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Well, I´m making "Cmoy amp" (Actually RA1 copy), but I've been stuck in the power management since I saw these:

Here the blogger says that the amp uses a real ground circuit. (And there's also a schematic diagram).

enter image description here

But below that diagram shows that the ground is Virtual ground, and really the only difference between the two schematic diagrams is that one has polarized caps.

enter image description here

I thought that in order to create a virtual ground you needed a couple of resistors. That made me realize that I really dont have idea of when a circuit uses virtual ground or not.

Please help!

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    \$\begingroup\$ "Virtual ground" means the process of making the inverting pin on an op-amp the same voltage as the non-inverting pin - en.wikipedia.org/wiki/Virtual_ground explains. At some point a fool came along and clouded the issue by talking about making a split supply from a single voltage source and there lies the problem. This question is flawed by its nomenclature and I hate the dumbing-down or reallocation of meaning due to idiots who think they know something. Unfortunately I have to vote to close the question on that basis or I wouldn't be me. It's not a dig at the OP BTW. \$\endgroup\$ – Andy aka Nov 26 '15 at 15:13
  • \$\begingroup\$ Please, if anyone can offer a new learning path for me then offer it. I'm quite willing to accept that "virtual ground" might have a double meaning but I want a decent and persuasive reason! \$\endgroup\$ – Andy aka Nov 26 '15 at 15:15
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There is no contradiction ! Ground is simply a reference point. The Cmoy amp needs +9 V and -9 V, the point "in the middle" is used as a reference and in this case also called ground. You could also choose the -9 V point to be ground and then the other points would be +9 V and + 18 V. But with the Cmoy amp you want to connect it to an audio source. The schematic shows how you need to connect the input signal of the Cmoy amp, the ground of the input signal (coming from the audio source) needs to be connected to the point between the two 9V batteries.

Virtual ground is a term used when an opamp is used in a feedback configuration with one input connected to ground and the other input is then a virtual ground since the opamp keeps it at the same voltage as ground but there is no physical connection to ground !

In the case of the Cmoy opamp there is no virtual ground ! I advise you not to worry too much about ground or virtual ground. It is not a big deal if you just want to build some design.

Just remember that ground is simply a reference point, nothing more than that. And virtual ground is when an opamp has feedback and one input is connected to ground, then the other input is called a virtual ground.

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  • \$\begingroup\$ Thanks! but I'm not going to "build just some design", that design will also work as Buffer for a JRC2068, that will make it almost an O2 amp. Here is the schematic diagram of the O2. drive.google.com/file/d/… And I´ll also use two 9V DC wallwarts to get a split supply, like here: electronics.stackexchange.com/questions/84893/… \$\endgroup\$ – Phoenix1 Nov 19 '15 at 8:17
  • \$\begingroup\$ The two 9V adapter solution will work but personally I would use a circuit like this: google.nl/… with a 15 to 18 V input voltage, that would give you +/- 7.5 V or +/- 9V at the output with using only one AC adapter. \$\endgroup\$ – Bimpelrekkie Nov 19 '15 at 8:24
  • \$\begingroup\$ Uh, that's a rail splitter. I can't use that, that would go completely against the O2's designer. (Nwavguy). I'm saying that because of this interesting article: nwavguy.blogspot.com/2011/05/… \$\endgroup\$ – Phoenix1 Nov 19 '15 at 9:17
  • \$\begingroup\$ @Phoenix1 If you're looking at the O2 amp, look at how it generates its power rails: with a low voltage AC plugpack and dual half-wave rectifiers. \$\endgroup\$ – Nick Johnson Nov 19 '15 at 10:42
  • \$\begingroup\$ Hmm, I don't see any proper technical arguments against using such a rail splitter nor why you should not use it. I have a Headphone DAC + amplifier which uses a rail splitter and I have no problems with it. From my engineer's standpoint 9V + 9V is the same as 18 V divided in two. I think you simply misinterpret the points Nwavguy is making. \$\endgroup\$ – Bimpelrekkie Nov 19 '15 at 10:50
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Well based on the schematic they use a different approach to get a split supply for the op-amp, like you mention using two equal sized resistors.

The reason why they do this (as they say) compared to the way it is done in your first link is, that the batteries can drain unevenly, in which the first approach (center-tapping the batteries) can lead to uneven +/- volatages, which might distort the audio.

With the resistor approach you always get a nice even split.


A virtual ground (in my thinking) is needed when you have a single supply op-amp but you need to amplify a signal which would go below the supply rail.

For example: you get a sine-wave with 1 V peak-peak, referenced to your ground, so it goes from -0.5 V to 0.5 V.

You want to amplify it to 3 V peak-peak, but the op-amp only has a supply of +5 V and 0 V. You are not allowed to apply a signal above and below the supply rails to the op-amp, so you have to shift the input signal to a point, where it will not go below the 0 V line.

In most cases (symmetrical signals) you'd go for a level in the middle of your supply rail. 2.5 V in this case. This is the virtual ground, because the signal you are interested in will be referenced to that value from now on.

So you couple the signal onto your 2.5 V virtual ground to get it up there, and the op-amp is now able to amplify it nicely to a 3 V peak-peak sine wave. But that 3 V peak-peak is referenced to the virtual ground.

If you measure the voltage from the "real" ground (the op-amps 0 V) the amplified sine wave will go from 1 V to 4 V (3 V peak-peak but riding on 2.5 V).


But really, ground is just a reference, in my example I said that the supply negative terminal is 0 V and the positive terminal is 5 V. So my ground was the supply negative terminal. I could have arbitrarily chosen -2.5 V and 2.5 V, but that would have made my 0 V not really available to the circuit. (and because of that the input signal would be on -3 V to -2 V probably)

Maybe you could say that a virtual ground has not the lowest impedance path to the supply (because it has to go through some resistors or op-amp if you have an active virtual ground) compared to the real ground. Or it's just the way it is commonly chosen.

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Two batteries in series would give a much stiffer "virtual" ground than a resistor divider would. I think that's why the first blog calls it a real ground.

The resistor divider that you mention (without diagram) is a way to bias opamps when using a single supply (i.e. a single battery). Some DIY audio community guy, from which you've got the 2nd image, indeed calls this "virtual ground".

enter image description here

The more professional EE documents on this like TI's, ADI's or Microchip's don't call a simple resistor divider "virtual ground". TI does call their TLE2426 "'Rail Splitter' Precision Virtual Ground". This does provide a fairly still "middle rail", i.e. a virtual ground. Note that it's basically just an buffer added after a resistor divider. (I'm guessing if TI's calls theirs Precision Virtual Ground you can argue that something less precise, e.g. a resistor divider as above is just "virtual ground"; semantics...)

You can make an argument that beefy enough caps after the resistor divider are good enough "virtual ground", but I've not seen this term commonly used for this kind of circuit outside of a certain the DIY community (at least not when referring just to a resistor divider). When EE pros design an opamp circuit that uses a single battery, they call it single-supply [most of the time]. Actually, I did find one article by ADI employees where they call it "false ground", including the case where it's just a well-bypassed (i.e. big caps after a) resistive divider:

enter image description here

And not to be less original, the LT guys call it 'artificial ground' (or supply splitter) but they seem to do so only when a buffer is used.

enter image description here

So, yeah, no shortage of variations in terminology on this concept...

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