I am developing an analog fully differential circuit that needs a tight capacitor matching to reach high performances. Because discrete capacitors are difficult (and expensive) to match for production I was thinking about using a calibration method based on a linear capacitance multiplier, using an operational amplifier with a digital potentiometer or a sample and hold circuit to store the multiplication factor. What are the drawbacks of such a method? I am thinking about the amplifier bandwidth (which is not a problem in this circuit), power consumption and, of course, the hardware overhead for both the calibration and the multiplier itself.

  • \$\begingroup\$ You could make a discrete capacitance DAC, instead of an R2R network, you could have a C2C network. You would need multiple switches, its a wild idea. \$\endgroup\$ – Voltage Spike Feb 1 '17 at 6:54
  • \$\begingroup\$ Pay for matching. State-variable filters have one resistor per capacitor, and trimming the resistors is sufficient to move every pole, so there ARE clever ways around a specific matching requirement, but no one can clever around the general problem. \$\endgroup\$ – Whit3rd Feb 1 '17 at 7:02
  • \$\begingroup\$ Get a cheap LCR meter. The absolute values might be a little off but you can match to a closer tolerance. \$\endgroup\$ – Robert Endl Feb 1 '17 at 7:26
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    \$\begingroup\$ analogue methods have noise and drift. It's difficult for me to conceive of an application that is so critical to be beyond discrete capacitor matching, yet can be matched with a digipot and will tolerate the noise and errors of an analogue multiplier. If the application depends on match, then it can measure its own mismatch, and tell you what shimming caps to add, or switch in shimming caps. A full specificiation of the requirements of the cap, long term drift, voltage rating, resolution, can it have one connection to ground, would help with suggestions. The application might help as well. \$\endgroup\$ – Neil_UK Feb 1 '17 at 7:31
  • \$\begingroup\$ Sounds like an XY problem. \$\endgroup\$ – Andy aka Feb 1 '17 at 8:28

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