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ADS1293: Question about the relation between Vref and DIVR

Part Number: ADS1293

Hi,

I would like to ask you following basic questions about the Vref, DIVR.

1) Vref(2.4V) has no relation with DIVR, right?

    Because DIVR is +/- 0.4V and Gain of INA is 3.5, that is to say, 3.5* +/-0.4=+/- 1.4V, not 2.4V.

2) Will you tell me the reason why Vref value is 2.4V?

3) The meaning of DIVR

    Let me understand the meaning of DIVR value correctly.

    DIVR = +Vin - (-Vin) = +/- 0.4V, is my understanding correct?

    It does not mean 0.8Vpp, right?

   Or +/- 0.4V means 0.8Vpp??

4) mid point of INA

    The INA output is +/- 1.4V, according to the specification.

    Will you teach me the mid point voltage of INA?

    Around VDD/2?

I am confusing a little to understand Vref and DIVR and I appreciate your support.

Best Regards,

  • Hello Takumi-san,

    1. There are 2 limitations with the input range of the device. First, the INA will not be linear if its output is larger than 1.4 V. That limit is not related to the reference voltage. The sigma-delta modulator after the INA will go unstable if its input voltage is larger than 0.74 x Vref. You can see in section 8.3.6.1 that the device will generate a flag if this happens. For you, it means that for any reference voltage you use, the output of the INA should not exceed 0.74 x Vref.
    2. This is an easy voltage to generate on a chip using a bandgap reference :-)
    3. +/- 0.4 V = 0.8 Vpp
    4. The INA is actually a differential in, differential out amplifier so the common-mode or "mid point voltage" of the INA output will be the same as the input. Equation 4 in the datasheet describes the limits to the inputs for proper operation.

    Regards,

    Brian Pisani

    **UPDATE 11/3/17: Although the modulator will remain stable for input voltages less than 0.74*Vref / Gain, distortion performance may degrade if the input exceeds the Recommended Operating Conditions of +/-400mV ( = 0.58*Vref / Gain). **