The virtual ground in operational amplifier circuits has exactly the ground potential only if the op amp is ideal and its differential gain is infinity. Under this condition when the non inverting input is at ground the inverting terminal will be a virtual ground. As the op amp output voltage is of limited value and the ideal op amp gain is infinite potential difference between the input op terminals will tend to zero.

Real op amps have very large gain but not infinite. Consequently, if one of the input terminals is at ground, the other terminal will have a potential that basically different from the ground potential and it is equal to the ratio of the output voltage to the differential gain of the op amp. Therefore the potential of the other non ground point will be small and can be considered zero. It is then said that this point is a virtual ground. Then, it is clear that the virtual ground has always a small potential , basically different from the ground potential at all output voltages different from zero.

The concept of virtual ground is introduced merely as an assumption to simplify circuit analysis, no more no less.

Since nodes with exactly the same potentials can be connect together with a conducting wire without disturbing the circuit and since the virtual ground is at different potential from the ground, one can not connect a virtual ground to the real ground other wise the circuit will be disrupted.

The virtual ground is merely an assumed approximate ground to simplify the circuit analysis no more no less.

Virtual ground is like as if 2 points are connected together with 0 resistance but NO CURRENT FLOWS through it. But in reality if we connect 2 points with a negligible resistance, the current will flow through it and hence its not allowed to happen. So the answer for your question would be NO, we cannot connect virtual ground to real ground.

Potential( Voltage) is measured with reference point, normally earth is considered as reference point(ground)at zero potential.If you take the point as a reference(at zero potential)but not connected to ground can considered as(flotting) virtual ground.Hence if your considered virtual grounds are at same potential, can be connected and if the both at zero potential can earthed solidly.

For my opinion, at first we have to discuss and clarify both terms. When is a ground potential "virtual" and under which conditions we call it "real"? I think such a definitions is necessary in order to avoid misunderstandings. Otherwise - thinking of the inv. opamp input - the answer would be: No, it is not allowed to connect both grounds because the amplifier would be dead.

Having thought about the question for a while my conclusion is as follows:

* In case the virtual ground potential is caused by fixed and independent energy sources (voltage or current) it is possible to connect this point with the real ground without changing the current/voltage distribution within the network.

* That means: If at least one dependent source (like opamp circuits with feedback) is involved, such a connection will alter the currents and voltages.

Prof.Cyril,

I think the answer for this question comes straight from our earlier discussions.

Refer Link:

http://en.wikipedia.org/wiki/Operational_amplifier

An operational amplifier (op-amp) is a DC-coupled high-gain electronic voltage amplifier with a differential input and, usually, a single-ended output. An op-amp produces an output voltage that is typically hundreds of thousands of times larger than the voltage difference between its input terminals.

The op-amp is one type of differential amplifier. Other types of differential amplifier include the fully differential amplifier (similar to the op-amp, but with two outputs), the instrumentation amplifier (usually built from three op-amps), the isolation amplifier (similar to the instrumentation amplifier, but with tolerance to common-mode voltages that would destroy an ordinary op-amp), and negative feedback amplifier (usually built from one or more op-amps and a resistive feedback network).

As the input signal connected to the inverting terminal appears across Rin connected between the inverting terminal which is at the Virtual Ground Potential and the Non-inverting Terminal which is grounded and hence is at actual ground potential Zero volts, if the inverting and non inverting terminals are shorted,the input terminal gets shorted . vide equivalent circuit of the op-amp given in the attachment and also in the link.

Hence the two terminals virtual ground and actual ground terminals should not be shorted.

Prof: P.S

Prof. P. Subramanyam,

your contribution is a good example for my desire first to define what we mean by using the phrase "virtual ground".

Is it "nearly" to ground (as in the opamp example) or is it exactly at ground potential, but "virtual" because of a missing wired connection ?

Prof.Lutz,

The inverting terminal is the Virtual Ground which is at near Ground Potential .The non-inverting terminal is actually connected to Ground which is at actual ground potential and is at zero voltage.

One end of the of the Resistance Rin in the equivalent circuit is connected to the inverting terminal which is at the virtual ground potential. The other end of this Resistance Rin is connected to the non-inverting terminal and is at actual ground potential.

The remaining part of the input signal circuit consisting of the signal source and its internal source impedance get connected between the inverting and non-inverting terminal on the other side through ground, thus completing the input signal circuit. This part of the circuit is not given in the equivalent circuit of Op-amp in the link or in the attachment sent by me.It is understood.

Now the input signal circuit consists of the signal voltage source .,its source impedance and Rin all in series and the circuit gets completed through ground.

We know that It is customary in Electronic Circuits to show the supply source by means of an arrow pointing towards + or - of the source voltage for eg.

the collector of of npn transistor is pointed towards +Vcc..The remaining part of the circuit or how the circuit gets completed is all understood or assumed.

Signal being an alternating voltage source,its phase terminal is connected to inverting terminal and its neutral terminal is actually connected to ground.Hence this neutral terminal of the source signal is at actual ground potential..Since the non-inverting terminal is also at the actual ground potential the input signal circuit gets completed through ground.

If this circuit is drawn separately in full it will be clear.

Thus the Signal Voltage appears across Rin..

Hence the inverting and non-inverting terminals should never be connected together to avoid shorting of the input signal.

Mr.Sanath Kumar has given the answer correctly, but his explanation is not properly brought out. I am sorry that Mr.Bhupender Desai's answer is ambiguous.He has not clearly mentioned whether virtual ground and actual ground terminals can be connected together..

Regards.

Prof.P.S.

Prof. P. Subramanyam,

thank you for your long explanation - however, there was a misunderstanding between us. Everything you have described is and was, of course, also my understanding. But again back to my last recommendation for a clear definition.

I think, when "virtual ground" is defined as described by you above ("near" by real ground) the main question at the beginning of this discussion ("can we connect...") is useless.

Of course, we are not allowed to connect because "near ground" - even if somebody may call it "virtual ground" - is not identical to the true ground potential - and as you also have mentioned, the differential input would be shorted.

Thus, I think the main question at the beginning (formulated by Prof. Cyril Mechkov) is related NOT to the inverting opamp input terminal because it never is a true "virtual ground". We only are allowed during calculation of gain, input impedance ,....to treat this node as if it would be a virtual ground. Am I right???

Virtual ground cannot be connected at true ground. I agree with Prof. Lutz comments as virtual ground is used to obtain the high value of gain

Prof.Lutz and Prof.Athar Azeez,

As already discussed Virtual Ground is not actual ground having zero voltage potential.Virtual Ground has a potential difference with reference to the ground,however small the value may be. Unless we want both the terminals to be at the same potential we should not connect them together.

Even in the case of negative feedback used for computation we are not actually shorting the virtual ground and real ground.We are only assuming that the voltage between these two points is so very small that the voltage at the inverting terminal is assumed to be at ground potential of zero volts for computation purpose only

As already explained earlier, virtual ground and actual ground points in the Op-amp should not be connected to avoid shorting of the signal applied to the inverting terminal since the signal voltage appears across these two points as the Op-amp is a dual input single output(DISO) Differential amplifier.

In short, we can say that no two terminals with different potentials can be connected,be they virtual ground and real ground or any two virtual ground points.They can be connected only if both of them have the same potential.

In Electrical Engineering Short circuiting means a Fault.

One of my colleagues,a professor himself, used to say jocularly that there are only two types of faults in Electrical Engineering namely open circuit fault and short circuit faults.Even the Ground Fault in a Transmission line is also a Short circuit between the conductor and earth.It is true for our Op-Amp or in general.

Pisupati.

May I come back to one of my previous comments (approx. 8 postings back).

I was pointing to the difference between independent and dependent sources.

Example: A chain of two resistors (10k and 5k) is connected between two ground referenced voltage sources (+10 V and -5 V). In the middle between both resistors there is what I call a "true virtual ground". This node can be connected with the common ground without changing the currents through the resistors.

This may illustrate my desire to have first a common DEFINITION of "virtual ground".

Of course, without any discussion it is obvious that the inverting terminal of an opamp must not be shorted. I am sure that was not the intention of the original question from Prof. Cyril Mechkov .

The virtual ground in operational amplifier circuits has exactly the ground potential only if the op amp is ideal and its differential gain is infinity. Under this condition when the non inverting input is at ground the inverting terminal will be a virtual ground. As the op amp output voltage is of limited value and the ideal op amp gain is infinite potential difference between the input op terminals will tend to zero.

Real op amps have very large gain but not infinite. Consequently, if one of the input terminals is at ground, the other terminal will have a potential that basically different from the ground potential and it is equal to the ratio of the output voltage to the differential gain of the op amp. Therefore the potential of the other non ground point will be small and can be considered zero. It is then said that this point is a virtual ground. Then, it is clear that the virtual ground has always a small potential , basically different from the ground potential at all output voltages different from zero.

The concept of virtual ground is introduced merely as an assumption to simplify circuit analysis, no more no less.

Since nodes with exactly the same potentials can be connect together with a conducting wire without disturbing the circuit and since the virtual ground is at different potential from the ground, one can not connect a virtual ground to the real ground other wise the circuit will be disrupted.

The virtual ground is merely an assumed approximate ground to simplify the circuit analysis no more no less.

I would like to add some examples of "real" and "virtual" grounds that may be tested if they can be connected together or to the real ground:

* a resistor connected to the real ground (the Lutz's idea above:)

* a voltage follower connected to the real ground creates a virtual "copy" of the real ground:)

* a resistive network consisting of two connected in series resistors ("biased voltage divider", "resistive summer" or "half-bridge circuit") that is "stretched" between the positive and negative rails

* the same resistive network but buffered with a voltage follower

* the same network but where the one ohmic resistor is replaced by a constant-voltage (voltage-stable) nonlinear resistor (e.g., a zener diode)

* a network consisting of two connected in series equal but opposite resistors ("positive" and negative) that is connected to the ground

* the common emitter point of a differential amplifier operating in a differential mode

* the middle point inside the load connected between the two collectors of a differential amplifier operating in a differential mode

* the middle point inside the common resistor connected between the two input op-amps of an instrumentation amplifier operating in a differential mode

* the inverting op-amp input of any op-amp inverting configuration with negative feedback

In the last classic virtual ground example, it is very interesting to see what happens if we connect two or more op-amp virtual grounds (inverting inputs) together.

You ask an excellent question; however, as a number of colleagues have noted, the idea of a virtual grround needs to be properly defined for the application at hand. You certainly can connect a virtual ground to ground, thereby bringing the virtual ground to ground. Good examples of virtual grounds occur in EMI measurements and in nerve cell, or electrophysiology, measurements. In modeling the conduction process in an axon, we typically assume that the extracellular space is ground, but, in fact, this is a virtual ground where the potential is time-varying and due to both dependent and independent energy sources.

The ideas given by Prof.Cyril Seem to be alrigt .It is also similar to the one we had been discussing with reference to negative feed back in op-amp earlier explained by me in terms of Potential Divider Principle.

People may be interested in going through some of the circuits given in Circuit- Fantasia link for which is given below.

http://www.circuit-fantasia.com/circuit_stories/inventing_circuits/virtual_ground/virtual_ground.htm

In addition one can refer to the link in which the circuit creates a solid, unbudgeable, virtual ground. It enables a DC power supply or wall adapter with a two conductor output to function as a split supply - with a three conductor output (i.e., positive, negative AND ground).

www.goldpt.com/virtual_ground.html

Some more circuits for creating virtual ground are given in the link below,some of them already known to us.

https://www.google.com/search?q=virtual+ground+circuits&hl=en&tbo=u&tbm=isch&source=univ&sa=X&ei=LrgbUcvPG4OsrAe-0YDIDQ&sqi=2&ved=0CD0QsAQ&biw=1440&bih=735

I hope that these links maybe helpful on the topic.

I am not well versed in nerve cell, or electrophysiology, measurements referred to by Prof.Savatore Mporgera,

Prof. P.S.

Lutz, would you clarify what you mean when saying "true virtual ground" in the example below:

"A chain of two resistors (10k and 5k) is connected between two ground referenced voltage sources (+10 V and -5 V). In the middle between both resistors there is what I call a 'true virtual ground'. "

It seems if there is a "true" virtual ground, we should observe a "virtual" virtual ground as well:)? Maybe, you consider the first as more natural and the second - as more artificial?

Regards, Cyril.

Probably Prof. Lutz means by the term "True Virtual Ground " a virtual Ground Point whose potential is also at exact ground potential of zero volts but away from the actual grounded terminal at zero voltage. It is different from a virtual ground point whose Potential is so very small that its value can be treated to be almost equal to exact ground potential of zero volts but still different from the actual ground as in the case of Op-amp with negative Feed Back.

Dr.P.S.

Thank you, Dr. P Subramanyam,

this was and is exactly the meaning of my contribution.

True virtual ground: As explained using the simple voltage divider

Near virtual ground: As can be found at the inv. terminal for real opamps with feeback.

Another suggestion - "static virtual ground" (created by a voltage divider) and "dynamic virtual ground" (produced by an op-amp)?

The concept of Static and Dynamic Virtual Grounds,raised by Prof.Cyril has some relevance, I feel.

When the signal is given to the inverting terminal of the Op-amp with negative Feed Back. what occurs is creation of a Dynamic Virtual Ground, as defined by Prof.Cyril,whose potential is approximately equal to actual Ground Potential.

Static Virtual Ground as created by Potential Dividers is always present without reference to Op-amp or Signals.

Creating a solid, unbudgeable, virtual ground enabling a DC power supply or wall adapter with a two conductor output to function as a split supply - with a three conductor output (i.e., positive, negative AND ground) is an example of Static Virtual Ground. The reference to it had been given by me earlier.

We have to contemplate (if it is possible, without shorting the Signal or in any other mode or device) about creating of a Dynamic Virtual Ground(Real Virtual Ground as defined by Prof.Lutz) since the Dynamic Virtual Ground created in. the Op-amp with negative feed back is an example of,if I can say, a Virtual or non -true Virtual Ground whose potential is approximately the potential of the Real Ground in the sense of Prof.Lutz's definition of Virtual Ground.

Dr.P.S.

...In this chain of thoughts, the "static virtual ground" cannot withstand the our intervention when "pulling" it down or up (by sinking or sourceng current) since it cannot "feel" and react to the disturbance. The "dynamic virtual ground" is "intelligent" and "clever"; it (the op-amp) reacts to our intervention drawing back. And the frequently used trick is to use the intervention as an input and the op-amp reaction as an output; this is the situation in an op-amp inverting amplifier...

Just like shifting of the operating point on a Load line by changing the biasing, and signal excursion with respect to the biasing point, I think the static Virtual Ground also can be changed by changing the the Supply Voltages to the Pots that create the static virtual ground and of-course the Dynamic Virtual ground in the case of Op-amp can dance according to the tunes of the signal as Prof.Cyril has pointed out.

Dr.P.S.

May I remind all RG members having contributed to this problem up to now that the original question was if a connection between real and "virtual" ground is allowed - of, course, without changing voltages and currents in the network.

About 6 days ago I have made an assertion regarding this point - and I am interested to hear about your opinion.

For the sake of simplicity I repeat my suggestion in the following:

* In case the virtual ground potential is caused by fixed and independent energy sources (voltage or current) it is possible to connect this point with the real ground without changing the current/voltage distribution within the network.

* That means: If at least one dependent source (like opamp circuits with feedback) is involved, such a connection will alter the currents and voltages.

Lutz, I second your assertion. These days, I have completely realized what virtual ground is and will expose my insights in a separate question. I hope it will be an interesting discussion.

Regards, Cyril.

Hi Cyril,

I forgot to mention that I strongly support your definition ("static" vs. "dynamic" virtual ground) you have proposed two days ago. I think, it is a very good and helpful definition that should find entrance into textbooks.

Regards

L.

Thank you, Lutz! To make the discussion more fun, here are two amusing examples of a static and dynamic virtual ground:

If there are 0$ in my credit card, this is an example of a "money virtual ground". But if I spend 1000$, it becomes -1000$ ("no virtual ground"). So, this is a "static virtual ground".

Another example of a "money virtual ground" is If there are 0$ in the credit card of my wife. In this case, if she spends 1000$, I will immediately pay in 1000$ and it will remain 0$. So, this is a "dynamic virtual ground":)

Have a fun, Cyril.

Prof. Lutz's statement " In case the virtual ground potential is caused by fixed and independent energy sources (voltage or current) it is possible to connect this point with the real ground without changing the current/voltage distribution within the network.

* That means: If at least one dependent source (like op_amp circuits with feedback) is involved, such a connection will alter the currents and voltages."

Prof.Cyril's example of credit card explains the phenomena of Static and Dynamic Virtual Grounds fairly well.

I may supplement it by having an initial balance of $500 instead of $0, for the same transactions the variation now will be with reference to the initial bias of $500 instead of $ 0. That explains my assertion of shifting of the virtual ground just like shifting of the operating Point in the load line by changing the bias and the excursion of the signal and amplified output around the new operating point.

As Prof.Lutz pointed out, this is a worthy contribution to be included / added to the topic of Virtual Ground and Virtual ground Op-amp with negative feed back.topic to bring out the difference between Static and Dynamic Virtual Grounds..

This Discussion throws light on only the Static and Dynamic Virtual Grounds.

However the original question of how to create Real Virtual Ground of actual Ground Potential in Op-amp with negative feed back still remains to be answered,for ; the virtual ground in this case has approximately equal ground potential and not exactly equal to ground potential.

At -least, to complete the current discussions. we have to establish whether such a task can be performed or not.

Dr.P.S.

Dear Pisupati, thank you for the contribution. Your thoughts are very useful for the progress of all the discussions in the area of electronics here. I cannot explain exactly why and how but it is true that you stimulate me to propose and develop a chain of questions; I am sure the other contributors will share my opinion as well...

In your comments, you implicitly pose for consideration various basic circuit concepts. Let me, following your suggestion, extract interesting questions from our discussions and pose separate new questions. I urge on you and the other contributors here to do the same. For example, it is interesting to consider the fundamental concepts of biasing, load line, shifted virtual ground, etc. proposed by you...

In the funny "money" analogies above I have used a credit card to be able to draw money even there is no money in the card:) IMO your suggestion to "shift up" initially the "money virtual ground" with $500 is more appropriate for the case of the debit card where I could not draw money if there is no money in the card. But you made me ponder whether it is better to change my wife's credit card with a debit one:)

Have a fun, Cyril.

Cyril, as you mentioned the creation of a "chain of questions" here comes such a question:

Imagine a simple differential amplifier made of two transistors (long-tailed pair) driven by two voltage sources with different sign. By proper selection of these sources and the bias point it should be possible to have a common emitter node with exactly ground potential.

For my feeling this could be a "static" virtual ground - and a connection to the real ground potential would not change dc currents or dc voltages.

However, as soon as we try to use this circuit as an amplifier (input signal at the base) the "static" virtual ground changes into a "dynamical" one, which must NOT be connected to the real ground. Otherwise the circuit could not work as desired, right?

Great idea, Lutz! I highly appreciate the topic about the long tailed pair, biasing, common and differential mode, dynamic load, op-amp input structure, cascode stages and all kinds of such odd and exotic circuits...

Regards, Cyril.

Lutz, your example (the long-tailed pair) is extremely interesting although we should consider it in a separate question. Yes, if we have chosen a constant appropriate common-mode input voltage (about 0,6V), the common emitter point can be set at 0 V and we can connect it to the ground; nothing will change.

Then, we can apply whatever differential input signal and the point will continue staying at 0 V; so we may connect it to the ground without any problems. In this differential mode, the long-tailed pair is not "interested" in the nature of the ground (virtual or real); it only "wants" some ground. Actually, there is no long-tailed pair; there are two separate common-emitter stages with (virtual or real) grounded emitters.

But we should not forget that this virtual ground is created and maintained by two voltage (emitter) followers whose outputs are connected in parallel. And they are circuits with negative feedback that endeavour to keep up steady common emitter voltage. So, if we connect this virtual ground to the real ground, they will not react since they are not disturbed. But if we connect this point (via a resistor) to some different voltage (e.g., to the power supply), they will react to this intervention by "pulling" the point to the opposite direction.; they will "cooperate" in this work "helping" each other. So, this virtual ground behaves as a "dynamic virtual ground" in the sense that it withstands any kind of disturbances.

But this configuration (with real grounded emitters) is not a true differential amplifier. If we change the common-mode voltage, the output collector voltage(s) will not stay constant; it (they) will change dramatically and finally the transistor(s) will saturate. This is the answer to your question, "Otherwise the circuit could not work as desired, right?"

The great idea behind the long-tailed pair is that the virtual emitter ground appears at the differential mode and dissapears at the common mode. Thus, in the first situation, the two transistors mutually "oppose" and virtually ground their emitters; as a result, they act as separated (by the common virtual ground) common-emitter stages. In the second situation, the two transistors mutually "help" and "cooperate" in the changing the common emitter voltage; they act as parallel connected common-collector stages (emitter followers).

As a conclusion: IMO your assertion is right (as usual:) But, as you have frequently said, we have to define precisely the term "static" and "dynamic" virtual ground. I suggest to consider a virtual ground "dynamic", if it is kept up steady by a negative feedback, not if it is simply variable. Maybe we should master more appropriate name? And maybe, the question should be, "As the virtual ground is a kind of ground, can we connect it to the real ground or (via a resistor) to a different voltage?"

I have an idea to discuss all these so interesting and important situations in separate questions. Regards, Cyril.

The question raised by Prof.Lutz and the answer given by Prof.Cyril are quite interesting.

As rightly pointed out by Prof.Cyril this question and its addendum can be posed as a fresh question because of the operations being different for Differential Amplifiers-common mode or Differential mode,with tail Resistance or without it or with tail Resistance having a parallel capacitor across it ,or Op-amps etc.

We have to limit the discussions on each question to sustain the interest of the participants,they being mainly students.

Hence I suggest that we can branch out to a new question.

Dr.P.S.

Yesterday, my students and I, all together, reproduced the Ohm's law in the laboratory to investigate the virtual ground appearing on a resistive layer. We made a movie (1 GB); here is the link:

https://photos.app.goo.gl/SDHk2iW3oK4RjneE6

The movie is closely related to the Circuit idea story about this experiment:

http://en.wikibooks.org/wiki/Circuit_Idea/Walking_along_the_Resistive_Film

Regards, Cyril.

Prof.Cyrils movies on Walking along the Resistive Film are Exciting and Wonderful. The Voltage Drop along the Resistor is normal as seen in any book,but the Virtual ground inside the Resistor is amazing.

I am curious about how these Visuals have been filmed. I really enjoyed the Films. Are they obtained by animation or by using Luminescent Material or by monitoring through a series of Voltage Sensors and Filmed continuously?It is really perplexing to me.I will be happy,so will be others,I suppose,if the Filming Process is explained at-least briefly. Long back in the Electronics Lab I remember to have seen some high wattage wire wound resistors on porcelain base glowing when passing current..

Thanks to Prof. Cyril for the Splendid work.

Dr.P.S.

Hi, Pisupati! Thank you for the admiration for my work. I created a series of such "real-time simulations" ("live animations") in the late 80's when I, being a young assistant, put in practice with a great enthusiasm this approach in the technical education.

The idea is simple: the computer controls (supplies, drives) the circuit (by DACs), continuously measures (by ADCs) the voltages of the basic circuit points (nodes), calculates the rest values and draws a respective picture (voltage bars, a voltage diagram, a mechanical or fluid analogy, etc.) thus representing the circuit operation in some attractive way. Actually, this is the educator's viewpoint at the circuit under investigation.

In this experiment, the computer supplies (by two DACs) a potentiometer by its two ends, measures (by three ADCs) the voltages at its ends and the slider, and draws three vertical bars which heights are proportional to the respective voltages. The position of the slider voltage bar is calculated. Then it connects the top of the three bars with straight lines representing the envelope of the voltage diagram. This setup is described in the section below of the Circuit Idea story:

http://en.wikibooks.org/wiki/Circuit_Idea/Walking_along_the_Resistive_Film#How_to_visualize_the_voltage_diagram_on_the_screen

I carried out this experiment with my students two days ago. They are so enraptured that, after the laboratory exersise, I decided to make this movie especially for you, my friends from ResearchGate-:) It was not so easy but I managed to do it. I held the camera with my right hand and struck the respective keys on the keyboard (W and Z for increasing/decreasing the left voltage, and I and M for increasing/decresing the right voltage), and also moved the potentiometer slider with my left hand to find the mystic virtual ground on the resistive layer.

Maybe you have seen the op-amp on the middle part of the laboratory setup. I used it in the end of the experiment when I, with a theatrical gesture, disconnected the right end of the potentiometer from the DAC and connected it to the op-amp output (I also connected the inverting input to the slider and the non-inverting input - to the ground). Thus the op-amp continued keeping up the virtual ground instead me... and it did this donkey work better than me... You guess this is a possible way of reinventing the ubiquitous op-amp inverting amplifier...

Best regards, Cyril.

I thank you Prof.Cyril for your immediate Positive Response and for opening the cave visited by Ali Baba with the Manthra "Open Sesame ".

Now We have access to the Vast Treasures of insight into the Circuit elements.

We have nothing else but Admiration for your Initiative and Dedicated Work.

Thanks a lot. to you and to your Devoted Team

Dr.P.S.

Having seen your link "Walking along the resistive film", this seems to be an excellent way to teach circuit concepts, with a visual demonstration. Also, getting the students to assemble the voltage sources, wire, and so on reinforces the learning experience. No doubt the classes are enjoyable too!

Here's a late answer to Lutz von Wangenheim's question posted Feb 17 2013. If you biased a differential amplifier (long-tailed pair) so that the emitter voltage is zero, and then hard-wired that node to a low impedance node at zero volts (that is, you connect the emitters to a 'real ground'), you are correct, the differential amplifier would not work.

To my mind, the explanation is, when the emitters are NOT connected to a 'real ground', the transistor emitter currents are constrained - the sum of the emitter currents must always be equal to the 'tail' current. So if the current increases in the left-hand transistor, it must decrease in the right-hand transistor, and vice-versa.

When you connect the emitters to 'real ground', you introduce another path for current flow at that node, and the constraint that caused the transistor emitter currents to be related no longer applies. You no longer have a defined 'tail' current, so the transistors are no longer obliged to 'exchange' current between one another.