0V in a schematic

[Rich]

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?

 

[Rich]

"Rich" <notty@emailo.com> wrote in message
news:6v81jaFiffjsU1@mid.individual.net...

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?

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?

 

[Tom Biasi]

"Rich" <notty@emailo.com> wrote in message
news:6v81jaFiffjsU1@mid.individual.net...

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?

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

Tom

 

[Rich]

"Tom Biasi" <tombiasi@optonline.net> wrote in message
news:498ed3b6$0$10182$607ed4bc@cv.net...

"Rich" <notty@emailo.com> wrote in message
news:6v81jaFiffjsU1@mid.individual.net...
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?


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."

 

[Tom Biasi]

"Rich" <notty@emailo.com> wrote in message
news:6v83asFific0U1@mid.individual.net...

"Tom Biasi" <tombiasi@optonline.net> wrote in message
news:498ed3b6$0$10182$607ed4bc@cv.net...

"Rich" <notty@emailo.com> wrote in message
news:6v81jaFiffjsU1@mid.individual.net...
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?


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

 

[Nobody]

On Sun, 08 Feb 2009 12:42:42 +0000, Rich wrote:

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?

"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.

 

[Rich]

I said:

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?

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?

 

[Rich]

"Rich" <notty@emailo.com> wrote in message
news:6v89tsFih6llU1@mid.individual.net...

I said:

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?

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.

 

[Greg Neill]

Rich wrote:

I said:

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?

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.

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

Like so?

Increasing the voltage will increase voltage at top end of R, but voltage
at
GND will always be GND.

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.

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.

Sure.

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.

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

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?

 

[Rich]

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)

 

[John Fields]

On Sun, 8 Feb 2009 15:59:44 -0000, "Rich" <notty@emailo.com> wrote:

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

 

[Rich]

"John Fields" <jfields@austininstruments.com> wrote in message
news:5j1uo41ft4cde7jckibqk3q28fnagelb82@4ax.com...

On Sun, 8 Feb 2009 15:59:44 -0000, "Rich" <notty@emailo.com> wrote:

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.

 

[Rich]

"Rich" <notty@emailo.com> wrote in message
news:6vamfpFj27vcU1@mid.individual.net...

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)

 

[Rich Grise]

On Sun, 08 Feb 2009 14:33:13 +0000, Nobody wrote:

On Sun, 08 Feb 2009 12:42:42 +0000, Rich wrote:

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?

"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.

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

 

[krw]

On Mon, 9 Feb 2009 12:39:35 -0000, "Rich" <notty@emailo.com> wrote:

"John Fields" <jfields@austininstruments.com> wrote in message
news:5j1uo41ft4cde7jckibqk3q28fnagelb82@4ax.com...
On Sun, 8 Feb 2009 15:59:44 -0000, "Rich" <notty@emailo.com> wrote:

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.

The differential amplifier is simply the combination of the two. Solve
them together and you won't be confused.

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.

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.

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",

By Jove, I think he's got it!

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.

....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.

 

[krw]

On Mon, 9 Feb 2009 14:16:43 -0000, "Rich" <notty@emailo.com> wrote:

"Rich" <notty@emailo.com> wrote in message
news:6vamfpFj27vcU1@mid.individual.net...

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)

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.

 

[Jasen Betts]

On 2009-02-08, Rich <notty@emailo.com> wrote:

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?

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.

 

[Jasen Betts]

On 2009-02-08, Rich <notty@emailo.com> wrote:

"Rich" <notty@emailo.com> wrote in message
news:6v81jaFiffjsU1@mid.individual.net...
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?

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?

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.

 

[Rich]

"krw" <krw@att.bizzzzzzzzzzz> wrote in message
news:nak1p45tgkbapiofvtch6d9pboc3rv8r8v@4ax.com...

On Mon, 9 Feb 2009 14:16:43 -0000, "Rich" <notty@emailo.com> wrote:


"Rich" <notty@emailo.com> wrote in message
news:6vamfpFj27vcU1@mid.individual.net...

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)

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.

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.

 

[Rich]

"Jasen Betts" <jasen@xnet.co.nz> wrote in message
news:gmrjh8$qoe$2@reversiblemaps.ath.cx...

On 2009-02-08, Rich <notty@emailo.com> wrote:

"Rich" <notty@emailo.com> wrote in message
news:6v81jaFiffjsU1@mid.individual.net...
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?

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?

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.

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.