# Why does non-inverting input of non-inverting amplifier need a path for DC to ground [duplicate]

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...?

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.

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

• 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. – George Herold Jan 6 '15 at 18:10
• 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. – Fizz Jan 6 '15 at 21:22
• 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. – supercat Sep 24 '15 at 18:48