External reference of 6363

[johndigsby85]

I have a further accuracy question.  For the PXI 6363 I am thinking of a configuration where I want to maximise the resolution over a narrow range of voltage.  For this, I think I could use the APFI 0 or 1 pins of the 6363 to provide a suitable external reference

But although I get accuracy specs for the internal voltages, what will affect the accuracy of my outputs driven using the external reference (except the accuracy if the reference itself)?  This might be asking what components lie between the APFI pins and the external output, but I'm just keen to understand if my resolution might be swamped by accuracy problems

[mdoherty]

Accuracy is determined by the bit measurement found on the manual for your hardware. This tells us exactly how precise it can get with measurements.

http://www.ni.com/pdf/manuals/377776a.pdf

[johndigsby85]

I don't know what you mean by "bit measurement", I'm afraid and this doesn't appear in the manual (which I have read previously in trying to answer the question)

If you mean the resolution, do you mean that the accuracy is simply divided by the full range resolution of the device?  i.e. the accuracy of the external reference divided by 2^16 in this case?

[mdoherty]

ADC resolution will determine the accuracy - in your case, this is 16 bits. So essentially it will divide the range you set by 2^16.

How to max this: https://knowledge.ni.com/KnowledgeArticleDetails?id=kA00Z000000P9BxSAK&l=en-GB

[johndigsby85]

Yes, I understand this, but you seem to be talking about precision/resolution, but I mean the accuracy.  For example, the accuracy of the 6363 is 1890uV at full scale 10V.  If I applied this across a potential divider to get a lower voltage, and fed it back into the APFI 0, the reference voltage would have an accuracy of some proportion with that 1890uV

My question is whether there are any additional gain/offset errors between the APFI pin and the AO pins that will increase the inaccuracy in the reference (or indeed any other sources of inaccuracy).

I think this is sort of answered by your link, so it looks like my inaccuracy would be similarly divided by the 16 bits?

[mdoherty]

Oh sorry about that.

Yes I would think that would be correct then.