This thread has been locked.

If you have a related question, please click the "Ask a related question" button in the top right corner. The newly created question will be automatically linked to this question.

FDC1004: FDC1004 - femtoFarad resolution - Queries for better understanding and possible application

Part Number: FDC1004


Hi, 

I'm now searching for a way to perform capacitive conversions of really small capacitors (20 fF - 50 fF) and I have two questions regarding the FDC1004 chip.

1) Why is the resolution 0.5 fF in the best cases? Theoretically, if the maximum value is 16 pF and we have 23 bits for the conversion (one is the sign bit so it does not count) the resolution should be 16 / (2^23) = 0.019 fF, pretty smaller than teh one in the datasheet.

2) For capacitance changes as the mentioned before (around 30 fF), is it recommended to your this chip? In case that yes, which considerations/recommendations would you give for obtaining the best performance? Afterall, parasitic capacities will be affecting the system...

Best regards.

  • 1.) The effective resolution specification in the electrical characteristics section is 16 bits. The 0.5fF number uses the effective resolution, so the calculation is 30pF/2^16 = 0.5fF
     
    2.) That is a fairly small amount, and it depends on your application. If you need the absolute value of the capacitance, then you will need a way to deal with board parasitic variation. There will be noise in the system too, partly depending on environmental conditions. You can use the EVM to see the approximate noise floor.

  • Hi Clancy,

    Regarding your answers:

    1) I still do not get it: why is 2^8 in the numerator and 2^19 in the denominator? In any case I realize I've made a mistake in my last post:

    Numerator: if the range of the converter is ±15 pF, it means the maximum value that the 24 bit can represent is +15 pF or -15pF so the numerator should be 15pF - (-15pF) -> 30 pF (length of the converter).

    Denominator: Quoting form section 8.6.1.1 : "The capacitance measurement registers are 24-bit result registers in binary format". So we have 24 bits to represent the values.

    Resolution: length represented / 2^(bits used) = 30 pF /  2^24 = 0.0018 fF.

    I realize this is a really small resolution, that's why I'm asking because I'm not able to see my error...

    2) I need to detect a difference of capacitance between two capacitors (small capacitors) so it should erase environmental noise. But see, a parasitic analysis will have to be done.

    I'll be awaiting your answer.

  • My apologies, I made a calculation error too! I am modifying my earlier post to have the correct answer (so people looking at this thread in the future are not confused).