THE INPUT QUANTITY OF ANY NEGATIVE FEEDBACK SYSTEM IS A DISTURBANCE; THE OUTPUT QUANTITY OF THE SYSTEM IS THE REACTION TO THIS DISTURBANCE.

Hello Cyril,

in the past, I was involved in control systems also. And the above sentence reminds me on two major cases we have to distinguish.

* Follow-up control (servo systems),

* Disturbance control (example: AGC).

And, in contrast to your sentence above, it is the task of a disturbance control loop to CANCEL the influence of disturbing signals so that they do not appear (or only with minor influence) at the output.

This kind of "disturbance control loop" must not be confused with the case as described by the quoted sentence above.

A short comment to your transimpedance example:

In case of follower operation the input current source delivers a current of 0 amps, right? In this case, we have 100% feedback and the follower output is -as we know from the classical follower operation - again at the output of the opamp, right ?(because there is no current through the feedback resistor).

Ismat and Lutz, thank you for your reaction to my disturbance-:)

Ismat, your stories are as always interesting and impressive... I will try to solve the problem of the "provocative" professor from your student years. First, assume G1 and G2 are going to infinite (op-amps). Then, applying the "reversal principle", the gain of the active feedback network is 1/H2. Finally, applying the "reversal principle" again, we obtain a whole transfer ratio H2. If H2 < 1, the whole system is an attenuator. This can be considered as an example of a "double reversal principle"-:)

You are right, I purposefully rearrange the inputs and outputs and assign new roles of the classic circuit building block. I do it with the purpose to make students and readers think, to wake their creative thinking and curiosity... since it is rare to see them nowadays...

I like your thoughts about the voltage difference at the op-amp input (the "error signal" or the "static error"). I have named the negative feedback circuit with grounded non-inverting input "zero voltage stabilizer" when we consider (use) the op-amp inverting input as an output; this is just a voltage follower with zero input voltage. This arrangement can be used for creating a virtual ground by copying the real ground. Indeed, it will differ slightly from zero (with the error voltage). It is interesting to see if it is possible to make it exactly equal to zero.

Thank you again, there are more interesting situations around the feedback phenomenon to discuss... Cyril

Lutz, thank you for all the comments; I will gradually answer them...

Maybe we have to define two kinds of disturbances - undesired and desired (useful)? Maybe, in one case we like some disturbance while, in another case, we hate it?

AGC (automatic gain control) is an interesting subject, PLL - also... BTW, is the virtual ground an example of a disturbance control (canceling the voltage at the op-amp inverting input)?

Sorry, but it is more than an hour after the midnight; I will comment the situation with the transimpedance amplifier tomorrow ... Cyril

Yes - I agree with Dr. I. Aldmour. As a general case, the gains in the forward path´s cannot simplified to be infinite.

However, I have another question related to this control loop. And this question touches the loop gain. This sounds as we would have only one single loop gain. Is that true?

I don`t think so - we have THREE different possibilities to open a loop (and to define a loop gain):

1.) Feedback loop remains closed

2.) "Large" loop remains closed

3.) No loop remains closed.

Question: Which of these three loops mostly defines the closed loop behaviour (for example: Phase margin, stability properties) ?

For my opinion, this has nothing to do with the question where the output of the system is defined. And therefore, we must not automatically follow our visual impression that the "large" loop dominates. What is the "large" loop?

I can redraw the system and there is no "large" loop anymore.

As far as I know - a final and general answer for such a multi-loop system cannot be found in books on control systems. But I know only a few of them.

Is there any answer?

Ismat and Lutz,

I would note that we (both teachers and students ) are not computers but human beings; so, besides precise formulae (at the stage of calculating) we need first to see what basic ideas behind circuits are. Computers do not need to understand what they calculate but people need. At the qualitative stage of understanding, we have to forget quantitative limitations and details; they prevent understanding.

That is why, I assumed an extremely high gain above to use (like Lutz in the question below) the simplified Black's formula. This allowed me, looking for the idea in the Ismat's feedback circuit above, to see two times the "reversal principle" - the unique negative feedback property discussed in the question below:

https://www.researchgate.net/post/Can_we_reverse_swap_the_input_and_output_of_any_unidirectional_device_by_applying_a_negative_feedback_If_yes_what_will_be_the_use_of_this_trick?

So, if H2 is a humble voltage divider, the local feedback system "reverses" (converts) it into a reciprocal amplifier, and the global feedback system "reverses" it back to the initial attenuator. The only difference is that the new attenuator is an active attenuator with low output resistance...

Regards, Cyril

"At the qualitative stage of understanding, we have to forget quantitative limitations and details; they prevent understanding."

Ohh Cyril - for my opinion a VERY dangerous claim!

Quantitative limitations and details prevent understanding?

I remember another discussion (in another forum) about the simple common emitter stage and its over-simplified gain formula: A=-Rc/Re.

And what happens if Re=0 or if Re is bypassed by a capacitor?

Such an oversimplification does not enable a better understanding - in contrary: it prevents a good understanding of the amplification principle. If such a simplification is applied it is absolutely necessary to state under which particular conditions it may be used.

And the same applies to all circuits containing amplifiers with feedback.

Shall the beginner believe that the overall gain is always the inverse of the feedback factor? No discrimination between inverting and non-inverting operation?

According to my experience, for a beginner it is necessary to know the "exact" formula (as "exact" as necessary) and to decide if - for a particular case - a simplification is appropriate.

IMO these are two dual cognitive techniques like the inductive and deductive approach...

About the simple common emitter stage and its over-simplified gain formula: A=-Rc/Re... The similar situation is in an op-amp inverting amplifier when R1 = 0...

Now more seriously... The common-emitter stage with emitter resistor, as every electronic device, has a limited operating region where the transfer characteristic ( A=-Rc/Re) is valid. Out of this region, the transistor is saturated (our case) or cutoff and the expression is not valid. A simple equivalent circuit can help us to realize ths situation and to explain it in a human friendly manner (for people, not for computers-:)

We can present the circuit as a network of three resistors (Rc, Rce and Re) in series. One of them (Rce) is variable; the other two (Rc and Re) are constant. The same current (neglecting Ib) flows through the network; so the proportion between the voltage drops across the two constant resistors is the same as the proportion between their resistances. Forced by the negative feedback, the transistor changes its present "resistance" Rce, accordingly the collector current, so that (Re/(Re + Rce + Rc)).VCC = Vin. So, when we begin decreasing Re (at constant Vin), the transistor begins decreasing its Rce to keep the assigned ratio... and when Re extremely decreases the transistor depletes all its available resistance, approaches the zero resistance limit and saturates... and the magic of the negative feedback ceases...

And to observe the tradition to be "provocative"-:), here is another fresh viewpoint at this traditional circuit - it can be thought as of a disturbed (by the resistor Rc) emitter follower. The transistor reacts to this disturbance by changing more vigorously its present resistance Rce and, as a result, transmutes from a follower to an amplifier...

Hi Cyril - perhaps we deviate somehow from the main subject of this thread.

However, one comment to your last example:

Of course, I know and accept that there are different approaches how to teach transistor basics. But I never would use such an equation based on the sum (Re+Rce+Rc). I think, it is very important - in particular for beginners - to realize that the BJT can be seen as a voltage-controlled current source.

Therefore, I never arrive at the simplified expression Rc/Re.

For my opinion, the most important transistor parameter which contains the amplification property cannot be neglected from the beginning. And that is the transconductance g.

Thus the gain expression is -Rc/[(1/g)+Re].

I think it is very important to see under which conditions 1/g can be neglected against Re. You remember: A small quantity may be neglected only if compared with another quantity.

Regards

Lutz

Hi, Lutz! Thank you again for supporting! It is a great idea to think about the possibly simplest things in this world. Unfortunately, there are few people doing it... or having a willingness to do it... I can understand it when talking about professionals... but I cannot understand in the case of teachers...

"The equation based on the sum (Re+Rce+Rc)" is the most universal way of presenting this phenomenon; it is valid for whatever implementation of this negative feedback system (tube, BJT, FET, "manual", op-amp, non-electrical, etc...) Ican't imagine a simplest way of presentation...

"BJT can be seen as a voltage-controlled current source" viewpoint is not valid in this negative feedback configuration; it is valid in the simplest BJT current source where the transistor acts as a constant-current nonlinear resistor:

http://en.wikibooks.org/wiki/Circuit_Idea/Group_65b#A_transistor_acting_as_a_current-stable_resistor

Here, the constant-current transistor is made act as a constant-voltage device (the emitter follower). The negative feedback as though has reverted its behavior; we have already discussed this remarkable property. So, here "a voltage-controlled current source" is the whole NFB circuit (the transistor + the constant emitter resistor), not only the bare transistor...

Interesting transformations... We take a constant-current element (the bare transistor); then we make it act as a constant-voltage device (an emitter follower) and finally, we use its current (i.e., we use it again as a current source)...

Lutz, I continue answering your questions... now about the transimpedance amplifier... You have said above:

"In case of follower operation the input current source delivers a current of 0 amps, right? In this case, we have 100% feedback and the follower output is -as we know from the classical follower operation - again at the output of the opamp, right ?(because there is no current through the feedback resistor)."

Well, let's toy with this so simple but obscure 1-resistor op-amp circuit (for convenience, I have again attached the diagram here). This arrangement resembles the biasing technique with injecting a current into the emitters of a differential pair (the inverting input corresponds to the common emitter junction). Here, we have to distinguish the two possible cases - with a perfect and with an imperfect input current source.

The situation when the current source is perfect (with high internal differential resistance) and it "delivers a current of 0 amps" means that there is nothing connected to the op-amp inverting input. So, as you have said above, if we apply an input voltage to the non-inverting input, both the "output" voltages - the op-amp output and its inverting input, will follow the input voltage. The current source will behave only as an extremely high dynamic resistor that will not interfere with the operation of the voltage follower.

When this perfect current source delivers some current (e.g., "sinks" 1 mA), it creates a voltage drop across the resistor acting as an additive disturbance for the op-amp. It reacts to this intervention by increasing its output voltage to keep the "virtual difference voltage" equal to zero. The same situation can be observed in the long-tailed pair when increasing the emitter biasing current at constant input voltages.

Now, if we apply and vary an input voltage to the non-inverting input, both the "output" voltages - the op-amp output and its inverting input, will follow the input voltage.... but they will differ with the voltage drop across the resistor. In this case, we have two outputs - disturbed (the op-amp output) and undisturbed (the op-amp inverting input). The same arrangement is used in the "shifting" output stages of op-amps (e.g., 709). We can also observe it in the long-tailed pair when increasing both the input voltages (a common mode of operation).

But the situation is quite different if the input current source is imperfect (Vin + Rin). There is no any problem if we vary the input current (Vin) since this imperfect current source is shorted by the virtual ground. But "in case of follower operation", when we vary the input voltage at the non-inverting input, the current and, accordingly the voltage drop across the resistor, will vary as well. Now the imperfect current source interfere the voltage follower operation... and it transmutes into a non-inverting amplifier...

Finally, let's return to the "disturbance as an input" idea. IMO both the input quantities - the input current injected into the inverting input and the input voltage applied to the non-inverting input, are disturbances for the initial elementary negative feedback system. Both they "invade the op-amp privacy" whose main purpose is to keep a zero difference voltage between its inputs (a "differential virtual ground"). So, it reacts to these interventions by changing its output voltage to restore the desired equilibrium...

It is obvious that all these considerations and mental experiments are a prelude to understanding the so interesting differential op-amp circuits...

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

At the end of this comment, I would like to ask some awkward questions:

I wonder if someone else will help us in this noble undertaking - to reveal the truth about circuits... or should we continue alone? Will we continue making our students memorize ready-made "explanations" about all these legendary circuits created with a lot of enthusiasm by beautiful minds... dull, boring and even hateful "explanations" that do not explain anything? Are there circuit designers here, in ResearchGate community... or we have to migrate somewhere else to find them? What is more important for these professionals - to make money for themselves and for their employers or to share their experience with all the people of this world? I wonder...

Cyril

Hi Cyril,

I will respond to two parts of your last contributions:

1.) Quote:"The equation based on the sum (Re+Rce+Rc)" is the most universal way of presenting this phenomenon; it is valid for whatever implementation of this negative feedback system (tube, BJT, FET, "manual", op-amp, non-electrical, etc...) I can't imagine a simplest way of presentation..."

I agree that this seems to be the simplest way of presentation. But - is this the only justification? For my opinion, there are over-simplifications that do not help at all.

I accept, of course, that different approaches in teaching are possible and even desirable. But - in this particular case, I have some doubts.

How do you find numbers for Rce and how do you arrive at a gain expression?

Of course, I know the meaning of Rce - it is the momentary DC resistance that swings around a fixed point. But - does this model help to understand basic transistor functions?

By the way: Also the transconductance g is used for BJT`s as well as FET`s and tubes.

2.) The last part of your last reply.

An answer to this part is somewhat more complicated.

"...to reveal the truth about circuits" - well, this sounds very good and challenging.

But, I think, we must be careful not to complicate things by trying to search more and other interpretations of an electrical circuit or device.

Don`t misunderstand me: To look onto a circuit from two sides (and to find two possible explanations/interpretations) can be very helpful and interesting.

That`s what I did, for example, by explaining some oscillator circuits, which can be seen as a negative-resistance device or as a closed-loop arrangement (one-port or two-port, respectively).

But in the present case (to consider the input signal of an amplifier as a disturbance) I am not yet convinced that it helps to "shed more light" on the system.

Nevertheless, such a task can be a good exercise to let the brain work, you know what I mean? (My english is by far not as good as yours). But that`s not your only intention, am I right?

(By the way: In another part of this forum I have asked for a solution or a comment to the loop gain problem for a multi-loop system. No answer up to now!).

Regards

Lutz

Hi Cyril,

I believe for any feedback system, the input has to be the signal which is followed by an output. This may include a deterministic or random signal. As long as the signal is merged from the external "universe", it should be considered as an input. The point of entry may be subjective or chosen based on a specific view-point, such as investigating the effect of noise at the output. As for the output, ANY signal except the ones from the external universe may be chosen as an output (I am sure people who cheat will hate this because they cannot pretend an "input" from somebody else as their own "output"). Therefore, my answer to your 2nd question is yes.

Cheers,

Masoud