CAV-444 Capacitance Measuring IC

Discussion in 'General Electronics Chat' started by shaun8567, May 6, 2011.

  1. shaun8567

    Thread Starter New Member

    Nov 24, 2010
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    0
    I'm currently working on my senior project for my BS in Mechanical Engineering and I'm using Analog Microelectronic's CAV-444 IC in my system, which converts the value of the capacitor to be measured into a differential voltage signal. My question is regarding the two different nodes that are labeled Vref, as shown in the image below (circled in red).

    [​IMG]

    Are these two nodes tied together (in other words, do I have to connect Vref on one end of R2 to Vref on pin 6)? If not, then is my differential voltage reading (Vout-Vref) between Vout and Vref on pin 6 or is it between Vout and Vref on the end of R2? The datasheet doesn't specifically address this, but my assumption is that the two nodes need to be tied together.

    Any input?
     
  2. Jaguarjoe

    Active Member

    Apr 7, 2010
    770
    90
    Yes, the two Vref points are to be tied together. Pin 6 is an artificial ground, set at Vcc/2 and supplies 2.5v (if you have a 5v Vcc). R2 is referenced to that artificial ground.
    If you take your output between pin 6 and pin 5, it will not be referenced to actual ground. It be offset by 2.5 volts+/- whatever the offset adjust provides . This will make it harder to feed A/D's or other circuits. If you take the output between pin 5 and actual ground, it will be easier to feed other circuits because they all share the same ground.
     
  3. shaun8567

    Thread Starter New Member

    Nov 24, 2010
    5
    0
    Just to clarify, I connect the output of pin 6 (labeled as Vref) and connect that to R2, and then I have my ADC compare between Vout and ground? I believe that datasheet for the CAV444 states that measurements are taken between Vout and Vref, like in the imagine shown below.

    [​IMG]


    In this case, would I connect pin 6 to R2, then measure between the connection point (between pin 6 and and R2) and Vout?
     
  4. Jaguarjoe

    Active Member

    Apr 7, 2010
    770
    90
    Pin 6 is a constant 2.5 volts. Yes, R2 goes to pin 6 (Vref).
    You are wasting an A/D input by measuring Vref. It is constant, 1/2 Vcc. If Vcc is stable, so will Vref be stable. So why even measure it? Any good design will have a stable, clean Vcc.
    Look at fig.4 in the data sheet. This is Vout referenced to ground. The signal starts at Vref (2.5v) and approaches Vcc. That's all you need. I don't know what the uC's A/D resolution is, but, just for talking, say the A/D is 10 bits and is refernced to Vcc (5.0v). Minimum capacitance will be (2.5/5) * 1023 = 512counts. At max cap, you'll be at 1023 counts.
     
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