakage current? Isn't their always leakage current
> on the pin,
> >regardless of whether you are sampling or not?
>
> Leakage current is not the primary error source here. The
> S/H capacitor
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2002\06\12@020713
by
Vasile Surducan
|
On Tue, 11 Jun 2002, Dwayne Reid wrote:
> At 01:34 PM 6/11/02 -0400, Bob Ammerman wrote:
> > >
> > > You have 2 potential problems: input impedance and acquisition
> > > time. Microchip specifies that your source impedance should be no higher
> > > than 10K - you have a worst case source impedance of 25K (when the pot is
> > > in the exact middle). The cure is simple - add a 0.1 - 1 uF cap from the
> > > analog input to gnd. The cap supplies the charge required when the input
> > > is sampled and restores full accuracy to the a/d input.
> >
> >This is _not_ a complete cure because, in addition to the CHOLD charge time,
> >there is leakage on the input pin that can pull the voltage away from where
> >the pot sets it.
>
> True. I did not mean to imply that the cap cures all the effects of too
> high a source impedance and perhaps I should have stated that in my initial
> post. But the predominant error source seems to be from the S/H cap and
> not leakage.
>
Right. The input impedance can be higher than 10k. But the SH capacitor
must have enough time to charge. The problem is not so evident using just
one ad input. But the required aquisition time must be at least doubled
when more than one ad input is used and imput impedance is 10...20K. Our
friend from Israel forgot to use the minimum delay of 20uS or so after
channel switching. He has no problem with the impedance for sure.
But may have problems with cumulative noises because the filtering cap at
high input impedances is not recommended.
Vasile
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2002\06\12@031200
by
Michael Rigby-Jones
|
> -----Original Message-----
> From: Dwayne Reid [SMTP:RemoveMEdwayner
TakeThisOuTPLANET.EON.NET]
> Sent: Wednesday, June 12, 2002 1:53 AM
> To: PICLISTEraseME
.....MITVMA.MIT.EDU
> Subject: Re: [PIC]:strange analog in problem
>
> True. I did not mean to imply that the cap cures all the effects of too
> high a source impedance and perhaps I should have stated that in my
> initial
> post. But the predominant error source seems to be from the S/H cap and
> not leakage.
>
The datasheet shows a typical leakage current of 500nA. With a 100K source
impedance this could give an error of 100k*500n = 50mV. On a typical PIC
using (5v) Vcc as a reference this would give around 10 counts of error
(around 3.3 bits). If this was constant then it may not be a problem, but
leakage current by their very nature are quite temperature dependant.
Regards
Mike
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2002\06\12@035857
by
Alan B. Pearce
>but I don't see how two resistors to two different
>sources would qualify as "parallel".
Go back to basic circuit theory and try the Thevinen (sp?) equivalent
circuit.
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2002\06\12@041657
by
Jonathan Starr
|
Put a 0.01uF tant across each analog input, I read somewhere in microchips app notes about an issue with more than one analog input.
Something like sampling one causes a current drop on the port, then when you do the other one the reference point is still changing _while_ you sample so it gets messed up, I can't remember exactly but it definitely said use a cap to cure the problem I'll try to find the app note for you..
Hope this helps.
Jonathan Starr
>>> Micro Eng <EraseMEmicro_eng
HOTMAIL.COM> 11/06/2002 17:14:35 >>>
Using the F877, I have PortA configured for analog inputs, and had 10K pots.
Seemed to work ok, I had a watch window set up, and a breakpoint right after
the read sequences (for 4 pots).
I've changed the pots to 50K now, and upped the aquisition time, but the
values seem to be way off now, and are affected by each other.
ie...
if I read just one, it is close...but not exact. If I put the second pot in
the loop to read, the first one goes way off scale. I've read the actual
voltages on the pins, and they are not changing.
Anyone ever seen where just reading a second port affects the value of the
first one?
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2002\06\12@042320
by
Vasile Surducan
On Wed, 12 Jun 2002, Alan B. Pearce wrote:
> >but I don't see how two resistors to two different
> >sources would qualify as "parallel".
>
> Go back to basic circuit theory and try the Thevinen (sp?) equivalent
> circuit.
Come on Alan ! ;-)
Don't send him to the basics, is much easy considering the gnd and the
Vcc with 0 impedance , gnd because is ground and Vcc because it cames from
a zero output impedance stabilizer, so you have two resistors connected in
parallel because vcc and gnd are equal to 0 ( as impedance and not as
values )
Of course the crude reality is a little different...
Vasile
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2002\06\12@043812
by
Alan B. Pearce
> Come on Alan ! ;-)
> Don't send him to the basics, is much easy considering
>the gnd and the Vcc with 0 impedance , gnd because is
>ground and Vcc because it cames from a zero output impedance
>stabilizer, so you have two resistors connected in parallel
>because vcc and gnd are equal to 0 ( as impedance and not as
>values )
So you have just explained the very basics that I was suggesting he go back
to. :)
After all this is how a Thevinin circuit configuration is arrived at, by
considering voltage sources to be a 0 ohm impedance.
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2002\06\12@063210
by
Bob Ammerman
2002\06\12@075453
by
Olin Lathrop
> If I recall correctly, the general idea is that (perfect) voltage sources
> have 0 impedance and (perfect) current sources have infinite impedance. If
> you consider the 0 impedance of the voltage source, the two sides are
> actually 'connected' in parallel.
>
> I hope I got it right. ;-)
Yes, you did.
*****************************************************************
Embed Inc, embedded system specialists in Littleton Massachusetts
(978) 742-9014, http://www.embedinc.com
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2002\06\12@103819
by
Micro Eng
|
Some more information....
after reading thru the posts.....seems to be a couple of avenues to go down.
The first is increase the setup time. Second is to add a capacitor to ground
on the pin. Haven't tried this, as it seems to be in discussion if this is
even a correct approach.
There is two problems still...accuracy and multiple channel reads.
I changed the code to read in just a single input,and modified the setup
delay values. I started with 255, ranged down to 10.
; xtal is 4.194 Mhz
AD_read
movwf ADCON0
call ADSetup
bsf ADCON0,GO ; Start first A/D conversion
btfsc ADCON0,GO
goto $-1
movf ADRESH,W ; Get A/D value and put in W reg
return ; return with value in W
ADSetup
movlw d'5'
movwf DelayAD
decfsz DelayAD,f
goto $-1
I put in a breakpoint right after the call to read, and a watch window to
look at the data.
The first time thru, after being reprogrammed, from an expected value of
d'178' for a voltage of 3.5V, the value is 178 with a delay of 10, or if the
value is '255' for the delay, the first read is 186. The second time and
all reads after that, read '187'. The interesting thing is from a range of
1.5V up to 3.5V, the expected value vs the final stable value all have a
difference of 7 bits. Someone mentioned the leakage current could produce
such an offset. I can deal with an offset if I knew it was constant, but
why the initial reading be closer to the actual value and the readings after
that be stable.
So, as a result, the setup delay doesn't seem to have a large difference,
although I would expect it to actually. Is there another delay that I didnt
account for?
And this still hasn't looked at the multiple channel reads either.
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2002\06\12@183857
by
Peter L. Peres
On Wed, 12 Jun 2002, Dale Botkin wrote:
>On Tue, 11 Jun 2002, Dave Dilatush wrote:
>
>> All this is related to a thing called Thevenin's Theorem:
>>
>> http://hyperphysics.phy-astr.gsu.edu/hbase/electric/thevenin.html
>
>Well, sunuvabitch. Guess this is one of those occasions I wish I'd had
>the opportunity to spend more than a semester in college. Happens a
>couple of times a year. Thanks for the link, that explains it nicely.
And then research Norton's theorem which is Thevenin spelled backwards, in
a way.
Peter
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2002\06\12@183900
by
Peter L. Peres
>And the computation below was done using 'typical' numbers. We all now we
>should never design to the 'typical', right?
Why, you mean things may be *different* in reality ? Shouldn't someone
*sue* them then for *misleading* the *public* ?
Peter
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2002\06\13@021352
by
Vasile Surducan
|
Have you considered first to suppress all analogical problem you may have
? Like analogic ground, the point where you connect the analogic input
voltage, the garbage on your Vcc supply ( assuming you still use
switching PSU for test) etc. 100nF and 10uF directly on PIC supply pins.
Usual no cap are required from ad input to gnd. If you still want them,
use a small signall input impedance and no more than 47nF.
And of course if your analogic signal is comming from elswere, then check
first how clean is.
Delay must be at least 20uS. A leackage offset is temperature dependent.
You have another problem there.
regards,
Vasile
http://www.geocities.com/vsurducan
On Wed, 12 Jun 2002, Micro Eng wrote:
{Quote hidden}> Some more information....
>
> after reading thru the posts.....seems to be a couple of avenues to go down.
>
> The first is increase the setup time. Second is to add a capacitor to ground
> on the pin. Haven't tried this, as it seems to be in discussion if this is
> even a correct approach.
>
> There is two problems still...accuracy and multiple channel reads.
>
> I changed the code to read in just a single input,and modified the setup
> delay values. I started with 255, ranged down to 10.
>
> ; xtal is 4.194 Mhz
>
> AD_read
> movwf ADCON0
> call ADSetup
> bsf ADCON0,GO ; Start first A/D conversion
> btfsc ADCON0,GO
> goto $-1
> movf ADRESH,W ; Get A/D value and put in W reg
> return ; return with value in W
>
>
> ADSetup
> movlw d'5'
> movwf DelayAD
> decfsz DelayAD,f
> goto $-1
>
> I put in a breakpoint right after the call to read, and a watch window to
> look at the data.
>
> The first time thru, after being reprogrammed, from an expected value of
> d'178' for a voltage of 3.5V, the value is 178 with a delay of 10, or if the
> value is '255' for the delay, the first read is 186. The second time and
> all reads after that, read '187'. The interesting thing is from a range of
> 1.5V up to 3.5V, the expected value vs the final stable value all have a
> difference of 7 bits. Someone mentioned the leakage current could produce
> such an offset. I can deal with an offset if I knew it was constant, but
> why the initial reading be closer to the actual value and the readings after
> that be stable.
>
> So, as a result, the setup delay doesn't seem to have a large difference,
> although I would expect it to actually. Is there another delay that I didnt
> account for?
>
> And this still hasn't looked at the multiple channel reads either.
>
>
>
>
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>
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2002\06\13@045643
by
Anand Dhuru
Along these lines...
What happens if I need to measure a resistance to the tune of 100K to 1M
using the ad on the 16F877?
I had been toying with the idea of using a fixed resistance and the variable
forming a voltage divider, with the common point being fed to the PIC AD.
Following this thread's indicated that the max. impedence for reliable
operation should NOT be more than 20K. Are there any suggestions?
The resisatnce to be measured is in fact a sensor, and therefore I do not
require the reading in ohms, as such. A look up table can give me the values
to be indiacted on the diaplay.
Regards,
Anand Dhuru
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2002\06\13@052423
by
Peter Betts
|
> Following this thread's indicated that the max. impedence for reliable
> operation should NOT be more than 20K. Are there any suggestions?
I'd consider signal conditioning circuitry.
Whenever sampling data it's a good idea to condition the signal before entering the ADC. Use it to amplify, clean up (filter), scale and impendace match the signal to the ADC inputs.
The cheapest and simplest item is an Op-Amp (High input impedance (so won't effect the sensor being observed) and low output impedance (so ideal for the ADC).
I've used op-amps to amplify the uV signals from a K-Type thermocouple probe before entering the ADC of a PIC. With correct gain setting of the ADC I was able to set it up so that the 10-bit number read from the ADC was the exact temperature. So no software scaling or look up tables required.
If you're interested in dynamic changes of the signal then you really should think about Nyquist theory. Limit the frequency bandwidth of the signal to 1/2 the sampling frequency, otherwise you'll get aliasing and a host of unusual effects. (Or rather... sample at twice the maximum frequency bandwidth of the signal of interest)
Simple R/C filters will do the trick for many systems, you can set gain and filtering with various op-amp circuits. Anything more and you should really be using a DSP as the PIC will begin to struggle.
Regards
Pete
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2002\06\13@111653
by
Pic Dude
For simplicity, could you not form a voltage divider
w/2 fixed resistors, and the large variable resistor
to be measured placed in parallel across one of the
fixed resistors? Of course you now need some math
since the voltage measured at the AD port is no longer
linearly proportional to the variable resistance.
Cheers,
-Neil.
{Original Message removed}
2002\06\13@114550
by
Michael Rigby-Jones
> -----Original Message-----
> From: Pic Dude [SMTP:RemoveMEpicdudeTakeThisOuT
spamAVN-TECH.COM]
> Sent: Thursday, June 13, 2002 4:17 PM
> To: EraseMEPICLISTspam
spamBeGoneMITVMA.MIT.EDU
> Subject: Re: [PIC]:strange analog in problem
>
> For simplicity, could you not form a voltage divider
> w/2 fixed resistors, and the large variable resistor
> to be measured placed in parallel across one of the
> fixed resistors? Of course you now need some math
> since the voltage measured at the AD port is no longer
> linearly proportional to the variable resistance.
>
> Cheers,
> -Neil.
>
The change in voltage at the A/D converter would be very small which could
reduce resolution to unacceptable levels.
Regards
Mike
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2002\06\13@121119
by
David Minkler
|
Hi,
The problem with these voltage divider schemes is that you will only be
able to use a comparatively small fraction of the ADC range thereby
limiting your measurement resolution. Adding additional "parallel (in
the Thevenin sense)" resistors to the network without scaling the
network to compensate will only make the problem worse. The voltage
non-linearity of these resistive networks may actually turn out to be a
benefit if the sensor changes its resistance in an exponential or
pseudo-exponential fashion (as in a thermistor) as the resistive divider
will (sort of) compensate in the reverse direction.
In order to provide a good solution to the problem we need to know more
about the sensor characteristics and the required measurement resolution
and accuracy. Tell us more.
Best regards,
Dave
Pic Dude wrote:
>
> For simplicity, could you not form a voltage divider
> w/2 fixed resistors, and the large variable resistor
> to be measured placed in parallel across one of the
> fixed resistors? Of course you now need some math
> since the voltage measured at the AD port is no longer
> linearly proportional to the variable resistance.
>
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2002\06\13@123317
by
Pic Dude
Good point.
How about adjusting the reference voltages to correct
for this?
Yes, I know there are better ways to solve this problem
-- I'm just in a Rube-Goldberg mood today. :-)
Cheers,
-Neil.
{Original Message removed}
2002\06\14@075055
by
Anand Dhuru
Peter, David, Mike, Neil,
Thanks for all the replies, folks. From what everybody's advised, I guess
the best way to do this would be to use an op-amp to match the impedences to
levels at par with the sensor and the AD input. If one is to do this, which
would be the best (cheapest, simplest to use) op-amp? The frequency response
is not an issue at all. And preferably with a quad configuration, as I need
to use this scheme on 3 channels.
By the way Neil, Mike was right about your suggestion. I ran the values
thru' a spreadsheet, and although the solution looks temptingly simple, the
changes in the effective value of the sensor would be too insignificant.
Regards,
Anand Dhuru
{Original Message removed}
2002\06\14@095408
by
Paul Hutchinson
|
LM324 is easy to use and very inexpensive.
Paul
{Quote hidden}> -----Original Message-----
> From: pic microcontroller discussion list
> [
RemoveMEPICLISTKILLspam
MITVMA.MIT.EDU]On Behalf Of Anand Dhuru
> Sent: Friday, June 14, 2002 7:48 AM
> To:
PICLISTSTOPspam
spam_OUTMITVMA.MIT.EDU
> Subject: Re: [PIC]:strange analog in problem
>
>
> Peter, David, Mike, Neil,
>
> Thanks for all the replies, folks. From what everybody's advised, I guess
> the best way to do this would be to use an op-amp to match the
> impedences to
> levels at par with the sensor and the AD input. If one is to do
> this, which
> would be the best (cheapest, simplest to use) op-amp? The
> frequency response
> is not an issue at all. And preferably with a quad configuration,
> as I need
> to use this scheme on 3 channels.
>
> By the way Neil, Mike was right about your suggestion. I ran the values
> thru' a spreadsheet, and although the solution looks temptingly
> simple, the
> changes in the effective value of the sensor would be too insignificant.
>
> Regards,
>
> Anand Dhuru
>
>
>
> ----- Original Message -----
> From: "David Minkler" <
spamBeGoneMinkSTOPspam
EraseMELUXTRON.COM>
> To: <
KILLspamPICLISTspamBeGone
MITVMA.MIT.EDU>
> Sent: Thursday, June 13, 2002 9:46 PM
> Subject: Re: [PIC]:strange analog in problem
>
>
> > Hi,
> > The problem with these voltage divider schemes is that you will only be
> > able to use a comparatively small fraction of the ADC range thereby
> > limiting your measurement resolution. Adding additional "parallel (in
> > the Thevenin sense)" resistors to the network without scaling the
> > network to compensate will only make the problem worse. The voltage
> > non-linearity of these resistive networks may actually turn out to be a
> > benefit if the sensor changes its resistance in an exponential or
> > pseudo-exponential fashion (as in a thermistor) as the resistive divider
> > will (sort of) compensate in the reverse direction.
> >
> > In order to provide a good solution to the problem we need to know more
> > about the sensor characteristics and the required measurement resolution
> > and accuracy. Tell us more.
> >
> > Best regards,
> > Dave
> >
> > Pic Dude wrote:
> > >
> > > For simplicity, could you not form a voltage divider
> > > w/2 fixed resistors, and the large variable resistor
> > > to be measured placed in parallel across one of the
> > > fixed resistors? Of course you now need some math
> > > since the voltage measured at the AD port is no longer
> > > linearly proportional to the variable resistance.
> > >
> >
> >
> >
>
>
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