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We are designing analog input with RC low-pass filter and we want to be able to disconnect the filter if needed. So we came up with these two options. 1st option has jumper at the right side and has twice as much inputs because you literally have to disconnect the wire of the input and connect it for example from IN1 to IN2.

enter image description here

The second design has only one input IN but we are concerned that if op-amp is used the wire above it might bring a lot of disturbances from the air... Also we are concerned about current which might flow in the op-amp output or (-).

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Well what do you think?

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    \$\begingroup\$ In the 2nd case you run straight into the cap. Better find a way to make that filter active and bypass the cap, which will make you filter drop an order. \$\endgroup\$ – a concerned citizen Apr 18 '18 at 11:15
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    \$\begingroup\$ What is the reason to not connect together in1 and in2 in your first schematics? \$\endgroup\$ – Martin Apr 18 '18 at 11:15
  • \$\begingroup\$ @Martin Oh Jesus... Thats right! Hahaha \$\endgroup\$ – 71GA Apr 18 '18 at 11:16
  • \$\begingroup\$ Should we also put zenner on pin 2 of the jumper in our first design? Might be good right? \$\endgroup\$ – 71GA Apr 18 '18 at 11:20
  • \$\begingroup\$ If you follow @Martin 's recommendation (which I second), adding a second zener is unnecessary. Whatever the jumper position, the input signal will be clamped by the zener. \$\endgroup\$ – dim Apr 18 '18 at 12:03
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1st option has jumper at the right side and has twice as much inputs because you literally have to disconnect the wire of the input and connect it for example from IN1 to IN2

Is there any reason why you shouldn't do this: -

enter image description here

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  • \$\begingroup\$ what is the purpose of having the zener diode in the circuit? \$\endgroup\$ – drtechno Apr 18 '18 at 16:18
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    \$\begingroup\$ @drtechno there may be a purpose but realistically it is the OP who made this decision. On the face of it, it looks like it might be some over-voltage protection incase IN1 got too high. Now it works also to protect the op-amp for the same reason. \$\endgroup\$ – Andy aka Apr 18 '18 at 18:33
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    \$\begingroup\$ It is an over voltage protection. If input current exceeds 3.3V zenner opens up and redirects current into GND. \$\endgroup\$ – 71GA Apr 19 '18 at 12:37
  • \$\begingroup\$ It does not assure a low (AC) impedance to the uP's ADC circuit, at J1-2, which many need (though I don't know Z of IN1). Nor does OP's of course \$\endgroup\$ – Henry Crun Apr 19 '18 at 21:15
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    \$\begingroup\$ because it would be asymetric because the signal swing the forward bias direction would clamp the signal at the junction voltage (0.6v or so). \$\endgroup\$ – drtechno Apr 21 '18 at 8:12
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What!!! Are you people made of jumper pins?

schematic

simulate this circuit – Schematic created using CircuitLab

This also provides a buffered input to the ADC for both settings.

Many uP on chip ADC require low AC Z (PIC, AVR) (e.g. 10k or XpF)

Further the uP is protected by U? in both settings.

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  • \$\begingroup\$ What are you trying to say? Are we missing anything? This design looks super simple! \$\endgroup\$ – 71GA Apr 19 '18 at 12:38
  • \$\begingroup\$ @71GA 2 pins is fewer than 3 pins. Oh, and it should be better as well as cheaper. 3 legs good, 2 legs better. Don't call me Muntz \$\endgroup\$ – Henry Crun Apr 19 '18 at 21:30
  • \$\begingroup\$ @71GA the first sentence of your question... you want to be able to disable the filter when needed. Well, this circuit might help because it acts as a buffer if SW2 is open. Or provides a filtered output when SW2 is closed. The possible disadvantage of this circuit is having a relatively high output impedance. I would use an active filter instead. \$\endgroup\$ – Rohat Kılıç Apr 19 '18 at 21:39
  • \$\begingroup\$ @RohatKılıç What's your simplest non inverting active filter? \$\endgroup\$ – Henry Crun Apr 19 '18 at 23:04
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    \$\begingroup\$ @Henry well, putting the opamp buffer after RC maybe (with keeping the switch, of course). Depending on the signal source's output impedance, we may require one more buffer so that the signal level is not decreased by dividing. I dont know the target's input impedance , but if it's in a few hundreds of kOhms or in MOhms range then the circuit above is well suitable. \$\endgroup\$ – Rohat Kılıç Apr 20 '18 at 4:00

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