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In a recent question I (rightly or wrongly) said in a comment that calling a mid-rail generator (two resistors and a smoothing capacitor) a "virtual ground" was wrong and that the term "virtual ground" is reserved for when using negative feedback on an op-amp; if the non-inverting input is at 0V (ground) then the inverting input is regarded as a virtual ground due to negative feedback and the op-amp's very high open-loop gain.

I looked on wiki to find that their construction definition is: -

A voltage divider, using two resistors, can be used to create a virtual ground node. If two voltage sources are connected in series with two resistors, it can be shown that the midpoint becomes a virtual ground if

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Wiki then goes on to show this diagram: -

enter image description here

And they have some describing words thus: -

An active virtual ground circuit is sometimes called a rail splitter. Such a circuit uses an op-amp or some other circuit element that has gain. Since an operational amplifier has very high open-loop gain, the potential difference between its inputs tend to zero when a feedback network is implemented. To achieve a reasonable voltage at the output (and thus equilibrium in the system), the output supplies the inverting input (via the feedback network) with enough voltage to reduce the potential difference between the inputs to microvolts. The non-inverting (+) input of the operational amplifier is grounded; therefore, its inverting (-) input, although not connected to ground, will assume a similar potential, becoming a virtual ground if the opamp is working in its linear region (i.e., outputs have not saturated).

In the first sentence they mention a "rail splitter" then they go on to describe pretty much what I (currently) understand as a virtual ground. I have problems with Wiki's explanation because they started with the words "rail splitter" and, from my viewpoint, a "rail splitter" means "mid-rail generator".

I've provided the wiki explanation (before asking my question) just to show I've done some research on this.

My question is this: -

Is a mid-rail generator (formed by passive components and sometimes an active buffer) a "virtual ground"?

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  • \$\begingroup\$ If we look at parts like the TLE2426 then even the chip manufacturers don't seem to make much of a difference. \$\endgroup\$
    – PlasmaHH
    Commented Jan 13, 2016 at 11:46
  • \$\begingroup\$ @PlasmaHH damn those marketing people for stealing "inappropriate" names!! \$\endgroup\$
    – Andy aka
    Commented Jan 13, 2016 at 11:59

2 Answers 2

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Yes. I've used the term "virtual ground" for a node used as reference for signals and that was generated by making roughly a mid point between two supplies. I think that's pretty common usage.

I also agree that "virtual ground" can be used to describe the negative input of a opamp in inverting amplifier configuration, but only when the positive input is tied to ground.

This is a loose enough term that it should be defined before use, unless it is clear from context. Talking to someone about virtual ground without context or a definition is wrong.

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  • \$\begingroup\$ (+1) Especially for pointing out that "virtual ground" is an overloaded expression and it is, confusingly enough, context-dependent. In a sense, in the case of the rail-splitter, the term "virtual" is particularly confusing: after all, ground (unless it is earth ground), it is just a point arbitrarily taken as the reference for voltages, thus the output of a rail-splitter is just the ground of your system (if it has just one) otherwise it is another ground! \$\endgroup\$ Commented Jan 6, 2018 at 14:45
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Yes.

Why: A Virtual ground is "any" (there are very few exceptions) reference point that is not directly provided by the voltage supply. That's all it means really. The reason that the non-inverting input to that inverting op amp is a "virtual ground" is because that's how the op amp equations work. The voltage difference at the inverting and non-inverting inputs is assumed to be zero, and the non-inverting input is grounded. That makes the inverting input always equal to zero. However, since it's not actually grounded it's "virtually grounded." Note that this voltage point isn't being fed to anything but the non-inverting input which, by assumption, should have no current input. That implicitly buffers that voltage. There are of course explicit buffers (linear regulators, op amps, discrete parts in a buffer formation), which is more of what they mean when they say "active."

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