Hmmm... another "provocative" question...

Cyril,

this is not a provocation, it is life :-))

I was wrong about this problem sometimes.

Fellows, 

Why    is   it   " Either /  OR " , 

Perhaps they are " Both "   Trans-Impedance ,

and perhaps  they are " Neither ".  

I think there can be more information forthcoming from Josef. 

(1) Depending on the circuit in which they are designed, 

both  may both be a Voltage controlled 

or Current controlled   ?  

( ET circuit designers are " imaginative & devious " in their methods ).

(2) Both have with high input impedance, and available current output. 

(3) Both appear to be linear

and to be usable with negative feedback. 

I thought at first they might be Trans-Conductance amps ( TCA ). 

However,  

(a) I see NO notation saying that either devices  

are  FET input op-amps that have very low input offset voltages.

( Any small input offset will be amplified into the output energy ).

which indicates both are not VCCS ( not TCA ) ,

(b) I see no "Ibias" inputs, ( so not TCA ) .  

still the apprentice, Glen 

In the Fig. a is used classical operational amplifier with great voltage gain (ideally infinite):

- the input impedances are infinite (voltage inputs);

- the inverting input voltage is created by feedback circuit - an error signal is difference of voltages;

- the output impedance is zero;

- low input impedance of the whole structure provides a feedback resistor – “not the amplifier”;

- the structure is “transimpedance” but not the amplifier itself.

In the Fig. b is used CFA amplifier with really current input:

- the noninverting input impedance is infinite (voltage input);

- the inverting input impedance is zero (current input);

- the inverting input voltage is created by input follower - sometimes named OTA - an error signal is inverting input current IX - converted on the output voltage by means of transimpedance;

- the output impedance is zero;

- the amplifier is “transimpedance” itself.

    The low inverting input impedance (and simple CFA construction) assures (for the same technology) better frequency and dynamic (high slew rate - 1000 V/micros) response of CFA compared to VFA (but worse DC properties of CFA).

The structure in Fig.a can be used to convert the current output of photo multiplier tubes, photo detectors and other types of sensors. Etc.

Some producers then write:

(SBOS313B − AUGUST 2004 − REVISED NOVEMBER 2004)

The OPA381 family of transimpedance amplifiers provides 18MHz of Gain Bandwidth (GBW), with extremely high precision, excellent long-term stability, and very low 1/f noise. The OPA381 features an offset voltage of 25 μV (max), offset drift of 0.1μV/°C (max), and bias current of 3pA. …..

But itself amplifier is still "voltage amplifier" – no transimpedance!!!

The confusion arises from the fact that the components are sometimes named "according to the application", rather than "by internal design" - by circuits principle.

Dear Josef,

I think everyone here should be grateful for this comparative analysis made with such care and understanding. With respect to me, it makes me think back on those "devilish current-mirror circuits":)

Regarding the confusion, I think it should be not a problem since we use a device (voltage amplifier) to build a circuit (transimpedance amplifier). The former is a component; the latter is a system built by this component...

Examples:

A transimpedance amplifier built by a voltage amplifier.

A non-inverting voltage amplifier built by a voltage amplifier.

An inverting voltage amplifier built by a voltage amplifier...

A bit laborious, but accurate :-))

Oh my God, Yosef! Why did you have to mention again this diabolical circuit (CFA)? I was living so quietly... but now my mental equilibrium is disturbed again:)

The problem is that I cannot agree that the inverting input is a current input. I continue to argue that it is a low-impedance but still voltage input!

Look at the CFA input stage. It is just an emitter follower classically biased from the side of the base (the non-inverting input)... and we want to control it from the side of the emitter (the inverting input). This is the well-known common-base configuration...

Now I would like to ask you, "How do we control a common-base stage - by voltage or by current?" Of course - by voltage, if we want to have a high voltage gain. If we drive it by a current (as in cascode circuits), there is no amplification (owing to the transistor) since the same as the input current flows through the collector resistor...

So, my conclusion is that both CFA inputs are voltage-controlled... as in any differential pair... but they have different input impedances. If you keep claiming that the inverting input is current controlled, you must agree that the common-base stage is also current controlled...

The input stage of CFA is just a transistor controlled by voltage both by the sides of the base and the emitter... it is a kind of a unbuffered differential pair - the imperfect prototype of the next perfect differential pair having two identical inputs...

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

Once accelerated, I can not stop to think on what is going on the common emitter point  (the inverting input)...

Well, let's use a system approach to clarify the philosophy behind this situation...

We can think of this stage as of a voltage stabilizer with negative feedback that keeps up the input/output voltage of the common emitter point equal to the input bias voltage (we suppose there is no input source connected to the non-inverting input).

If we apply some voltage to the joined emitters... or inject some current into this point... we actually disturb this servo system... and it reacts to our intervention trying to restore the equilibrium. The change in the collector currents of Q1 and Q2 represents this reaction...

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

Cyril,

I believe that we must apply the "global perspective" (black box point of view). No current (to the inverting input) - no the output voltage. The alone voltage at the inverting input is nothing. The current must flow! From this point of view is the inverting input (of CFA) a current input.

Josef

Actually, when there is no current injected to/drawn from the inverting input, the output voltage is zero since the bias voltage at the non-inverting input is zero (assuming there is no input voltage applied or it is zero).

Then, if we apply zero voltage at the inverting input, no current will flow again... and the output voltage will continue staying zero.

But if we slightly change the applied voltage at the inverting input, the system (emitter follower) will react to this intervention by vigorously changing the collector currents... and some voltage will appear at the output...

So, a difference between the emitter voltage and the applied voltage is necessary so that some current to flow.

Well, let's for example consider the case when we try to "pull up" the common emitter point by injecting a current into this point. As the base voltage is steady, and the emitter voltage tries to raise, the system will react so that to restore the emitter voltage - Q1 will close and Q2 will open more. Accordingly, the Q1's collector current will slightly decrease, and the Q2's collector current will increase...

Quote Josef: "I believe that we must apply the "global perspective" (black box point of view). No current (to the inverting input) - no the output voltage. The alone voltage at the inverting input is nothing. The current must flow! From this point of view is the inverting input (of CFA) a current input."

In principle, I would agree to this "global perspective".

I think, all of us agree that the input transistors of both types of amplifier blocks are - of course - voltage controlled (as mentioned also by Cyril: "I continue to argue that it is a low-impedance but still voltage input! "). However, can this physical aspect justify the view that also a transimpedance amp (CFA or realized via opamp) should be regarded as as "voltage-controlled"? I don`t think so.

I think, the term "transimpedance amplifier" is strictly application-oriented and describes nothing else than the fact that the transfer properties of such a circuit can best be expressed by a voltage-to-current ratio (in accordance with the intended use!) .

Of course, we can measure a signal voltage at the low-resistive inverting CFA input as well as at the opamps inverting input (in transresistance applications).  And we could define a corresponding voltage amplification value. But for which purpose? This definition would not be in accordance with the intended use of the circuit (it is more or less a parasitic property of the circuit).

Fazit: From the application point of view - the common CFA as well as the voltage opamp (if wired as a transimpedance amp) can and should be described as transimpedance amplifiers (in spite of the fact that the input transistors, physically, are voltage controlled devices).

Lutz, please be reasonable... This gadget (when driven from the side of the emitters) is just a humble common-base stage! Will you say that a common-base stage is a transimpedance amplifier?

Cyril,

Are you saying that the humble 741 could be wired 

to become anything that an imaginative engineer could dream ?  

If so, you are adopting the "System Perspective" ...  a good thought ! 

So,

if the 741 could be wired and function as a

CFA, VFA, TransImp, TransCond  circuit ( system ) ,

then it would be named according to the functional circuit . 

To me, it seems that the "system" perspective 

resolves many of the above mentioned problems. 

I think that Josef uses this approach, called "global".  

To me,

it seems that each of our engineers

is adopting a different "basis of comparison"

or some mixture thereof, 

thus generating confusion. 

In comparing various entities,

the "basis of comparison"  must be  

the "same basis" for all entities.  

This is a method rule (logical requirement) 

in statistics and accounting (finances).  

Lutz, please be reasonable... .Will you say that a common-base stage is a transimpedance amplifier?

Cyril, why do you classify my view as not "reasonable"? Didn`t you realize that I strongly support the application-oriented view?

So I repeat: If a circuit with a low-resistive input (of course, "low-resistive" must be defined in comparison to the source resistance) is intended to be used as a current-to-voltage converter it should be called "transimpedance amplifier". This is my recommendation because the exploited transfer properties of the circuit are best described using the term "transimpedance ".

Back to your remark: If it is the purpose of a common-base stage to transfer an input current (input voltage with a source resistance much larger than 1/gm) into a corresponding output voltage (must be bufferd - otherwise it cannot be called "voltage source) - yes, such an arrangement in my view is best described as "transresistance amplifier".

In short: All transistor types are (physically) voltage-controlled devices - however, they can be used - together with other active or passive parts - in electronic circuits having an input-output relationship that can best be described with "current in" and "voltage out". That`s what we call "transresistance". 

Lutz,

your "In short" is very well.

Josef

Here is an interesting observation: The performance of a voltage-opamp with resistive feedback used as a current-to-voltage converter (low input resistance) ) is best if the opamps input resistance (at the inv. inpiut node) is as large as possible. 

Glen,

Thanks for your wise observations about the 741 op-amp and system approach... Your participation is crucial for the success of our discussions... Thank you again!

Lutz,

You are right... and I completely second you! I am sorry that I had to ask this "provocative" question about the nature of the common-base amplifier... but the result is ultimately positive and benefit to all of us is obvious.

So the conclusion is that we need more often to ask such "provocative" questions... and respond in this brilliant way:)

Now about the nature of the topic...

You really helped me enrich my notion about the odd common-base stage with an additional perspective. Let me "philosophize" a little bit because you managed to excite my imagination... and make me go back in the role of optimist and not a pessimist as I (temporally) was above:) By doing this, we probably will gradually uncover the secrets of these bizarre circuit solutions inside OPA, OTA, CFA, etc. ... jealously kept by circuit designers:)

Well, we can drive a transistor from the side of the emitter in two completely different ways - by voltage or by current:

• The first is incorrect since there is a "conflict" between two "fighting" voltage sources - the input voltage source and the output of the emitter follower (another voltage source). However, although it is obviously incorrect, this configuration is widely used - in the classical common-base voltage amplifier, differential pair...
• The second is correct since there is a "cooperation" between two heterogeneous but "helping" sources - the input current source and the output of the emitter follower (a voltage source). However, being correct, this configuration is more rarely used - let's see where and how...

....................

Cyril - please be patient with me, but I do not understand the last two parts of your reply (regarding the common-base stage). Why do you speak about an "emitter follower" - the emitter potential follows which voltage? And what means "the first is incorrect"? What is "incorrect" and why?

Dear Lutz,

When talking about circuits, I am infinitely patient because, for me, this is one of the greatest pleasures in my life... but usually others lose patience to follow my reasoning:)

Well, in the so-called common-base configuration, the emitter voltage follows the base voltage... so this is an emitter follower. But since the base voltage is kept constant, this emitter follower is a voltage stabilizer (with negative feedback)... i.e., a constant voltage source.

Then we connect another (input) voltage source and begin slightly varying its voltage. But the first source does not like to change its voltage:)... and reacts to our intervention by vigorously changing the current... thus trying to restore the previous voltage. So there are two voltage sources in parallel... and each of them tries to set its voltage at the emitter.

Note that, in this configuration, the input voltage source is stronger than the transistor voltage source... and manages to establish itself.

This arrangement is incorrect since we know from Physics and Basic Electricity that it is a bad idea to connect in parallel two voltage sources with different voltages (they are like shorted). By the way, another "bad idea" is to connect in series two current sources with different currents... but this is another circuit story...

Thus I figuratively depicted the operation of the common-base stage when voltage driven. Besides this case, we can see this situation in all sorts of emitter-coupled circuits (long-tailed pair, emitter-coupled logic, emitter-coupled amplifier...)

Lutz

the propertis of the current to voltage converter see below, derivation in Annex, theory in

https://www.researchgate.net/publication/281480600_Theory_of_electronic_circuits?ev=prf_pub

Book Theory of electronic circuits

I thank Joseph for the kindly provided theory... and, if you let me, I will continue with the no less useful and necessary intuitive explanations of the second part of my reply...

Cyril - thank you for the prompt response. However, I must admit that I have problems with your "transistor voltage source". More than that, you speak about negative feedback. Feedback means that a signal is fed back to the input thereby reducing the input signal. So - where does this happen in common base? I really do not understand.

My simple understanding of the common-base configuration is a s follows:

The base node is kept at a constant potential and the emitter node acts as an input for an external signal. Hence, the voltage Vbe is altered resulting in an increase of the emitter current Ie. In contrast to a common emitter configuration, the signal source must deliver the base current as well as the emitter current. This means: The signal source works against a much lower input resistance r(in) which can be calculated as r(in)=1/gm (gm: transconductance).

Yes, I see Lutz... and will explain with pleasure my "negative-feedback viewpoint" at this configuration (although I have already done it).

Strictly speaking, if the input voltage source is absolutely perfect (ideal), really there is no negative feedback... but this is the extreme case... Practically, it always has some internal resistance... and the "transistor voltage source" also has some resistance (re)... so that the two sources are connected through some resistances to the common point (like R1-R2 network in the op-amp inverting amplifier).

I suggest to you to make such a simple experiment with the common-base stage:

https://www.researchgate.net/post/Can_we_see_the_negative_feedback_principle_in_the_operation_of_the_common-base_stage_Can_we_think_of_it_as_of_a_disturbed_common-collector_stage

 Cyril,

You present new ideas. 

This is how I visualize a Common Base Transistor circuit : 

Cyril - I am afraid that I am the only one who does not see (1) that we have an internal "transistor voltage source" and (2) why we have negative signal feedback in the described common-base stage.

Quote, item (3) .....and the emitter voltage will (almost) restore to its previous value.

I don`t think that this observation can be explained with feedback. For feedback, a portion of the output must be back-coupled (with a minus sign) to the input. Is this the case in your example? I rather think, it is simply the exponenetial function Ie=f(Vbe) that keeps the emitter voltage almost to its previous value (0.65...0.7 V).

More than that, it was not clear to me from the beginning if you speak about a configuration with or without Re (see glen`s circuit above).

Dear Lutz,

The "internal transistor source" in a common-base stage consists of the power supply, the bias circuit (a voltage divider) that drives the transistor base, and the very transistor that follows the base voltage at its emitter. As the base voltage is constant, the emitter voltage is constant as well... and the whole combination acts as a constant voltage source... or as a voltage stabilizer, whose terminals are the emitter and ground.

This "voltage source" is loaded with an emitter resistor Re... so the voltage across this resistor is the source's output voltage. It is kept constant by means of the negative feedback (emitter degeneration) as follows.

The output emitter voltage (across the emitter resistor) is subtracted from the constant bias voltage at the base... and the difference drives the base-emitter junction... and accordingly, the collector current, so that to eliminate this difference.

When we try to change (somehow) this output voltage of this system with negative feedback (voltage follower), it reacts to our intervention (disturbance) by changing its (collector) current... we convert this current reaction into voltage one (by the collector resistor) and use it as an output signal.

So, from this "negative feedback viewpoint", the common-base stage is an emitter follower that is disturbed at its output by the input voltage source.

Cyril,

Are you thinking about zero drive into the Re

as  the  'idle'  or   'quiescent'   starting point ? 

Are you then applying a variable  V  to the Re   as the 'disturbance' ? 

Dear Cyril - I think the core of the disagreement between us is your sentence "As the base voltage is constant, the emitter voltage is constant as well"..

Of course, this is true as long as there is no input signal at the base or at the emitter (trivial insight). However, as soon as we "disturb" this quiescent system with a signal at the emitter (direct or via an emitter resistor Re) - and that`s we are speaking of - the emitter voltage will NOT be constant. Otherwise we would not observe a current change at all (because Vbe=const). However, it is a small change only because of the steep slope of the exponential relation Ie=f(Vbe).

I think, to describe this phenomenon, we do not need any "internal transistor voltage source".  What is wrong in my explanation?

Dear Glen and Lutz,

I have attached a practical AC common-base amplifying stage to remove the uncertainties about this circuit...

Regarding the need to see an "internal transistor voltage source" I would say some words to justify it.

Yes Lutz, we can explain this circuit without seeing such a device.. and without introducing the fundamental negative feedback principle... but this means "not to see the forest for the trees". When understanding and explaining a new circuit, I always try to group the particular components into functional blocks... to see already known sub-circuits into the new circuit... and the basic principles on which they are based...

As you have noted, this is a DC voltage source that, when there is no AC input signal applied to the emitter, produces constant voltage (a copy of the bias voltage). Then we come and connect (through a decoupling capacitor) our AC input voltage source to the output of the source (the "quiescent system"). This capacitor quickly charges and equalizes the difference between the zero voltage level of our voltage source and the initial quiescent voltage of the "transistor voltage source". From now on the two sources are connected in parallel...

Please, note the difference with the Glen's textbook circuit diagram - here the input voltage source is connected in parallel, not in series to Re. This is the classical "voltage-controlled common-base stage"... while the Glen's digram represents the "current-controlled common-base stage"... which is the object of our lively discussion here... and which we have yet to discuss...

Now I want to digress a little from the topic to diversify the discussion...

A well known paradox is that poor people are more generous than the rich. So that I have only "naked ideas"... and generously give them away to people:) Here is another powerful idea that could serve you in understanding transistor circuits...

My eccentric idea is that the serial negative feedback (emitter degeneration) is an inborn trait of the transistor since its input and output are internally joined to the common emitter terminal. So, the "naked transistor" can be considered as an elementary negative feedback "system" consisting of two parts - a comparator C and regulator R (see the attached picture). 

As a result of this connection, the transistor always tries to make its emitter voltage equal to the base voltage... even when it is impossible.

Here are some examples:

• In the common-collector stage (emitter follower), the transistor successfully changes its output emitter voltage when the input base voltage changes
• In the common-emitter stage, the transistor tries to change the emitter voltage... but failed... since the emitter is firmly fixed to ground
• In the common-base stage (our case), the transistor tries to keep unchanged the emitter voltage when the input voltage source "brutally" changes it... and succeedes or fails depending on the internal resistance of the input voltage source

Cyril - you may remember, that I have asked you for patience. Now - based on the diagram you have provided - I can imagine what you mean ("two voltage sources in parallel").  But - is that really true? Two voltage sources in parallel? I am afraid, the whole thing was a big misunderstanding. Let me explain:

1.) Let me remind you on the classical opamp adder: Two voltage sources are combined using two resistors which meet at the common opamp input. I am sure you know what I mean. Now - are we speaking of two parallel voltage sources (which would be forbidden)? No - we realize that both sources drive corresponding currents through the resistors - resulting in a certain voltage at the common node wher both currents meet (representing the sum of both parts).

2.) Now - look at the emitter node in your diagram. The DC current IE creates a certain DC voltage VE at the emitter node - and the signal AC voltage drives another current (AC) into this common node thereby causing a superposition of the two currents (DC and AC).  As a result, we have a DC voltage with an additonal AC voltage at the emitter node. That`s exactly what we want to see.

But where are two voltage sources in parallel?  We just have added two currents in a common node according to Kirchhoffs current law.

EDIT (PS): I think, one of the reasons for the misunderstanding was the fact that you spoke about an "internal transistor voltage source" - but, in fact, you had the complete circuit (including RE) in mind.  And the low-resistive output at the emitter node did you consider as "voltage source", correct?

Dear Lutz,

Now I ask you to be patient since I will be absent till Sunday:)

Cyril

Each amplification is always a gentle (more or less) sweep of a quiescent working point only. If the AC part is small enough, we can use small signal models - and superposition theorem too.

The signal source (AC - for the CB connection) must be able to add "whole" emitter current.

The signal source (AC - for the CE connection) must be able to add base current only (which is much smaller than the emitter current, ideally zero) - the same voltage B-E.

From this is evident that input impedance of the CB connection is small, while the input impedance of the CE connection is great (ideally infinite).

Hi Josef - your first sentence (..."sweep of a quiescent working point..) leads me to another question - away from the main subject of our discussion.

Question: Can a "quiescent point" be swept? Is it then still a quiescent point? Or is it more correct to say that we have a signal that swings around the quiescent point (which, of course, remains fixed) because - in my view - the Q point is determined by DC quantities only.

I would agree that this question seems to be rather "academical" - on the other hand, I remember that Prof. Erik Lindberg very often has used the term "moving quiescent point". And, sometimes, this expression has caused some disagreement and/or confusion on my side.

And anyway, CB involvement can be considered a non-inverting transimpedance amplifier. Its output resistance is, however RK :-))

See Annex

Lutz

It is a philosophical question directly. It would be many things emerged from the graphical solution. But it is no longer "fashionable".

Josef

"It is a philosophical question"

Josef - yes, I agree. But does this mean it cannot be answered?

I am going to think about it :-)

Lutz

I sketched a little. I think it is really possible to say:

"Each amplification is always a gentle (more or less) sweep of a quiescent working point only".

We can use a DC +AC source (of UBE) or an AC component to superimpose on the DC component (via capacitor mostly).

Josef

Josef, perhaps this question is not worth to be discussed further - nevertheless, I see a contradiction in the term "sweeping the quiescent point". What then is the meaning of "quiescent" ?  The excursions in your drawing (green colour) are not "quiescent".

To me - we are using DC voltages/currents to fix a point that remains "quiescent" - and a superimposed signal causes periodic excursions around this point. Don`t you think that this is a much more better description of what really happens?

Hi Lutz.

It seems you are right.

Yes, if we will use the term "quiescent" only for the situation "no signal", then:

"Each amplification is always a gentle (more or less) sweep of a current (voltage) around the quiescent working point only".

Josef

Any amplifier circuit that converts current input into voltage output is a transimpedance amplifier with gain in ohm.

Josef - thank you. That`s logical.

Hi everybody! I am already here... and what my eyes see... again a very interesting and in-depth discussion about my favorite topic - biasing! I think that only "academic people" involved in the field of technical education can penetrate so deeply into the matter ... for others it is, most frequently, just a waste of time...

Well, then let me temporarily deviate from so interesting topic about the transimpedance amplifier... and try to generalize, once and for all, what biasing is:

• Biasing is a summation of two input quantites - variable input (AC) and constant biasing (DC). As a result, a third quantity is obtained - the sum of both (AC + DC)
• This summation is implemented by a summer - serial (according to KVL) or parallel (according to KCL)

So, in whatever circuit that is somehow biased, we should always be able to see some sort of a summer.

Another problem is whether we can see it:)

To defend my thesis, please offer me various biased circuits... and I will show what is the summer there...  and what kind it is...

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

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

Regarding the Erik's "moving quiescent point" ("moving something immovable":), the explanation is extremely simple - he talks about the resulting quantity (AC + DC) at the summer output... the biased quantity...

So, to avoid ambiguity, we must clearly distinguish between the three quantites - input (AC), biasing (DC) and biased (AC + DC).

Biasing is a summation of two input quantites - variable input (AC) and constant biasing (DC). As a result, a third quantity is obtained - the sum of both (AC + DC)

Hi Cyril - welcome back.

May I ask you: Is the above YOUR definition or do you think it is based on some kind of common understanding? I am sure, it is no surprise for you that I cannot follow this philosophy: What has an ac input signal to do with "biasing"? To me the term "biasing" is exclusively connected with a DC determined bias point (I repeat: POINT!).

Hi Lutz, it is always interesting with you:)

The written above is my definition based on the common sense...

For itself, the biasing implies only the DC ingredient - we can bias the device (transistor, op-amp, etc.) without applying an input AC signal... But this "only biasing operation" does not make sense becausе, in this case, this "biasing" quantity wil be not biasing but simply a DC input quantity...

The biasing only makes sense as a prerequisite for an AC input signal...

So, even in this particular case, we need a summer... but with a zero AC ingredient at its second input... and when we want, we apply the AC input signal at this input... Thus this arrangement always includes a summer.

The biasing only makes sense as a prerequisite for an AC input signal...

Cyril - we should not mix different aspects. I did not ask WHY we perform biasing or under which conditions it would make sense.

The question is simply: Do the electrical quantities which are connected with the term "biasing" are DC values only - yes or no?

Again: According to my understanding (and, as far as I know, also in accordance with common understanding) we bias an electrical circuit because we want to select one particular operating point on a non-linear input-output characteristic. And because this is a time-independent "point" (mostly a certain voltage-current pair) we are using only DC quantities for this purpose. 

Of course, we do this because this well-biased circuit will be used for some specific operations (amplification, waveform shaping, ...) - as a "prerequisite" (as you say). However, using this biased circuitry to process any signal (voltage, current, temperature,...) is another step and cannot  be seen as part of biasing.

Indeed, there is a special and unlikely case when the input voltage source is "internally biased", i.e., it wiggles its output voltage around a DC level. For example, my function generator (an instrument) has an adjustment of the DC level... so I can connect it directly to a transistor base...

Lutz,

I will repeat it again... The bias quantity is an initial "accessary quantity" that implies another "true input quantity"... for which it is made...

The "only-biasing quantity" is just a humble input quantity. If you simply applied it to the amp input, how will you later apply the input quantity to this input? For example, if it is a perfect voltage source, you cannot directly connect it to the amp input... you can do it through a summer.

Cyril, in order to avoid misundestandings - I did not argue against using a summing circuitry, how could I?

My only point was your sentence "Biasing is a summation of two input quantites - variable input (AC) and constant biasing (DC)."

I think, the process of "summation" must not be confused with "biasing". Biasing is - as you wrote - a kind of "prerequisite" (a precondition) to allow desired operation of the whole circuit. Therefore, biasing is the result of DC quantities only!

Consider the classical opamp. Using negative feedback the opamp is well biased at a good bias point in the linear region. Then, it CAN be used to amplify an input signal. However, even without any input signal it is still a well-biased unit.. 

Lutz, I agree with you that, strictly speaking, the biasing is only the  DC part of the whole AC + DC signal applied to the amp input. But I think our mission here is not to write strict definitions as for an encyclopedia; it is rather to understand what actually stays behind this term.

Maybe we need some more general word for the whole operation (e.g., "summing") that combines the two sub-operations - "biasing" and "wiggling". But I think the key point here is the summing circuit ... and therefore it is interesting to see it in various applications.

The op-amp biasing system is an interesting topic to discuss. There the input differential pair is biased by the side of emitters by sicking/injecting current from/into the common junction... while the current caused by the input voltage is imposed on the initial bias current.

So, the input quantity (base voltage) is not directly summed with the biasing quantiry (emitter current). Instead, the collector current caused by it, is summed with the bias current.

https://www.researchgate.net/post/Can_we_reveal_the_brilliant_ideas_behind_the_741_op-amp_circuit_solution_of_genius

Cyril,

I must be on Lutz's side here,

although writing with different words,

which come from my different world. 

I hope you and Lutz can come to agreement, 

and I can better struggle to follow the original question/solution. 

Yes Glen, I agree that there is such a need...

Regarding the usage of "bias", although I am not a native English speaker, I think that still there is a difference between "biasing" and "biased". I repeat it again, terms are not the most important thing here... the most important is to understand what lies behind them. But you see how difficult it is implemented in our discussions... what efforts cost me to bring the discussions to that direction... and how reluctantly my detailed explanations are met...

Regarding your "picture" above, I would add that " biasing (DC) " is also an input, most frequently - internal. In this connection, here is another my insight about this arrangement:

Amplifiers are actually 2-input summers.

Cyril,

Agreed :  "Amplifiers are actually 2-input summers."  

   Interesting clarification. 

Agreed :  "biasing (DC)" is more complex than I stated. 

But, Those clear ideas were not the point of my post.

Is the discussing about variations of bias in relation to VFA / CFA  , etc ? 

Hi Glen - I must admit that I am "guilty". I don`t think that the question of bias has a direct relation to to the question VFA/CFA. Two days ago I have replied to Josef`s sentence "Each amplification is always a gentle (more or less) sweep of a quiescent working point only".

I was not happy with the expression "sweeping of a quiescent point". I think this has been clarified meanwhile between Josef and myself.

However, at this moment - and with respect to your last contributions - I think, there is no fundamental difference between (bias point" and "quiescent point", right?

Cyril - I do not recommend to continue the discussion of bias methods and alternatives within this thread. However, from your contribution one can derive that you consider the opamp as "current - controlled", is this correct?

If you think, this could/should be dicussed in more detail, we could open a new thread?

Quotes:

"the most important is to understand".

"the current caused by the input voltage is imposed on the initial bias current."

"Cyril - I do not recommend to continue the discussion of bias methods and alternatives within this thread." 

Why not Lutz ... if we have something more to say? But maybe we should limit to these aspects that are related to the Joesef's question... and to redirect the common bias questions to my question about biasing (however I do not want to abuse the hospitality of Josef here... I feel that it is running out:)

Glen,

I admit honestly that whether there is a connection between the biasing idea and these gadgets named VFA, CFA and OTA, is the last thing that would interest me here. The most valuable for me is to get an interesting and frank discussion, in which creatively thinking (and well-minded) participants generate chains of powerful ideas... and each of them encourages the other... thus accelerating its creativity... and if once this miracle happened, to squeeze everything out of it... But as you can see, always some wise guy will break it at its most interesting part... and the miracle vanishes...

Instead of giving an answer to your question, I suggest yourself to establish whether there is such a connection by answering a few questions:

Is there a voltage summer in the circuit of CFA below that sums the input and bias voltage? If so, what is it? How does it sum the voltages - in a serial or parallel manner? What are here the "biasing" and "biased" (i.e., your "operative" ) quantities?

"Why not Lutz ... if we have something more to say?"

OK - why not? And your expexted answer to my last question has a certain relation to the problem VFA/CFA. Therefore, again my question:

Does your statement ("the current caused by the input voltage is imposed on the initial bias current.") mean that you consider the opamp as a current-controlled unit?

Sorry Lutz, I have not forgotten about all your questions... and think over them... and intend to answer them...

It is an interesting question, which brings us back to the topic about the transimpedance amplifier closely related to the Josef's question... Here are my first observations on this issue (for simplicity, I mean here an ordinary long-tailed pair with emitter current source):

• Regarding the base inputs (for the input voltage sources), the op-amp is, of course, voltage controlled
• Regarding the emitter input (for the bias current source), it seems it is current controlled

Here is another my insight about this differential arrangement:

Differential amplifiers are actually 3-input devices.

I believe that the CFA (transimpedance) rather uses a voltage follower and summed together the currents in the current input.

But I think our mission here is not to write strict definitions as for an encyclopedia; it is rather to understand what actually stays behind this term.

Yes - I support this sentence by 100%.

Therefore, my comment to one of your last statements ("Regarding the emitter input for the bias current source, it seems is current controlled"):

To me, it is always the voltage between base and emitter (VBE=VB-VE) which controls the collector currents. This holds, of course, also for the classical long-tailed pair.

In detail: A  voltage change at the left transistor base causes a voltage change at the common emitter node which - at the same time - is the input  voltage for the most right transistor (constant base voltage).  Hence, both collector currents change (different directions).

Cyril I do not think that we can use a term the emitter - input for this amplifier (case).

We'd already investigating analog multipliers :-))

See for example

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?ev=prf_pub

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

Lutz,

I was talking above about the "bias adjustment" when we change the current of the bias current source. This is the third operating mode besides the other two - differential and common.

In this situation, when the emitter current changes, the transistors cooperate and simultaneously change their base-emitter voltages so that to adjust the sum of their emitter currents equal to the whole bias current. So, the situation is as follows: the emitter current changes -> the base-emitter voltages change -> the partial collector (emitter) currents change. You may decide what a control this is - current or voltage. 

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

https://en.wikipedia.org/w/index.php?title=Operational_amplifier&oldid=436147005#Biasing_circuit

Cyril, yes - I agree. We have to discriminate between the "naked" transistors (only VBE matters) and the circuitry consisting of the two main transistors and other elements (like a current  source). For the whole circuit, we can speak of current-biasing because that`s what we can determine (the current in the common emitter leg).

OK - misunderstandings solved.

But now we have another problem ( because Josef has brought the multiplier cell into the discussion):

When we consider the transistors T1 and T2 - as seen from the emitters - as current-controlled, which signals then are added (using the adder model as mentioned by Cyril and Josef) ? Voltage signals at the base nodes and currents from the other sides? 

Here is how the discussion begins to accelerate to unprecedented rates:)

I thought to start this topic... but I did not dare because there are great specialists in this field... and whatever I would say, it will seem silly:(

But I might try... whatever happens...

Lutz,

"""there is no fundamental difference between (bias point" and "quiescent point", right?"""

Right !   by the old usage of these terms. 

Well... let's begin talking about the long-tailed pair acting as a multiplier. I will try to reveal the idea of this application in a simple and intuitive way.

It is interesting that the legendary AD 7520 digital-to-analog converter, when used in bipolar applications (2-quadrant multiplier)... and a popular in the past trick when weighting by mechanical scales... are based on the same idea. Let's see what it is...

In this differential arrangement, we get an initial reference quantity... divide it into two equal parts... and subtract them from each other. Thus the inital result (output quantity) will be zero.

Then we begin moving entities from the one part to the other. As a result, the result will change from zero to positive or negative direction reaching the maximum - the reference value.

Thus the reference value determines the steepness of the transfer characteristic and the operating range of these bipolar devices... and the output value is proportional to the product of the reference value and the input value... so such devices are named multiplying...

In a long-tailed pair with emitter current source, the bias emitter current serves as the initial reference quantity... It is divided into two partial collector (emitter) currents (initially equal)... and they are subtracted from each other. Thus the inital result (output voltage) is zero.

When we begin changing the input voltages in a differential manner, the one partial current will increase... and the other - decrease... while the common bias current will stay unchanged. Or, figuratively speaking, the current is redirected (steered) from the one leg to the other.

In the ordinary long-tailed pair, the bias emitter current is set constant while in the analog multipiers, it is used as a second input quantity. Our task here is to see its role as a steady bias quantity...

I think that the fact that there is a multiplication here, is not of interest for us. We see only the initial quiescent state of the system... and begin varying differentially the input voltages. Thus we will add or subtract current from the initial bias value... will increase or decrease it...

So, I think at first glance, it seems we have again the same biasing arrangement where the input signal is imposed over (summed with) the bias quantity... the input signal current is summed with the bias current... there is a current summation...

"We see only the initial quiescent state of the system... and begin varying differentially the input voltages. Thus we will add or subtract current from the initial bias value"

Cyril, this sounds as if you are going to subtract a current from a voltage.

No Lutz... I have said it...

First we convert the input voltages into partional collector currents... and the bias voltage into a bias emitter current...  by transistors acting as transconductance amplifiers... then sum the currents... and finally convert their difference into an output collector voltage...

Cyril - again, please be patient with me, but it is not yet clear to me ( "then sum the currents"). Please be detailed - which currents are added ? Emitter and collector currents?

On the other hand, I think, we have clarified this point already.

Let me summarize again (diff. amplifier with BJT`s):

(1) Considering the whole circuit with a sufficient large emitter resistor RE we CAN TREAT (we can, but we are not forced to do it) the BJT as if it would be driven from the emitter side with a current (current-controlled). (But this view cannot be used when we try to apply the summing model.)

(2) However, if we try to use the summing model - as proposed by you some days ago - we have no other choice than to add/subtract two VOLTAGES. And this is very simple because the BJT itself is already a diffferential amplifier (reacting upon the difference VB-VE=VBE).  Hence, any change in VB or VE or VB and VE at the same time will influence the collector current(s).

"Please be detailed - which currents are added?" Well, the situation is as follows:

The whole bias emitter current stays unchanged as it is set by a constant current source. This current is divided into two partial emitter (collector) currents. These half bias currents are equal when the input base voltages are equal (zero differential input signal)... so the whole differential system is fully symmetric.

When we begin varying the input differential signal, an amount of the one partial current "moves" (is steered) to the other partial current, i.e., adds to it. Simultaneously, this amount of current subtracts from the first partial current.

So, if we observe only the one collector, we will see an increase/decrease of the collector current above/below the quiescent half value set by the emitter current source.

My idea is that in this differential current steering system, actually the signal current is not added/subtracted to the bias current... it is taken from one place and is given to the other place... it is "moved" from the one to the other leg.

(I want to ask you to be more indulgent with me, because these thoughts are still new to me... and they occurr to me at the time... i.e., they are not yet established and may have been wrong...)

Let me now return to the so interesting topic about the transimpedance...

Sorry that I had forgotten that the common-base stage is a transimpedance amplifier:) I have even asked such a question about the similarity between the transistor common-base and the op-amp inverting amplifier where I have noted that both devices have a middle point with constant voltage (virtual ground). Here is my insights about the similarities and differences between the operation of these circuits.

Both they are circuits with voltage type negative feedback that makes them keep up a constant voltage (virtual ground) at the middle point. So, when we begin to draw (more or less|) current from the left side of this point, its voltage begins decreasing... and they should react to our intervention to restore the equilibrium - the previous voltage of this point.

Regarding the voltages, they do it in two different ways:

• In the common-base stage, the transisor decreases its "resistance" connected in series; as a result the voltage of the middle point  (the emitter) increases and restores to its previous value
• In the op-amp inverting amplifier, the op-amp increases the magnitude of its output voltage; as a result the voltage of the middle point (the inverting input) increases and restores to its previous value

Regarding the currents, both circuits adjust the currents flowing through their right parts, equal to the input current. To see how, imagine that their right parts consist of two "things" in series - resistance and voltage... that constitutes "current sources":

• In the common-base stage, the resistance of the "current source" is changed
• In the op-amp inverting amplifier, the voltage of the "current source" is changed

https://www.researchgate.net/post/Is_there_any_relation_between_the_common-base_amplifier_and_op-amp_inverting_amplifier

(I have edited my previous post at the end of Page 8)

Looking at both circuit diagrams, I try to figure out why the "CB involvement can be considered as a non-inverting transimpedance amplifier"... as Josef claims at Page 4. What I see is that both input voltages are positive... both input currents are "sucked " from the inputs... and both the output voltages are positive... We draw currents from both circuits... and they react by producing positive output voltages... Then why the one is non-inverting and the other - inverting?

Maybe we should take into account the direction of changes instead the polaritiy of the signals? If so, when we increase (draw more) current, the collector voltage changes down... while the op-amp output voltage changes up...

Maybe we should take into account the direction of changes instead the polaritiy of the signals? If so, when we increase (draw more) current, the collector voltage changes down... while the op-amp output voltage changes up...

Cyril - according to my understanding it is a commonly used method to "take into account the direction of changes". That is the only reason for the gain of a common-emitter stage to be defined as "negative".

Regarding the comparison of the two gain stages (common-base vs. inverting opamp): Yes - the visual similarities are obviuous. However, the opamp output is a voltage and the BJT output is a current (which is defined as positive if it goes into the collector). And - a positive opamp output gives a current out of the device (in contrast to the BJT).

Lutz,

"The BJT output is a current" but CB output is a voltage (the current converted by the collector resistor into a voltage)...

Cyril

But, output resistance is only resistance in the collector.

Josef

Yes, Josef... I have not forgotten your remark about this fact. I agree with you that this is a significant difference between these circuits... and it is interesting to see why this is so... I know we know why:) but it could be interesting for young readers... although, as you already said, such topic is not so "fashionable" nowadays...

Well, let's try to explain it in a more colorful way to make it more interesting (compared to sterile textbook explanations)... hoping that, when reading it, our imaginary students will not yawn as during a classic PP lecture:)

The problem is how to convert the current into voltage... Here are three ways:

1. First, we can simply pass the input current through a resistor... and take the voltage drop across the resistor as an output. The output resistance of this possibly simplest passive current-to-voltage converter is determined only by the resistor... and only imagine what will happen if connect a low-resistive load.

2. Then, we can draw the input current from the emitter of a transistor with fixed base voltage (i.e., common-base configuration). In this more sophisticated arrangement, we make the transistor adjust its base-emiter voltage so that to make its collector current (almost) equal to the input current... and then convert this collector current into voltage by the same humble (collector) resistor. So, again the output resistance of this CB stage is determined only by the (collector) resistor...and a low-resistive load will cause the same trouble as above. Then why to do it in this way? Maybe because the input current source will see (almost) zero load resistance?

3. After, to be in fashion:) we can include the ubiquitous op-amp in this "game". So, we can make it copy the voltage drop across the resistor... and then use this inverted copy instead the original voltage. Now the humble passive current-to-voltage converter is already buffered... and this way it is transformed into an active current-to-voltage converter... that can be loaded with a relatively low-resistive load. In addition, the input current source sees zero load resistance (virtual ground).

These were three solutions with increasing complexity ("only resistor" -> "resistor + transistor" -> "resistor + op-amp")... but note that all they contain in themselves, as a "germ", the humble Ohm's resistor acting as a passive current-to-voltage converter.

Hmm, as usual, when I begin to develop my most interesting ideas, the discussion dies... instead to heat up even more... An interesting psychological phenomenon, which makes me think again about the secrets of human soul...

But our favorites is likely to fall asleep sweetly (such as during a dull PP presentation in a semi-dark classroom:)... so we need to attract their attention with something new... What could it be? Aha, I guessed:

4. We can combain 2 and 3 by attaching the modern op-amp to the old common-base stage. For this purpose, we connect its inverting input to the transistor collector and use the op-amp output as a varying power supply for the common-base stage... and, of course, as a buffered output. Thus we will invent a new circuit topology that can name with the pompous name "improved common-based stage":)

.....so we need to attract their attention with something new... What could it be?

Perrhaps the question if it is really true (physically) that "Ohm's resistor is acting as a passive current-to-voltage converter." (quote Cyril) ?

Can a resistor really convert a current into a voltage? (I remember we had a similar discussion already). How does this work?

Yes Lutz... we have discussed this topic many times...

My last understanding of this situation is that the current creates an (output) voltage drop across the resistor... with the caveat that this current was created by another (input) voltage...

So, in this arrangement, there are in total three "cascaded" quantities (voltage -> current -> voltage)... where the current serves as an intermediate quantity... In other words, the voltage creates a voltage by means of the current...

If we see only the second and third quantity, we can conclude that the current creates the voltage...

I guess these my thoughts will wake up our favorites from their sweet dream:) ... and we can continue to consider our main issue...

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

"If we see only the second and third quantity, we can conclude that the current creates the voltage..."

Yes - that`s the common view and it works. But - is it the physical truth? If the current within the resistive body consists of moving charged carriers under the force of the existing E-field, this current cannot be the cause of the field (resp. the corresponding voltage).  

Hmm... you perfectly know what I will ask now, "Then what the hell is this current? Or it is just a side product like the base current of the BJT?"

My notion is that the current serves only as an energy carrier... as a kind of a "transmission" connecting two points - initial and final... It does not create the voltage at the final point... it only carries the energy to this place that creates the voltage...

But these are only intuitive guesses... Frankly, the physical side of the phenomena does not interest me as much as the fundamental circuit ideas that can be easily figured out by means of various analogies and situations from our human routine...

I am not sure if this topic is interesting for our favorites... rather I have the feeling they are bored... and will again start dozing off:)

So I suggest again to awaken their interest with the next clever question:

5. Why do not buffer the common-base output by a non-inverting (emitter, op-amp...) follower instead by an inverting follower (transimpedance amplifier)?

(The thesis that a transimpedance amplifier is an inverting follower will have an additional refreshing effect... and will further encourage their thinking:)

Josef,

While you meditate on my questions above:) I would like to discuss your circuit diagram representing a common-base stage with a bias current source in the emitter and an AC input current source...

First, I think there is no much use to include a constant current source in the emitter of the CB stage since the emitter voltage is fixed... and a simple resistor would work well... This would make sense in a case of the common mode of a differential pair... where the emitter voltage varies...but here?

Moreover, the low input resistance of the CB stage shunts this current source (they are in parallel)... and the input current source sees low resistance... so the bias current soutrce can be replaced again by an ohmic resistor...

Then why to use a bias current source instead a bias resistor? 

Then, I ask you again, what is the point of using an input current source instead a voltage source?

In this case, the transistor is somehow "unused"... since it only conveys the input current  to the collector resistor... and it does not amplify... The effect of this arrangement is just as much as if the transistor was not there... and the input current source was passing its current directly through the collector resistor...

Then why to use an input current source instead a voltage source?

Of course, one reason to drive the CB stage by a current source, can be to beat the Early effect (of the input transistor) ... as in the case of the cascode configuration...

Also, I wonder about the role of the input decoupling capacitor... Is far as I know, it should decouple (isolate) the biasing circuit from the input source.

But here they, being current sources in parallel, are decoupled by the fixed emitter voltage... Then is this capacitor needed?

I use the current source to get the big "RE" - ideally infinite. See attachement once more.

I see Josef... and really "RE" is almost infinite... but I am not sure if we need it here...

We will need such a high dynamic resistance to keep the current constant if the (emitter) voltage across this resistance varies... but it is relatively constant here... So, in this case, a constant ohmic resistor is enough to keep the current constant...

Cyril - just a small remark regarding your answer 1 day ago (cascode): I suppose you have mixed-up the Early effect and the Miller effect, correct?

Lutz,

In my opinion, both effects are beated in the cascode configuration - the Early effect in the lower transistor (fixed collector voltage), and the Miller effect in the upper transistor (no capacitive negative feedback between the collector and the base).

Once we began talking about the cascode configuration, let's say some words about what it is...

In this arrangement, two heterogeneous "controlled sources" are connected to each other... or two converters (amplifiers) are cascaded:

• a voltage-controlled current source (a transconductance amplifier) and
• a current-controlled voltage source (a transimpedance  amplifier)

A cascode configuration consists of cascaded transconductance and transimpedance amplifiers.

Now let me share my last insight about the odd cascode configuration. As usual, it is based again on the ubiquitous negative feedback principle...

From this viewpoint, in this configuration, we can see two nested circuits with serial negative feedback (emitter degeneration). The first of them sets the current, and the second sets the voltage of the common point (between the collector of the lower transistor and the emitter of the upper transistor). Look again at the nice Josef's picture to see them...

The lower transistor acts as a circuit with serial negative feedback - emitter follower, which keeps a constant voltage across a constant emitter resistor. As a result, it keeps a constant emitter (collector) current... and it acts as a constant current source with negative feedback (current stabilizer).

This circuit with serial negative feedback is inserted in the emitter of the same circuit with serial negative feedback - another emitter follower implemented by the upper transistor. Like its lower "brother":), it tries to keep a constant voltage at its emitter (voltage stabilizer). As a result, it adjusts its emitter (collector) current equal to the collector current of the lower transistor (current follower).

So, let's repeat it again: in the cascode configuration, there are two interacting NFB systems - the lower sets the current, and the upper sets the voltage of the common point.

Each of them provides ideal load conditions for the other: the upper provides zero load resistance for the lower; the lower provides infinite load resistance for the upper.