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Ground for 5v card suplied voltage

Hi,

im using a 6023e to sample rotaions in a simple armature. Im using single-ended analog inputs that use the +5v line on pin 14 that are connected directly to a potentiometer that is then connected to an AI channel.

I have tried to use pin 13 DGND as ground by connecting it to AISense or AGND but I only ever get fixed readings at either 4.8v or 0.06v on all channels. This is when I use NRSE or RSE AI mode.

Is this the correct pin to use for a ground from the card for use with the +5v line?

Or have I made some other mistake

thanks

Leo
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Message 1 of 8
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Leo, I would try using the RSE configuration. In this configuration you would leave the AISENSE pin unconnented, and connect the DGND for the 5V source and the ground of your sensor circuit to AGND. You need to make sure you configure the Analag input for RSE.

Note: you might make sure the current you are using from the 5V supply of your card does not exceed specification.
Message 2 of 8
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Hello Leo,

When using the AI channels in differential measurement mode you need to use the AIGND pins (32,29,27,24,56,59,64, 67). AISense is for use when measuring in the Non - reference single ended measurement mode and the DGND is obviously the ground for digital signals.

The 5volt line can supply up to 1 amp of current and can be used to power what ever peripheral device you require, or can provide voltage/current excitation for a sensor. Obviously you need to use this pin within the spec detailed in the manual.

Without looking at your full circuit diagram I cannot comment to see if the measurements you are reading are correct. Can you verify them using a scope or a DMM at each AI node? It maybe that depending on your circuit connected to each
AI channel that you are saturating the AI channel?

Hope this helps.

Can you post a circuit diagram to help visualise your equipment setup?
Message 3 of 8
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Thanks for all the responses!

Steve, I think I can describe what Im doing.

I am running a lead from pin 14 (+5v) connecting it to a (variable) resistor and then connecting that to one of the AI channels so that i can read it. In effect I could be running a wire directly from pin 14 into one of the AI channels and trying to read that, if the resistance was 0.

Not having an electronics background (im a software engineer), I assumed that the 6023e would automatically test the voltage on the input channel against 0 volts and this would be the basis of the reading that i would get later in software. This i now realise is incorrect, but I need to adapt what i have and make it work.

The problem im having is that I dont see how I can provide a gro
und as i dont seem to have one!

and there was me thinking that my circuit was SO simple that even I couldnt get it wrong!!!

All suggestions gladly appreciated
thanks

Leo
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Message 4 of 8
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Hello Leo,

In order to measure a voltage that varies using this variable resistor you need two resistors in series from the +5v pin to the AIGND pin to create a potential divider. Take the node in between the two resistors to an AI channel. You should then be able to change the voltage at this node using the variable resistor which should then be read by the AI channel.

This network works simply as a ratio of the second resistance over the total resistane multiplied by 5 volts should be what you are reading at the channel.

Hope this helps.

Thanks for your reply, I thought you had other components on the circuit.

Steve
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Thanks for that - I stuck some 100k resisters on the end of the curcuit and jammed them into ground so I now get a reading. I'm a little concerned that the reading is logarithmic (the pots themselves are linear - well thats what i ordered) - is this right? The problem becomes more apparrent when I use a 3k pot.

anyway - the worst of the problems has been dealt with - so thanks alot for the help

cheers

Leo
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Message 6 of 8
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Hello Leo, as this is a ratio of the pot over the total resistance the value of the resistor you apply will scale/attenuate the response I.e will act as a negative gain. If your pot response is logrithmic then the network response will be logrithmic with respect to the attenuation factor so I think your pot is not linear. So to answer your question this is normal.

Kind Regards

Steven Bird
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Hi Leo

Assuming that your potentiometer is a three-terminal device, you have a wiper terminal, and terminals at each end of the resistive element. You should connect your analog high input to the wiper. You should connect one end of the potentiometer resistive element to the +5 and the other end to the digital ground. If you're using differential connections for your analog input, connect analog low input to the digital ground. These connections will give you voltage readings that are matched to the taper of the potentiometer.

Regards,

Bob
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