# Why does an opamp buffer output voltage with no input?

I have an opamp (specifically a TL072) I'd like to use as a high impedance buffer for a digital filter circuit I'm building. When I power the opamp with +5/0 on the rails configured as a unity gain buffer - with nothing connected to V_in, V_out = 1.4V~. Even if I model this in circuitjs, I'm almost seeing exactly the same behaviour - so what am I missing? Where does this floating voltage feedback voltage come from?

• Well, thus it is not "floating" because in the real world your capacitor may not be ideal. It may have some leakage. Thus, knowing what an OPAMP is doing when really not connected to anything won't help you much. This is why it is advised to explain what you are trying to achieve in the first place. In that application it may be a good idea to fix the Vin offset voltage externally using very high (but not too high) resistors. Commented Apr 19, 2021 at 13:11
• You can't run a TL072 on a single 5V supply (at least not in any way that's useful for much). If you Refer to the Fine Manual you'll find that the input & output ranges are limited to a couple of V from either supply rail. Commented Apr 19, 2021 at 14:35
• If the input is floating, what voltage do you expect on output and why? Isn't 1.4V just as good as any other voltage? Commented Apr 19, 2021 at 15:53
• Floating is floating. Understanding why this specific op-amp goes to 1.4V is an academic curiosity, but from an engineering perspective the only thing to take particular note of is that the output voltage with a floating input is undefined. A different op-amp would not necessarily behave this way because there is generally no deterministic requirement for the floating input to produce any particular output. This is basic Garbage-IN/Garbage/OUT.
– J...
Commented Apr 19, 2021 at 17:35
• Also bear in mind that the simulation model is only a model and may not be accurate in certain situations, floating input being one of them.
– Frog
Commented Apr 19, 2021 at 21:19

An opamp in a circuit according to your schematic (with nothing connecetd to IN+) will drift to the input voltage point where its input bias current crosses the zero axis.

Example from the OPA145 datasheet:

It crosses the zero at pretty much mid supply, but other amps are different and would bias differently, or even not at all (go to rails).

e.g. for your TL072 it looks like it would not bias at all:

But when practically building circuits with pA input bias current, there will be also leakage on your board, which will move the datasheet curve around slightly. All the leakage current into/out of the V_in node taken together has to sum up to zero, to keep its potential steady. The opamp input bias current is one contribution, but other contributions could be leakage through esd diodes, through capacitors and even board surface creepage to nearby metal parts at different potential. Therefore, the self-bias point in a real circuit is often unpredictable. If you get the same value in a simulation and real circuit from a JFET input opamp, I would guess it is coincidence.

• What do you mean by "such an amp?" Commented Apr 19, 2021 at 12:20
• @AlexTurner I meant it as in "an amp as drawn in your schematic with nothing connected to In+" Commented Apr 19, 2021 at 12:29
• Ah thank you - so as per your answer, the OPAx145 (if configured as per my question with +5/0 on the rails) would bias to 2.5V with a floating input? Commented Apr 19, 2021 at 12:41
• And the leakage of the capacitor connected to V_in... Because V_in is connected to a cap as the OP added in a comment. Commented Apr 19, 2021 at 13:13
• @AlexTurner theoretically yes. But for different supplies, even the same Opamp like the OPA145 might bias differently. And in a real product leakage on the pA level is often unpredictable. So this is never a reliable way to bias the working point. Commented Apr 19, 2021 at 15:11