Confirming readADC input impedance requirement

wapo54001

Senior Member
This is a pretty primitive question, but I need to confirm this.

I have a sensor output with specs that say 2ma max into a minimum 2K load.

I want to run this output through a variable pot and then a fixed resistor to ground, and take the output of the pot wiper as the input to a picaxe 08M, taking into consideration the input impedance requirement of the 08M.

The basic circuit is below (if my formatting survives).

Question is this: How do I figure the input impedance in this scenario? Given the input impedance requirement of the 08M, what are the largest values I can safely use for R1 and R2, given R1 is roughly 1/10 the value of R2? Are 1K and 9K reasonable values here, or can I go with 10K and 90K? I'd like the higher values, if they will not affect the readADC10 accuracy.

Help greatly appreciated.


<code><pre><font size=2 face='Courier'>

sensor output
-------|
\
/
R1 \&lt;------ 08M input
/
|
\
R2 /
\
/
|
GND

</font></pre></code>
 

BeanieBots

Moderator
In a potential dividver arrangement, the overall impedance that will be seen by the &quot;load&quot; is given by R1*R2/(R1+R2).
So, for 10k and 90k that would be 9k which would be fine for an 08M ADC input.

That does of course assume the output impedance of the sensor is zero. namely a voltage source. If it has a high impedance, then you need to add it to the top part of the divider resistance and include it in the calculation.

Edited by - beaniebots on 03/09/2007 19:53:34
 

sghioto

Senior Member
Question? Why are you using a resistor in series with the pot? The arrangement shown will only use about 10% of the range of the sensor. Eliminate R2, just use the 10K pot unless that's your intention. Just a thought.

 
 

wapo54001

Senior Member
BeanieBots,

The sensor is an active device requiring 5vdc to operate, and the output varies from .5vdc for zero to 4.5vdc for full scale. I do believe it's a voltage source.

sghioto,

The purpose of the pot is to be able to trim the input up or down so as to be able to compensate for potential mismatches between devices downstream that use separate power supplies to set their reference voltages.
 

boriz

Senior Member
I have found I get much better, more accurate response from the ADC when using an op-amp impedance buffer. The circuit is simple. This image was drawn in response to another thread, but you get the idea. <A href='http://www.zen86415.zen.co.uk/buffer.gif' Target=_Blank>External Web Link</a>

(Mentally replace the words &#8216;peak detector&#8217; with &#8216;pot wiper&#8217;)
 

BeanieBots

Moderator
wapo54001, sounds like it's the output of an op-amp or similar (which explains the 2mA current limit) so you should be OK with a 10k pot and 90k fixed. I assume it's that way round from your description of only requiring a small trim.
If was high impedance, the divider would load it and you wouldn't get what you expected. If that is the case, then you would need to buffer it the way boriz suggests.

Is there no way you could do the 'trim' in software. eg feed in known voltage, push button or on power up and have it self calibrate or maybe use another input looking at a reference voltage (eg diode drop or LED voltage)?

Edited by - beaniebots on 03/09/2007 21:42:30
 

wapo54001

Senior Member
BeanieBots,

Really, this is the best way to do it for several reasons.

One, it's an 08M with very limited code space, and I've pretty much used it up.

Two, There will be times when the circuit is reset when the input is not at zero. Also, users must be considered to be not skilled. Portends a variety of problems. Simpler is better.

 

BeanieBots

Moderator
It was just a thought. I have a particular dislike of pots and try to avoid them whenever possible. Users love to fiddle!
 

womai

Senior Member
Opamps are a problem if your input signal covers the whole range from 0V to Vcc. If you run the opamp from the same supply (0/Vcc) as the Picaxe then they become nonlinear when the signal approaches either end. (Yes there are so-called rail-to-rail opamps but they don't guarantee nonlinearity, they simply ALLOW input signals over that range.) So either restrict the signal to less than Vcc and more and 0V, or use separate supplies &gt;Vcc and &lt;0V for the opamp.

Wolfgang
 
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