Cyril - yes, I share your point of view.

However, for the long-tailed pair I do not like too much the combination of different equivalent resistors in equivalent circuit diagrams which can be found in many text books.

I am interested simply in the current which is drawn out of the small-signal source that is connected to the base of one transistor (Q1). This current defines the corresponding input resistance at this point (and,yes, referenced to common ground).

The situation (computation) is complicated because this current (hence, the input resustance) depends not only on the DC conditions (quiescent operational point) but also on the conditions (input signal) at the other input (Q2).

Oh, Lutz... to confess I asked this question in secret from you:) as an addition to your question about the input resistance of the differential amplifier:

https://www.researchgate.net/post/What_is_the_input_resistance_of_a_differential_amplifier

I secretly hoped that someone pro from ADI will take a position on this issue by demonstrating that such internal resistor really exists...

My view of the problem is contained in

Method Supplement to "The second journey: from the linear OPA model...

I am convinced that we can "work" with the term input resistance - as described.

Hello, Josef !

Thank you for providing us with your supplement (regarding the diff. input resistance).

However - I have the following question:

Your result for the resistance rD depends on the voltage differenec Vd only - and not on the corresponding input voltages which form this difference.

As a consequence, the result will be the same for the two following cases:

(1) Vd=10mV with Vin1=5mV and Vin2=-5mV

(2) Vd=10mV with Vin1=10mV and Vin2=0.

But we know that in both cases the value of rd will be different by a factor of "2".

Or - was there any misinterpretation on my side?

Thank you and regards

Lutz

Dear Josef - I am very sorry, I was too fast with my comment.

And - yes - it was a severe misunderstanding on my side.

Only now I have realized that the resistance rd is just a device in your model with two open nodes. Therefore, the input current - and with it the input resistance - at one of these two nodes (against common ground) depends on the voltage at the other node.

Finally, I completely agree with your derivation and your results, which can be also written as (neglecting the common mode influence):

rin1=rd*(1+Vin2/Vd)

which simplifies for the special case Vin2=-Vin1 to

rin1=rd/2.

Hi Lutz, I am pleased all is OK :-)

Josef

The concept of rd is very useful, see for example results in:

https://www.researchgate.net/publication/283329017_The_fifth_journey_-_a_short_look_in_unreliable_country_of_R_filters

The issue here lies with the word "exist." Does the input resistance of any electronic device exist? Can you pick up this resistance with your fingers and connect it to another circuit? So, the "differential input resistance" is just a resistance that exists in the model of your device.

I rather think that "the issue here lies with the two words" Resistance and resistor". The resistor is a part which can be picked up with my fingers - and it has a property which is its "resistance". And it does "exist".

As far as the input resistance of a circuit is concerned, you are right that it cannot be icked up, but does it exist? Of course it does exist because it can be measured. The corresponding electrical quantities (voltage and current) do both "exist", do they not?

A really interesting topic to discuss...

Hi Cyril,

I am glad that may to read your words again.

Josef

Hi Josef, Lutz and other participants in this discussion. As usual, my creative thinking wakes up just before I go on vacation...

I tend to think the essence of question is not whether this thing labeled Zdiff means "resistance" or "resistor"... but rather where it is located... what are these two points to measure the voltage across and current between them.

I think the ADI picture is misleading because it gives the impression that there are two types of resistances ("resistors") Zcm and Zdiff, located in different places; Zcm is between the (one) base and ground, and Zdiff - between the two bases. Thus, it is likely that the inexperienced reader will (correctly) connect the input voltage source between the base and ground to measure Zcm... and (incorrectly) between the two bases to measure Zdiff... like in the question below:

https://www.researchgate.net/post/Can_we_connect_a_floating_voltage_source_between_the_two_inputs_of_a_differential_amplifier_If_so_at_which_conditions

(the problem of the last connection is that there is no path for the input biasing currents)

In fact, the input source is always connected between one of the bases and ground... and the resistance "seen" by it is always between that base and ground. However, in common mode, the emitter voltage follows the base voltage, while in differential mode it is constant or changes in the opposite direction. So, the resistance is different (high in the first case and low in the second one)... but the place is the same.

Indeed, if the one of bases is grounded and the only one voltage source is connected to the other, we can think that the input source is connected between the two bases... but this is only a special case...

Thus, even when working in differential mode, the differential pair is a 2-input device. So it is always driven by two single-ended (grounded) voltage sources... and they "see" two differential input resistances between each base and ground.

If we want to drive it by only one floating source, we split its voltage (by a voltage divider) into two equal but oppositely changing voltage drops. This voltage divider is formed by the two input resistors connected between the bases and ground.

In this chain of thoughts, we can note the term "differential input resistance" may be misinterpreted.

"Differential" can be thought of as "dynamic", "small-signal", "AC" (not static)... but here it is used in the sense of "differential mode". Principally, both resistances - at differential mode and at common-mode, are "differential.

So, the correct name of this kind of resistance should be "differential-mode input resistance".

Of course, the most interesting (and important) is to explain why the differential input resistance is low at differential mode and why is high at common mode.

А fresher explanation is to see the Miller arrangement here - a resistor connected between the input voltage source and an additional ("opposing" or "helping") voltage source. The "resistor" is the base-emitter resistance rbe; the additional voltage source is assembled by the collector-emitter junction of the transistor and the common emitter resistor.

At common mode, the transistor "opposes" the input voltage source by changing its emitter voltage in the same direction. According to Miller theorem, this decreases the input current and virtually increases the input resistance.

At differential mode, the transistor "helps" the input voltage source by changing its emitter voltage in the opposite direction. According to Miller theorem, this increases the input current and virtually decreases the input resistance.

Cyril - for explaining the mode-dependend input resistances you have used the MILLER theorem as well as a more "practical" view (...the transistor "helps"...).

To me, the variation of the input resistances can best be explained using the concept of negative feedback (which, of course, has some relations to the MILLER theorem).

Lutz, the only problem with this negative feedback approach is that there is no negative feedback at differential mode.

A negative feedback is introdiced by the emitter resistor (emitter degeneration) only at common-mode. This resistor is shunted by the low output resistance of the other transistor at differential mode. So, the trick here is to suppress the amplification only of common-mode signals by introducing a negative feedback.

A few circuit phenomena (Miller effect, virtual ground, bootstrapping, true negative resistance) are based on the same idea (including a varying voltage source in series to the input one) and generalized by the Miller theorem. They do not necessarily require negative feedback but can be observed in some cases when the input voltage source and the amp output are connected by the feedback network. Op-amp inverting circuits are typical example of this configuration.

Cyril, I do not understand where you can see "a problem" with the concept of negatibe feedback. We have to analyze three basic cases:

No feedback, small feedback and large feedback.

Quote: " A negative feedback is introdiced by the emitter resistor (emitter degeneration) only at common-mode. . "

Yes - and such a common-mode signal does exist also for Vin1=Vs (signal) and Vin2=0 (unsymm. diff. mode).

In this case, we have, of course, negative feedback (the emitter of Q1 sees the small input resistance rin2=1/gm of the second transistor (in paralle to the very large resistance re in the common emitter path.).

This (small) negative feedback effect reduces the gain in this mode to A=(1/2)gm*Rc (for symm.diff. mode with vin1=-vin2 the gain is A=gm*Rc.).

At the same time, the input resistance at the base of Q1 is inceased (due to negative feedback) to rin1=2*h11 (symm. differential without feedback: rin1=h11).

Summary:

1.) No feedback: Vin1=-Vin2 with rin1=rin.2=h11

2.) Small feedback: Vin1=Vs and Vin2=0 with rin1=2*h11

3.) Very large feedback (pure common mode): rin1=rin2=h11+h21*2*re

Lutz, I agree with you. The "problem" was in the first case (true differential mode) where you were talking about negative feedback... but actually there was no feedback...

👍😉

Lutz, I really agreed with you. Your explanation, and especially your summary at the end, are alluringly simple and straightforward for the reader who does not want to go into the depths of things. They would do (and probably did) a very good job in a university textbook on circuit design. I would say that you have explained the phenomenon from the point of view of a scholarly and conscientious lecturer by relying on the classical theory of systems with negative feedback. Probably, your explanation would appeal to modern people who do not want to waste their valuable time getting into circuit details.

But in such a discussion, where "old circuit stagers" came together:), I allow myself to explain them from the perspective of an inventor and unconventional teacher, relying more on my imagination than on my logical reasoning. I am aware that my explanation is more verbose and requires more effort and imagination... but this is how inventors think.

I want to justify the existence of circuit techniques above (Miller effect, virtual ground, bootstrapping, true negative resistance...) and their generalization - Miller theorem, because obviously there was some reason to introduce them ever.

Note although these techniques are most commonly observed in negative feedback circuits, they are not necessarily implemented in this way. With the same success, they can be implemented without negative feedback. I think they are introduced specifically for the cases where we can observe or we want to make artificially changed resistance (impedance). That is why, I think they are the most appropriate means of explaining the modified input resistance in various operating modes.

As I said above, they are based on the same idea - connecting an additional varying voltage source in series (contrary or in the same direction) to the input voltage source. In circuits with Miller effect, virtual ground and S-shaped true negative resistance (VNIC), the additional voltage source is connected in the same direction (travelling along the loop) to the input voltage source that leads to an increase in current and virtual decrease in resistance. In circuits with bootstrapping and N-shaped true negative resistance (INIC), the additional voltage source is connected contrary to the input voltage source that leads to a decrease in current and virtual increase in resistance.

Thus, adding/subtracting additional varying voltage to/from the input voltage, we change the common current by changing the total voltage but this creates the illusion that we changе the resistance. We see resistance but in fact it is resistance plus voltage… and this combination is virtual resistance.

So my conclusion is that it is more accurate to say that not the negative feedback but the Miller arrangement decreases/increases the input impedance. The main point here is not that we returned a part of the output voltage back to the input (feedback) but that we inserted the output voltage in series to the input voltage (Miller idea)... and this is only true for some types of feedback. In these circuits, the output voltage depends on the input voltage according to the negative feedback principle… but in the case of the true differential input voltage it is independent.

Very exciting - I have managed to make a connection with another our discussion:

https://www.researchgate.net/post/Is_the_low_output_resistance_of_the_emitter_follower_a_result_of_negative_feedback

It was about the role of negative feedback in achieving low output resistance, particularly in the case of the emitter follower. Here I will generalize this technique and will show my new insight that actually it is again the ubiquitous Miller idea.

I have illustrated my explanations with the attached picture. It shows a typical circuit with negative feedback - an op-amp inverting amplifier, driven by constant input voltage Vin. So the circuit output voltage Va is constant as well. The internal op-amp output resistance is represented by the resistor Rout; so, the op-amp output and circuit output are different.

The circuit output resistance, for some reason, is less than Rout… and we want to know why. To measure this resistance, we connect a testing voltage source to the circuit output and vary its voltage Vtest. This source can have some internal resistance Rtest (a), i.e. it is a real voltage source… or it can have not internal resistance (b), i.e. it is an ideal voltage source.

In the first case (a), the testing voltage Vtest is summed with the op-amp output voltage... then, in a similar way, with the input voltage Vin… and the resulting voltage drives the inverting input. So the negative feedback mechanism is in action.

In the second case (b), the testing voltage Vtest is directly applied to the circuit output and completely determines the circuit output voltage. But a part of this voltage determined by the R1-R2 voltage divider (the negative feedback network) is summed with Vin and drives the op-amp input. The op-amp output voltage changes in an opposite direction, the current increases... and the resistance virtually decreases.

Now, the insight… In both cases we can see the same Miller arrangement - a main input voltage source (Vtest), additional varying voltage source (the op-amp output) and resistance Rout between them. The additional voltage is summed to the input voltage… so the current increases... and the resistance virtually decreases...

Thus in both cases there is a connection (negative feedback network) between the op-amp output and inverting input... but the negative feedback mechanism functions only in the first case. So, the connection between the op-amp output and input (the negative feedback network) but not the very negative feedback principle is what causes the low output resistance… and this is the Miller arrangement.

So, the negativ Feedback principle makes use of thr Miller Theorem. Okay?

Cyril, I think thereis no fundamental difference between both effects. The Miller effect is identical to the effect of voltage controlled curent feedback.

Why do you prefer one with respect to the other one? Just different names for the same effect!

....as far as the input resistance is concerned....

Lutz, indeed, in terms of input resistance, both concepts are the same… and are represented by the same expression - Rin = R/(1 - K). I just want to remind that it is a great illusion to think that such a simple expression can explain things in depth; it can help calculating. For the purposes of explaining, it is rather convenient in cases where we do not want to do it but would like to finish with the "explanations" as soon as possible (for example, at the end of a lecture:)

I will illustrate how we can explain in a fully intuitive way, without any formalism, what the Miller's and negative feedback principle are... and then see what makes use of what…

First, I want to make one clarification. Miller's theorem is a generalization of the Miller effect. It was not invented by Miller but was later named after him as an appreciation for his original idea. In itself, it is a mathematical expression. But since we, engineers, deal with material electrical circuits, it might be more correct to talk about Miller arrangement, Miller configuration, Miller circuit, Miller idea...

To explain the Miller idea, let's first connect a voltage source to a resistor with resistance R1 and begin varying the source voltage V1. The current will vary according to Ohm's law - I = V1/R1, and the voltage source will "see" the genuine resistance R1 = V1/I. In this arrangement, the current depends only on one variable - the voltage V1 (the resistance is constant).

The generalized Miller idea is to connect a second voltage source in series to the resistor and simultaneously varying its voltage V2. Now the current depends on two variables - V1 and V2, and will vary, as before, according to Ohm's law - I = (V1±V2)/R1. It will be changed - smaller or larger depending on the second source polarity. So the first source will “see” changed resistance R = (V1±V2)/I - decreased, zero, increased, infinite, negative…

The "trick" here is that the first voltage source "thinks" the resistance R1 is changed since it does not "see" the second voltage source. It is not simply loaded by a resistor but by a network consisting of a resistor and a voltage source in series. So the first source "sees" not only the resistance R1 but the total resistance R1+R2 of the network. But what is this R2? Where is the "resistor" representing R2? The voltage distribution can help us...

The total voltage across the network is divided into two parts - the voltage drop VR1 across R1 and the voltage (drop) VR2 across the second source. The former is obvious… but what is the latter?

If V2 is connected in such a way that its voltage is subtracted from V1, it acts as a voltage drop. And since it is proportional to the current, it acts as another positive resistance added to R1. So the first voltage source "sees" increased total resistance (a sum of two positive resistances). An example is the common mode of the differential pair.

If V2 is connected so that its voltage is added to V1, it acts as voltage (electromotive force). And since it is proportional to the current, it acts as another but negative resistance subtracted from R1. And since it should be less than the fully positive resistance R1 (?), the result is decreased total positive resistance (

Cyril, of course, I agree to everything.

In short, we can state that the current through a resistor depends on the voltage difference between both nodes of the resistor. Hence, nothing else than Ohms law for the special case that one of the voltages depends on the other one....

Yes, Lutz... and this is the case of the deep negative feedback where V2 is proportional to V1. And if a resistor is connected between them, its resistance will be virtually modified.

Only, the Miller arrangiment does not necessarily require V2 to depend on V1; they may be changed simultaneously and independently of each other (for example, V2 can be obtained from V1 by a fixed-gain amplifier... or simply V2 can be the inversion of V1 as in the case of the differential mode of the differential pair). So, there is no feedback in this case.

The conclusion is that, "when the input resistance is concerned", the negative feedback arrangement is a special case of the Miller arrangement... it is its possible way of implementing. That is why I so stubbornly insist to mention the Miller's idea when considering the "modified resistance" side of circuits...

(Hmm... I just saw now that the edit & delete button is already at the bottom, right of the button Share)

Dear respected colleague Cyril,

Welcome!

Hope you are well.

The quantities are measured or calculated as they are defined. With reference to a differential amplifier, the amplifier has two inputs the noninverting and the inverting inputs. The differential input resistance is defined by dVdi/dIi. Which means that one applies a small signal voltage voltage source between the inverting and the noninverting inputs dVdi and sees the current dIi flowing into the non inverting input and coming out of the inverting input of the differential amplifier.

Best wishes

Dear respected colleague Abdelhalim,

Regarding your last contribution...I am not sure if the definition of the diff. input resistance BETWEEN both inputs is of common use. I am afraid, it makes not much sense because in most cases this definition is not in accordance with most applications. As far as I know, the diff. input resistance is measured and defined at one of the inputs only. In this case, of course, this input resistance depends on the operational conditions (input voltage) at the other input......this effect was the background of my former question regarding a formula which takes this second input voltage into consideration.

Regards

Lutz

As far as I know, producers measure the differential input resistance between two inputs but the second one is grounded. This actually means that the differential input resistance is measured between the one input and ground.

In this arrangement, the current exits the one input, flows through the input source and enters the ground (not the other input).

Greetings fellows,

I have enjoyed so much this imaginative analysis by Cyril.

It is good to read Dr. Z's erudite response.

Prof. Lutz reminds me of one of my prof in school, always inviting a discussion.

I first encountered the OpAmp in 1976, and they have always been intriguing.

I have developed another audio filter, using radical phase filtering

based on OPA All-Pass filter stages.

When I have gathered more explanatory data,

then I plan to present the project as a "Question" on RGN.

A few of my measured observations appear to be counter-intuitive ( illogical )

and I am sure there will be good comments from this forum.

Hi Glen, I'm glad to hear from you again. Josef

We say resistance where the source is DC.