operational-amplifiervirtual-ground
I understand what a real ground for a Opamp is.It is a terminal which is physically connected to ground or earth which acts as the reference point for the entire circuit.
But what do we mean by a virtual ground?
And how do we explain it with the help of opamps?
And finally is my understanding of real ground correct or is there something more specific that i can add.
The purpose of this circuit is to make sure the circuit has sufficient phase margin (does not oscillate). It's a particular problem with MOSFETs and not BJTs. 100nF is a very large capacitor- 1nF would work as well here (10nF is a good value), but maybe they wanted a bit of a LP filter or just wanted to be sure.
The problem is that the MOSFET (with such a low sense resistor) represents almost a purely capacitive load on the (exceptionally wimpy in this case) op-amp output. That produces a phase shift with the (relatively large) open-loop output impedance of the op-amp. In the case of the MCP6002 the maximum capacitance you can safely put on the output is less than 100pF with G=1. The Cgs is relatively low on that MOSFET (31pF typically, 46 max) but Miller capacitance comes into it too. Fortunately, with an LED load it's almost looking like a cascode arrangement, so you may be out of the woods.
You'd have to do a bit more calculation or simulation to be 100% sure- maybe try feeding a square wave to the non-inverting input and look to see how much overshoot you get in the current waveform. Varying when someone touches it sounds like it might be oscillating!
It's poor form IMO to do this in general- the second circuit above is the right way to do it. Even if you conclude it's working well enough, be careful that in production the MOSFET does not get replaced with something with substantially more capacitance. For example, the inexpensive AO3418 has a Cgs of 235pF (typical).
It's in the datasheet:
The outputs are floating. You can connect any one of them to ground (but normally we'd connect the 0 V / Common to ground to give a symmetrical supply.
The voltage you're seeing on the output is due to stray capacitance between the mains side and the output side. There isn't enough to provide any useful power so, for example, if you were to connect a lamp between any one of the outputs and mains earth there wouldn't be enough power to light it. Your multimeter input impedance is high enough to give a reading without loading the stray voltage.
Best Answer
In the context of an op-amp circuit, ‘real’ ground is the common zero-volt referenced to the power supplies and rest of the system.
‘Virtual’ ground is a reference used for op-amp signals that represents a signal zero-volt reference.
Virtual ground comes up mainly when using a single-ended supply with an op-amp. In order to represent signals that have negative and positive swing, the op-amp’s input and output signals need to be offset so that they fit between ‘real’ ground connected to -V, and positive supply connected to +V.
So most often, a low-impedance midpoint reference is used as ‘ground’ to give the desired offset to the op-amp.
In fact the virtual ground can be any offset with respect to real ground (including no offset). What makes it virtual is that it’s a driven, low-impedance signal reference and not a direct connection to any ground or supply voltage.