What is the purpose of the C2 capacitor in the feedback loop?

What is the time-constant of the R4/R5/R6/C2 net?

opamp schematic

This is for a pH-probe.


  • \$\begingroup\$ For high frequencies C2 has a low impedance, that will short the output and - input of U1. It will then behave as a voltage buffer (gain = 1) for high frequencies. For low frequencies R4, R5 and R6 determine the gain. \$\endgroup\$ – Bimpelrekkie Aug 14 '19 at 12:15
  • \$\begingroup\$ so, it's an unity buffer for low frequencies and null gain for higher frequencies? the time constant of the filter isn't (C2 x (R4+R5))? why do you think it depends on R6 too? \$\endgroup\$ – Vinlar Aug 14 '19 at 12:23
  • \$\begingroup\$ it's an unity buffer for low frequencies and null gain for higher frequencies? No unity gain (gain = 1) for high frequencies and gain > 1 for low frequencies. "null gain" doesn't mean anything. gain = 0 means that nothing comes out, whatever the input signal. If you want the actual transition frequency, do the analysis yourself, you'd have to choose a position for R5 as it is a potmeter. Do the analysis and see how the value of R6 matters. \$\endgroup\$ – Bimpelrekkie Aug 14 '19 at 12:29
  • \$\begingroup\$ I was hoping you'd link not only to the original schematic (which you really presented well in your question, but to the context, i.e. the project in which it's used \$\endgroup\$ – Marcus Müller Aug 14 '19 at 12:31
  • \$\begingroup\$ yes it depends on R5 position, but not on R6, I'd rather say the time constant is C2 x (R4+R5pos) \$\endgroup\$ – Vinlar Aug 14 '19 at 12:32

C2's purpose is providing negative feedback for higher-frequency signals, thereby attenuating them.

In other words, U1's configuration works as an active low-pass filter. Not by any means a steep one.

I have several problems with the shown schematic, though. They range from aesthetic (GND should always point downwards, full stop) to the functional (the input R1/C1 is a 15Hz wide Low-pass filter. What do the other stages do?) to the component choice (why use an offset-nullable, strobable high-bandwidth opamp, and then use none of the features nor more than 15 Hz of the bandwidth? The extreme input impedance means higher noise than necessary in this configuration; why a JFET as the second stage? Why are the used opamps more than 40 years old? Is this a 1970's design? Do the restrictions that led to that design still apply?). It's always good to evaluate the quality of your circuit sources.

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  • \$\begingroup\$ I am surprised too about the 4.7M resistor, should be a lower value as input to reduce noise (and higher value cap) \$\endgroup\$ – Vinlar Aug 14 '19 at 12:20
  • \$\begingroup\$ @Vinlar but that might load whatever you attach there. It really depends on what this circuit is for – which I don't know. \$\endgroup\$ – Marcus Müller Aug 14 '19 at 12:21
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    \$\begingroup\$ and please edit your question to include the link to the original schematic! \$\endgroup\$ – Marcus Müller Aug 14 '19 at 12:28
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    \$\begingroup\$ it's still an unnecessary source of problems, and EMI – You'd normally really just connect your probe to the non-inverting input of a low-noise, low-offset opamp to not burden the sensor, see analog devices' example; as said, these are 1970's components, maybe this was a great circuit for the available technology at the time... \$\endgroup\$ – Marcus Müller Aug 14 '19 at 12:30
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    \$\begingroup\$ yes, but that happens by conducting as much current as needed to keep the voltage at 1.25V. So, if a surge happens, the zener just conducts that current to ground, nothing happens. \$\endgroup\$ – Marcus Müller Aug 14 '19 at 12:44

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