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'[PIC:] ADC source impedence'
2004\05\27@092911 by Mike Hord

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How does one calculate the source impedence for input to
an ADC?

More particularly, given a resistor divider, how do you find
the Rs at the midpoint?  I'm looking at using a 1M-2M
divider in a low power circuit to provide a quick and dirty
battery voltage estimate for a 9V battery driving a 3.3V
system.  Obviously, this VASTLY exceeds the 2.5k max
the datasheet on the PIC18F2320 I'm using recommends,
but since I'm only using one channel of the ADC, I can
leave that connected all the time, allowing for
tremendous acquisition times.  I don't need a lot of
accuracy, just a ballpark figure (say, tenths of a volt,
+/- 50 mV) to let the user know when to change the
battery.

I REALLY don't want to add another opamp to this
circuit, and most of the discrete component voltage
followers that immediately spring to mind would consume
rather more current than I want to spend.

Thanks!
Mike H.

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2004\05\27@105047 by Nigel Duckworth

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I've done this several times to save on components and unnecessary battery
drain by putting 100nF adjacent to the ADC input pin.
Potential divider charges the 100nF in slow time, charge is transferred to
the ADC sampling cap when the channel is selected,
100nF acts as a low impedance source to the ADC. Only works for
slow-to-change analogues though (like battery volts).

I've got a horrible feeling someone's going to say I shouldn't do it but it
works for me :)

Nigel Duckworth




{Original Message removed}

2004\05\27@110255 by Alexander Rice

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On Thu, 27 May 2004 08:28:56 -0500, Mike Hord <spam_OUTgaidinmdTakeThisOuTspamhotmail.com> wrote:

> How does one calculate the source impedence for input to
> an ADC?

It's easy but don't bother, just put a .01uF or so cap from the a2d input
to ground, the cap provides the necessary low impedence to charge the 20pF
capcitor in the a2d that does the measuring.

Regards

Alex Rice

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2004\05\27@111537 by fred jones

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===================================
I don't need a lot of
accuracy, just a ballpark figure (say, tenths of a volt,
+/- 50 mV) to let the user know when to change the
battery.
==================================

I have a question.  At what voltage do you read from the 9v battery do you
determine that it should be changed?  I am building a device using some 9v
alkaline batteries and I am putting a low battery indicator circuit to light
an LED but not really sure at what point you should have it trip and light
the LED to tell the customer it's time to change the battery.
Thanks,
FJ

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2004\05\27@113859 by Bob Axtell

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Several battery manufacturers specify the EOL (End Of Life) point of
their batteries in their documentation. Eveready ("Energizer Bunny"
folks have published several books on the subject; Duracell does too.

Google for 'em.

--

I too have had extremely good luck with a 0.01uF to 0.1uF cap right on
the ADC input, but keeping the the analog input enabled for aquisition
all the time places the input impedance error into the calculation.

To use this correctly: turn on the analog input only a few uS ahead of
the sample time, then read it, then turn it off immediately. You will
notice a tiny movement on the scope at the acquisition time when the
0.1uF charges the ADC input cap. You'll obtain a very accurate reading
this way.

--Bob




fred jones wrote:
{Quote hidden}

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2004\05\27@115107 by David VanHorn

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At 08:28 AM 5/27/2004 -0500, Mike Hord wrote:

>How does one calculate the source impedence for input to
>an ADC?
>
>More particularly, given a resistor divider, how do you find
>the Rs at the midpoint?  I'm looking at using a 1M-2M
>divider in a low power circuit to provide a quick and dirty
>battery voltage estimate for a 9V battery driving a 3.3V
>system.


It's the parallel combination of the two resistors.
AC impedance can be lowered by adding a cap, but that won't help you with fighting input offset or leakage on the ADC pins.

An emitter follower will though.  You aren't really interested in the exact voltage anyway, you can easily compensate for the error introduced by the EF.

Now your input will be in the 10's of ohms, dependent on the gain of the transistor used.

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2004\05\27@132325 by Herbert Graf

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{Quote hidden}

       Using standard circuit theory you short all voltage sources and open
circuit all current sources, and then solve for R.

       Another solution I see instead of the huge acquisition times is to put a
cap on the line, it will limit how fast the value can change, but if it's a
battery voltage it shouldn't be changing quickly anyways. TTYL

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2004\05\27@150217 by Mike Hord

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>Using standard circuit theory you short all voltage sources and open
>circuit all current sources, and then solve for R.

Sigh.  Classic Thevenin equivalence problem, right?  Now I feel like a
dope.

>Another solution I see instead of the huge acquisition times is to put a
>cap on the line, it will limit how fast the value can change, but if it's a
>battery voltage it shouldn't be changing quickly anyways. TTYL

This is an interesting idea; I like it.  I'll probably use it in the future.

Thanks to everyone who responded!

Mike H.

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2004\05\28@080117 by Russell McMahon

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> > I have a question.  At what voltage do you read from the 9v battery do
you
> > determine that it should be changed?  I am building a device using some
9v
> > alkaline batteries and I am putting a low battery indicator circuit to
> > light

> Several battery manufacturers specify the EOL (End Of Life) point of
> their batteries in their documentation. Eveready ("Energizer Bunny"
> folks have published several books on the subject; Duracell does too.

That's the best answer - but as a rule of thumb 1v per cell = 6v for a 9v
battery is a reasonable end point. An alkaline cell is still going at 0.9v
but the end cometh soon. Depends on how much reserve you want. Temperature
extremes will also affect this.



       RM

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2004\05\30@112701 by Mike Hord

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Will the battery continue to trickle even down to, say .7 or .6V per cell?
I would assume that for uA range currents, even a very low voltage
would be sufficient, since the problem is one of increasing internal
resistance as the stored energy decreases.

I'm strongly suspecting that I'll get to a point where the battery can no
longer power the product to do anything, but can keep the data safe.

Mike H.

>That's the best answer - but as a rule of thumb 1v per cell = 6v for a 9v
>battery is a reasonable end point. An alkaline cell is still going at 0.9v
>but the end cometh soon. Depends on how much reserve you want. Temperature
>extremes will also affect this.

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2004\05\30@193551 by Russell McMahon

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> Will the battery continue to trickle even down to, say .7 or .6V per cell?

Yes. But you are probably off the bottom of anyone's spec sheet and the
capacity would be getting tiny.
This may keep am memory alive BUT if you have a voltage regulator in there
anuthing could happen. Some regulators shut down below a certain Vin and
some draw substantially more quiescent current in dropout (and some no doubt
do both :-) ).

> I'm strongly suspecting that I'll get to a point where the battery can no
> longer power the product to do anything, but can keep the data safe.

If you care about data backup then it would be advisable to have a secondary
power source that never powers the main circuitry but provides backup power.
Can be very small and may even be "sealed for life" as far as user is
concerned.




       RM

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2004\05\31@225205 by Mike Hord

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>If you care about data backup then it would be advisable to have a
>secondary
>power source that never powers the main circuitry but provides backup
>power.
>Can be very small and may even be "sealed for life" as far as user is
>concerned.

Interesting...like a hard-soldered lithium battery?

That could be an interesting thought.  Data backup isn't crucial, though.
I can afford to lose a little bit.

I think I am going to have to put in some kind of voltage follower, since I
really should check the battery voltage during the flash write cycle.  The
power draw peaks up near 35 mA during that...

Thanks for the ideas!

Mike H.

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