I can see that when you see a voltage statement on a schematic, that is a statement of the voltage at that point with reference to some other point in the schematic. It's not saying necessarily that the stated voltage is across the component.
Okay, is 0V on a schematic meant to be zero volts above ground potential, or can it be any arbitary voltage above ground potential? Is it an absolute statement that the point is at zero volts with reference to ground?
Lets say that one end of PSU and resistor are both connected together and to GND. Other side connects to a resistor or to a divder network.
Increasing the voltage will increase voltage at top end of R, but voltage at GND will always be GND. R sees itself as being connected to GND one one of it's ends. The source sees itself as being connected across R, which of course it is.
Lets say we now have a voltage divider, two 10 Ohms resistances, R1 and R2 which is connected the GND. When voltage rises it will do so across top end of R1, and the middle tapping point.
Does R1 see itself as being connected to GND? No. Why? Because with reference to ground it's lower end increases or decreases. R2 sees itself as connected to GND. The source sees itself as being connected across the divider network, because it is.
Lets say that now we don't connect one end of R to GND, but to a virtual earth.
If the voltage with respect to GND increases (it does because source is connected at one point to GND) voltage with respect to GND will increase at the top end, but will it increase at the bottom end? I don't think so. I think it has to always remain at a constant potential no matter what voltage is put on R's top end. Does the voltage at the bottom end of R really have to be at GND potential? Or is it sufficient it remains constant?
If a source is connected to GND and if it is feeding into a resister R, whose other end is connected to a virtual GND, the potential of that virtual GND must be 0V. If it's not then it will see a load that is nor R. So, a source only sees the full resistance of a load, if the other end of the load, or terminating input resistance, is at 0V with rspect to GND.
Yes, I later realised the math was wrong. I keep switching between inverting amplifier and non inverting amplifier. I should concentrate on each type of circuit individually.
That diagram I drew was to show *to myself*, how in principle we obtain a virtual ground at V- for the inverting amplifier. And why we get Vout as we do for a given combination of Rin and Rf.
I never knew what virtual ground was until now. And now I know, that as long as one end of a circuit compont is held at 0V with reference to ground by an "opposing voltage", for all intents and purposes it's akin to being connected to GND, or 0v. As long as there is somewhere for the current to flow.
I mean, current must find it's way back to the source, so, to the source, it looks like it's terminated in Rin.
V- must be 0V above grund or it would look to the source as if it was terminated in a different resistance value than Rin. If for instance Vin was
10V, and Rin was 10K, and V- was at a potential of 5V above ground, the voltage across Rin would be 5V, not 10V. So the source would see an input resistance of 20K, because current flowing in Rin would be 5V/10K = 0.5mA., whereas it ought to be 1mA. I think I'm getting it. :c)
The differential amplifier is simply the combination of the two. Solve them together and you won't be confused.
Yep. I(Rf) must be equal to I(Ri) (Zin- = infinite), therefore Vo/Vi = -Rf/Ri (since V- is a virtual ground).
Now add a second Ri and you get Vo = - (V1*Rf/Ri1 + V2Rf/Ri2), or the (negative) sum of V1 + V2 times their respective gains. V1 is a virtual ground but it is also called a "summing point" because it can be used to add signals.
By Jove, I think he's got it!
...and enough of it. When the output turns out of "head room" it can no longer supply the current so the "virtual ground" isn't.
it's a benchmark to measure other voltages from, in an isolated device (like bettery powered pocket radio) there is no earth connection and Ov serves to simplify the exporession of the voltages present.
Well, what I'm saying is that the input source is connected to GND, or a point that we call 0V. The source is connected to one side of Rin. It's essential that the other side of Rin is either connected to GND, or 0V for the source to see it's being connected across Rin. With a virtual ground arrangement, if the "earthy" one side of Rin got to be at 5V, then source will not see Rin, but some other terminating value. That's what I understand.
The thing is, GND, if strictly interpreted, is a statement that the point is at zero potential with respect to the actual ground.
But, we all know that in many cases GND is non other than a circuit reference point from which voltages are measured and stated, and that actually GND may not quite be at 0V with respect to true ground.
As to a virtual ground, well, that again does not necessarly mean the point is really at 0V with respect to ground.
Usually virtual ground is generally at the same potential as the 0V point in a circuit, often labelled GND, but it does not have to be. For instance, with the inverting amplifier what matters is that V- is at the same potential as the common connection between source and output. The cicuit goes from source, thru Rin, through Rf, through the output cicuit, and the output connects to the other side of the source. I guess there is no reason the common connection needs strictly be at 0V. See diagram.
But I think in the vast majority of circuits V- will almost invariably be at
Whatever the actual voltage is at V-, one thing is for sure, virtual ground in the context of Op Amps is practically a stable value, and almost invariably at 0V.
Not entirely sure what the midpoint of a voltage divider is, when both resistors are the same value. A lot depends on whether for some reason someone wants to declare it as 0V or GND. Not sure if the center point has some aspect of "virtual" to it or not. Possibly I suppose.