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the role of Fin Select of FDC2214

Takao Yamamoto
Genius 4135 points
Other Parts Discussed in Thread: FDC2214

Dear all,

I think that Fin Select 1,2 is shown as '÷m' at Figure 12 at data sheet p.13.

What role is this?

*Sorry...I could not understand from data sheet...

Best regards,

  • 0
    TI__Expert 8050 points
    Hello Yamamoto-san,

    The divider is to make sure that the sensor frequency does not violate any timing constraints. Please refer to Table 1 on page 14.

    Regards,
    Yibo
  • 0 in reply to Yibo Yu
    Genius 4135 points

    Dear Yibo,

    This EVM is constructed as single-ended sensor configuration.

    So, must we set to '2'?

    *attached file is GUI view

    Fin Select.pdf

    And, what role is this function?

    Why does this device devide by 1 or 2?

    Best regards,

  • 0 in reply to Takao Yamamoto
    TI__Genius 9430 points

    Dear Yamamoto-san,

    Yes, it is recommended to set Fin_Sel to "2" for single-ended sensors (as referenced by Table 1 in section 9.3.1).

    The role of this function is to divide the incoming oscillation frequency to get it into a smaller range. (please see figure 12). Note that CH0_FIN_SEL (0x14) is the divider we are changing.

    Yibo, correct me if I'm wrong here, but I believe the purpose is simply to make sure that the max oscillation is within range of the frequency counter. In single ended mode, you can get relatively high capacitance readings. Dividing the input clock allows for higher capacitance readings before saturating the frequency counter. (notice also that we have FIN_SEL set to 2 when max oscillation frequency approaches 10MHz for both differential and single ended). I assume there is some minor trade off of resolution at lower frequencies to extend the maximum input.

    Make sense?

    Hope this helps,

    Nate

  • 0 in reply to Nate Enos
    Genius 4135 points

    Dear Nathaniel Enos,

    Thank you for your reply.

    I would like to know about your question, additionally.

    ■Question 1,

    Why can we get relatively high capacitance in single ended mode?

    ■Question 2,

    Could you teach me following point of view? Lower frequency = extending the maximum input? Trade off?

    ' I assume there is some minor trade off of resolution at lower frequencies to extend the maximum input.'

    Best regards,

  • 0 in reply to Takao Yamamoto
    TI__Genius 9430 points

    Dear Yamamoto-san,

    1) you get a relatively high capacitance in single ended mode because you are not subtracting anything out of the single like in differential mode. In differential mode, you are only looking at the difference between two sensors so the absolute value of that difference can be smaller. This isn't always true, it depends entirely on the application (which is why it is still recommended to use the divide by 2 when the differential sensor is large (refer to previous table I mentioned).

    2) Sure, I'd be happy to. When you divide the input frequency you are effectively halving the full scale input. If noise is constant in the system then your effective resolution goes down (this goes back to our previous conversation about effective resolution). For example, let's say 5Mhz is the most the digital core of the FDC2214 can handle (just for ease of calculation, please see datasheet for specifics), if the noise of the device is "x" then your resolution is log2(5Mhz/x), but if you divide that by two, then your resolution is now log2(2.5Mhz/x), or 1 bit less. But if your signal was 10Mhz, you must divide it down so that it is within 5Mhz so the digital core can process it. Make sense?

    One thing to note is that most of the noise comes from the sensor and not the device. The sensor noise will also half when divided. It is only the device noise that will be similar, so the total effect will likely be small because the sensor noise is dominant. If you are concerned about this you can always change the divider on the EVM and see how much of a difference there is.

    Hope this helps,

    Nate