ADC Input Impedance

[Fowkc]

I've been trying to work this out myself, but I want to make sure. I have the circuit below measuring a voltage level from 10 to 15V divided down by three equal resistors A, B and C. What sort of values should they be to make sure the ADC works, in terms of input impedance? Ideally I'd use an op-amp buffer here, but I don't have the space on my PCB, due to my stupid on-the-fly designing and not thinking enough.

<code><pre><font size=2 face='Courier'>
------------------------ 10~15V
|
|
---
| |
| | A
---
|
|
---
| |
| | B
---
|__________ 28X ADC Input
|
---
| |
| | C
---
|
|
--------------------
</font></pre></code>

(ASCII art not really necessary, but it's kind of fun).

 

[Technical]

Total resistance should be about 3-5k, so 1k or 2k2 on each one would be fine.

 

[Michael 2727]

Voltage Divider Def:

Two resistors in series connected across a voltage source.
The voltage at the common junction is a fraction of the total applied voltage,
determined by the resistor values.
<code><pre><font size=2 face='Courier'>Pos Neg &lt; side connected to.
eg: 10K + 10K = 1/2 supply voltage.
2k + 8K = 3/4 supply voltage.
8K + 2K = 1/4 supply voltage. </font></pre></code>
All measured between junction and Neg/Zero supply rail.
You can have more than two resistors, three resistors for example will
produce a window voltage eg: 20V supply, 10K + 5K + 10K will produce
&quot;Four Volts&quot; across the 5K resistor. 2/5 + &quot;1/5&quot; + 2/5 of total supply.
But this voltage has no direct reference to either supply rail, it is floating
somewhere in between.

There are plenty of resistor calculators,
voltage divider calculators on the web -
Use Google.

Edited by - michael 2727 on 02/05/2006 15:04:03

 

[Technical]

Michael, he wasn't asking what a pd is, the principle is obviously already understood from the orginal post! The most appropriate resistor is the uestion - with relation to the input impedance used with the ADC pins! e.g. using 3 100k resistors would not be appropriate.

 

[Michael 2727]

Not having delved into the data sheets quite that far.
Has anyone done any real testing on what the
Max impedance can be used with reasonable result ?

 

[BeanieBots]

As per the microchip spec, the source impedance should be less than 10k.
If you consider a potential divider made from resistors A and B (in your example A is the top two resistors and B is your bottom resistor).
The impedance that will be seen by the PICAXE is given by A*B(A+B).
(parallel resistor combination, also look up &quot;Thevenin equivalent voltage source&quot
Hope I spelled it correctly!

To use your exact example:-
((A+B)*C)/(A+B+C) must be less than 10k

I am really pleased to see that someone is actually making an effort to take this part of voltage divider design into account. Well done.

 

[BeanieBots]

Michael 2727,
In real practice the results depend on how often you take a reading, what voltage (and associated impedance) is connected to other inputs. A VERY big influence is any negative current that makes it into the chip substrate.
With the download cable NOT connected, impedances as high as 50k can probably be &quot;got away with&quot; for 8-bit readings but not for 10-bit.
Try a 100k+100k divider using readadc10 with and without the download cable to see just how bad things can be.
Then try with 10k+10k.
Then add in the enhanced download circuit.
How much does your multimeter make the 100k+100k drop by? (this is also something that many forget about).

Edited by - beaniebots on 02/05/2006 15:28:59

 

[Michael 2727]

That text was borrowed from an article I wrote
for an OP Amp project published in the US
a couple of years back.
I have my lazy moments,,~ ;o)

 

[hippy]

<i>As per the microchip spec, the source impedance should be less than 10k. </i>

I believe there is some variation in what the datasheets say and it differs on PICAXE variants, but I don't off-hand know which is what. I'll go and take a look and report back later.