Hi Cyril - nice to hear from you again.

Here is my short answer to your question: I think, your explanation in the last part of your contribution is nothing else than again a desription of the Miller effect with your own words. In fact, it IS the Miller effect which "apparently" lowers the load resistance. However, only "apparently" because speaking about load or input resistances we reference them to ground.

We allocate the current to one voltage source only - but in reality there is another voltage source which "sucks" a part of the current (50%) out of the common load.

The Output Voltages of the Two Amplifiers are Opposite in Sign and the Load is connected in between the Two Voltage Sources. Thus  the Two Voltage Sources are connected in Series aiding mode and connected in series with the Load. Thus Twice the Voltage Output is Supplying the Load resulting in Twice the current. This is Equivalent to having a single voltage source with half the load impedance to have Double the Current.

Hence each Amplifier sees half the Load impedance. This Explanation is from Circuit Point of View.

P.S.

Dear Cyril,

my preferred explanation is yours :-)

"the amplifier change the output voltage and the current doubles compared to the value referred to the attached load because the other terminal of the load experiences an opposite voltage change".

But rather than to connect it to the Miller theorem I prefer to use the Bartlett theorem on symmentrical/antisymmetrical circuits, applied  to the circuit in the attached Figure.

Simone - great idea! Congatulations!

A cluster of brilliant ideas, my friends! Thanks for the instant response!

Lutz, a striking explanation! Exactly, it is the Miller effect that makes this illusion. As you have noted ("speaking about load or input resistances we reference them to ground"), our delusion comes from the fact that we think of the whole network (speaker + output) as of a load. Indeed, it is a load, but a kind of a more complicated... composite... active load... consisting of a humble passive load (the speaker) and an attached voltage source destroying a part of the load resistance... And, as you have noted, in this way, we have transformed the floating load with a grounded "load" with reduced resistance... what is the great idea of the Miller theorem...

I would only add some interesting observation to your Miller effect viewpoint. Usually it is associated with circuits having negative feedback... but notice that there is no any feedback here...

Pisupati, nice explanation! Really, both voltage sources (amp outputs) are connected in series, in the same direction when travelling the loop, so that their voltages are summed... and we can think of them as of a single voltage source with a doubled voltage.

Simone, it is relly a great idea to use the simmetry when explaining circuits! I remember that this technique was used by one great Bulgarian popularizer of  electronics - Atanas Shishkov (his books in the field of transistors rival such bestsellers as The Art of Electronics; it is only regrettable that they were not translated into English). There he spoke about the "axis of symmetry" in differential circuits, which obviously was the Bartlett theorem...

We can see also the ubiquitous virtual ground at the middle point inside the load (speaker coil)...

But it is interesting to ask another question related to your suggestion: "What will happen if the circuit is not so symmetrical? What will 'see' each of the outputs in this case?"

Dear Cyril,

regarding your answer to Lutz and the attached picture, what it shows is only an approximation that assumes that A is not changed by Z.

Only if you assume that A is the gain of the yellow two-ports, loaded by Z you obtain the axact solution and the appearance of feedback.

Dear Cyrill,

regarding the asymmetry of the circuit I think your answer reported in the picture is correct.

For example the amplifier with the greater gain sees a greater load:

- r2/r1 = V2/V1 = - (G2Vin) / (G1Vin) ;

G2 / G1 = r2 / r1 ;  r2 = G2 / (G1 + G2) * r;

Yes, I think so... and the virtual ground will move inside the load towards the output with the lower voltage.

Dear Cyril,

satisfy my curiosity, how do you make such amazing graphs by hands ? Do you use a CAD with a  graphics tablet ?

cheers

S.

No, Simone... this was the well-known old "technology" - color pens and paper:)

Many color pens, rules, natural light from a big window ... your desk is that of a graphic designer !

Prof.Cyril,

I relish your idea of Input Impedance including the Active Sources ,Prof.Lutz's reference to the Popular Miller's theorem and and Prof.Simone Orcioni's reference to Barlett Theorem.

Your Idea can be correlated with Barlett Theorem,with Active Sources and Input Impedances interchanged. If one is Symmetrical the other one can be treated as Anti symmetrical,I think.

Nice Explanations.

Best Regards.

P.S.

Dear Cyril,

Let us assume that the upper amplifier has an output voltage Vo, the lower amplifier wil have an output voltage of  -Vo. The current pushed by the upper amplifier will be 2Vo/ RL. Which can be put in the form Vo/ (RL/2) = Vo/ RL effective. Which results in RL effective= RL/2 as required.

More important is the concept of changing the impedance of a port by controlling the the voltage of the other terminal which is used to increase the input impedance by an amplifier by feeding a positive voltage to the other terminal.

Assume that the lower terminal of the load is at Vol and the upper is at Vou, then the current I= (Vou-Vol)/ RL= Vou/RLeffective,

Then RLeffectivve=  RL / ( 1- (Vol/Vou)). So RL effective can be controlled by the ratio and sign of the ratio Vol/Vou. For this ratio equals -1 as in our case RLeffective= RL/2.

In conclusion the impedance at a terminal and ground will be lowered by negative voltage and raised by a positive voltage on the other terminal.

Thank you for your question

best wishes

Very interesting thoughts, Abdelhalim! Your calculations are fully consistent with Miller theorem where K = -1:

RL1' = RL/(1 - K) = RL/(1 + 1) = RL/2 (RL seen from the side of OUT 1)

RL2' = K*RL/(K - 1) = -RL/(-2) = RL/2 (RL seen from the side of OUT 2)

(https://en.wikipedia.org/wiki/Miller_theorem#Explanation)

So this is a particular "symmetric" case of the Miller theorem.

Thanks.

I would only add that we can observe the same phenomenon in an op-amp inverter (an inverting amplifier with R1 = R2, K = -1), in the input stage of a 3-op-amp instrumentation amplifier, in the ordinary transistor differential amplifying stage (long-tailed pair)... and maybe in many other symmetric circuits.

Prof.Abdelhalim's Observation, "the impedance at a terminal and ground will be lowered by negative voltage and raised by a positive voltage on the other terminal." is Really interesting.The Observation made is to have RL/2 or 2*RL. I think by changing the values of V1and V2  We may be able to get any multiples of RL ,a wonderful Observation. We have an Impedance Converter without a Transformer and tap changing for the purpose !

P.S.

As mentioned already - phase inversion between both driving voltages is identical to the input referenced Miller effect. Can we say that the opposite - equal phase - is what we call "bootstrap effect"?

Dear Pisupati, I have examined in detail the six possible combinations, which can be obtained from the two possible source connections (in the same and in the opposite direction) and the three proportions between their voltages () in the Wikipedia article about the Miller theorem:

https://en.wikipedia.org/wiki/Miller_theorem#Applications

(by the way, this was one of the powerful ideas on which my thesis was based)

Dear Lutz, thank you for your interest and patience shown by you to this topic that we have discussed many times. I know that you are too far into this subject; so I appreciate your comments very high.

Yes, the "bootstrap effect" is the case when the two voltages have equal values and polarities in regard to ground (but actually they are opposite when travelling the loop):

https://en.wikipedia.org/wiki/Miller_theorem#infinite

I want to thank you for the welcome greeting at the beginning of this discussion. I withdrew for a time from this society since I experienced some stress a few months ago; I was in something like a depression and I was withdrawn from the usual creative activity. Some people here, in my work, and other circumstances contributed to feel a sense of doubt about the value of what I do. I tried to join another popular (US) site, but he was very prosaic and pragmatic... and that finally pushed me away... Only now I start to recover towards life and all these friendly comments help me a lot...

Thank you again,

Cyril.

Prof.Cyril,

I hope you are completely back to Electronics, getting rid of your Stress Fully.

I have been exploring other Areas in which I have some knowledge and am happy to be Associated with R.G.Also it helps in leaning many things new to us.

At times when I am Busy with the Research Work of my Students and Preparation of Lesson notes for M.Tech, Power Systems,Power Electronics and Computer Science, I do not contribute any thing to R.G. Platform except glancing through it.

I am also associated with Academia but not deeply.R.G. is an interactive site with Students,Research Scholars,and Experts like you,Prof.Lutz ,Prof. Abdelhaleem Zekry etc.

To reduce the stress I contribute to Facebook, writing poetry in Telugu,my Mother Tongue, Tamil,which I studied and writing captions,passing Comments (in English also) etc.Cartoons are also good.

My Special thanks to you as you are the First Person who encouraged me in Academic Discussions in R.G.

Best Regards.

P.S.

Cyril,

I think you are a wonderful scientific constructive critical person. I admire your patient and persistance to make things glass clear by intuition and feeling. This is your style.

Relying on the analysis, mathematical tools may help much.

We are born different to enrich the life.Thank you Pisupati for including me in your friends circle.

Wish you the best in the life. 

Hi Cyril - I repeat: Welcome back to R&G.

I cannot add anything to the last three contributions from valuable R&G members.

I only can say: I second every sentence and I wish you all the best for the next weeks and months.

Perhaps you can help me to find an answer to a new question which I have posted some minutes ago?

https://www.researchgate.net/post/What_is_the_definition_of_resonance?_tpcectx=qa_overview_following&_trid=55aa5b615f7f71be538b45a0_

Kind regards

Lutz