Hello Cyril - I am looking forward to the first contribution ...

Thank you for the permission, Lutz...it gives me confidence to start... Some of my previous successes in uncovering the secrets of famous circuit solutions (e.g., the exotic Widlar&Fullagar op-amp 741 input circuitry, ECL and other related to this topic) also encourage me to start:

https://en.wikipedia.org/wiki/Operational_amplifier#Internal_circuitry_of_741-type_op-amp

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

You understand it is a great challenge to reveal the clever ideas of famous circuit designers and to explain them to people... and even to very circuit designers:) The problem is that if you (I) are "nobody", people do not (want to) beleive you... they would beleive if you are "somebody":)... and this is my greatest tragedy:)... But let's ditch these unhappy thoughts and start our imaginary excursion in the past...

1. UNDER-USE OF THE "DIFFERENTIAL" INPUT STAGE.

It is logical to start creating our amplifier with a choice of the input stage. The classical circuit solution is the fully symmetrical differential stage to which we can reach in three steps (I use this "building scenario " to present this circuit to my students):

• First, we can control the transistor from the side of the base (keeping the emitter voltage steady)... thus " inventing " the common-emitter amplifying stage. It has moderate input resistance because of the small base current drawn from the input source. If we allow the transistor to "self-change" its emitter voltage (floating ground), we will obtain the common-collector stage (emitter follower) having an extremely low input current because of the neutralizing emitter voltage (emitter degeneration).
• Then, we can control the transistor from the side of the emitter (keeping the base voltage steady)... and so we get the common-base amplifying stage. It has extremely low input resistance because of the big collector current "pushed" by the transistor into the input source.
• Futher, we can control the transistor both from the side of the base and emitter... and so we get the simplest differential amplifying stage. But it is (ß times) asymmetric with respect to the input currents. So, to solve the problem,
• Finally, we buffer the emitter input by another buffer (emitter follower, common-collector stage) thus obtaining the fully symmetric differential (long-tailed) pair.

This structure has two opposite (inverting and non-inverting) inputs. So, we can use both to measure "floating" (ungrounded) voltages, i.e. as a differential amplifier. But we can use only one of them as an input where to apply the input voltage, and the other - as an "internal input" (e.g., for a negative feedback)... i.e., to use the differential amplifier not as differential but as a single-ended amplifier... like all the op-amp inverting and non-inverting circuits... We can even do not use (ground) the "unnecessary" input as in the case of the inverting amplifier...

Then the second buffer (emitter follower) becomes redundant since, in these applications, we do not need precisely balanced inputs... so we give it up... and use the emitter of the transistor just as a second input...

So, the first idea behind the CFA is the refusing to make the second (inverting) input perfect (low-current) - when we need a high impedance input, will use the first (base, non-inverting) input; and when we can meet or want a low impedance input, will use the second (emitter, inverting) input.

Well - good start. May I continue?

We want to use both inputs (base node and emitter node at the same time) for differential applications. But there is a dc voltage between both nodes acting as an unwanted "offset". Consequently, we drive the base of the npn transistor with another emitter follower , which however is realized as pnp transistor (offset cancellation). This results in a differential two-transistor stage with zero dc offset and two inputs (one high-resistive and the other one low-resistive).

Lutz, we have started really a unique discussion... like brainstorming or synectics sessions in the past... It would be wonderful if we can continue in this vein... And since you raised the issue of biasing, let me first consider a fundamental difference between the CFA and classic op-amp with respect to biasing... and then continue with your suggestions that raise new interesting issues for discussion about this exotic arrangement...

2. CONSTANTLY APPLIED BIASING FROM THE SIDE OF THE BASES.

It is clear that transistor amplifiers require biasing to compensate "the dc voltage between both nodes acting as an unwanted 'offset'" (the voltage drop across the base-emitter junction). Actually, the bias voltage is just another input signal which is summed with the main input signal. Like above, we can apply a "stiff" bias voltage both from the side of the base (common-base amplifier) and from the side of the emitter (common-emitter amplifier)...since we have two circuit inputs (the base and the emitter) that can take the two input voltages. But the differential amplifier needs in total three inputs - two (inverting and non-inverting) for the input voltages, and one for the bias voltage.

In the input differential stages of the ordinary "voltage feedback amplifiers" (VFA, e.g. classic op-amps), this problem is solved by replacing the "stiff" bias voltage source with a "soft" bias current source. It injects the bias current directly into the common emitters of the two transistors forming the long-tailed pair (they adjust their base-emitter voltages to pass the bias currents through the collector-emitter junctions):

https://en.wikipedia.org/w/index.php?title=Differential_amplifier&oldid=441555401#Biasing

Thus the both bases remain vacant for the two input signals and the "soft" emitter current source allows the emitters to follow the base variations. Note that this biasing is prepared in advance but it comes into play only when we connect sources to the inputs and the bias currents begin to flow through them. This means that we should not leave the inputs of a VFA op-amp unconnected because it will saturate.

Unfortunately, in the input differential stages of CFA, the common emitters are occupied (fixed) by the one input voltage and we can not apply there the bias voltage. So we go back to the classical bias arrangement and apply (sum) it from the side of the bases along with the other input voltage. In contrast to VFA (classic op-amps), here the biasing is constantly applied ("by birth") to the bases. So, we can leave the inputs of a CFA op-amp unconnected because there is always an input bias voltage applied to its non-inverting input that causes a zero "output" voltage at its inverting input...

Now about your speculation above that we introduce another emitter follower with the purpose to cancel the base-emitter voltage... and "this results in a differential two-transistor stage with zero dc offset and two inputs (one high-resistive and the other one low-resistive)".

I am rather inclined to believe that the problem of the base-emitter "offset" is solved by connecting the bias circuit to the bases of the two transistors... while the second emitter follower (the transistor pair) solves another problem - the compensation of temperature variations...

So, I see the next two great ideas here - "dynamic load" and "passive compensation". Let me consider them.

3. "DIFFERENTIAL" DYNAMIC LOAD.

The input quantity of the transistor is the base-emitter voltage changed from the base, from the emitter, or from both. The output quantity is the collector current; so we have to convert this current into a voltage by passing it through some kind of a resistor (in the simplest case, a linear ohmic resistor). To obtain a maximum possible gain, we choose to pass the current through a dynamic constant-current "resistor" (a constant-current source); another transistor can serve as such a constant-current "source". Thus the two transistors are oriented with their collectors to each other and act as opposing current "sources". As a result, a conflict appears and the voltage in the middle point between them (the common point of the collectors) vigorously changes:

http://www.circuit-fantasia.com/my_work/conferences/cs_2005/paper1.htm

We can improve this simple dynamic idea if we make the input voltage control the two current "sources" simultaneously but in opposite directions (differentially); thus the voltage gain dramatically increases...

http://www.circuit-fantasia.com/my_work/conferences/cs_2005/paper2.htm

As we have said above, the two transistors should be oriented with their collectors (drains) to each other to act as opposing current "sources". A good example of this configuration is the classic CMOS stage where the two transistors are connected with their sources to Vdd and Vss (ground) and their drains are joined.

In the classic emitter-coupled (long-tailed) pair, the collectors of both transistors (T1 and T2) are turned up. So, we should turn only the one collector (the left in the attached picture) back down to stand against the other collector. We can do this "magic" by one simple current mirror (T3-T4)...

https://en.wikibooks.org/wiki/Circuit_Idea/How_to_Reverse_Current_Direction#The_problem:_current_direction_inverter

Unfortunately, the transistors of our CFA input stage are turned with their emitters (not with their collectors as we would like) to each other...

So, we have to "rotate" both the transistors (currents) back to make them stand with their collectors against each other. To do this "magic", we need two simple current mirrors...

4. PASSIVE COMPENSATION.

Maybe the nicest feature of any differential pair (especially in microelectronics) is to compensate all sorts of undesired disturbances affecting simultaneously and equally both the transistors. We can see this great idea in the operation of our CFA input stage.

Assume a zero input voltage. The equal base-emitter bias voltages make the transistors produce equal collector currents. They are reversed by the current mirrors and made oppose to (subtract from) each other. As a result, this fully symmetric circuit is well balanced and the output voltage between the collectors of the output transistors Q4 and Q6 is equal to the (zero) input voltage.

Now imagine the temperature changes. The two base-emitter voltages and, as a consequence, the two collector currents, equally change... but the output voltage does not change...

Cyril, please excuse me, but - without large detours - I would like to come back to our main topic: Internal structure, working principles and main application areas for Current-feedback amplifiers (CFA). In my first reponse to your invitation regarding CFA I have mentioned - as the first topic - the input stage consisting of two emitter followers. The attached picture shows how this input stage is completed by a current mirror and an output stage (The picture is from a Sergio Franco contribution in "Analog Circuit Design", EDN series, edited by J. Williams).

Please note

* that the circuit is completely symmetrical,

* if the high impedance node (without the compensating C) is externally available, we have a Current Conveyor (CC).

PS: Unfortunately, I don´t know how to include the figure directly (to be seen).

Lutz, but everything I wrote above is intended just for the CFA! This is my attempt to reveal consistently, step by step, idea after idea, the design and the implementation of the main ideas in this circuit solution. I still have not told the most important things that I realized actually this morning...

I would like to say these are not trivial circuit solutions... they were invented with a lot of imagination and fantasy... they are just brilliant... and can not be explained by standard trite "textbook" phrases... they deserve more...

Okay, let's look at this more detailed figure and comment some details in it (I have copied it from the screen by SHIFT+PRTSCR, then placed it in the window of the Windows Paint program and cut). I prefer to use the more simplified but conceptual WP picture above but it is interesting to break for a moment our "historical excursion" and look at the details inside the Sergio's picture...

Well, let's give due attention to your insertion above about the inclusion of additional emitter followers at the base inputs:

"We want to use both inputs (base node and emitter node at the same time) for differential applications. But there is a dc voltage between both nodes acting as an unwanted "offset". Consequently, we drive the base of the npn transistor with another emitter follower , which however is realized as pnp transistor (offset cancellation)."

What can I say? A great idea... another excellent illustration of the general idea of "passive compensation"!...

(Lutz, I only need to interrupt about half an hour for dinner and will continue my commentary later...)

We discuss here how the biasing is particularly implemented (both in input and output complementary stages). Here is my story about biasing...

Figuratively speaking, the problem is how to "pull up" the base of the upper n-p-n transistor and how to "pull down" the base of the lower p-n-p transistor with Vbe... or how to "straddle" the bases with 2Vbe...

We can realize it simply by a bare resistive voltage divider as in the conceptual picture under the question. The problem of this simple solution is that the bias voltage will vary when the input voltage varies...

...and there is another problem: Monolithic integration of 4 resistors.

Yes... so we can improve this simple bias arrangement by replacing the resistors R2 and R3 with diodes... Then the bias voltage will vary slightly... and even more slightly if we replace the resistors R1 and R4 with current sources...

Quote: ...but everything I wrote above is intended just for the CFA! This is my attempt to reveal consistently, step by step, idea after idea, the design and the implementation of the main ideas in this circuit solution"

Cyril - I know that all this stuff belongs to the story. However, I am not convinced that it will be possible to "reveal...step by step...the design and the implementation of the main ideas".

Or do you think that I am too quickly and superficially in the discussion (because I have presented already a simplified scheme of the CFA principle) ?

But still the bias voltage drop across the diodes will vary under the influence of the temperature... and these variations will not perfectly match the variations of the voltage drops across the base-emitter junctions...

... So we can replace the diodes by base-emitter junctions...

... and finally, by emitter followers... to obtain an additional advantage - input buffering...

Lutz, you are the perfect interlocutor I have ever seen... Of course, it is not absolutely necessarily to follow a strict "scenario"... and actually there is not such... We can deviate from the main topic when we meet another interesting idea...Maybe we will not find the absolute truth (if it exists), but we must try:)

And walking towards it, we will explain many important circuit concepts... I frequently say to my students that not final and perfect circuit solutions are so important for understanding... the path to them, the evolution of circuit ideas, are much more important... But we nave not still answered what "to understand a circuit" means:

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

Whatever it means, I am firmly convinced that we must see and retrieve gradually, one after the other, the fundamental circuit concepts behind the concrete circuit solution. So if you let me, I will continue keeping them numbered and labeled with capital letters...

5. CASCODE IDEA.

The biasing emitter followers are loaded by current sources... and the pair "emitter follower + current source" is actually the so called "cascode". What is it and what is its idea? This is another great circuit idea that deserves a special question... but let's begin talking about it here...

In this arrangement (we can see it also in the classic long-tailed pair at common mode), a voltage source (the emitter follower) is connected to a current source; as a result, they provide each other ideal loading conditions - the current source provides an extremely high resistance load for the varying voltage source (our case); the voltage source provides zero resistance load for the varying current source. Figuratively speaking, like people in life (not like RG members in our discussions:), they "help" each other:

http://www.circuit-fantasia.com/my_work/eng_review/ing_review_11_99_a.htm (in Bulgarian)

Let's see how they do this magic in our circuit...

6. CONSTANT-CURRENT DIFFERENTIAL RESISTOR (NOT CURRENT SOURCE)

First at all, let's clarify that the current source in the emitter is not a source since it does not produce energy... it is just a resistor... but not the ordinary linear ohmic resistor... it is a nonlinear resistor with a horizontal IV curve (a constant-current differential resistor)... Such an element is simply implemented by the collector-emitter junction of a transistor with fixed base voltage... or more sophisticated, by the output part of a current mirror... Let's consider the operation of the upper cascode (Q3).

Think of the cascode pair as of a voltage divider consisting of two "resistors" - the dynamic emitter resistor (the "current source") and the static ((instant), chordal) emitter-collector resistance of Q3. Now imagine the input voltage Vp increases. The transistor Q3 should adjust its emitter voltage equal to the new input voltage; so it begin increasing its chordal emitter-collector resistance to "lift" the emitter voltage. But this decreases the common current and the dynamic emitter resistor reacts to this "intervention" by decreasing its chordal resistance to keep the current (it is still a constant current "source")... thus it "helps" the transistor to increase its emitter voltage... As though there are two connected systems with negative feedback that interact ("help") each other...

So, the use of the emitter dynamic resistor ("current source") is that the collector current, accordingly the base current and, as a result, the base-emitter voltage of the biasing transistor Q3, stay relatively constant when the input voltage varies.

7. USING UNDESIRED EFFECTS FOR USEFUL PURPOSES

The base-emitter "shift" Vbe is an undesired effect in the input CFA emitter follower acting as a voltage drop (a loss). It is interesting that we eliminate it by another equivalent but desired base-emitter "shift" acting as a voltage source (gain).

This is a passive compensation since the compensating element does not follow the undesired voltage drop by means of negative feedback. We rely only on the identical IV curves of the two elements (even when the temperature varies)...

We can see the same biasing arrangement in the CFA output complementary emitter follower (Q5 and Q6). Even here we can see how the current "sources are made (by Q12 and Qi6). The only strange thing is that they are not so constant "sources" since they copy Q1 and Q2 collector currents. It is interesting to see why...

Hi Cyril - you contributions numbered with 5,6 and 7 give a good overview over (about?) important principles applied for analog IC´s. Perhaps you could collect all relevant information you have in a corresponding document ("Classical principles in analog circuit design") for everybody (pdf format) ?

Hi Lutz! I finally managed to come back here after a long break:)

Thank you for the great idea! It is yet another confirmation that you and I think alike on the most important concepts of circuitry!

The idea of creating such a hierarchical system of fundamental circuit principles (like Altshuller's 40 inventive principles) arose in my mind somewhere in the middle 90's. I presented it with great enthusiasm at a conference on electronics in Sozopol... this was in the distant 1998... But instead of admiration, I met a disapproval... and this on the part of my reputable professor since my student years... By the way, his (and later, of some my colleagues and "friends") compensatory reactions gave me the idea on how to convert the simple follower into a negative feedback amplifier and how to explain the odd common-base stage:)

In 2002, I uploaded an interactive classification of circuit principles on circuit-fantasia.com:

http://www.circuit-fantasia.com/collections/circuit-collection/menus/old-menus/evolving-followers.html

In 2004, I added a weekly column about basic circuit principles

http://www.circuit-fantasia.com/tutorial/intro/list_of_principles.htm

In 2007, I began creating a "book about great circuit ideas" - Circuit Idea wikibook, and arranged its content in a form of a set of basic principles instead of dull conventional circuit names:

https://en.wikibooks.org/wiki/Circuit_Idea#Table_of_contents

As you can see, under the title of each Circuit Idea story (web page) there is a short description of the respective basic circuit idea (in red).

Since then, I do just that - extract general principles from specific circuit solutions and use them to explain other circuit solutions...

So, what is the main idea behind the CFA? If we manage to reveal it, we can include it into the principle collection...

"So, what is the main idea behind the CFA?"

I think, it is the desire to find an amplifier concept which overcomes some of the main limitations connected with the VCVS principle for classical opamps:

* Necessity to compensate the frequency response of the open-loop gain Aol (because of stability reasons),

* Slew rate limitations caused by the (internal) compensating circuitry.

As a result, the loop gain of CFA amplifiers is no longer determined by the feedback factor (which, at the same time, defines the closed-loop gain Acl).

Instead, the loop gain (and the stability properties) for CFA-based amplifiers is determined by a single feedback resistor Rf only, which therefore must have a certain minimum value (defined by the manufacturer).

Hence, no internal compensation is necessary to ensure stability for heavy feedback.

This allows fantastic slew rates of several thousands of volts/µsec.

As usual - the gain is determined by the feedback factor which can be selected independent on the element Rf.

One restriction exists: No feedback capacitor is allowed and, hence, some applications rerquire some special considerations (filters, integrators).

"...the loop gain (and the stability properties) for CFA-based amplifiers is determined by a single feedback resistor Rf only"... but why?

Because the input resistance at the inverting input node is very low (max. 50 ohms) we consider this as a "current input" - and the feedback "factor" is defined using the current "ir" which is fed back into this node.

Hence, the feedback factor Hf=ir/Vout=1/Rf.

(Now the feedback factor is not a "factor" but a conductance)

What does "low" mean - "permanently low" or "dynamically low"? Since the gain (Rf + Rg)/Rg is exactly the same as in the case of the standard "voltage feedback amplifier" (VFA) where the input impedance is high...

So, I will ask a (very) "provocative" question:) Is the current feedback amplifier really an amplifier with a "current feedback"? Is it not the same "voltage feedback"? Or, in some moments it is a current type while in others - a voltage type feedback? Or simply, it is a "low-resistive voltage feedback" in the steady-state?

It may seem quite daring to oppose the so authoritative sources...

http://cds.linear.com/docs/en/design-note/dn46fa.pdf

http://www.ti.com/lit/an/sloa066/sloa066.pdf

http://electronicdesign.com/analog/what-s-difference-between-voltage-feedback-and-current-feedback-op-amps

http://www.embedded.com/design/other/4006793/Back-to-the-basics-Using-current-feedback-op-amps-for-high-speed-designs

http://www.analog.com/static/imported-files/tutorials/MT-034.pdf

http://www.ieee.li/pdf/viewgraphs/current_feedback_vs_voltage_feedback_amplifiers.pdf

... but I want to stay true to yourself and find the whole truth about this great circuit solution...

Cyril, In response to your "provocative" question, I think it is correct to speak about "current feedback" because the output is a voltage that is related to the current that flows out of or into the inverting input.

(a) This current is the superposition of two currents: A current caused by Vin and the feedback current (caused by Vout); both currents meet in a commmon node (inv. input);

(b) This current is internally used as an input for a current mirror that feeds the next stage.

This in contrast to the classical voltage feedback opamp, where we also have the superposition of two currents. However, it is the voltage produced at this node which feeds the input stage (the corresponding current into the amplifier is neglected and is considered as parasitic).

To continue my answer: As a logical consequence, the open-loop "gain" is given in the data sheets as open-loop "transimpedance" Z(tr) and the loop gain is A(loop)=Z(tr)/Z(feedback).

On the other hand, I agree that each voltage-current relation can be interpreted also as a current-voltage relation because each current is connected with a corresponding voltage.

For my opinion, the terms "voltage feedback " or "current feedback" are chosen with respect to the most simple and most logical explanation/interpretation of the circuit´s working principle.

One well-known example - very often discussed and sometimes a subject of controversial debates - is the working principle pf the BJT. Very often the BJT is treated as a current-controlled device, whereas in reality it is controlled by the base-emitter voltage.

(To me, it is surprising that even in some textbooks the BJT is described as current-controlled. Just because we are using the simple relation Ic=beta*Ib ?)

Oh Lutz, believe me, I feel like a donkey between two haystacks:) Well, as if this question seems more intriguing... I will start with it...

Lutz, why have you stepped away from the "voltage-controlling" idea? The input buffer is a composed common-base stage... and you know very well that the common-base stage is driven by applying a voltage to the emitter(s)... This voltage is produced by the output buffer and is applied through the low-resistive voltage divider Rf=Rg to the output (the joined emitters) of the input buffer... that actually serving as an inverting input... So it is the attenuated output voltage...

8. BOOTSTRAPPING. In the feedback loop, we can see again the ubiquitous Miller arrangement - now an opposing voltage source (the output of the input buffer) virtually increases the resistance seen from the side of the input voltage source (the output of the output buffer acting through the voltage divider Rf-Rg):

https://en.wikipedia.org/wiki/Miller_theorem#Applications_based_on_subtracting_V2_from_V1

Really, the bootstrapping effect exists in the non-inverting configuration implemented by the ordinary voltage-feedback amplifiers (op-amps) where it virtually increases the differential input resistance. There this effect is not so noticeable because the input resistance is high "at birth"... but here the situation is quite different...

Cyril - I am with you that my reasoning contains a kind of conflict. But remember my attempt to solve the conflict:

"The terms "voltage feedback " or "current feedback" are chosen with respect to the most simple and most logical explanation/interpretation of the circuit´s working principle."

And in the present case, we have the situation that

(a) there is a current out of (or into) the low-resistive node , and

(b) this current is internally mirrored and processed further.

Therefore, in this case, I think it is appropriate to follow the terminology commonly used and speak about "current feedback" (although it is the voltage at the emitter node which is the physical cause of the effect).

Dear Lutz,

Аfter one year, the mint&sprite season come again and this sharply increased my "artistic" activity:) Another reason for this rise is the fact that in the middle of the week I go on vacation in the lands of the ancient Thracians among a lot of sun and mineral water... and that is why I hasten to reveal the ideas behind this gracefully circuit solution...

In my opinion, the input resistance of the inverting input is not only (permanently) low. Sooner it is dynamic - sometimes (during the transition) it is extremely low... but sometimes (during the steady-state) it is extremely high. Thus the so-called "current feedback" is not permanently "current feedback"; during the transition it is really a "current feedback"... but during the steady-state it is a "voltage feedback".

I will illustrate my speculation by a 19-sentury example - the famous compensation method for a voltage mesument, to show, for the umpteenth time already, that "the new is well-forgotten old".

https://en.wikipedia.org/wiki/Potentiometer_(measuring_instrument)#Constant_current_potentiometer

This is а legendary idea, since it was first proposed by Baron Munhauzen in the 17th century... then it is realized electricall in the 18th century... and finally, in an electric form - in the 20th century... Unfortunately, teachers in the field of measurement do not consider this technique so generalized and as a result, students cannot see later the connection between this pure electric technique and the more sophisticated electronic circuits based on the series negative feedback and "bootstrapping"...

In this "manual servo system", we change the attenuated voltage Vout to make it equal to the input voltage V(-) and the indicator shows zero. The problem is that the indicator (it was the so-called galvanometer) and the input voltage source shows extremely low resistance; so during the transition the whole current is diverted from Rg and it flows entirely through the input source (the inverting input or the common emitter point, in our CFA circuit).

So, the first conclusion is: DURING THE TRANSITION, THE INVERTING INPUT OF CFA EXHIBITS EXTREMELY LOW RESISTANCE.

Actually, the output of the voltage divider is shorted... the lower resistor Rg is bypassed... and a current Vout/Rg flows through the feedback network; thus the name "current feedback".

So, the second conclusion is: DURING THE TRANSITION, THE CFA FEEDBACK IS A CURRENT FEEDBACK.

I know that it sounds strange but I will say that a huge current flows between the two sources - the common emitters (what is an input as well!) and the voltage divider output VRg, since there is a voltage difference between them (the strange is that they are joined... but they have different voltages:)

Well... the negative feedback makes the system "move" towards its equilibrium. The difference between V(-) and Vg decreases... and finally becomes almost zero. The "output voltage" V(-) of the inverting input (how strange it sounds!) is almost equal to the "input" voltage Vg. Seen from the side of Vout, the inverting input has extremely high resistance (open circuit)... it is not really but virtually increased input resitance... As a result, the whole current is diverted back from V(-) and it flows entirely through Rg.

So, the third conclusion is: DURING THE STEADY-STATE, THE INVERTING INPUT OF CFA EXHIBITS EXTREMELY HIGH RESISTANCE.

Actually, the output of the voltage divider is (virtually) opened... and the whole unaffected output voltage Vrg = Rg/(Rg+Rf).Vout of the unloaded voltage divider is applied to the inverting input... so, the feedback is "voltage"... not "current".

So, the fourth conclusion is: DURING THE STEADY-STATE, THE CFA FEEDBACK IS A VOLTAGE FEEDBACK.

We are ready to make final conclusions:

• THE INVERTING INPUT OF CFA EXHIBITS DRASTICALLY CHANGING RESISTANCE.
• THE CFA FEEDBACK DYNAMICALLY CHANGES FROM CURRENT TO VOLTAGE ONE.

Cyril - one question:

Suppose, we are using this CFA device as a non-inverting ac amplifier for an input signal of 1kHz (closed-loop gain Acl=2).

Now - with respect to this example: Which time periods belong to "steady state" and "transition" conditions, respectively?

Hi Lutz,

I have said it already, I will say it now again... How wonderful in this world is to have at least one but quality interlocutor (this applies to friends... and colleagues too...)

Your insertion is completely appropriate. I myself can not fully realize this idea... how it really happens in practice... but one I am absolutely convinced - the behavior of the CFA (the inverting input resistance, (the input impedance of the inverting input, current and voltage in the circuit of feedback...) is time variable...

Here is another typical and well-known example - charging a capacitor through a voltage divider, which is really time variable.

During the transition, the capacitor bypasses Rg and diverts all the current I = V/Rf through itself. Seen from the side of V (via the voltage divider), the capacitor has extremely low impedance (short circuit).

During the steady-state, the voltage across the capacitor is almost equal to the unaffected output voltage of the voltage divider VRg = (Rg+Rf)/Rg.V... and the whole curent flows through Rg. Seen from the side of V (via the voltage divider), the capacitor has extremely high impedance (open circuit).

Lutz, I continue thinking about your question where is the "transition" and where - the "steady-state"... I am inclined to believe that simply the first is valid at sufficiently low frequency, while the second - at very high frequency...

" I am absolutely convinced - the behavior of the CFA (the inverting input resistance, (the input impedance of the inverting input, current and voltage in the circuit of feedback...) is time variable..."

I think - strictly spoken - you are correct. In a non-linear circuit input impedances are time-variable (voltage dependent). However, here we have discussed terms and circuit properties for the CFA like gain, input impedance, feedback factor and stability, which all are small-signal parameters. And we apply rules and formulas (at least in our brains) which can be applied for linear, small-signal and time-invariant conditions only. Hence, I think we are always in the steady-state mode during discussion of the above mentioned parameters.

Well Lutz, but how do we explain the fact that, if we are always in the steady-state mode, the negative feedback is both voltage (determined by Rg/(Rg+Rf and Vout) and current type (determined only by Rf and Vout)? Does a current enter/exit the common emitters? If so, the feedback is current; if no, it is a voltage type...

I would like also to repeat that the bootstrapping effect exists in the non-inverting configuration implemented by the ordinary voltage-feedback amplifiers (op-amps) where it virtually increases their differential input resistance. But the range of variation is different - while there (VFA) the input resistance is increased by the bootstrapping from "very high" to "extremely high", here (CFA) it is increased from "extremely low" up to "extremely high". So, while there the input current is originally small and can not decrease more than that... here it can significantly vary... There practically there is no current... here we have two cases - there is no current or there is current...

Well Lutz, this was only one (but interesting) side of this exotic circuit solution that we should gradually clarify. But the most important thing here is to reveal the main idea...

This is a brilliant circuit solution (like Widlar's op-amp input differential stages)... so it deserves a brilliant explanation as well... and if someone can do it, his/her explanation deserves at least one (positive) vote or, more probably, many... many negative votes:) Well, I will try it first...

Imagine we are in 80's... and we must invent this circuit (our bosses wish to do it:) Here is one possible "scenario" to get out of this critical situation...

First at all, we know that there are various stray (mainly, temperature) influences that we should compensate... We know the remedy - we should make them oppose and, as a result, eliminate each other (the differential idea)...

We are already familiar with the classical long-tailed pair... but to impress our bosses (and to deserve our $$$:), we decide to make a more exotic arrangement... For this purpose, we take two inverse (complementary) emitter followers and join their inputs (bases) and their emitter voltage outputs . Then, to make this arrangement even more sophisticated and exotic... and, of course, to get a high gain and only one single-ended voltage output, we reverse their collector current outputs (by direction inverting current mirrors)...then join them... and finally buffer this output since it is extremely high resistive...

Then we connect two equal bias voltage "sources" (forward-biased diodes or base-emitter junctions)... they are "floating" since we have already joined the two bases and there is no way to apply grounded bias voltages. These two common-mode initial input voltages make the emitter followers "fight" each other (like a "tug of war") - the upper n-p-n transistor "pulls up" while the lower p-n-p transistor "pulls down" the common emitter point. As a result, the emitter voltage does not "move" at all and remains zero... The "reactions" of this "fight" (collector currents) equally increase and their difference, accordingly the output voltage, remains zero...

When the temperature varies, the situation is the same since the temperature influences both the upper and lower parts of the circuit...

Thus we obtain a beautiful, fully symmetrical differential pair, where we can observe two conflicts: the first is a "voltage conflict" causing vigorous current variations; the second is a "current conflict" causing vigorous voltage variations...

Of course, we have already observed these phenomena in the classical differential pair with dynamic load... but there are some essential differences between the two great solutions...

In contrast to the long-tailed pair, in our "invention" the two bases are joined... so we can apply only one signal to ground (like any complementary buffer); it is single-ended but it acts as the diferential input signal in the classic long-tailed pair. Figuratively speaking, it urges the upper n-p-n transistor to "pull up" while the lower p-n-p transistor to "loosen" the common emitter point. They do this by changing differentially their collector currents ("reactions") that, after the reversal, are subtracted in the output point thus producing the output voltage (the dynamic load idea)...

So, to make a real differential input, we have to find another input... an opposite (inverting) input... We begin thinking... and finally, feeling the success (and the possible good consequences for us:), we come to the so absurd and misunderstood in circuitry... but so widespread in our life idea - to intervene in the "conflict" between two "forces" as a third "force"... Thus we help one of them against the other... and they "react" to our intervention trying to maintain the "status quo"... The only problem is that we have to endure their reaction...

An interesting fact is that, in this conflict, we stand on the side of (help) one force... but it does not help us; in this case the two forces help each other against the common enemy:)

So, the situation in our CFA circuit is as follows.

Two emitter followers (a complementary emitter follower) keep up the common emitter voltage (almost) equal to the input common base voltage. We connect another emitter follower to this point and try to change its voltage. The two followers change their collector currents trying to keep the existing voltage... and, as a result, the output voltage vigorously changes... The only problem is that our emitter follower has to endure the reaction of the two followers...

This was my fancy story about the creation of the so-called "current-feedback amplifier"...

Still regarding the CFA concept:

"Does a current enter/exit the common emitters? If so, the feedback is current; if no, it is a voltage type..".

As I have mentioned (and as we know) there is not only a current into (out of) the common emitters, but this current is mirrored and further processed. This fact leads to the amplifying propertes of the CFA which are described using the term TRANSIMPEDANCE.

Regarding the current mirror based diff. stage:

"Thus we obtain a beautiful, fully symmetrical differential pair"

I think, it is not yet "fully symmetrical". Both collector currents differ by the base current of the upper left transistor.

However, this can be healed in the next stage (connected to "out"), which needs a base current. By proper designing this stage (with the appropriate base current) we can establish "fully symmetrical" properties.

Cyril and Lutz,

I have read through this discussion, as best I can ,  

regarding the Truth about CFA,  

I sense an "either"  vs "or"  goal. 

Eiher "Voltage" or "Current"  as the controlling entity.

From Philosophy, I have come to see the possibility 

that neither inifinity nor zero actually exist. 

From Math, I have come to see that 

we have "agreed" on methods of handling infinity and zero. 

In practice I have come to see that 

both Voltage "and" Current are always present ( OHMs Law ), 

so "both" are always parts of a controlling entity. 

This has been a very interesting presentation, 

and cooperative interplay of perspectives, 

and certainly the whole story has not ( possibly never ) been told. 

O my God, Glenn! You can not imagine how much you moved me with your positive votes! Of course, I could thank you in the usual way for the scientific community - by sending an internal message. But I want to do it here to send a few key messages to the participants in this forum.

You made me read in one breath and remember this exciting circuit "tale" that I wrote with such imagination and enthusiasm last summer (I have not looked at it since then). Then I felt that I had revealed the great secret of this bizarre circuit by seeing in it and showing many other basic circuit ideas... and the little thing I needed was at least what you made a year after that - a simple vote.

Now I ask myself a chain of simple questions, "Why people do not admire when they see something created with such enthusiasm, but rather try to downplay it? Why they do not comment on any perspective offered there but instead deny everything written arguing with general phrases?" The result would be much better, impressive and useful for all of us - discussions will become more passionate, interesting and alive, as they should be...

People are different according to their way of thinking, interests and realization in society. Some of them are interested in underlying ideas while others - in their implementations. So it should not oppose ideas to their implementations...

Glen - I agree with you, of course.

Let me reply as follows: In electronics nothing is ideal. That is an old wisdom and, of course, also applies here. No "sinusoidal" signal is a pure sinus without harmonics and no "linear" amplifier really is linear. More than that, there is no clean voltage feedback and there is no clean current feedback. We always have a mixture of both - but we know which property is dominating - and we select a proper corresponding name. Regarding "current feedback" we assume that the input impedance of the feeedback node will be zero, which is never the case.  But everybody who has a certain background will know under which circumstances we are allowed to speak about "current feedback" - with respect to other uncertainties and error sources.

Do you know what I mean?

Cyril,

De Nada.  

It has been the great pleasure of this geeky geezer 

to read the thinking of men more talented than himself ,

and to comment without consequence. 

Lutz,

Point well made.  

We casually speak of the "dominant" entity. 

I only felt that "Rule Number Zero"  

needed to be brought to the front

... if only for a moment. 

I think that is another story...

And I also think that after all is preferable to use the name of the transimpedance amplifier. Potential of current input is defined by voltage input - always, even without feedback. This is a significant difference compared to conventional OPA - if there no feedback nothing work at all. :-))

Yes, Josef... "potential of current input is defined by voltage input - always, even without feedback" since the bias voltage is preliminary applied by the side of the bases while, in the case of the conventional OPA, it is applied by the side of the emitters... and the external input circuit must close the biasing loop.

Do you share my "provocative" assumption that the current-feedback amplifier is a voltage-feedback amplifier at low (normal) frequencies and a current-feedback amplifier only at the maximum frequency?

Cyril, please can you explain the following claim?

... the current-feedback amplifier is a voltage-feedback amplifier at low (normal) frequencies and a current-feedback amplifier only at the maximum frequency?

I do not know how the frequency can play such a role (and what is the "maximum" frequency)?

Dear Lutz,

Thanks for returning to the discussion that we started with a great enthusiasm last summer. Regrettably, the mint&sprite season is over and my "artistic" activity is a little decreased:) But I will try to get into а good creative condition by a little Greek Metaxa:)

Now a little more seriously... My explanation is simple:

At low enough frequency, the "input" voltage VIN- of the inverting input manages to follow the output voltage of the voltage divider RG-Rf. So, the "input" impedance is high (due to the mystic bootstrapping effect), and the voltage divider is practically unloaded... it has the illusion that it is not connected to anything:) This explains why the gain is determined by the voltage divider... and this is the well-known voltage-feedback amplifier.

At high enough frequency, the "input" voltage VIN- of the inverting input does not manage to follow the output voltage of the voltage divider RG-Rf, it is "stiff". So, the "input" impedance is low (there is no the fancy bootstrapping effect), and the voltage divider is practically shorted. This explains why in this case the gain is determined only by the resistor Rf... and this is the so called current-feedback amplifier.

Cyril - I am sorry, but I cannot follow you in this respect. The input impedance at the inverting input of a CFA is always "low" - this input terminal is conncted to the common emitter node of a npn-pnp combination. There is no "bootstrap effect".

Both transistors have the same current - unless there is an external current (provided by external feedback circuitry). Now - the produced current difference is sensed by two current mirrors - and their outputs are used as an input for a current-voltage converter.*** For these reasons, I think it is correct to use the term "Current-Feedback Amplifier" because - indeed - it is a current difference which is evaluated internally of the device.

*** I just have discovered that your first posting in this subject shows a typical block diagram of a CFA.

If 'the input impedance at the inverting input of a CFA is always "low"', how do we explain the gain formula (K = RG + Rf)/RG)? Is there no the low input resistance to bypass RG in the case of the non-inverting amplifier?

I think, we all have problems to "explain" why the non-inverting gain of an opamp with feedback is (1+Rf/Rg). We need some knowledge about the principle of feedback - and we need formulas showing the relationship between the several state variables. And then we can proof that - assuming ideal opamps - that the gain indeed is (1+Rf/Rg).

And the same applies to CFAs. We cannot "explain" the resulting gain expression. We need to derive the gain formula based on current-voltage relationships - under the assumption that the transimpedance Z(s) of the CFA device is approaching infinite. This derivation was done long time ago (can be found elsewhere) - and the result is indeed (1+Rf/Rg).

Vasile - I am afraid there is a misunderstanding:

YOU are speaking of a classical voltage opamp that CAN be used as a transimpedance amplifier (current input).

In contrary, WE are speaking of another device - a Current-Feedback amplifier that has one low-impedance input (inverting). This CFA device has - from each nature - an open-loop transimpedance (in contrast to the opamp which has an open-loop gain). 

Hi Cyril

I must agree with Lutz,

I did a little" homework": non-inverting amplifier with CFA, see appendix.

Josef

Josef, can you explain your results more physically? What happens in this circuit below and above the 3 dB frequency?

Lutz,

I think that one thing is to change the emitter voltage of an emitter follower when the base and accordingly, the emitter voltage, is fixed (as in the ordinary common-base configuration) and other - when the base and emitter voltages change in the same direction... The latter is a sort of a common-base stage with varying base voltage and we can see here the effect of bootstrapping...

So, in the last Josef's picture, the voltage of the inverting input depends on (follows) the voltage of the non-inverting input... Here the inverting input is not independant as in the ordinary op-amp; it depends on the non-inverting input. So, in this non-inverting configuration, the inverting input is an output that follows the non-inverting input:)

For example, imagine the input voltage of the non-inverting input increases... the emitter voltage also increases... the output voltage of the buffer increases.. and the output voltage of the voltage divider also increases... The result should be a decreased input current and accordingly, an increased input impedance, of the inverting input... All this is valid at low (below 3 dB) frequency.

Vasile,

For basic ideas it does not matter whether something is old or new. Great ideas are immortal... only their implementations die...

For me, it is difficult to imagine anything more exotic than this circuit...

An excellent material... but I need something more...

Thank you Lutz for clarification. Can you point please to a datasheet of such device?

Hello Vasile - here are some CFA types:

AD844 (extensively applied), AD8012, LT1223, LT1227/28, MAX3970, LM6181, OPA691/695, OPA2691/2694,...

It is important to note that the stability properties (as a result of negative feedback) do NOT depend on the feedback ratio (as is the case for voltage opamps) but on the feedback resistor Rf only. Hence, using a suitable value for the feedback resistor Rf the circuit does not need any frequency compensation - and, hence, offers excellent slew rate values.

Quote C. Mechkov: "An excellent material... but I need something more..."

Cyril - here comes another "explanation":

For deriving the gain formulas for classical opamps we assume infinite input resistances. Likewise, for CFA`s we assume a ZERO input resistance at the inverting node. Based on this condition, we can say the following:

1.) Assuming a large open-loop gain, the classical formula from H. Black for the closed-loop gain reduces to Acl=-Hf/Hr (Hf=Forward function; Hr=Return function). This applies to each feedback system - even if the functions have a dimension (other than volt/volt).

2.) In case of CFA we have to consider the current id (out of the inv. low-resistive node) as the relevant input signal which controls the output voltage. This current id is the DIFFERENCE between the currents i(p)  (produced by the input voltage) and i(n) (produced by the output voltage). This difference is the equivalent to the VOLTAGE difference in case of classical opamps. This property gave reason to call the device "Current Feedback amp".

3.) For a CFA with feedback (Rf, Rg) we have Hf=id/Vin=1/(Rf||Rg) for Vout=0 and id=current out of the inv. terminal. 

4.) The return function is Hr=id/Vout=-1/Rf (for Vin=0).

5.) Therefore, the ratio is Acl=-Hf/Hr=(Rf+Rg)/Rg.

6.) Hence, assuming ideal open-loop devices, for a CFA we arrive at the same closed-loop gain as for classical opamps.

Hi Cyril

Hi all

I try explain my results more physically, see below.

Josef

Hi Josef - thank you for the illustrative diagram.

Just for clarification: The circuit diagram shows the internal (principle) structure of the device AD844, which resembles a Current Conveyor (CC) together with an output buffer (I-V converter). This is the only CFA device which provides external access to the high-impedance node. The first part (CC) sometimes is called "diamond transistor" because it shows some similarities to the BJT: One high-impedance and one low-impedance input and one high-impedance output node.

Dear Lutz and Josef,

Your last comments made me seriosly think... and, as a result, this evening I finally realized how different we are... I have never told this to anyone... but I decided to share it with you now, here, and in the context of this circuit...

You try to express and explain circuits by means of expressions and laws... while my main goal is to grasp the basic idea behind the specific circuit implementation and to express it by words.

For this purpose, I try to see well-known configurations, devices, analogies, principles in the unfamiliar circuit... to reduce it to a new combination of old building blocks... Formulas do not help me almost nothing to achieve my goal .... I do not need them... they even hinder me obscuring and watering down what is achieved by other means... Maybe it is because formulas show the quantative side of the circuits... but, at this stage, I need qualitative means... I need something else that is not written in sources...

Simply put, I realized that we speak different languages, and nobody listens to what the other said. I wrote huge masses of explanations in all these nine pages, but no one wants to think that way. You also wrote detailed explanations, but I do not want to use them...

I think my way of thinking is inherent rather to inventors than engineers and scientists, and if I was at a suitable place, perhaps I really would invent devices... but now only analyze and comment them...

But puzzling to me is that I noted (from the first time, in 90's) my circuit insights irritate and even enrage (even not conventionally thinking) professionals. What is most interesting, they do not refute my specific allegations but deny them by conclusions of a general nature...

But let's finish this "lyrical retreat" and return to our circuit...

To illustrate the written above about my way of thinking, I will say that I have obtained a perfect notion about the operation of the CFA non-inverting amplifier by the help of the humble 18th century electromechanical analogy - measuring a voltage by a compensating voltage. On the page 5 of this dicussion I have written something like this:

"In this 'manual servo system', we change the attenuated voltage VRG to make it equal to the input voltage and the indicator shows zero. The problem is that the indicator (it was the so-called galvanometer) and the input voltage source shows extremely low resistance; so during the transition the whole current is diverted from Rg and it flows entirely through the input source..."

And then, "This 'manual negative feedback' makes the system move towards its equilibrium. Finally, during the steady-state, the input voltage is almost equal to the unaffected output voltage of the voltage divider VRg = (Rg+Rf)/Rg.V... and the whole current flows through Rg. Seen from the side of V (via the voltage divider), the input voltage source has extremely high impedance (open circuit)."

So, I had managed to realize the circuit idea by thinking of it as of a kind of a servo system.

Vasile,

Can you find something more exotic than the input part of the CFA shown in the attached picture below? An input voltage Vn is applied to the output of the amplifier x1?!? It would be interesting to see the look on the face of a student who is introduced for the first time to this circuit...

Your last comments made me seriosly think... and, as a result, this evening I finally realized how different we are................

Simply put, I realized that we speak different languages, and nobody listens to what the other said.

Hello Cyril - at first, thank you for your criticism, your frankness and openly expressed feelings. However, I do not hink that we are so far from each other as you may believe.

Let me explain (without any formula):

We speak about amplifiers with feedback. Hence, those amplifiers needs a differential input, OK? Now - we have two sorts of amplifiers:

1.) Voltage Feedback Amplifiers (VFA) reacting upon a VOLTAGE difference at the input, and

2.) Current Feedback Amplifiers (CFA) reacting upon a CURRENT difference at the input.

In both cases, it is the feedback signal that reduces this difference.

That is the whole story as far as the basic principles of both devices are concerned. And if you carefully look again into the text of the various contributions you will find these simple statements.

To finalize these statements and for answering your own question (Why is the CFA gain (Rf+Rg)/Rg ?) it was necessary to develop some formulas. So - what is the problem?

Remember: The question in your title is "Is it really a current-feedback device"?

I think, this question has been answered, don`t you think so?

With all my respect

Lutz

Hi Cyril, do not be vexed.

We can not all be the same. Your way of thinking is good at inventing circuits - "what happens if ...?"

When analyzing circuits I personally prefer a useful routines - such as general voltage analysis, as You can see in my "works". Enough it make things easy. But before I start with that, I try to grasp the essence of the circuit, always.

This path leads to a brief (and accurate) descriptions like:

CFA (transimpedance amplifier) is a current controlled (inverting input) voltage source (Output) - voltage of the current input is defined by means of a voltage input (non-inverting input).

A person which is familiar with the basics of the art is able from this definition to derive the ideal characteristics. As: infinite input impedance of the non-inverting input, zero input impedance of the inverting input, zero output impedance and relation

Uout = Z x I-, where Z is transimpedance, etc.

Once you begin to explore the real properties of the structures, the language of mathematics is needed because is concise.

Josef

"So, my suggestion is to try unveiling the clever secrets behind CFA as we go back in the past and try to recreate, step-by-step, the way its creators (David Nelson, Kenneth, Saller Bill Gross etc) have thought when inventing it. If you allow me, I can start the discussion by exposing below my story about the possible reasons for originating this clever circuit solution."

Hi Cyril, as a further response to your last contribution I will try to partly answer another of your earlier questions (see the quote above).

I think, the "clever solution" in form of a CFA was born much earlier. As we have seen - the first part of a CFA is nothing else than a so-called "Current conveyor (CC)". And the first generation of such devices (CCI) was introduced as early as 1968 (!). This first generation was eqipped with two low-resistive current inputs (and one high-resistive current-output).

Much more versatile was the second generation (CCII, introduced around 1970) having one high-resistive voltage input (non-inverting). That is the device which was named (I think it was the manufacturer Burr-Brown) as "Diamond Transistor" (perhaps primarily an advertizing gag). 

If you are interested in the development process here is a source of information:

Sedra, Roberts, Gohh: The current conveyor: history, progress and new results.

(IEE Proc.-G, vol 137(1), 1990, pp 78-87)

Dear Lutz,

I would like to respond to your previous comment where you have explained in two points ("without any formula") what is CFA. BTW some flu caught me and what is most interesting is that it excites my imagination and creativity ... affecting me like MX******* (you can guess what I mean:)

IMO if "current" in "current-feedback amplifier" means "current flowing between the negative feedback network and the inverting input", this name is misleading because no current flows (excluding the small erroneous current). In steady state, what is the normal condition of the circuit, the feedback network (the voltage divider) produces following voltage that is equal to the emitter voltage; so, (almost) no current enters/exits the inverting input. This is a unique property of any bridge circuit including the "manual servo" above where no current flows through the bridge element (regardless of how it is low-resistive), and is named "bootstrapping". The manual servo (I have attached it again for convenience) is an excellent illustration of this phenomenon - if you vary the input voltage and I keep the compensating voltage equal to it, no current will flow through the galvanometer as much as low-resistive it is... as though its resistance has become infinite...

So, in the steady state, we have a voltage-type negative feedback, implemented by a classical voltage-to-voltage converter (voltage divider) that continuously adjusts its output voltage R1/(R1+R2)*Vo to make it equal to the input (and emitter) voltage VIN. As a result, the output voltage Vo is (R1+R2)/R1*Vo... and this is exactly the same as in any "voltage-feedback" op-amp configuration...

Thus we have every right to call this device "voltage-feedback amplifier".

If, for some reasons, the output voltage Vo cannot exactly follow the input (and the emitter) voltage (I suppose this will happen above the cutoff frequency), the voltage-type negative feedback will become a current-type... the voltage-to-voltage converter (the voltage divider) will transmute into a voltage-to-current converter (the resistor R2). In our manual servo, this situation means that I cannot follow you and a significant current begins flowing through the galvanometer... the role of the lower resistor RG is less (it is shunted) by the low emitter resitance... and the voltage divider on the right acts as a simple resistor RS...

Now we have every right to call this device "current-feedback amplifier" since a significant current flows between the feedback network and the emitters.

By the way, I (being a junior assistant) realized one of the first "CFA":) in the middle 90's, when trying to show how the bootstrapping effect increases the input resistance of op-amps. For this purpose, I connected a resistor between the 741 op-amp inputs and even a low-resistive (bad) voltmeter... and the op-amp continues working...

 To "CF ideal OP amp model"

Hi Cyril,

But comparable beauty is mirroring the current flowing via "follower output" (current input of the all structure). And this beauty is lost in a simplified "Thevenin model", which is of course OK. And I can also use it.

It is clear that in different situations we use various means for description of reality. However, we must always see to it that they are correct.

Josef

Quote: IMO if "current" in "current-feedback amplifier" means "current flowing between the negative feedback network and the inverting input", this name is misleading because no current flows (excluding the small erroneous current).

Cyril - for a voltage opamp there is no voltage between both input terminals (excluding the small erroneous differential voltage). Do you know what I mean? Of course, there is ALWAYS a voltage - the small erroneous voltage is the only driving voltage for the opamp.

And in case of a CFA? The "small erroneous current" is the only signal which is driving - via current mirrors - the I-V converter and, thus, the output voltage.

And - yes - you are right that the voltage between both CFA input terminals disappears, but this happens NOT because of feedback but because the connection between both terminals consists of a unity gain buffer.

Therefore, I again cannot follow your explanations.

Here is a short and good introduction to the CFA principle (Analog Devices):

http://www.analog.com/library/analogDialogue/archives/30-3/ask.html

A dramatic discussion:)

Well, Lutz... let's understand more deeply, with the power of our imagination, what happens in the input stage... to see if "this happens NOT because of feedback but because the connection between both terminals consists of a unity gain buffer", as you said... Let's do it in a human friendly, figurative and colorful way that can be not found in all these excellent sources.

For concreteness, imagine the input voltage changes in a positive direction; figuratively speaking, the input source begins to "pull" the transistor bases up. Transistors react to this "disturbance" by trying to restore the equilibrium. For this purpose, the upper transistor "pulls" the common point between the emitters while the lower transistor "looses" this point. This is the usual mechanism of the series negative feedback acting from the side of the emitter (emitter degeneration) in an emitter follower, which in our case is a complementary pair...

But in this case, another device - the output buffer, joins the "game", and "helps" the upper transistor by "pulling" the emitters up... This is another negative feedback acting from the side of the collector in parallel to the conventional emitter degeneration. This negative feedback is more perfect than the inherently existing because of the huge gain of the output current mirror dynamic load.

(see also my explanations on Page 6 of this discussion)

Now another insight about the nature of this circuit...

The absurdity of this name "current-feedback amplifier" is that actually there is a current when there is no negative feedback - when applying the input voltage directly to the inverting input while the non-inverting input is fixed (i.e.,  a common-base configuration)... and v.v., there is no current when there is a negative feedback (as in the case of the non-inverting amplifier). Looking from the side of the output buffer, the last configuration corresponds to a common-base stage with changing in the same direction base voltage.

....there is no current when there is a negative feedback (as in the case of the non-inverting amplifier).

Of course, there is a current - the most important one: A very small erroneous current which drives the current mirrors.

Yes Lutz, there is some current... but this current is negligible if we compare it with the current that would flow if we apply the input voltage directly to the inverting input (without negative feedback).

I would like to add that, you and all here, understand that we conduct this discussion not for the very name; it is the less important thing here. We want to clarify the basic idea and the circuit operation... and this can be done in the course of such heated conversations where we try to look at the circuit from all the possible sides. I think everyone here will agree with me that now we know much more about it than before the start of this discussion. So I thank you and all here for the assistance.

Yes Cyril - I agree to your conclusion.

It seems that our discussion is finished, but I still can not stop to admire this beautiful fully symmetrical circuit topology!

Yes Cyril, it is very pretty circuit, simple and useful.

Josef

Yes, I agree to the last two conclusions - and to summarize the surprising facts:

For our calculations of systems with feedback we are allowed - in most cases - to neglect the most important parameter : The driving signal for the amplifier, which is the differential input voltage for the VFA and the differential input current for the CFA.

But actually, we drive the CFA by voltage... really some current flows but we apply voltage as an input quantity... This voltage is determined by two cascaded "voltage devices" - the buffer output and the voltage divider while the current is a undefined erroneous quantity that tends to zero...

Cyril - I think we have discussed this subject earlier, didn`t we? There is a current coming out of the buffer (driven by the signal source) and another current going into the buffer (driven by the feedback network). And both are superimposed and form the (small) driving current for the current mirror. That`s the whole story.

Of course, you are right that both currents are driven by corresponding voltages - but this applies to all currents in such networks. For my opinion, what really matters is the question: Which quantity (voltage or current) is used for further processing. And in case of CFA it is the "erroneous" current which contains the feedback information and which is used as an input for the next stage (current mirrors).

Perhaps this discussion is nothing else than two look from two sides of the same coin ?