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FDC1004: Increase channel count for Out-of-Phase measurement

Part Number: FDC1004

Hi,

I'm wondering if it is possible to use the CAPDAC as reference instead of CIN4 like you descripe in the Application Note (https://www.ti.com/lit/an/snoa925/snoa925.pdf).

As explained in section 3, the FDC1004 measurements are configured as follows:

  • Meas1 = CIN1(CHA) – CIN4(CHB). CIN1 is set as the positive input channel, and CIN4 is set as the negative input channel.
  • Meas2 = CIN2(CHA) – CIN4(CHB). CIN2 is set as the positive input channel, and CIN4 is set as the negative input channel.

Can I replace CIN4(CHB) to CAPDAC(CHB) above and add for example CIN3 as follows:

  • Meas3 = CIN3(CHA) - CAPDAC(CHB). CIN3 is set as the positive input channel, and CAPDAC is set as the negative input channel.

If in this configuration the differential measurement still works, I guess switching the Shield for the OoP technic would allow me to use CIN3 and CIN4?

Hence, Shield 2 is the Guard for CIN3 and CIN4 and Shield 1 is the Out-of-Phase counterpart. --> the oposite as descriped in the application note where Shield 1 is the Guard for CIN1 and CIN2 and Shield 2 is the Out-of-Phase counterpart.

Hope you can follow my thoughts and understand my attempt to explain it.
So far it's just a mind experiment.

Best regards,
David

  • Hello David, 

    Since the CAPDAC doesn't work on the same excitation as the channels do, the out of phase operation won't quite be the same. But you might be able to still achieve similar results using the CAPDAC instead and setting up the shielding as you mentioned. 

    Best Regards, 

  • Hello Justin,

    thanks for your quick answer.

    Do I understand correctly, the shield drivers of SHLD1 and SHLD2 are still active and "Out-of-Phase" / 180° phase shifted? If that's the case, I'm gonna test it on a prototype.

    Just the CAPDAC runs on its own "clock"/excitation? Therefore the results may differ from the original "Out-of-Phase" setup.

    I'm interested only in a relative change, as long as the change in capacity is not smaller compared with the "original OoP". I guess that, I have to find out with the prototype..

    Best Regards,

    David

  • Hello David, 

    Yes your understanding is correct. This is shown in the functional block diagram in the datasheet (section 8.2) if you want to look at that as well. 

    Best Regards,