The main advantage of the "differential pair" is "floating voltage" of emiters. This enables large changes of common mode voltages. This propertie has not "differential circuit", its emiters are connected fixedly to zero.

The main (great!) idea of the differential (long-tailed) pair is not only the "floating voltage" of the emitters... but sooner the ability of this voltage to be either "floating" or "fixed" depending on the operating mode:

• At the common mode (Vin1 and Vin2 simultaneously vary in the same direction), the emitter voltage is "floating" - it simultaneously "moves" with both input voltages. Thus the three voltages - Vin1, Vin2 and Ve, simultaneously "move"... and both transistors do nothing... their collector currents do not change... and, as a result, the collector voltages do not change as well...
• At the differential mode (Vin1 and Vin2 simultaneously vary in opposite directions), the emitter voltage becomes "fixed" - it does not "move" at all... and the common emitter point behaves as a virtual ground... Now the transistors are all-powerful... their collector currents vigorously change... and, as a result, the collector voltages also vigorously change...

Thus the differential pair solves a fundamental problem of the analog circuit design - how to set the desired collector (quiescent ) current from the emitter without losing control from the base. Or, in other words, how to make the emitter voltage "soft" for the undesired influences and "hard" for the useful input voltage applied to the base. And this problem is solved by connecting another emitter follower (transistor) in parallel to the first that keeps the emitter voltage constant when needed...

Note the advantages of this "long-tailed" configuration are most significant in the case of a single-ended output (when taking the output voltage only from the one of the collectors). In this case, the left "differential circuit" is absolutely unfit...

Another example of this trick is an AC amplifier with emitter degeneration where a bypass capacitor is connected in parallel to the emitter resistor to suppress the slow DC variations and amplify rapidly changing AC input variations. See my explanations in the Wikipedia talk page about ECL:

https://en.wikipedia.org/wiki/Talk:Emitter-coupled_logic#...from_the_side_of_the_emitter

You can distil both answers into one statement. Controlled degenerative feedback. Circuit on the left, not present, circuit on the right, has it.

I must admit that - up to now - I never have seen such a comparison:  The classical diff. pair to be compared with two separate transistors (without any DC stabilization).

When I like to amplify a differential voltage I ask myself: Is there any simple amplifying device which works with a voltage difference? Answer: Yes, the BJT amplifies the voltage difference between the base and the emitter node (Ic=f(Vbe).

The only disadvantage is the low-resistive input at the emitter. However, not a big problem - at the emitter node we use in addition an impedance converter (common-collector stage, unity gain stage). And the result is: The classical differential pair.

Dear Lutz,

The circuit on the left is not so bad if we do not apply a common-mode input signal (or if we slightly vary it) and if we take a differential output signal...

I fully agree with your scenario of "inventing" the long-tailed pair... and I use it to introduce the basic idea to my students... But the next key point in this arrangement is to replace the emitter resistor with a current "source"... and we should explain its purpose...

Regarding your remark that "you never have seen such a comparison", I will share that, when I was a student, and I had to design a differential amplifying stage for a measuring device, I was wondering why the two emitter resistors were combined into one common resistor...

Alan,

Could you explain what does "controlled" mean in "controlled degenerative feedback"? Maybe that in one situation (common mode) there is a degenerative feedback while, in other (differential mode), there is no feedback?

We may also add "common" to this definition (see my student wonderment above)...

Hi Huan,

Your concern regarding the attached slide could be explained as follows:

When Vin1 [say, sin(wt)] and Vin2 [-sin(wt)] signals do not have a well-defined CM DC level, i.e. input CM level (Vin, CM) changes, therefore the bias current of M1 and M2 changes as well, hence varying both the gm of the devices and output CM level ( plz take a look  the upper and lower signal diagrams of fig b). When I/P CM level is very low (compared to the upper signal diagram), the min. values of Vin and Vin2 eventually turn off M1 &M2, leading to clipping at the output. Therefore, It is important the bias currents of the devices (M1, M2) have minimal dependence on the input CM level, and introducing differential pair, or long-tailed pair or Source-coupled Pair is the best solution to get rid of it.

It is really "the best solution to get rid of it"... but why?

Controlled degeneration, a bit of play on words, however consider the circuit on the left. The Transfer Function is of the form of an exponential while on the right, a hyperbolic function.

The presence of an impedance in the emitter, ranging in scope from a R (re) to that of a constant current source has a profound affect on the operation of these two differential circuits. Depending on how I control or implement this degeneration, I affect the common mode gain, the nature of the transfer function and eventually the amplifier degree of linearity.  

In the left circuit, the degree of degeneration is minimal, the common mode gain is large and undesirable. However, consider the constant current "sink" in this case, the right side circuit and the topology. The degeneration to the common mode signal is (ideal sink) infinite, while the differential gain is not affected by this degeneration. The current sink is essentially transparent and so is the degeneration.

I will rephrase my first comment saying that not so much the solitary emitter (source) degeneration is important but sooner the common emitter degeneration. I will show it by converting the left differential circuit into the right differential pair... and this can serve as another (besides the Lutz's) scenario of possible "inventing" the famous circuit:

• In the left figure, M1 and M2 are firmly grounded... and we control them by applying input voltages from the side of the gates
• But we want to bias them to set the desired quiscient drain currents... and we decide to make it from the side of the sources by the mechanism of the serial negative feedback ("degeneration")
• For this purpose, we insert current sources in the emitters and adjust the drain currents to the desired values
• Unfortunately, we lose control from the side of the gates since M1 and M2's sources are already "floating"
• Finally, it comes to mind to make M1 and M2 virtually grounded by joining their sources so that each source keeps the other "immovable" (the right figure)

In this arrangement, there are two independent inputs - a differential voltage input (the gates of M1 and M2) where we usually connect two single-ended input voltage sources, and a current input (the common sources) where we connect the biasing current source:

• When we vary the source bias current, both source followers cooperate and keep the common source voltage constant (approximately equal to the common-mode input voltage).
• When we differentially vary the input gate voltages, the source followers mutally oppose and keep again the common source voltage unchanged... as defined by the common-mode input voltage.
• When we simultaneously vary the input gate voltages (common-mode), the source followers cooperate again and easily change the common source voltage. At the same time, the bias current source keeps the drain currents unchanged and the output drain voltages do not change as well.

As a whole, in this arrangement, there are three interacting sources - two voltage sources and one current source, that depending on the situation, "help" or "oppose" each other...

Another interesting idea should be to control the "biasing" current source by another input (voltage) signal. Thus the common source current will depend only on the third input voltage... but the partial drain current of  each transistor will depend also on the differential input signal (since the common current is steered between the two transistors)... and this is the clever idea of an analog multiplier with controlled transconductance...

But I think there are more competent participants in this forum who could tell us more about this story...

Hi Cyril, it requires Mr. Barrie Gilbert :-))  and his "cell".

Josef

Dear all,

Thank you very much for the helps.

I am just a student and have a lot of questions now.

I will spend some time study it carefully. There are some points hope you could make it more clear.

@Prof. Lutz:

"The only disadvantage is the low-resistive input at the emitter."

Could you explain what do you mean by this? Does this mean that the impedance from emitter to ground is low?

I think that it not what you meant because emitter is connected to ground in this case.

@Prof. Josef:

"The main advantage of the "differential pair" is "floating voltage" of emiters."

I think you meant emitter varies together with gate(base) voltage so the current is constant when common mode voltage varies. Is this right?

@Prof. Cyril:

Your explanation is great!

"And this problem is solved by connecting another emitter follower (transistor) in parallel to the first that keeps the emitter voltage constant when needed...

"

I don't get why it is parallel here. Could you explain more?

"Regarding your remark that "you never have seen such a comparison", I will share that, when I was a student, and I had to design a differential amplifying stage for a measuring device, I was wondering why the two emitter resistors were combined into one common resistor... "

I also wondered about that. Could you share the result, why are they combined?

"Finally, it comes to mind to make M1 and M2 virtually grounded by joining their sources so that each source keeps the other "immovable" (the right figure)"

I still can't see why join these sources work as expected? If we join two sources, if one increases the other will decrease the same amount and this keep the emitter votlage constant?

Question: Does this mean that the impedance from emitter to ground is low?

I think that it not what you meant because emitter is connected to ground in this case.

My answer was not related to the shown figures. It was a general remark how to realize an amplifier for a diff. input voltage. The output current of a transistor (FET or BJT) is controlled by the input voltage (VGS or VBE). In both cases, the controlling quantity is the DIFFERENCE between two node voltages (G-S or B-E). Hence, by its nature, the transistor IS already a diff. amplifier - however, it is a problem to use it as such in reality because the signal input at the gate (corrected: source) resp. emitter node is very low (in both cases: r,in=1/transconductance gm). Therefore, we use an additional impedance transformer (source resp. emitter foloower) to have a larger input resistance. Do you understand now?

Dear Huan Nguyen

I recommend You to read

https://www.researchgate.net/publication/282705038_The_shortest_way_to_OPA

https://www.researchgate.net/publication/282778766_The_second_journey_from_the_linear_OPA_model_to_the_basic_principle_of_Gilbert_cell_via_large_signal_OPA_model

https://www.researchgate.net/publication/282862862_The_third_journey_from_the_BJT_input_OPA_model_to_the_FET_input_OPA_model?ev=prf_pub

Research The shortest way to OPA?

Research The second journey: from the linear OPA model to the basic p...

Research The third journey: from the BJT input OPA model to the FET i...

Dear prof. Lutz,

There is only one point I still not sure. Why the input impedance at gate node is 1/gm?

The input current is zero so should the impedance be infinity?

Dear prof. Josef:

Thanks for the reference. I will study it and ask if there is any problem.

There is only one point I still not sure. Why the input impedance at gate node is 1/gm?

Sorry - this was, of course, a typing error. Please read "source" instead of "gate".

Dear Huan,

Your questions are so well targeted, clever and even a little "provocative", that I begin to doubt whether you really are just a student:) In my observations, today's students are interested in different aspects of the topic being studied...

So (within a joke:) I begin thinking if you are not a young specialist from a well known company who is tired of blindly implementing boring tasks set by his boss... and wants not only to do... but to deeply understand circuit ideas:)

Now you have two altenative ways to understand the great idea behind this bizare configuration figuratively named "long-tailed pair" (fortunately, our ancestors have had a good sense of humor... and I am happy to continue this good tradition:)

You can try to see the idea either in Josef's "journeys" filled with math... or in my (or, to some extent, in Lutz's or else's) intuitive explanations based on well-known everyday situations... or humble electrical concepts... or in both... The choice is yours...

Regarding "I don't get why it is parallel here. Could you explain more?", I could say:

If we think in terms of electrical equivalent circuits, we will see that in the common source point, as though three electrical sources are connected in parallel - two voltage sources (the outputs of the source followers M1 and M2) and one current source (the drain of the biasing transistor)...

For simplicity, you can think of the combination of both voltage sources as of a only one (composed) voltage source. In such an arrangement, the voltage sources will determine the common voltage... and the current source will determine the common current...

The problem of the voltage sources is that they are two... and in some situation (common-mode) they cooperate and help each other (what is even more interesting, in this situation the current source also cooperates and helps them... and, as a result, in total all the three electrical sources do the same work - change the common source voltage)...

In other situation (differential mode) the voltage sources are at odds and opppose each other. As they are equal, nobody can win this "fight" (as in the game of arm wrestling)... and as of a result, the voltage does not change. In this case, the bias current source passively "observes" the "fight"... and does not do anything... it works at ideal load conditions (short circuit).

Dear Colleagues,

Dear Huan,

I am very happy to see this rich discussion and the many answers of  Huan question. I would like to greet Huan for her persisting discussions with my well known colleagues. I would lie to see the thing from the perspective of straight forwrd circuit analyses. I think it is the most convincing.

 It is so that the source coupled pair with an ideal common source  current source is the most suitable circuit as a differential amplifier because  from simple circuit analysis the common mode voltage will not affect any change of the bias currents of the transistors. It just shifts the common source node to the common mode voltage.Where as the the differential voltage will see the emitter as a virtual ground and the circuit works as a common source amplifier, where the the drain current of one transistor increases with gm Vid/2 and the other decrease with the same value. So,ideally this circuit amplifies the difference and will not be affected by the common mode voltage. So, its common mode rejection will be ideally infinite.

The other circuit in principle is not be compared by the source coupled circuit as it is just a two common source circuits, one works as a reference for the other. 

The major difference is the source  tail.  Without this tail the common mode voltage will change  the operating point of the transistors and there by its gain. For example as a demonstrative case say that the common mode voltage is high so that it can then saturate the amplifiers and its small signal gain will tend to zero. If it is too small the transistors will work  near cut off.

The best operation of this circuit may be to bias it as a class A amplifier and couple the signals by a coupling capacitor. In this case a common mode Dc voltage will not affect the transistor biasing.

In conclusion the source coupled pair is superior for common mode rejection and therefore can be direct coupled to the input signal sources.

Best regards

Abdelhalim,

You have probably said "The best operation of this circuit may be to bias it as a class A amplifier and couple the signals by a coupling capacitor" about the differential circuit on the left?

Cyril,

Yes! exact !

Thank you for your comment.

Best wishes 

Dear professors,

Thank you very much for the helps.

@Prof. Cyril:

"You can try to see the idea either in Josef's "journeys" filled with math... or in my (or, to some extent, in Lutz's or else's) intuitive explanations based on well-known everyday situations... or humble electrical concepts... or in both... The choice is yours."

I would like to try both! First by intuitive understanding and then formalize with math.

Thanks for the intuitive explanation. In the part "differential mode" I have a bit problem. 

Three sources are in parallel (two voltage sources and one current source) and they are joined by a common node, say Vc (please see the attached image).

In differential mode, two voltage sources are equal in magnitude and opposite in phase. So we have:

V1 = Vid/2

V2= -Vid/2

But in this case two voltage sources are in parallel so their voltage should be equal.

V1=V2=Vc that results in Vid =0.

So the voltage at the common mode should be zero. Is this understanding correct?

@Prof. Lutz:

The only disadvantage is the low-resistive input at the emitter. However, not a big problem - at the emitter node we use in addition an impedance converter (common-collector stage, unity gain stage). And the result is: The classical differential pair.

Could you give a picture about the classical differential pair? I tried to to draw the picture from what you described above. First stage common emitter, then at the emitter add another stage but I can't make a common collector here.

@Prof. Abdelhalim:

Thank you for the detailed answer. 

I would like to discuss about the point how the common source node voltage is possible to shift here. In this case, can we said that there is a negative feedback that causes the shift here? There is a relationship between Id and Vgs. So for common mode, Id is fixed and equal to a half of the ideal current source. For that reason, Vgs should also be fixed. That results in voltage at the common source node shifts together with the input common mode voltage. 

Is my understanding correct?

 It is so that the source coupled pair with an ideal common source  current source is the most suitable circuit as a differential amplifier because  from simple circuit analysis the common mode voltage will not affect any change of the bias currents of the transistors. It just shifts the common source node to the common mode voltage.

Well Huan... I will answer your reasonable question... but first try to find the answer in this Wikibooks story...

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

... and particularly, in this its part:

https://en.wikibooks.org/wiki/Circuit_Idea/Walking_along_the_Resistive_Film#Varying_both_V1_and_V2:_a_resistive_summer

If you manage to do this, you will succeed to make a connection between our 2016 with the distant Ohm's 1826... and you will be convinced in the power of "pure" ideas...

Dear prof. Cyril:

I read the link and in picture 19, there are two opposite voltage with a resistor and the voltage at the middle point is zero. So I think this is what you meant to say.

Do you consider the BE junction is a resistor in differential mode as in the picture below? If so I think I have no problem with it.

Dear Huan, 

You are a good electronic assistant. Yes,what you said is okay except that there is no feed back. It is just the properties of the current source and the device performance which dictates this condition where the current source  forces constant current between its terminal irrespective of the voltage between its its terminal that can vary in a wide range while the current is constant. It is so that the excess voltage over the required VGS to support the drain current will be absorbed by the constant current source.

I expect you a bright future with electronics.

Wish you success

Could you give a picture about the classical differential pair?

Hi Huan,

the enclosed figure shows what I mean:

The emitter of the left transistor is considerd as a low-impedance input (remember the common-base configuration). Therefore, we are using a unity gain common-collector stage (with its high-resistive input) for injecting an input signal. This procedure immediately results in a two-transistor circuit with common emitter resistor RE.

Finally, implementing the collector resistor RC2 makes the circuit fully symmetrical. Of course, due to the value of RE the common-mode properties of the circuit need to be improved. Therefore, we replace RE by a very large dynamic resistance re to be realized using a transistor-based current source instead of the ohmic passive part RE. (re is identical to r,out of the current source).    

Further questions?

Dear Abdelhalim,

Let me disagree with your claim that there is no feedback. IMO, in total, three circuits with negative feedback are connected to the common source point - two voltage followers with serial negative feedback, and one current stabilizer (if it is implemented by a source degeneration)... and they mutually interact. For simplicity, we can think of the current source as of a dynamic resistor (what it usually is)... and in this case we have two negative feedback systems acting at a common dynamic load.

When both input voltages simultaneously change, (the so called common mode), e.g., increase, they try to increase the common voltage by decreasing the total resistance in the circuit thus increasing the common current. As a result, the voltage Ve = Ie.Re will begin increasing since Ie increases.

But, at the same time, the bias current source (dynamic resistor) begins increasing its static resistance Re thus acting in the same direction as the voltage followers (now the voltage Ve = Ie.Re increases since Re increases).

So, in the Ohm's law - Ve = Ie.Re, the voltage is created by (depends on) both the variables - the current Ie and the resistance Re. If the current source is perfect, the voltage followers have nothing to do and their VGS stay unchanged... the current source does the "donkey work"...

Huan,

In Fig. 19 there is no common-mode signal; there is only a differential signal. So better look at Fig. 18, where both exist...

BTW these terms (common-mode signal and differential-mode signal) are artificially created... Actually, in the case of DC amplifiers, there are (we see) only one signal that is a sum of both.

Similarly, when talking about DC and AC part of a signal, there is only one signal (as in the WB story, where students produce these signals simply by varying the input voltages).

Another example can be presenting a complex signal as a sum of sine waves (in Fourier analysis).

Dear prof. Abdelhalim, prof. Lutz, prof. Cyril:

Thank you very much for the explanations!

@Prof. Abdelhalim:

I thought about feedback because I thought only with feedback Vgs can be influenced by Id.

(Normally Vgs is controlling quantity so it is Vgs that determine Id.)

For example, in MOSFET with diode-connected configuration (drain and gate are connected together), the input current Id is converted into its own Vgs.

I think this circuit can be called current to voltage converter.

However, I was mistaken. In the case with differential pair above, Vgs actually is constant. It is just Vg and Vs compared to ground changes with the same rate.

@Prof. Lutz:

It is interesting way to look at it.

I think I get it now.

Please see also the attached image if I got it right.

I have never thought about input impedance for this circuit so it is a bit new to me.

My understanding is just the bjt transistor will take the difference between V1 and V2 and convert it into collector current.

However, based on reading your comment, I think I got why emitter node low input impedance has effect on circuit performance.

From the picture, there are two input voltage sources, so we can use superposition here.

So, the circuit is divided into two subcircuits (common emitter and common base).

The common base has low input impedance so its output collector current will be not good.

For that reason, an emitter follower (which has high input impedance) is added to inject the signal.

I have two questions, could you clarify:

1. Why RE is added? I think it is for biasing only.

2. As you said, the second stage is common collector or emitter follower. So shouldn't the collector is connected directly to Vdd or Rc2 is equal to zero?

Remarkable thoughts and hand-drawn pictures, Huan! Go on in this direction...

I would comment your insights and answer your interesting questions... but will wait for colleagues first to respond...

Have you realized why the common source voltage is not zero... and how much it is?

Dear prof. Cyril:

Thank you for kind words. I think I got it. In common mode, the current flows through each branch is constant and equal to a half of the source current.

With collector current is fixed, the base-emitter voltage of bjt is also fixed according to  Shockley diode equation.

So when the input common mode voltage increases, to keep base-emitter voltage fixed, the voltage at common source node should also increase with the same rate.

For differential mode, the voltage at bases of two bjts are equal and opposite in phase. So its effect on the common source node will cancel each other and so in ac the node acts as a virtual ground.

Bravo, Huan!

As I can see, you have realized the great idea behind the long-tailed pair where three electrical sources interact ("help") each other:

In the bias mode, when we set the bias current, the upper voltage sources provide for the lower current source perfect load conditions (short circuit or artificial "ground"). As you have noticed, by the help of the serial negative feedback mechanism, the transistors adjust their input VGS voltages so that to pass the bias current. So VGS significantly cnahges.

In the common mode, the lower current source provides for the upper voltage sources perfect load conditions (open circuit or "nothing"). Now the transistors just start to change their VGS... but since the current source changes its dynamic resistance, they stop. So VGS does not cnahge.

In the differential mode, the one voltage source provides for the other perfect load conditions (short circuit or artificial "ground"). In this case, the transistors do all the "donkey work" vigorously changing their VGS; the bias current source does not do anything... it only passively contemplates:)

So we both have seen this truth... but how do we make conventionally thinking professionals see it?

Regarding "In common mode, the current flows through each branch is constant and equal to a half of the source current", the sum of the drain currents is constant but they can be in any proportion relative to one another.

Then regarding "For differential mode, the voltage at bases of two bjts are equal and opposite in phase", it is true only about the differential parts...

And finally, regarding "So its effect on the common source node will cancel each other and so in ac the node acts as a virtual ground", only the differential parts will be cancelled.

The "virtual ground" is equal to the common-mode input signal (the arithmetic average of the two input voltages).

Quote Huan: So when the input common mode voltage increases, to keep base-emitter voltage fixed, the voltage at common source node should also increase with the same rate.

Well Huan, this applies under IDEAL conditions only. I think your understanding can be improved if you realize that a FINITE resistance in the common emitter path provides negative feedback (with an ohmic RE=re or a large current source with re) according to the gain formula for common mode signals:

Acom=-Rc/(2re + 1/gm)

This gain formula clearly shows the influence of re (Ideal current source has infinite re).

Huan, let me try to answer some of your questions/remarks:

My understanding is just the bjt transistor will take the difference between V1 and V2 and convert it into collector current.

Yes - and the main question we are dealing with is: How does the transistor complete this task?

However, based on reading your comment, I think I got why emitter node low input impedance has effect on circuit performance.

Be specific! Why do we want a large input resistance? Which kind of "effect on circuit performance"? Answer: Each voltage amplifier should have a large input resistance because otherwise the (mostly unknown) internal resistance of the source (and in reality it is NEVER zero) must be taken into account - unless it is much lower than the input resistance of the amplifuier. This is the reason we do not use the emitter node as input

From the picture, there are two input voltage sources, so we can use superposition here.

Yes - we always may apply superposition for two independent signal sources.

So, the circuit is divided into two subcircuits (common emitter and common base).

No - as seen from one of the inputs the whole circuit can be treated as common-collector stage in series with common-base. This is the simplest way for calculating the diff. gain for the circuit.

The common base has low input impedance so its output collector current will be not good.

Excuse me, but this sounds not "good". This is no technical language. In electronics we never should use terms like "good" or "not good" without sufficient explanation.  The disadvantage of a  low input impedance was explained above.

Finally - Huan, feel encouraged for asking further questions.

Dear Lutz,

I would only note that your comment before last (about the finite Re), although being useful, is not closely related to the Huan's quote (about the voltage at common source node). It does not  enrich the main idea but rather deflects the thought from it. Huan is talking about the common source voltage, not about the particular drain voltages.

Cyril - not closely related to Huans question?

Here is his question again:

Quote Huan: So when the input common mode voltage increases, to keep base-emitter voltage fixed, the voltage at common source node should also increase with the same rate.

I think the mentioned increase "with the same rate" applies to IDEAL condidtions (re infinite) only!

For finite re values the emitter/source note does not change with the same rate. Thus, we have a CHANGE in VBE resp. VGS and, hence, in the collector/drain currents. This the only reason for implementing a current source (unfortunately not ideal) within the emitter/source path. Am I wrong?

I think, each beginner should know WHY we use a third transistor in the common leg. And the formula shows why a large re is desired.

Dear prof. Lutz:

Thank you very much for the detailed answers.

Be specific! Why do we want a large input resistance? Which kind of "effect on circuit performance"? Answer: Each voltage amplifier should have a large input resistance because otherwise the (mostly unknown) internal resistance of the source (and in reality it is NEVER zero) must be taken into account - unless it is much lower than the input resistance of the amplifuier. This is the reason we do not use the emitter node as input .

Sorry for being unclear. Your explanation is also what I meant too.

No - as seen from one of the inputs the whole circuit can be treated as common-collector stage in series with common-base. This is the simplest way for calculating the diff. gain for the circuit.

For the common emitter and common base, I was refering to my picture attach in the post (the bjt with two input signals at base and emitter).

For the differential pair with three transistors, I agree with you.

Relating to the differential gain calculation, I am familiar with the method haft-circuit using virtual ground.

I will try the method you said. It seems also interesting way.

There still remain two questions in my previous post.

These question I am refering to the classical differential pair you attached.

I have two questions, could you clarify:

1. Why RE is added? I think it is for biasing only.

2. As you said, the second stage is common collector or emitter follower. So shouldn't the collector is connected directly to Vdd or Rc2 is equal to zero?

Dear prof. Cyril:

Thank you very much for the comments.

I need some time to digest them.

First about perfect load condition. I think here is what you mean.

1. For a voltage source, perfect load is current source (infinite resistance).

2. For a current source, perfect load is voltage source (zero resistance).

I also see that you divided the operation into three modes.

In the bias mode, you said that the Vgs significantly change.

Do you mean that it changes significantly compared to the case without the bias current source?

Hello Huan!

1. Why RE is added? I think it is for biasing only.

Why do you think RE would be "added"? It is the most important part of the emitter-follower as shown in my hand drawing.

2. As you said, the second stage is common collector or emitter follower. So shouldn't the collector is connected directly to Vdd or Rc2 is equal to zero?

Yes - you are right. And, indeed, there are such unsymmetrical diff. amplifier with only one Rc. However, because of symmetrie we can/should add this additional resistor Rc2. And this makes the second (right) transistor to a common-base stage. Otherwise, there would be no resistor transferring current into voltage.

In summary:

* As seen from the left input the left transistor works in common emitter configuration with the gain-determining resistor Rc1 (with signal feedback RE||r,in2);

* At the same time, the right transistor (seen from the left input) works in common-base configuration (gain is determined by Rc2) and is feeded by the left transistor which now works as an emitter follower.

* The same considerations can be made from the input node at the right side (vice versa).

* Consequence: It is NOT possible to say transistor 1 or 2 would work in a particular configuration. Instead, it depends which of the two inputs and which of the two outputs we are looking at (which output is referenced to which input).

I hope, I could expreess myself clearly. Do you understand my explanation?

Dear Lutz,

Maybe you are right at this particular situation (imperfect voltage follower) but my comment is about the causality in the general perfect arrangement.

You know this is a fundamental idea that is widely used in circuitry (in various current sources with negative feedback, transistor amplifiers with emitter degeneration, non-inverting amplifiers, etc...)

The idea is to make the voltage across a resistor equal to the input voltage... and to use the current through it as an output quantity.

My thought is that in this arrangement, the voltage across the resistor is set not by the current source but by the input voltage... and the current source sets only the current...

To convince you in this statement, I can connect an op-amp before each transistor and take the negative feedback after the transistor (from the source). As a result, the voltage across the emitter resistor will be practically equal to the input voltage.

The resistance of this resistor will determine the current, not the voltage. When we change the input voltage, the current will also change according to Ohm's law. But if we replace it by a dynamic resistor (current "source"), the current will not change...

This is our situation - the drain current, accordingly, the drain voltage, does not change when the voltage across the emitter dynamic resistor changes since the resistance changes as well...

Hi Huan! Nice to see your questions!

Regarding "In the bias mode, you said that the Vgs significantly change. Do you mean that it changes significantly compared to the case without the bias current source?", I will say:

Vgs always will vary significantly regardless of the type of the bias element - a humble ohmic resistor or the more sophisticated constant-current dynamic resistor (current "source")... since they both do the same work - change the current, while the voltage across them stays constant.

"Vgs always will vary significantly regardless of the type of the bias element - a humble ohmic resistor or the more sophisticated constant-current dynamic resistor (current "source")... since they both do the same work - change the current, while the voltage across them stays constant."

Cyril - typing error?

"This is our situation - the drain current, accordingly, the drain voltage, does not change when the voltage across the emitter dynamic resistor changes since the resistance changes as well..."

Dear Cyril - sorry, but again I must ask you for clarification.  I don`t understand the bold part of the sentence. The resistance changes??? That`s new to me.

Hi all,

I think that the best explanation goes out from "superposition teorem" for desribed circuits. It is easily applicable methode and gives us very good results.

Josef

Dear Lutz,

It is interesting for me to get to know what a notion you have about this 2-terminal element connected between the common emitter point and the negative supply...

Quote: "Dear Lutz, It is interesting for me to get to know what a notion you have about this 2-terminal element connected between the common emitter point and the negative supply.."

Cyril - is it really necessary to answer this question? In any case, the element connected between the common emitter point and the negative supply is, of course, a constant resistance (either ohmic RE=re or differential re). And the common mode gain is Acom=-gm*Rc/(1+gm*2re).

This expression corresponds with a small-signal emitter current (for a common mode signal):

i,e=Vin,com/(2re + 1/gm).

Is something wrong? Why do you think that this resistance re changes its value in common mode operation?

Lutz,

Frankly, I have heard about "ohmic", "static", "chordal" resistance... I have also heard about "dynamic resistance", "dynamic load", "differential resistance"... But  I have never heard about "constant differential resistance".

Such a "constant" differential resistance (graphically represented by an inclined line that does not pass through the coordinate origin) is just an illusion. Actually, it is varying static resistance...

I could demonstrate to my students how this trick is implemented in the long-tailed pair by connecting a humble 19-th century rheostat as a source (emitter) current-stabilizing element... a kind of a manually-controlled current "source":)

First, I will not move the rheostat's slider when both input voltages vary (common mode). So, in Ohm's law - I = Ve/Re, the current will vary since Ve vary but Re = const. This would be an example of the simple long-tailed pair with source (emitter) resistor.

Then, when both input voltages increase (common mode), I will move the slider so that to increase the resistance, and v.v. Thus, in Ohm's law - I = Ve/Re, the current will not vary since both Ve and Re vary in the same direction and rate. This would be an example of the more sophisticated long-tailed pair with source (emitter) current "source".

Finally, I will only say to my students that this work is too boring for me (a thinking but laisy human being:)... and I intend to assign it to a "stupid" but "loyal " transistor... and will do it...

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

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

https://en.wikibooks.org/wiki/Circuit_Idea/Negative_Differential_Resistance

"Why RE is added?"

The source (emitter, cathode) follower is a device with a voltage output. To control the output voltage, at least two elements are needed - a "pull-up" element and a "pull-down" element, that are connected in series between the supply rails. They "pull" the common point to opposite directions like people in the game of "tug of war":)

In the long-tailed pair, two "pull-up" elements are connected in parallel to the positive rail and one element - to the negative rail. The two input variables (voltages) control the "pull-up" elements, and the bias variable - the "pull-down" element.

"But I have never heard about "constant differential resistance"..."

Hi Cyril - don`t you know what I mean? The explanation is simple:

A differential resistance (re) is identical to the slope of a non-linear V-I characteristic. However, in case the slope is constant within the region of interest (as is the case for the transistors output characteristics we are speaking of) this resistance (re) remains constant. This is the basic working principle of a transistor based current source as it is used also for differential amplifiers. Agreed?

But what about the main question: Do you still think that the resistance value in the common emitter path changes in common-mode operation? (I have tried to answer all your questions). I think, Huan needs clarification - otherwise he would be confused.

Cyril, 

Up-Vote for your diagram!

Electronics is very much a "Balancing Act".

( So also the rational world)

Your picture is worth a thousand words in school class. 

I have used a three person "tug-of-war" cartoon

to communicate (1) Voltage Pushing, (2) Resistance Impeding (3) Current Flowing

... with Resistance squeezing the Watts

out of the Amperes flowing through.  

Three little men each doing his thing. 

About the Re being dynamic, 

I have always thought of it

as being the product of dynamic forces.  

the apprentice Glen

"But what about the main question: Do you still think that the resistance value in the common emitter path changes in common-mode operation?"

Yes Lutz, the static (chordal, instant, ohmic...) resistance changes... if only I move the slider:)... or the transistor appropriately opens/closes...

But what we see, is just an illusion - a line representing the trajectory of the moving operating point...

Glen, thanks!

I have searched in Google Images for a 3-way tug of war but I have not found...

But I have found a 4-way version. Perhaps it can serve to invent a multiple input differential circuit where four sources interact:)?

Cyril - I do not move any slider; instead I try to discuss properties of a real diff. amplifier. 

It seems there were a severe misunderstanding between us because in earlier posts you spoke about "dynamic resistances".

Example 1: "...but since the current source changes its dynamic resistance,"

Example 2:  "..the drain current, accordingly, the drain voltage, does not change when the voltage across the emitter dynamic resistor changes since the resistance changes as well...

This was my only point, because this "dynamic" resistance (I have named it "re" in my formulas), of course, remains constant. There is no reason for a change.

Now you are speaking about another "resistance" (chordal, ohmic) which is a ratio of corresponding DC values. And this  - fictive - resistance changes of course. But this resistance plays no role for common-mode operation. 

I think, this fact should be clearly stated  - otherwise we would cause additional confusion for the questioner Huan.

Lutz,

Saying "dynamic resistance", I mean "varying resistance"... or more precisely, "self-varying resistance" (since it is 2-terminal)... This name only express the fact that this resistance changes when we change the voltage across it...

I don't think it is fictive... Conversely, I think it is real but varying...

And exactly this varying resistance plays the main role for the common-mode operation since dV/dR = I = const.

At differential mode, this resistance does not vary since the voltage across it is constant... it is shunted by the virtual zero output resistance of the voltage followers... and really,  it plays no role in this case...

Finally, I should clarify that all my thoughts above served the process of understanding... and formally, you are right about these terms, Lutz... Really, the differential resistance has a constant value... and the static resistance changes its value...

But my purpose was to understand and explain what happens there, in this "long tail"... to reveal the mechanism... not to be correct regarding the terms... Now, if you want, we can place all in its place... but it is not interesting for me... I prefer to "move sliders" in my imagination...

Cyril - thank you for clarification.

Finally - Huan, everything clear now?

Gentlemen, coming back to Huan`s original question (Diff. circuit vs. diff. pair) presently I am concentrating on the signal input resistances. Of course, in case of two separate stages we can use standard formulas.

However, for the BJT-based diff. amplifier with a common leg the situation is a bit more involved since the input resistance at T1 seems to be not independent on the signal input at T2. 

Therefore, I expect that we have r,in (T1)=f(v,in2).  As a result of my first considerations, there are three special cases, which are relatively easy to analyze: (1) Common-mode operation and (2) diff. mode (v,in1=-v,in2) and (3) v,in2=0.

But what about the general case with two arbritrary voltages for v,in1 and v,in2 ?

Up to now, I didn`t find any contribution dealing with this problem. Do you have an answer?

And what about this (from https://www.researchgate.net/publication/282778766_The_second_journey_from_the_linear_OPA_model_to_the_basic_principle_of_Gilbert_cell_via_large_signal_OPA_model)

Research The second journey: from the linear OPA model to the basic p...

Hi Josef - thank you for replying.

Sorry, but i forgot to make clear that I am, of course, interested in the (differential, dynamic) SMALL-SIGNAL INPUT RESISTANCE (r,in).

Your expression gives the "momentary" static DC ratio (ohmic resistor, Vin/Iin). 

For three special cases we know:

* r,in1 (v,in2=0) = 2*hie=2*r,be

* r,in1(v,in2=-v,in1) = hie=r,be .

* common mode: r,in1=r,in2 = hie + 2*hfe*r,e (hfe=beta, r,e=eff.emitter resistance)

I expect a formula r,in1=f(v,in2) which contains the above three special cases.

Maybe the Miller theorem can give an answer to this question... since this is a typical Miller arrangement where two sources are connected by a resistor "bridge"...

Hi Lutz

and what this way

Hi Josef, when I have time I will try (try!!) to check your calculation - because I think it cannot be correct. But at the moment I don`t know where the error is.

Inserting small signal parameters in the final expression which you have shown (for Vd=0) the result would be  r,in=2hie=2h11 (because IQ/2Vth=gm=beta/hie).

However, this is the known result for the case Vin,2=0.

In case of Vd=0 we have commom-mode operation with a much larger input resistance (that is the purpose of requiring a large Re resp. re in the emitter leg).

EDIT: Josef, I think, I have found the error:  In the paper you have derived formulas using the expression Vd=Vbe1-Vbe2.  However, for calculating the input resistance you have used Vd=Vin,1-Vin,2. However, both expressions are identical for a constant emitter potential only. And this is the case for Vin,1=-Vin,2 only.

Correction: I am afraid, my comments above under EDIT are not valid. The emitter potential is not required to be constant, but IDENTICAL for both transistors. And this is, of course, always ensured. So, please forget the text under EDIT. 

Updated explanation: For my opinion, we cannot use at all the exponential expressions as a function of Vd. The reason:  This function contains only the difference between both input voltages but  NOT the level of this difference (which gives the common mode portion).  Hence, resulting from the differentiating process we get only the slope of the tanh function IB=f[tanh(Vd)]

Hi Lutz

I'm still thinking about it. It seems that there are still some "philosophical problems" :-) Josef

B y the way: Simulations confirm the results for the mentioned three special cases (V,in1=-V,in2; V,in2=0 and V,in1=V,in2).

I have tried to calculate the input resistance r,1 at the base node of T1 - dependent on a varying input voltage v2 at the base of T2. In accordance with the common practice  no phase difference - other than 180 deg - was assumed between both input voltages v1 and v2.

Result:        r,1=r1,o*[1+(v2/vd)].           (1)

with:

r1,o = input resistance for v2=0;

r1,o=2hie=2h11=2rbe;

vd=v1-v2.

_____________________________________________________

Expression eq.(1) confirms and satisfies the following special cases:

(1)  v2=0: r1=r1,o=2hie=2h11=2rbe;

(2)  v2=-v1=-vd/2:  r1=0.5*r1o=hie=h11=rbe;

Comment: Both results can be easily found also directly from the circuit.

(3)  v1=v2 (vd=0): r1 infinity. 

Comment: This result is a simplification and is valid for an ideal current source in the emitter leg only. . During derivation of the given expression eq. (1) the common-mode portion for the input current was neglected if compared with the much larger diff. portion of the current.

From the BJT circuit we can directly calculate for case (3):  r1=hie+hfe*2r,e=h11+beta*2r,e ( r,e=resistance of the common emitter leg).

This is a very large value and confirms the introduced simplification.

__________________________________________________________

As another confirmation of the derived expression eq. (1) we can analyze the input resistance for a system (diff. amplifier) with open-loop gain A and negative feedback.

In this case: v2=k*v,out and v,out=vd*A (k: feedback factor).

It is easy to show that in this case eq.(1) gives:

r1=r1,o*(1+k*A).

This is the known expression for the input resistance of a system with feedback. The input resistance without feedback (r1,o) is increased by the classical factor (1+loop gain).  This result again confirms eq.(1).

Dear Lutz von Wangenheim

Here, the comparison is intended to emphasize the effect of input common mode voltage on the working of a differential circuit. I believe, this slide tries to differentiate between the use of two separate circuit vs using the differential circuit. in fact, in the case of the latter, the tail current source absorbs the input common mode voltage and hence leaves the circuit immune from common mode variation. while in the first circuit, the variation of input common mode will be transmitted to the output by the gain of the input devices.

@Nima Shahpari, And how does "the tail current source absorb the input common mode voltage"?

@cyril Mechov, when the input common mode changes, the gate of input devices change. If the tail current is ideal, the source current of input devices cant change. therefore, the Vgs of the input devices is kept constant and all the input common mode variation will be absorbed by the ideal tail current source.

@Nima Shahpari, And how does this magic happen - the input common-mode voltages change but the common source current does not change? I need some more realistic explanation to this phenomenon...

Not a magic, simple KCL from Electrical circuit theory.

@Nima Shahpari, I have a simple explanation of the behavior of the element included in the emitters. It behaves like a self-changing (dynamic) resistor. For example, if the gate voltage increases, the source voltage also increases... but this element also increases its resistance. As a result, the current does not change - I = V/I = Vinc/Rinc = const.

Hi Cyril - may I give a short comment...

I can follow your view ....but for my opinion, your explanation (better: artificial-theoretical idea) is NOT a "simple" one. If it would be simple, we would have no problem to explain the function of such a "self-changing" resistor.

I think, this problem results from another "artificial-theoretical" view: The introduction of an ideal current source symbol.

Such an element is just theoretical and does not exist (like the "self-changing" resistor, which also does not exist.).

In reality, we have a fixed large resistor (static or a differential one) in the common emitter leg - and what happens for a common-mode input? We have maximum negative feedback as is the case also for the classical emitter follower.

The resistor remains fixed, of course, but the voltage follows the input and we have a very small change of current only (leading to a correspondngly small common mode gain).

Dear @Lutz, By the concept of "dynamic resistance" I explain the more abstract concept of "differential resistance".

For each point lying on the horizontal part of the IV curve, there are two types of resistance - static and differential... and for me the former is more primary. I understand and explain what the differential resistance is through a changing static resistance.

In my opinion, there is no such thing named "differential resistance"... there is only changing static resistance. When we vary, with some step, the voltage across the nonlinear element (current "source"), we actually see different static resistances at each step.

Oh no! Certainly, there is something like a "differential resistance" rd. And this resistance is another quantity than a "changing static resistance" Rs. For a varying operational point on a non-linear V-I-characteristic (example: pn diode) we can always (for each pair V-I) find the ratio Rs=V/I.

But this value Rs will certainly not be identical to the slope of the curve wich is rd=d(V)/d(I).

@Lutz, You made me think again about this phenomenon... and I should prepare another answer... But before, I would ask you what you mean when (sometimes) saying "differential (dynamic) resistance"? What do you mean in this case by "dynamic resistance"... because the differential resistance has a constant value?

Hi Cyril - I must admit that - sometimes - I am using the term "dynamic" because I can find this term in some books - perhaps because it is a good contrast to "static". But I agree that it sounds a bit misleading - the term "differential" is much more correct. In the future, I should restrict myself to the term "differential".

Thanks, @Lutz...

I have new insights about my answer but need some time. For now, I would say:

The two resistances are different but related quantities; the differential resistance is a constant resistance that is obtained by a dynamic static resistance.

Cyril, here is an example (concerning "dynamic"): "Electrical Engineering (A.L.Hambley): "Notice that the slope has the units of inverse resistance. Hence, we define the dynamic resistance of the diode as rd=...."

As you consider the changing static resistor as "primary": But I suppose you are using also the h- resp. y-parameters, don`t you? And all the small-signal equivalent diagrams are also in terms of diff. quantities... But I must admit that I do not use anymore these equivalent diagrams (I suppose the same applies to you?).

Quote: "...the differential resistance is a constant resistance that is obtained by a dynamic static resistance."

Is it really "obtained"? How? For a very small change of the operating point, the difference between the two static resistances is very small (nearly zero) - however, the slope of the V-I curve can be pretty large.

@Lutz, More precisely speaking, "can be obtained"...

"More precisely speaking, "can be obtained"..."

May I ask - HOW ?

…

@Lutz, I will illustrate it by a picture...

@Lutz... I suggest you (as I have done before) to demonstrate the operation of the "dynamic resistor" through the simplest possible experiment shown in the attached figure. I give you the main role of a varying voltage source V, and I take on the secondary role of a varying resistor R.

This is an imaginary experiment, but it is so simple and convincing that it hardly needs to be implemented. But if necessary, it can be easily carried out. We can even imagine that it could have been realized in the 19th century, for example by Ohm and Kirchhoff- :)

The operation of this arrangement is extremely simple:

When you increase the voltage V by setting successive increasing values V2, V3, V4 and V5, I increase the resistance R by setting successive increasing values R2, R3, R4 and R5 so that the ratio V/R remains constant - V2/R2 = V3/R3 = V4/R4 = V5/R5 = I = const. As a result, you (the voltage source) see the infinite differential resistance Rd -> infinity. But this is an illusion since actually you see the varying static (ohmic) resistance R. So the variable resistor R acts, with my help, as a self-varying “dynamic” resistor.

This is illustrated graphically in the attached figure. When you increase the voltage V of the voltage source, its vertical IV curve moves horizontally (translates) to right. Since, at the same time, I increase the resistance R, its inclined IV curve rotates clockwise and the intersection (operating) point moves along a horizontal line (positions 2, 3, 4 and 5) thus forming the IV curve of the current-stable differential resistance dR…

What conclusions can be drawn from this 19th century experiment?

It shows:

• how an element with infinitely large differential resistance can be made by a resistor with varying static (ohmic) resistance...
• how the operation of an element with infinitely high differential resistance can be explained by a resistor with varying static (ohmic) resistance...
• the two resistances are different but related quantities; the differential resistance is a constant resistance that is obtained by a dynamic static resistance.
• the Early effect not taken into account because it is unrelated to the concept of dynamic and differential resistance.
• the differential resistance is not consided in some point (classic small-signal view); it is considered along a long region of the IV curve (large-signal view).

Cyril - thank you for all your effort to show me something....it is a nice graphical visualation of Ohm`s law (first figure) resp. the definition of a static DC-resistor in case of a nonlinear characteristic I=f(V) for a varying operational point.

However - can we really deduce from the drawing that "the differential resistance is a constant resistance that is obtained by a dynamic static resistance..." ?

For my opinion, when one quantity is "obtained" (derived) from another quantity, there must be a certain numeric relation between both quantities. Let me give an example:

When in your drawing the voltages V2...V5 would have other values, the corresponding static resistors (defined by the crossings with the horizontal line) would also have other values.

How can you say that the diff resistance (unchanged, still infinity in your example) would be "obtained" by these static resistors?

As you can see, I have some problems to follow the logic of your steps...

Hi, Lutz... Thanks for the contribution.

As you can see, I have not taken into account the Early effect because it is unrelated to the concept of dynamic and differential resistance. But if you want, you can consider the case of high (not infinite) differential resistance. Then the horizontal section of the green IV curve will have some small slope... but note it will be straight line again... and all explanations will be valid again!

It is the interesting point here that I do not consider the differential resistance in some point (classic small-signal view); I consider the large-signal differential resistance along a long region of the IV curve (between points 2 and 5).

So, we are talking about two kinds of linear resistances having IV lines - static and differential.

Hi Cyril, sorry to say that - for my opinion - the two main problems I have are not yet clarified:

* Why do you consider the "changing static resistor" as "primary" (if compared with the diff. resistance of a non-linear characteristiv,

* How can you say that the diff resistance (small-signal or "large-sigmal") would be "obtained" by these changing static resistors? (My example has shown that there is no relation).

* By the way: Up to now I never have heard about a "large-signal differential resistance". What is it - in contrast to the classical diff. resistance (slope of the curve)?

Hi @Lutz,

* I consider the "static" resistance as something material, tangible, physical... not only as a "bodiless", abstract ratio between two quantities.

* I have said (and shown) that (and how) the more abstract and sophisticated "differential resistance" can be obtained by varying the simple, humble, well-known 19-century ohmic, static, constant resistance...

*The differential resistance is introduced to define a local resistance at a given point of a non-linear IV curve; so it is different at different points of the curve. But note that in this case (constant current "source") and also in many other cases (eg., negative resistance circuits) the differential resistance is the same along a long section of the IV curve... it is not only a local resistance. So, we have obtained a linear IV section on the diffetential resistance IV curve by a fully linear IV curve of a static (but changing) resistance.

--------------------------------------------------------------

(By the way, can I ask you something off topic? Do you use your phone to edit your comments? I ask because I haven't been able to do it for many years and this is a big problem for me here...)

Quote: "I have said (and shown) that (and how) the more abstract and sophisticated "differential resistance" can be obtained by varying the simple, humble, well-known 19-century ohmic, static, constant resistance..."

I must admit that I do not consider the the slope of a function as "abstract" and "sophisticated". We are using such quantities many times for calculating input/output resistances and gain values for a variety of active electronic circuits.

More than that, in your drawings, the 5 static resistors are defined by the SLOPE of the straight lines only - NOT by the length. Hence, I cannot see how the they can serve to "obtain" any differential resistance rd.

You could convince me if you would describe the relation between both resistances (varying static and differential, resp.) with WORDS or with a formula.

Quote: "...the differential resistance is the same along a long section of the IV curve... it is not only a local resistance."

So - both are the same (small-signal and "large signal") for a constant slope ; why then make a difference?

Regarding your last question, I do not understand what you mean with "edit your comments" using a phone. In the past, I think, we have communicated already by email?

@Lutz, I frequently work in the bed, on my smartphone. My browser is Chrome. But I cannot edit my answer.. since there is no such option below my answer when I click to open the drop down menu. There is only "Delete" but I need "Edit". For comparison, there is such option on my laptop.

@Lutz, I really cannot understand why we cannot agree on such a simple concept... It is just a variable resistor supplied by a variable voltage source... and this arrangement is a dynamic version of the Ohm experiment. And because I change the current resistance of the resistor simultaneously with the input voltage variations, this effect occurs - the voltage source "sees" different (higher, infinite, negative) resistance. In the case of my picture (infinite resistance), it "sees" a horizontal line in the region between points 2 and 5... but it can see slightly inclined or even with a negative slope line... it depends on me (how vigorously I will change the resistance)...

Cyril, regarding the "edit" function: I never use the phone for writing in forums. I need a true keybord with large symbols....

Regarding our technical problem:

Lets take a simple example: I have an unknown non-linear two-pole.

Three measurements for a changing static resistor ("dynamic version of the ohm experiment") for 3 different voltages:

R1=1 kOhm, R2=1.6 kOhm, R3=2 kOhm.

How can you "obtain" the corresponding differential resistance of the test device from there measurements?

@Lutz, You are a happy person then-:) I frequently write quite late in bed and sometimes I even wake up and look at the phone screen...

Regarding your "technical" question, I need to know the values of the three different voltages... and then I will determine the three operating points... and the "large-signal differential resistance". But I am not sure if the three points will be located on a straight line (it depends on your resistance choice).

Cyril - slowly we are approaching the heart of the problem.

Of course, you need the corresponding voltages - and then you will calculate the corresponding currents (since the resistor values are given). From this, it is a simple task to find the ratio of the differences and the corresponding slope delta(V)/delta(I). Hence, you are doing nothing else than to use the definition of the "diffreential resistance".

However, this works only under the assumption that the real non-linear curve does not change its slope between the two test points.

Do you now understand why I could not agree to your statement that "the differential resistance is a constant resistance that is obtained by a dynamic static resistance." ?

The static resistors do not help at all - you always need the corresponding exact values for the current/voltage pairings (and not only the ratios).

Regards (and good night).

(I am stopping for today).

@Lutz, Do not forget that my goal is to explain the phenomenon in an intuitive way... to reveal and show the idea, principle, concept... not to calculate, to find the exact value... the relation... I do not know how the nonlinear element "calculates" the needed coordinates of the moving operating point but I know the final result - the varying voltage source "sees" a linear IV curve with a modified slope. This is the phenomenon that concerns me. Good night.