Very Elementary Question

[IanB]

As I understand grounding means designating one node to be the reference node relative to which other voltages are measured.

No, this is not what grounding means. Grounding means making a physical connection from some part of a circuit to an external point of known and stable potential (commonly the earth, but not always), such that all voltages in the circuit are predictable relative to that known potential.

This can be done for safety, circuit protection, or to eliminate unwanted noise signals in the system.

To give a contrived example:

You could have a battery powered signal source like a microphone with preamp, feeding a battery powered amplifier. Suppose now you attach the microphone to your nylon shirt that has a static charge of 10,000 V on it. The microphone is now sitting at a potential of 10,000 V induced by the nearby static charge. It doesn't care of course, because it is isolated and floating and does not know what the outside world is doing.

Now suppose we plug the microphone into the amplifier, which is sitting on a desk and not charged up to 10,000 V. As soon as we plug in the microphone there will be a voltage difference of 10,000 V between the microphone and the amplifier. This might be bad for the amplifier, since its input circuits are only expecting to see a few mV. The amplifier could get fried.

One solution to this is to establish a common voltage reference between the mic and the amp, so they are both sitting at the same potential. This is usually done with the shielding braid on the microphone cable. Both mic and amp  physically connect their circuit grounds to the common braid and now both of them share the same ground potential.

[smoothtalker]

May i ask a Very Elementary Question too? i'm confused with mains earth reference. i'm a newbie too.

Why does grounding a circuit makes it reference to that grounded voltage? For.eg, the 230V AC line. The neutral is grounded. i know it doesn't affect the circuit connected to it. we will see 230v rms between hot and neutral when it's open, but why do we still see a 230 rms between hot and neutral when the circuit is loaded? isn't the AC sine wave also running in the neutral?

i don't understand how grounding affects a circuit.

thanks

[IntegratedValve]

As I understand grounding means designating one node to be the reference node relative to which other voltages are measured.

No, this is not what grounding means. Grounding means making a physical connection from some part of a circuit to an external point of known and stable potential (commonly the earth, but not always), such that all voltages in the circuit are predictable relative to that known potential.

This can be done for safety, circuit protection, or to eliminate unwanted noise signals in the system.

To give a contrived example:

You could have a battery powered signal source like a microphone with preamp, feeding a battery powered amplifier. Suppose now you attach the microphone to your nylon shirt that has a static charge of 10,000 V on it. The microphone is now sitting at a potential of 10,000 V induced by the nearby static charge. It doesn't care of course, because it is isolated and floating and does not know what the outside world is doing.

Now suppose we plug the microphone into the amplifier, which is sitting on a desk and not charged up to 10,000 V. As soon as we plug in the microphone there will be a voltage difference of 10,000 V between the microphone and the amplifier. This might be bad for the amplifier, since its input circuits are only expecting to see a few mV. The amplifier could get fried.

One solution to this is to establish a common voltage reference between the mic and the amp, so they are both sitting at the same potential. This is usually done with the shielding braid on the microphone cable. Both mic and amp  physically connect their circuit grounds to the common braid and now both of them share the same ground potential.

You are referring to "earth" ground. What we call reference node (ground) is relative and can be any node. However I'm wondering if there's one and only one node in the circuit that all currents sink into it or lets say "best" "ground" candidate.

[IanB]

As I understand grounding means designating one node to be the reference node relative to which other voltages are measured.

You are referring to "earth" ground. What we call reference node (ground) is relative and can be any node. However I'm wondering if there's one and only one node in the circuit that all currents sink into it or lets say "best" "ground" candidate.

No, I am not referring to earth ground particularly, I am referring to grounding in general. I am trying to correct your misunderstanding about grounding.

Grounding has nothing to do with reference nodes in circuit analysis. Picking a "ground" reference on a circuit diagram is purely algebra, nothing more. It works like this: all circuit voltages are differences in potential between two points. Let's say we have three points on a circuit diagram, call them A, B and G. Let's reference the voltages to point G. So the voltage at point A referenced to G is V_A - V_G. And the voltage at B is V_B - V_G. Therefore, the difference in voltage between points A and B is:

  (V_A - V_G) - (V_B - V_G) = V_A - V_B

That's all it is. A pure numerical convenience to avoid unnecessary arithmetic and keep the degrees of freedom straight. There is no node where all currents "sink into" (since currents don't source or sink anywhere, they just flow round and round in circles). There is no "best" candidate other than the one that is most convenient for humans looking at the circuit.

Just keep in mind that grounding a circuit happens physically, in the real world. It can be protective earth grounding, or signal grounding, or combinations, but it is always a practical matter with a functional purpose. Do not confuse it with reference nodes in circuit simulations which have no functional purpose except computational convenience.

[IntegratedValve]

Ok then how does a circuit know I used a specific reference node to get it working if it's not connected to some earth point? What about battery powered devices like cellphones?

[Len]

Ok then how does a circuit know I used a specific reference node to get it working if it's not connected to some earth point? What about battery powered devices like cellphones?

It doesn't know! The circuit doesn't care which node you choose to put your black DMM probe.

It's like measuring distance, the world doesn't know or care where you choose to put the end of your ruler. There is no "centre of the universe" that everyone starts measuring from, and there is no "universal zero voltage" point in your cellphone. Any voltage measurement is measuring the "distance" between two points chosen by you. (That's why voltmeters have two probes, not one.)

[IanB]

Ok then how does a circuit know I used a specific reference node to get it working if it's not connected to some earth point? What about battery powered devices like cellphones?

Again, you are mixing up two different concepts. Reference nodes in in circuit simulations have nothing to do with connections to earth points.

Reference nodes in simulations are to make the simulator work. They have nothing to do with whether or how the circuit works.

Earthing points in the real world have nothing to do with circuit simulations.

Therefore:

The circuit doesn't know or care what reference node you used to simulate it. The circuit will do whatever it does regardless. The circuit will work even if you don't simulate it at all and just go ahead and build it.

Battery powered devices like cellphones work because earth points are not needed for circuits to work. Currents flow in loops. If there is a closed loop current can flow and the circuit can work.

Consider a circle. Which side of the circle is the bottom?

Consider a battery circuit. Which part of the circuit is the ground?

See? The question doesn't really make sense.

Reference nodes in simulators are there to make the computer simulation have a numerical solution. They have no bearing on the function of the circuit being simulated.

Try this question:

Point A is 4 m higher than point B.
Point B is 2 m lower than Point C.

What is the elevation of point C?

[IntegratedValve]

This is one battery cell used to dual-supply Op Amp. Anything wrong with this design?

[tszaboo]

This is one battery cell used to dual-supply Op Amp. Anything wrong with this design?

Yes. You did not decouple the virtual ground. The power supply voltages depend on the load current of the virtual ground.

[IanB]

This is one battery cell used to dual-supply Op Amp. Anything wrong with this design?

With the partial circuit as drawn you can remove both resistors and the virtual ground and it will be exactly the same circuit. At present the resistors serve only to drain the battery.

In case you are puzzled by this, observe that the two resistors are not connected in any way to the op amp unless you complete the circuit by making some connections to the op amp inputs. Once you have wired up the inputs in this way and made all necessary connections, you will discover that the ground symbol you have drawn is not important to the operation of the amplifier.

Did you reach a conclusion on the question about the elevation of point C?

[IntegratedValve]

Did you reach a conclusion on the question about the elevation of point C?

Relevant to what point? It's lower than A by 2 m and higher than B by 2 m.

This is an inverting amplifier configuration, I tried calculations using nodal analysis and I still get same results if were calculated without considering power supply. Where did I screw up?

[IanB]

Did you reach a conclusion on the question about the elevation of point C?

Relevant to what point? It's lower than A by 2 m and higher than B by 2 m.

It's relevant to voltages at nodes in circuit simulators and why you need to add a ground symbol somewhere when simulating a circuit.

Consider:

There are two equations in three unknown variables:

A = B + 4
B = C - 2

Since there are three variables and two equations there is one degree of freedom. It follows that we can assign any elevation to C that we wish. If, for example, we assign an elevation of 100 m to C, then B is at 98 m and A is at 102 m.

The same situation happens with voltages at nodes in circuit analysis. There is always one degree of freedom that must be arbitrarily satisfied. It is a convention to pick a node, any node, and arbitrarily assign it a voltage of zero. This is done by attaching the ground symbol to that node in the circuit. The circuit in real life has no physical device corresponding to that ground symbol.

This is an inverting amplifier configuration, I tried calculations using nodal analysis and I still get same results if were calculated without considering power supply. Where did I screw up?

When you do your nodal analysis you are considering the amplifier to be a "black box" and not including what is inside it. As a black box, the amplifier is assumed to be self powered and to have an infinite range of possible voltages. As such, the actual power supply (and any limitations it may impose) is not part of the analysis.

Therefore, if you erase from your circuit diagram the battery and the two resistors and the connections to +Vcc and -Vcc, you will still have the same ideal circuit. Furthermore, if you now connect each of the ground symbols with a line and erase all the ground symbols, you will still have the same circuit. The ground symbol is just a shorthand for "all these points are connected together".

[IntegratedValve]

Correct me if I'm wrong, are you saying the ground between the two resistors of the voltage divider is not necessary? Lets consider the typical two battery supply connection presented all textbooks, it shows the node between the two batteries is "grounded", is this necessary? What type of ground they are referring to? Is it connected to OpAmp external circuit "ground"?

[IanB]

Correct me if I'm wrong, are you saying the ground between the two resistors of the voltage divider is not necessary? Lets consider the typical two battery supply connection presented all textbooks, it shows the node between the two batteries is "grounded", is this necessary? What type of ground they are referring to? Is it connected to OpAmp external circuit "ground"?

Yes, I am saying that, in a way. For the ideal black box op amp schematic the virtual ground is not needed. For the real op amp in a physical circuit it is needed if you want a reference voltage mid-way between the power rails. When it is needed, it is needed because the op amp output can source or sink currents and these currents need a return point to get back to the power supply.

But I see you are still getting stuck on this idea of "grounding". It would probably help you to start drawing circuit diagrams without any ground symbols, since the ground symbols seem to be causing confusion. Instead of using ground symbols, just draw a line on your schematic between points that share a common connection. Do that for a while and see if things become clearer.

[Galaxyrise]

Correct me if I'm wrong, are you saying the ground between the two resistors of the voltage divider is not necessary?

It is not necessary that you called it ground, or that you designated the connection with a net symbol instead of solid lines.

Lets consider the typical two battery supply connection presented all textbooks, it shows the node between the two batteries is "grounded", is this necessary? What type of ground they are referring to?

Just the "I hereby declare this to be 0V" ground in those simple examples.  You could just as easily let -VCC be 0V; it's not necessary to declare 0V to be between the batteries.  The circuit will function just fine whether you call one of its voltages "ground" or not.

And in your circuit, the voltage from VCC to "ground" will vary based on the output of the opamp, which is probably not the intent of the design. This is what NANDBlog was referring to. It is a common beginner mistake (I'm certainly guilty of it!) to design a part of the circuit in isolation (in this case, the resistor divider) and then use it in a way that makes it work very differently (by attaching the output to it.) And that's why the textbook example is two batteries instead of a battery and a resistor divider

[IntegratedValve]

I think I got it now.

Thanks you all for your patience and informative help.

Conclusion:

1- The point between the two batteries has to be connected to the OpAmp external circuit so that the internal circuitry swings the output voltage between the two rails.

2- In case of using one battery of double voltage (2 x Vcc), although the voltage difference between +Vcc and -Vee OpAmp connections is the same, but connecting one end to the OpAmp external circuitry will cause the swing between 0 and 2xVcc?

3- the voltage divider method does not work because when it's loaded the voltages across each dividing resistor will change based on load current, which in turn will affect the normal operation.