0V in a schematic

Also other related questions:

If I have two equal resistances in a voltage divider, is there any special significance to the tapping point (in between the two resistors) as a voltage refrence?

Is it ever considered to be a "natural" 0V point, or GND? Could you ever call that point a virtual ground? Or is the virtual groubnd concept something entirely different?

Its indicating that it is the reference point for the other voltages. Common and ground may not be the same thing.

Tom

I reckon then that when you see the GND or chassis symbol, that's usualy to be taken as:This is the reference point for the voltage statements in the schematic - unless otherwise stated."

It should be noted on the schematic the reference point for voltage readings given.

Tom

"virtual ground" normally refers to a point which is maintained at ground potential by feedback, e.g. the inverting input of an op-amp whose non-inverting input is tied 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?

I think I've worked it out.

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.

Virtual GND achieved by way of feedback.

special

to

] a ] +----+----[R]---------+ ] | | | ] | O | ] | GND | ] |+ [R2] ] [ V ] | ] |- | ] +---------------------+

Like so?

at

Voltage with respect to what? If GND is your reference, the voltage at point *a* is fixed with respect to that reference at GND potential.

Sure.

R2

end

Draw the circuit so we can be sure of what you mean.

as

at

voltage

Okay, I'm now beginning to get this and I'll be able to understand OP Amps.

When you look at an Op Amp circuit one part of the input source will be connected to GND, the other to say V-. There is an output Vout.

V- is meant to be at 0V with rspec to GND, not that it has actually be connected to ground. There is a clever way of making V- 0V with respect to GND without actually connecting it to GND.

Here it is:

+--------------------------+ | | | +ve | [Vin] 10V | | -ve R1 5R | | | | V- |----GND +---------0V | | Virtual GND | | | +ve 30V R2 10R [Vout] | | | | | +--------------------------+-- V out

V out = R1 + R2 ------- * Vin R1

V out = 5 + 10 ------ * 10 = 30V

Amplification factor: 3.

It works because of negative feedback and making sure V- goes to 0V with respect to GND.

Of course V- is an input to the Op Amp. The V- input has a very high input resistance.

Just me explaing it all to myself.

All quite simple when one can look at a good diagram. :c)

--
That\'s a _horrible_ diagram, and your math, well...



Here\'s a good one:

.V+>---------------+
.                  |   10R
.     10V      +---|--[R2]--+
.    /    5R   |   |        |
.Vin-----[R1]--+--|-\\       |
.                 |  >------+-->Vout
.GND>-------------|+/
.                  |
.                  |
.V->---------------+


Since you\'ve drawn an _inverting_ amplifier:


             R2 * Vin     10R * 10V    
     Vout = ---------- = ---------- = 20V
                R1            5R 


Gain is simply:


           R2     10R
     Av = ---- = ----- = 2
           R1     5R

or:

           Eout     20V
     Av = ------ = ----- 2 
           Ein      10V

So that checks out.
    

Pretty stiff opamp, too,  since to drive the virtual ground to 0V will
require 2A out of the opamp. ;)

JF

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)

I had an engineer once (I'm the tech), that was looking through a schematic with me, who asked, incredulously, "What the heck is a zero-gain inverter for?" ;-)

To the OP, it depends mostly on what else the circuit is connected to.

It's most likely just the reference point for measurements in the circuit, and/or the DC return of the power supply.

If the whole thing (including the power supply) is isolated, then it really doesn't matter.

Cheers! Rich

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.

No, if I understand the above, the input impedance is still Rin since the '-' input is still a voltage source. It's no longer "ground", but ground is just what *you* choose to call ground.

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.

when dealing with some circuits it may be convenient to call that point a ground especially if the midpoint of the divider is only lightly loaded.

"virtual ground" has a special meaning - it may be better to refer to the divider midpoint as a synthetic ground.

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

0V, GND.

+--------------------------+-- +12V | | | +ve | [Vin] 10V | | -ve Rin 5R | | | | V- +---+2V +--------- +2V | ^ | ^Common Connection | "Virtual GND" | | | +ve 20V Rf 10R [Vout] | | -ve | | | +--------------------------+-- -18V Vout

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.

Above, just my take on things.