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So this circuit is from ICL7106 datasheet, dual-slope ADC ASIC, what I want to understand is the circuit connected to IN HI and IN LO. So to understand at what my confusion is I need to tell what I understand first:

  1. I fully understand how the ICL7106 works, and how dual-slope ADC works in general.

  2. This ICL7106 works by charging the INT capacitor with (REF_HI - REF_LO) for a fixed amount of time and then the INT capacitor is discharged by (IN_HI - IN_LO) until empty. This discharge time is measured and from those we can get the ratio (REF_HI-REF_LO) and (IN_HI-IN_LO)

  3. The circuit attached on IN_HI and IN_LO is used to get AC input and get the RMS of the AC after the whole process of integration. Meaning Ideally the output of that circuit after integration should get the same result as Integral(VAC^2 dt)

What I don't understand:

  1. How does the circuit attached to IN_HI and IN_LO works? Are there any technique names for changing AC to RMS like this?

  2. How to analyze the circuit?

  3. What does "FOR OPTIMUM BANDWIDTH" below 100pF capacitor means?

  4. How to start tweaking this circuit?

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  • \$\begingroup\$ IN_LO becomes the 'virtual earth' to the op-amp. The op-amp is a non-inverting amplifier and the rest of the circuitry is a 'precision rectifier' and filtering. The RMS conversion doesn't look to be 'true-rms' but rather assumes a sine wave and scaling to suit. Shove the circuit into a simulator - the 7106 isn't required. In the simulator you can tweak to your heart's content and see what effect the 100pF cap has. \$\endgroup\$
    – Kartman
    Oct 10 at 9:49
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    \$\begingroup\$ It doesn't compute RMS except when a sine wave is applied. It's a circuit that falls into the category of precision rectifier. \$\endgroup\$
    – Andy aka
    Oct 10 at 9:58
  • \$\begingroup\$ Thank you for the comments guys, @Kartman by it is not "true-rms", or "doesn't compute the RMS" do you guys mean it really is just precision rectifier? So that the integral should be abs(sin(v)) dt instead of sin^2(v) dt \$\endgroup\$
    – Bun
    Oct 10 at 13:20
  • \$\begingroup\$ @bun, the actual transfer function will be a little more complex due to the filtering so, yes, closer to abs(sin). \$\endgroup\$
    – Kartman
    Oct 10 at 20:43

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