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ADS5483: Vcm with TSH4130

Part Number: ADS5483

Dear Sir,

I use the TSH4130 FDA to provide the ADC input. The ADC works in external reference mode. The Vcm (3.1V typ.) is generated by the ADC and supplied to the FDA. The Vocm FDA input absorbs few hundred of uA (confirmed both in simulation and measure) but, since the FDA Vcm output is buffered by a 2 k resistor, the Vocm voltage of the differential signal is about 2.8V. But inside the input "Analog Input Circuit" (data-sheet @ pag. 28) of the ADC the Vocm appears applied to the common node of the differential signal.
I wonder if the applied internal voltage (reported in "Analog input circuit") is downstream or upstream of the buffer resistor. Because if is upstream of the buffer resistance, a difference voltage of about 300 mV will be present between Vcm_signal and Vcm. On the contrary, a voltage of about 2.8V will be the Vocm of the system.
Which is the case? Depending on it, does the system works well?
Thank you
Davide

  • Davide,

    We are looking into this.

    Regards,

    Jim

  • Davide,

    Figure 60 of the data sheet does not show the VCM output circuit. It only shows the VCM internal circuit. These are two different things. Since you are using an external reference, refer to Figure 66 for the expected voltage coming from the VCM pin with respect to the applied external reference voltage.

    The applied internal voltage is before the buffer, as shown in the circuit. What is not shown is the circuit used to drive the VCM pin. This should work fine for your application.

    Regards,

    Jim 

  • Dear Jim,

    thank you for your prompt reply but, sorry, I would like to be sure that I have explained correctly my doubts.
    The circuit used to drive the VCM pin is not shown, ok, but as stated on pag. 29 of data-sheet:  "The VCM output is buffered with a 2kΩ series output resistor".  On this 2k resistance, there is a drop voltage due to the current flowing in it, absorbed by the Vocm pin of the FDA. Thus, the Vocm voltage input of the FDA is reduced to 2.8V instead of 3.1V and this is the value of the common-mode voltage of the differential signal at the input of ADC, not anymore 3.1V. Is this condition ok?

    Also, I would take advantage to ask you another related question.  

    When the ADC is powered on in the external reference configuration from the power-off state, the VCM voltage reaches the plateau value after a time constant of the order of ms due to the capacitor (200 nF in my case) placed at the input pin Vocm of the FDA. In this configuration, the wake-up time in the light sleep mode is dominated and determined by the Vocm time constant which is greater than the value of 0.6 ms reported in table 5, is correct?

    Thank you

    regards

    Davide

  • Davide,

    If the common mode is at 2.8V instead of 3.1V, you will lose some dynamic range. Each input must swing no more than 3.1V +/ 0.75V. Since you have this common mode offset, as the input approaches 2.4V p-p, the device will start clipping in one direction. This is not a problem if you keep the max swing below this value. You can then use the external reference to apply gain to the signal if needed. When using external reference mode, make sure PDWNF and PDWNS are connect to 3.3V per Table 5 of the data sheet.

    Per the data sheet,

    "However, the wakeup time from light sleep enabled to external reference mode is dependent on the external reference voltage and is not necessarily 0.6 ms, but should be noticeably faster than deep sleep wakeup".

    So to answer your question, I would say yes.

    Regards,

    Jim