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ADS1213 Analog Signal Range

JACR
Intellectual 270 points
Other Parts Discussed in Thread: ADS1213
Hi people
I want to use an ADS1213 in order to convert 4 analog signals into digital codes.  Here's the thing, the sigma-delta converter is differential and my signal is unipolar, so i think i should just connect one of the input pins of the converter to the analog ground.  If i do such a thing, i will lose some dynamic range, but i will take advantage of some other things like oversampling and digital filtering so im guessing the tradeoff would be justified.
But... here's the problem; if i make the suggested connection (Figure 12 of the datasheet), in order to have a +/- 10V- or +/- 5V input voltage range i think i will have a lot of problems when my voltage signal is below the tens of milivolts range  (problems in the voltage divider with resistances, even if im using 0,5% tolerance)
So, i want to know what is the best design approach with this device,  should i use with the sugested circuit?  Or it will be better if i use a voltage level shifter and change my +/-5 V signal to a 0-5V and then buffer it to the input pins of the ADC.
I also want to know, if possible, the exact formula for the input voltage range using Vbias (i tought it was just {[Vin*R3/(R1+R3)] +Vbias}, but that is not it, at least thats is not what im getting when i make the measurements using a fluke)
Thank you very much for your time and best regards
  • 0
    TI__Guru** 113765 points

    JACR,

    I'm a little confused.  At first you state that your input is unipolar, but then you talk about converting your +/- 5V signal.  So what exactly is your input range?

    With the ADS1213 you can apply 0-5V directly.  This is the maximum input range of the ADS1213 inputs.  You can apply a larger input voltage as shown in Figure 12, but as you pointed out, this adds some inaccuracies due to part tolerance as well as be susceptible to drift. So with a direct connection from 0-5V, one LSB is positive full-scale/2^23, or 596nV per code.  Due to noise within the converter, the best effective number of bits you can achieve is 22.  This means that you can resolve better than you can measure, so I doubt that you will achieve better performance by changing the resolution.  What degree of resolution are you hoping to achieve?

    The formula for the inputs for figure 12 is:

    [(Vin - Vbias)/(R1+R3) * R1] - Vbias

    which is the voltage at the input of the ADS1213.

    Best regards,

    Bob B

  • 0 in reply to Bob Benjamin
    Intellectual 270 points

    Hi Bob..

    I made a mistake when i wrote my question.... what i meant was that my signal is single ended instead of differential.  However, it is bipolar (+/- 5V), so im going to ground one of the differential inputs in each of the analog input ports.

     

    So, the question will be.... is it better to use the circuit on figure 12?, or it will be better to use a level shifter in order to convert my +/-5 to a 0-5 and then use it as an input for the ADS1213.

     

    Thank you very much and best regards

     

    P.S Im trying to achieve something close to 16bits of  ENOB

  • 0 in reply to JACR
    TI__Guru** 113765 points

    JACR,

    It is hard to say not knowing your application, but the end result must be such that the analog inputs do not exceed 5V or go below ground.  The simplest solution is the Figure 12 circuit.  However there are some advantages when using an op-amp to level shift as it allows you to also add some filtering and possibly increase the input impedance of whatever you are trying to measure.

    So the answer is...it depends.  If you match the resistors, the Figure 12 circuit will work quite well.  So you need to look at your requirements, board area, cost, etc. and determine what will work best for your application.

    On another note, all signals applied to the ADC inputs are differential (or pseudo differential) as the return code is relative to a difference between the inputs as referenced to the AINP input.  So, if AINP is +5V and AINN is at 0V, the difference is +5 and is positive full-scale.  If AINP is 0V and AINN is 5V, the difference is -5V and the result is negative full-scale.  The ADS1213 is a little confusing as it has different options for expressing the resultant code.  It also refers to unipolar and bipolar measurements that measure the same way, but only expresses the result in the selected manor. 

    My understanding of what you are trying to do is have one output of your sensor that will be at ground and the other output could be up to the extremes +5 or -5V (like a function generator output.)  If that is correct, then to get the maximum range for whichever method you choose you will want to divide your input in half and center the common mode point at 2.5V instead of ground.  You can correct your reading back to the original scale within software.

    Best regards,

    Bob B