Analog pin problems

The crosstalk between channels happens when switching between channels in a high-impedance circuit, shouldn't be a problem if low-impedance voltage source connected.

Thanks but what does low impedance voltage source mean?

according to my book :

when output impedance is small a relatively large output current can be drawn from a devices output without significant drop in in output voltage.

When the output impedance is large a relatively small output current can be drawn from the output of a device before voltage at the output drops substantially.

A rule of thumb for efficient signal transfer is to have an output impedance that is at least 1/10 of the loads input impedance to which it is attached.

Now i do not know what half of that means or how one would go about measuring this impedance, it seems like in a very simple simple way its the resistance "outward" being that high impedance lowers current output.. so yeah maybe you can decipher that better then i.

UPDATE: dam! i think it just clicked, so if i have a million things in a circuit all taking a chunk out of its current then furthermore if i try to output something it will be at a very low current since it is all being used up... interesting

It turns out to be pretty easy to measure source (or output) impedance. You need only a variable resistor (like a pot) and a volt/ohms meter. Measure the unloaded output voltage from the mystery source. Then attach a variable load (your pot) across the output and increase the load until it is 1/2 the open circuit (unloaded) voltage which you previously measured. Then disconnect the pot and measure the resistance. That resistance is equal to your source impedance.

The principle is that you are forming a voltage divider. The "top" resistor in the divider is the mystery source impedance, and the "bottom" resistor is your pot. When the voltage is 1/2, then the two resistors are of equal value.

Of course, if you are measuring something at the extreme ends of the spectrum, conventional parts and techniques don't apply. So measuring the source impedance of a guitar pickup or a car battery can't be done using this simple scheme.

if i could pay you for posting that i would, thanks for that.

The voltage sources are trimpots (see the link I posted earlier) connected to the VCC pin.

Fantastic! Will give it a try! Thank you for all your help, guys!

Wow, that worked! However pin 3 reads nothing, even when connected but I guess that must be a contact trouble or breadboard issue. Guys, thank you sooooooo much!

BTW, 100 K potentiometer and resistor was the solution! Again, big thanks!

Since my last post I've been experimenting with different pull-down resistor and pot values. Unfortunately, whatever resistor I use the connections effect each other. I came to the conclusion that I have to break the connection between the pull-down and the analog pin the moment I connect a voltage source. Alas, I tested the concept on the breadboard and it works (I simply removed pull-downs from the connected analog pins). I wonder, is there a plug socket that breaks the existing connection between the pull-down while creating a new connection? Or should I add a transistor switch driven by the voltage source inserted?

When switching analog inputs quickly, SOP is to sample the new pin twice and ignore the first read.

alkopop79:
I wonder, is there a plug socket that breaks the existing connection between the pull-down while creating a new connection?

Yes. Many mono and stereo phone jacks (like for headphones) have a switch which opens when you plug in a cable. In normal use this disconnects the speaker(s) when you plug in headphones but can be used for your pull-down resistors.

Brilliant! Thanks!

Unfortunately, whatever resistor I use the connections effect each other.

That is because the values you are using are too large.
If the impedance of anything fed into the analogue inputs is greater than 10K, there is not enough time to charge the sample capacitor before the sample is taken.
So can you lower your impedance?
Your solution of shorting it out will only result in you not getting the right value the first time you take the reading although you might miss this.

I've tried 1 ohm to 1 M reistors and some in between. No succes. What are you suggesting, Mike?

Grumpy_Mike:
Your solution of shorting it out will only result in you not getting the right value the first time you take the reading although you might miss this.

I guess if I add a short delay, it would should be all right.

[quote author=Grumpy_Mike link=topic=72289.msg597190#msg597190 date=1321357387
That is because the values you are using are too large.
If the impedance of anything fed into the analogue inputs is greater than 10K, there is not enough time to charge the sample capacitor before the sample is taken.
So can you lower your impedance?
[/quote]

Sorry Mike but I'm still struggling with the idea of impedance. Can you give me an example? And what does that mean that the analog inputs have great impedance? What are the implications of these?

If the analog input has high impedance, what happens when you add a low and a high impedance source?

If an analog input has high impedance (whatever it means), does it mean it allows more current through than a low impedance input? Any help to shed some lights on this issue would be appreciated!

BTW, does it help if I set the internal pull-ups? Are they any better or reliable than the external ones?

Are they any better or reliable than the external ones?

No.

An impedance is like a series resistance to the voltage source. In your case the voltage source is a resistance so it means the value.

I guess if I add a short delay, it would should be all right.

No... the delay you need is between the switching of the multiplexer and the grabbing of the sample. This can only be acheaved by taking two readings of the same panel.

I've tried 1 ohm t

If you truly have then you have something else that is seriously wrong. Can you provide a schematic of what you have.