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I'm just building my first separate ADC circuit, after deciding Arduino was limiting me too much / it is just time to start understanding how to do things properly.

I currently am using a LT1290 ADC with a LT1027 5V reference. I have essentially stuck to exactly the example circuit given in the data sheet to get started, it is what I need anyway. I am currently sticking with single-inputs.

I was just wondering if by re-scaling my input signal from 0to5V to -5Vto+5V, and feeding this in as a differential input, is this an easy way of doubling the resolution (2 x 4096)? Or is there a high price to pay in accuracy compared to the increased complexity of the needed circuit?

Thanks

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    \$\begingroup\$ The first question to ask here is why do you need more resolution? Do you really need it? And second, what kind of signal are you sampling? Give us a clear picture of what you want to do so we can better assist you. As it looks, you might have the wrong hardware. \$\endgroup\$ – Mister Tea Nov 20 '16 at 23:48
  • \$\begingroup\$ I'm sampling a new strain sensor that I haven't calibrated yet using a standard voltage divider. I know how to choose the correct resistor values for that, and I think I'm clear on how to use op-amps correctly to make sure I'm sticking to the full range of available input values. I usually have voltage divider across 5V into analog input on Arduino, and leave it at that. \$\endgroup\$ – WamboBen Nov 21 '16 at 0:13
  • \$\begingroup\$ I might not need more resolution, you're right. It's very possible, the sensor is just too noise anyway, I was querying this as wanted to know if I had understood correctly, so I know if this might be an option in the future. \$\endgroup\$ – WamboBen Nov 21 '16 at 0:16
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    \$\begingroup\$ Why do you think I have the wrong hardware in the first place, that worries me a lot more than my actual question? \$\endgroup\$ – WamboBen Nov 21 '16 at 0:16
  • \$\begingroup\$ Well, if the example circuit is what you need, then adding a negative rail to use an unnecessary differencial input is a high price to pay. Whatever cost/benefit you application will have cannot be speculated on with the information currently present on your question. \$\endgroup\$ – Wesley Lee Nov 21 '16 at 5:49
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No, you will get 12-bit resolution in either case.

The codes for unipolar operation

enter image description here

and bipolar operation can be found in the datasheet

enter image description here

And you can see that for bipolar operation the LSB changes from 1.2mV to 2.4mV. So there is no increase in resolution, just a different scaling.

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  • \$\begingroup\$ This is great information, thanks! I suspected that something along these lines might be the case, but just didn't know how /where the error might show itself, I'm still at a point where it is hard to check stuff, I just don't know quite what question to ask! This is very clear, thankyou! \$\endgroup\$ – WamboBen Nov 21 '16 at 12:17
  • \$\begingroup\$ I'm too lowly to actually upvote you yet sorry! \$\endgroup\$ – WamboBen Nov 21 '16 at 12:19
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When you talk about the resolution of an ADC, you are talking about the number of bits actually. So a 12 bit ADC is a 12 bit ADC and unless otherwise stated in the datasheet, it will remain a 12 bit ADC whatever your input range is.

However, by doubling the input range, you are actually doubling the step size of the ADC.

I'm not sure if this is a good or bad thing for you because this really depends

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  • \$\begingroup\$ Great, thankyou, doubling the stepsize, really helpful seeing people explain it slightly differently. \$\endgroup\$ – WamboBen Nov 21 '16 at 12:18
  • \$\begingroup\$ I'm too lowly to actually upvote you, sorry! \$\endgroup\$ – WamboBen Nov 21 '16 at 12:20

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