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This question already has an answer here:

Wikipedia says that, but doesn't give any explanations.

The non-inverting input of the operational amplifier needs a path for DC to ground; if the signal source does not supply a DC path, or if that source requires a given load impedance, then the circuit will require another resistor from the non-inverting input to ground. When the operational amplifier's input bias currents are significant, then the DC source resistances driving the inputs should be balanced. The ideal value for the feedback resistors (to give minimum offset voltage) will be such that the two resistances in parallel roughly equal the resistance to ground at the non-inverting input pin. That ideal value assumes the bias currents are well-matched, which may not be true for all op-amps.

As far as I've concerned, it says just like it really must have this path in case the signal source does not supply a DC path. Then, it says about a consequence of making this path such as bias current which produce extra input offset voltage and so on.

Some people say the input does need to have DC path to ground. They say in case there is no path to ground and the non-inverting pin "hang in the air" for some unwanted unexpected reasons, the op-amp may go into saturation or something like this.

Other people say it doesn't need for the input to have this path. Opinions divided.

What do you think?

UPD: Added the schematic to avoid misunderstood. The schematics a little bit differs from the schematic in the question Use of 100K ohm resistor along with 0.1uF capacitor? So then maybe my question is not duplicated...?

schematic

simulate this circuit – Schematic created using CircuitLab

The values of R2 and R4 (as well as value R1, though) are default cause they don't matter in this scope. We discuss the role of R1.

UPD:

I've found out one more opinion that seemed for me to be suitable.

They say, that this resistor (R1) can help to attenuate some noise which source has large internal impedance. Source of noise and this resistor forms some kind of voltage divider and this divider attenuates the noise significantly.

It seemed to be possible explanation.

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marked as duplicate by Dave Tweed Jan 6 '15 at 18:15

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • \$\begingroup\$ A simple opamp model says the inputs draw no current. But this not right. There is bias current for the input stage and you have to provide a path for that current. \$\endgroup\$ – George Herold Jan 6 '15 at 18:10
  • \$\begingroup\$ This was indeed a duplicate question. But it should not be deleted because the other/prior one has a much less well chosen title. I'm also giving the asker an upvote for some level of prior research. \$\endgroup\$ – Fizz Jan 6 '15 at 21:22
  • \$\begingroup\$ Figure that every input may have a high-value (perhaps gig-ohms or tera-ohms) parasitic resistance to some unknown and perhaps arbitrarily-changing voltage. If an input has a 100Gohm resistor to the positive rail and no other DC path, what will happen? What will happen if one adds a 10M resistor to ground? If the parasitic resistance is large (as it often is), almost any deliberate resistance will dominate its effects but if there's no deliberate resistance, parasitic effects will become noticeable. \$\endgroup\$ – supercat Sep 24 '15 at 18:48
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As a general rule, all inputs of any kind need a DC path to ground. If you don't have a DC path to ground, the DC offset of the input can float to any voltage. You have capacitive coupling to the AC input, but you also have (weak) capacitive coupling to the rest of the circuit, the power lines, your body, etc. All of those can influence the DC component of the input voltage. A resistive path to ground (or a DC-coupled input signal) overpowers these other effects and forces a well-defined DC offset.

Why do you need a well-defined DC offset? If you don't have one, your input voltage could easily drift beyond the supply voltage range, saturating the op amp.

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    \$\begingroup\$ This answer pertains more to FET-input operational amplifiers which have a very high input Z, and low bias currents. BJT input op-amps need a path to ground simply so that their bias current has somewhere to go. If the only place it can go is a coupling capacitor, it will charge it. No stray fields are needed. \$\endgroup\$ – Kaz Jan 6 '15 at 18:19
  • \$\begingroup\$ Adam, why exactly DC offset can float to any voltage? Is there any severe explanations? How does the frequency response of the input circuit change after we add the resistive DC path to ground? \$\endgroup\$ – konstunn Jan 24 '15 at 10:56
  • \$\begingroup\$ Kaz, Adam, couldn't you please clearify, what do you mean under the term "coupling capacitor"? - Is it the capacitance of the p-n junction of the input transistor of opamp? There is some kind of an unwanted high-pass filter which blocks the DC component, right? \$\endgroup\$ – konstunn Jan 24 '15 at 11:10
  • \$\begingroup\$ This resistor (why so low?) is necessary only if an input coupling capacitor is used which does not allow the necessary (small) input dc bias current. In many cases this small dc currrent (nA range) is allowed to flow through the signal source. In this case, you should not use such a resistor because it unnecessarily degrades the input resistance of the whole circuit. \$\endgroup\$ – LvW Jan 24 '15 at 12:36

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