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ADS1262: ADS1262 differential input range

user5834775
Prodigy 220 points

Part Number: ADS1262

Hello

I have a question related to the differential input range of the ADS1262. 

I have a system where it is used a load cell of 100kg with a Wheatstone bridge configuration. It is excited with Vexc(+)=5V and Vexc(-)=0V (GND) and it has a sensibility of 2 mV/V; so the differential output of the load cell is:

- (10 mV) when it is applied a traction effort. In the output terminals there is a voltage aprox. of IN(+) = + 2.5V and IN(-) = + 2.49V.

- (-10 mV) when it is applied a compression effort. In the output terminals there is a voltage aprox. of IN(+) = + 2.49V and IN(-) = + 2.5V.

Connected to the load cell in ratiometric connection is the ADS1262. The ADS1262 is supplied with AVDD=5V and AVSS=0V, and VREFP=5V and VREFN=0V. If it is programmed a gain for example of 32V/V, the differential input range of the ADS1262 is of ±156mV. And the question is, why the ADS1262 can measure a negative differential input signal, as the - 10mV, if it is supply with a positive signal (AVDD=5V and AVSS=0V), that is, of unipolar way? Is not it supposed that supplying in a unipolar way should only be able to read positive values in the input?

Best regards.

  • 0
    TI__Mastermind 44820 points
    Hi,

    Let me know if my other E2E reply at e2e.ti.com/.../2961985 is able to answer this question with sufficient details.

    A negative "differential" voltage in this case (with a unipolar supply), just means that the ADC's negative input is "more positive" than the positive input pin. If you are unsure about the ADS1262's input votlage range, do take a look at the www.ti.com/.../ADS126X-CALC-TOOL design calculator, which will tell you if a given input condition is outside of the ADC's linear measurement range.
  • 0 in reply to Christopher Hall
    Prodigy 220 points

    Hello

    Thanks for the answers. I understand the concept of the two answer, they resolved a little my doubt, but I continue a little confused, because one thing that I wonder is:

    ((With a VADD=5V, VASS=0V, Vref(+)=5V, Vref(-)=0V and PGA = 32V/V))
    For example, in the differential input of the ADS1262 there is a voltage of -10mV (VINP=2,49 and VINN=2,5), these -10mV are amplified by the PGA by 32, so in the differential input of the ADC´s ADS1262 are -320 mV, but if the ADC is supplied with a unipolar supply (5V), how and why it can read this negative value? Should not it read values only between 5V and 0V? Because a unipolar ADC only can read positive voltage in its inputs.

    I know that the ADS1262 can do that because I do that and it works correctly, but I want to understand, how and why it can?

    Thanks again.

    Best regards.

  • 0 in reply to user5834775
    TI__Mastermind 44820 points

    Hi,

    Your question is a good one. I've worked with Delta-Sigma ADCs for awhile now and the idea of measuring negative differential voltages seems natural to me, I've forgot that this can seem odd at first sight.

    If you refer to a basic block diagram of a delta-sigma modulator (see Figure 1 in this blog post), you'll notice that it consists of a summing junction, an integrator, a comparator, and a 1-bit DAC in the feedback path). You can think of the summing junction as a typical amplifier with a positive and negative input, the more positive input tends to steer the output either towards the positive or negative rail; however, all of these voltages can be positive.

    In the feedback path, we can either add or subtract from the input voltage, since this is typically a switched capacitor circuit. In the case where the feedback subtracts from the input, the reference inputs can be swapped so that the negative input to the summing amplifier sees a more positive voltage (i.e VREFP).

    It's hard to find good reference material that shows a Delta-Sigma modulator with a differential input (most simplify the summing junction and only show a single-ended input signal), but if you can mentally replace the summing junction with a difference amplifier, then you can see how negative differential voltages are measured. An input signal of -VREF would tend to output all 0's on the modulator's output bitstream, so internally all circuit nodes can remain at a positive voltage and we can represent negative voltages as having a 1's density of less than 50%.

    Best regards,
    Chris

  • 0 in reply to Christopher Hall
    Prodigy 220 points

    -In the feedback path, we can either add or subtract from the input voltage, since this is typically a switched capacitor circuit. In the case where the feedback subtracts from the input, the reference inputs can be swapped so that the negative input to the summing amplifier sees a more positive voltage (i.e VREFP).

    So it works by swapping the reference inputs of the summing amplifier?

  • 0 in reply to user5834775
    TI__Mastermind 44820 points

    Yes, typically + or - Vref (the differential voltage) is used in the feedback path, but VREFP and VREFN are both >= 0V (assuming AVSS = 0V).

  • 0 in reply to Christopher Hall
    Prodigy 220 points

    okey, thank you for resolve my doubt.

    Best regards.