Common Base Amplifier Confusion

My answer to SE EE question Common Base Amplifier Confusion

This story is another case of rude administration in StackExchange EE. I tried to reveal in detail the operation of the legendary common-base stage by illustrating its operation with the help of voltage bars and current loops. But this did not become easy and required multiple edits to reach perfection. For some reason incomprehensible to me, however, the moderator did not like it and at one point he locked my answer. The problem is that I improved Fig. 2 as I drew the paths of charging currents and I want to replace the figure. Also, I found Cyrillic letters in the captions under some figures and I want to correct them. I asked the moderator to unlock my answer (see the last comments below)... but so far I have not received an answer...

Answer

Common emitter and common base are formal terms; common collector is even meaningless. For the last time, I came to this conclusion during yesterday's laboratory exercise dedicated to these basic transistor circuits.

Looking for a meaningful classification

Transistor setup

To explain to my students what lies behind these names, I used my favorite transistor setup - Fig. 1.

Fig. 1. A setup for investigating basic transistor stages.

It was extremely simple - two voltages (Vb and Ve) were produced by 1 k potentiometers (P1 and P2) and applied to the base and emitter of the transistor... and a LED was connected in series to the collector resistor Rc. The voltages were measured by voltmeters (V1 and V2) and the collector current (transistor state) was indicated by the LED. In addition, in the colorful circuit diagram drawn on the blackboard, the voltages were represented by voltage bars with proportional height (in red)... and the currents - by current loops with related thickness. To see what makes sense in the meaningless classification, we conducted a few experiments varying the one, other or both voltages.

Experiments

1. Base-driven (common-emitter) stage. We set P2 wiper at zero position (Ve =0) and began carefully moving the P1 wiper from the zero position (Vb was increasing). From one moment on (Vb = 0.6 V), the LED gradually started to light up (aha ... the transistor started to conduct current in active mode). We even wiggled P1 slider in a sinusoidal manner around these 0.6 V (bias voltage). Eventually the LED began to glow constantly (clearly, the transistor was "saturated").

Then we moved the P2 wiper at some nonzero position (Ve = const) and repeated the experiments above...

2. Emitter-driven (common-base) stage. Now we swapped the roles - set P1 wiper at some nonzero position (Vb = const) and began carefully moving the P2 wiper around Vb - 0.6 V. Now the LED (collector current) was behaving in the opposite way... but the collector voltage in the same way (OP's question). It was amazing that Vc did not reach the ground...

3. Base-driven with negative feedback (common-collector) stage. Now we disconnected the upper P2 end from Vcc and the transistor began itself producing Ve by passing its collector current through the lower part of P2. The students were impressed with how the transistor itself was "moving" the voltage of its emitter so that it would follow Vb.

4. Base-emitter driven (differential) stage. Finally, we decided to vary both input voltages - in an opposite direction (differential mode) or in the same direction (common mode). In this way, my students got an idea of how the simplest comparator and differential amplifier can be made.

Comparison

So, what is the difference between the two classifications?

Actually, there is always only one input voltage Vbe that is applied between the base and emitter (across the base-emitter junction). In most applications, it is floating... but, as a rule, in circuits, input voltages should be grounded (single ended). That is why, for greater versatility and flexibility, we form Vbe as a difference between two input voltages (Vbe = Vb - Ve)... and then vary the one, other or both. We have two ways to name these configurations:

In the classic centuries old classification, we give the name of the configuration according to which of the transistor terminals is with a fixed voltage (AC grounded, passive, neutral). But it is the less important terminal, especially in the case of the so-called "common collector" (the bare fact that the collector is AC grounded does not mean anything since the input voltage is not applied between the base and collector). In the differential stage there is no such AC grounded terminal at all.

In the suggested (just fabricated) classification, we give the name of the configuration according to which of the transistor terminals is driven by the input voltage (active)... implicitly assuming that the other terminal is with a fixed voltage (AC grounded).

Making a connection

Now, when my students ask me, "What is actually the common-emitter stage?", I will answer them: "This is a stage where we drive the transistor from the side of the base while keeping the emitter voltage constant"...

... common-base? -> "...driving the transistor from the side of the emitter while keeping the base voltage constant"...

... common-collector? -> "...driving the transistor from the side of the base while it changes its emitter voltage so that to keep it equal to the base voltage"...

... differential amplifier? -> "...driving the transistor both from the side of the base and emitter"...

Why CB stage does not invert

Short explanation

Think of the output circuit as of a voltage divider of two resistors - constant (Rc) and voltage-controlled variable "resistor" (Rce).

CB stage: When we increase Ve, Vbe decreases (since Vb = const) -> Rce increases -> Ic decreases -> Vce increases. So, CB stage does not invert the input voltage.

CE stage: When we increase Ve, Vbe increases (since Ve = const) -> Rce decreases -> Ic increases -> Vce decreases. So, CE stage inverts the input voltage.

Detailed explanation

It is so amazing how such a simple electrical phenomenon can be explained in such a formal way (excluding to some extent LvW's explanation)... There is a need for a simple qualitative explanation here. It can be done through a simple electrical equivalent circuit of a voltage divider consisting of two resistors - constant and variable (rheostat). The latter can be made by a potentiometer by using the partial resistance between the wiper and one of the end terminals. As early as the 19th century it was known that if we increase the variable resistance (moving the rheostat's wiper up), the voltage across it will increase and v.v., if we decrease the resistance (moving the wiper down), the voltage will decrease. If we use the other end terminal, the relation will be reversed - if we move the wiper up, both resistance and voltage will decrease and v.v., if we move it down, they will increase.

Now it only remains to replace the manually-controlled rheostat with a "voltage-controlled resistor" (BJT, FET) and we will obtain today's amplifying stages. For some (unknown for me) reason, all they behave like a "reversed rheostat" - when we increase the input voltage, their resistance decreases and v.v.

Indeed, they have non-linear resistance (seen from their horizontal output IV curves) but this is irrelevant for the purposes of our question. The only important thing here is that this is a resistance controlled by the input voltage…

Is it a current source?

Contrary to the conventional representation of the transistor as a current source, I will answer:

No… it is not a current source in the literal sense of the word (device producing electricity)... and in the ordinary human sense. Indeed, it is an element keeping up a constant current… but how does it do this magic? How can be a current changed in a circuit supplied by a constant voltage? The only way is by changing the resistance. So the main property of this element is "resistance" (in the broad sense of the word)...

Visualized operation

To truly understand this circuit, you need to imagine where the currents flow and what voltage drops they create for three typical cases of the input voltage - zero, positive and negative. For this purpose, the currents are visualized by full closed paths (loops) with related thickness (in green) and the voltages - by voltage bars with proportional height (in red). The current directions and voltage polarities are real.

For simplicity, I have replaced the bias voltage divider R1-R2 and the decoupling capacitor Cb with a voltage source Vb. The coupling capacitors Ce and Cb have large enough capacitances so the voltages across them do not significantly change when the currents flow through them. The input voltage VIN is represented by a variable battery and the load - by a resistance RL.

Vin = 0 (charging the "batteries"). In terms of DC mode, the CB stage is a CE stage with current negative feedback (emitter degeneration) driven by a constant input voltage Vb - Fig. 2. In this way, the needed initial (quiescent) emitter Ve0 and collector Vc0 voltages are set.

Fig. 2. CB stage at zero input voltage.

When the power is turned on, the capacitors, like "rechargeable batteries" begin to charge: Ce charges through the input voltage source (+Vcc -> Rc -> Tce ->Ce -> Vin -> -Vcc); Cc charges through the load (+Vcc -> Rc -> Cc -> RL -> -Vcc). Note the charging is not shown in Fig. 2 above; only the final result is shown. This process requires both the input source and load to be "galvanic" (DC conductive)... i.e., the input source is a "piece of wire" and the load has a low enough resistance. Thus, at the end of this initial part, the capacitors have copied the according initial DC voltages (VCe = Ve0 and VCc = Vc0).

Vin > 0. During the positive half wave, the input voltage is added in series to the voltage of the input coupling capacitor Ce and the total voltage Vin + VCe of this "composite battery" exceeds the quiescent emitter voltage Ve0 - Fig. 3. So, an input current Iin begins flowing through the input capacitor in a direction +Vin -> Ce -> Re -> -Vin.

Fig. 3. CB stage at positive input voltage.

Since the capacitance Ce is large enough, the capacitor is slightly discharged and the voltage across it does not change noticeably; so the positive input variations appear at the emitter (figuratively speaking, the charged capacitor "shifts up" the positive input voltage). As the base voltage is constant, the transistor is driven in an opposite manner compared to the CE stage, and the collector voltage changes in the same direction. Since it exceeds the voltage VCc across the output capacitor, a load current IL begins flowing through the output capacitor in a direction +Vcc -> Rc -> Cc -> RL -> -Vcc. The capacitance Cc is large enough... the voltage Vc0 across it does not change noticeably... so the positive collector variations appear at the load. Figuratively speaking, the charged capacitor Cc "shifts down" with Vc0 the positive output variations of the collector voltage.

Vin < 0. During the negative half wave, the input voltage is subtracted from the voltage of the input coupling capacitor Ce and the total voltage VCe - Vin is less than the quiescent emitter voltage Ve0 - Fig. 4. Now, an input current Iin begins flowing through the capacitor in a direction +Vcc -> Rc -> Tce -> Ce -> Vin -> -Vcc. The capacitor is slightly additionally charged and its voltage remains almost unchanged; so it "shifts up" the negative input voltage variations.

Fig. 4. CB stage at negative input voltage.

As above, the collector voltage changes in the same direction. But now it is less than the voltage VCc across the output capacitor and something interesting happens - the charged capacitor Cc begins acting as a source that supplies the load. The load current IL begins flowing through the load in a direction +Cc -> Tce -> Re -> RL -> -Cc. The capacitance Cc is large enough; so the voltage Vc0 across it does not change noticeably... and the positive collector variations appear at the load as negative variations (the charged capacitor Cc "shifts down" with Vc0 the positive output variations of the collector voltage so the voltage across the load is negative).

Generalization. As can be seen from the pictures above, the coupling capacitors Ce and Cc play the role of "voltage shifting elements". Ce "shifts up" the input voltage variations while Cc "shifts down" the output voltage variations. Thus the transistor stage is properly biased - its emitter and collector voltages are positive, while the input and output voltages wiggle around the ground.

Concepts

It is a well-known fact that people with practical and inventive thinking understand and explain circuits using knowledge not so much about the physical nature of active devices but mainly about their behavior in circuits. They manage to see basic ideas (concepts) extracted from their previous circuit experience and life. This generalizing thinking allows them to make connections between seemingly different circuit solutions... explain more and more circuits... and even invent new circuits. Here are the concepts I used to explain the CB stage:

Voltage division: R1 and R2 form an ordinary 2-component voltage divider that "produces'' a constant base voltage Vb. Re, Rce and Rc form a 3-component varying voltage divider producing the quiescent DC voltages Ve and Vc. Rce and Rc form an AC voltage divider that produces the AC output voltage Vc (Re is shunted by Vin through Ce).

Voltage "shifting": The coupling capacitor Ce and Cc "shifts up" the input voltage Vin; the output capacitor Cc "shifts down" the collector voltage variations.

Voltage fixing: The decoupling capacitor Cb fixes the R1-R2 divider's output voltage Vb.

Voltage following: Re introduces a current-type negative feedback ("emitter degeneration") that keeps up the DC emitter voltage and collector current constant thus stabilizing the quiescent point.

My comments

Imagine an emitter follower with a resistor inserted in the collector (connected between the collector and Vcc). It is a "disturbed emitter follower" but still follower. Is it a "common-collector stage"? Is the collector connected to the AC ground? No, it is connected through a resistor...
Agree that "Rc is irrelevant for CC gain"... and it is interesting to explain why. IMO it is because of the begative feedback. Rc is a disturbance for the emitter follower and the transistor does what is necessary to neutralize this disturbance - it decreases its static resistance Rce and accordingly, its voltage Vce. As a result, the collector current does not change... and the emitter voltage as well. But the transistor will more easily saturate. Regarding the Horowitz book, the scenario is the same for me - I start reading something and begin thinking deeply... and give up reading...
You can write such a book by collecting your answers and comments here... Hmm, this is a good idea that can be enlarged - someone could gather and organize all the valuable EE writings in one big "book" under the name "The SE art of electronics" or "The wisdom of SE EE" :)
I understand all this... but from my life experience as an "understanding", "explaining" and "inventing" human being, I have come to the conclusion that basic ideas must be explained with ideal models, even though they do not exist. If we start with the details, we kill the ideas. The details are for later when we decide to implement these ideas and then we remember that we live in a real world in which there is Re, Rin, etc. In this case (CB stage) the Re role is crucial for the explanation: if Re =0, it is voltage controlled; otherwise it can be considered as current controlled...
"It" means "CB stage input" or "emitter follower's output acting as an input" or "emitter with AC grounded base" or something that is "AC virtually grounded". It is "current-controlled" in the usual functional sense of the word - Vin and Re form a simple imperfect current source. But here, it works at ideal load conditions (AC shorted), so it acts as a perfect current source (i = vin/Re). A transimpedance amplifier (or op-amp inverting amplifier) is a good analogy.
The problem is that the more you go into detail, the more the main idea is lost... and you get what can be seen in most of the answers here... Therefore, at the stage of understanding, one should work with "ideal" elements in order to grasp the principle, the concept ... The details impede the understanding... This is a basic principle of the creative thinking...
I mean that the "ideal" voltage follower should keep up a constant output voltage when the input voltage is constant (acting as a voltage stabilizer). So, it will have zero output resistance. The emitter follower is quite far from that but the op-amp follower is quite close.
I understand the concepts behind the imperfect (non-ideal) transistor stages by the perfect ("ideal") op-amp circuits. Similarly, we consider the Zener diode IV curve vertical (ideal voltage stabilizer) and the transistor output characteristic horizontal (ideal current stabilizer).
... The gain is Rc/Re if the source is connected in series to Re. But in this case it is connected in parallel and the gain is really gm.Rc. Also, can I ask you to explain such obvious things more simply and clearly? Not everyone here is a circuit designer... at least OP...
Rc/Re case can be easily seen if we see the connection between CB stage and the op-amp inverting amplifier (R2/R1). I have asked such an RG question. In this case, there is a negative feedback in the CB stage. When a low-resistive input voltage source is directly connected to the emitter, there is no negative feedback. Analogical situation in the inverting amplifier is when the input voltage source is directly connected to the op-amp inverting input (in parallel to R1)...
"Increasing Ie externally" is not true. The input voltage source connected in parallel to Re does not increase the emitter current Ie; it "want" to increase the current through Re... which is something different. "Emitter current" means "current coming out of the emitter"... and here only two internal currents (Ic and Ib) leave the emitter. Iin is another (external) current that should be added to them... but they decrease because Ve increases and Vbe decreases...
Why should I read the simulation or look at the scope screen if I can imagine the circuit operation in my mind? Let it be read by one who does not understand what is happening and has no imagination to see it. But he will not find there the explanation he needs because this is a work of the human mind. The (output part of) a transistor behaves as a current source; so the current flowing through it cannot be directly changed. It can be done only by the side of the emitter since, in this case, the input source affects Vbe that, in its turn, changes the output current...
A typical example of this arrangement is the differential pair with a current "source" in the emitter. The current source cannot directly change the emitter current; it can do it only by the help of Vbe. So, it changes Vbe of both transistors thus making them adjust their emitter currents so that the total emitter current becomes equal to the current source current. This is the mechanism of a negative feedback...
Here is a virtuoso explanation of your Rc/Re arrangement that your simulator can not give (unless it is eventually equipped with artificial intelligence:) Due to the negative feedback, the collector-emitter part of the transistor acts as a "differential negative resistance" that is equal to the positive Rc; so the result is zero AC emitter resistance seen by Vin via Re. As a result, your real input voltage source (Vin with Re) is virtually shorted and it sinks a current Ie = Vin/Re from the emitter. It flows through Rc as well and creates a voltage drop VRc across Rc; hence K = Rc/Re...
I think the problem was linguistic and for me it has already been clarified. Once we opened this topic, it is interesting for me to know if you see a negative feedback here and resemblance to the op-amp inverting amplifier. I have drawn the two circuits one below the other in this picture so that they can be easily compared.
With the "virtuoso explanation":) above, I have explained the transistor function in CB stage by the concept of "differential negative resistance" (Rc is neutralized by an equivelant Rce resistance). With another "virtuoso explanation":) now I will explain the op-amp function in the inverting amplifier...
Here, the op-amp output acts as a "true negative resistance" (varying voltage source) that is equal to the positive R2; so the result is zero input resistance seen by Vin via R1. So, your real input voltage source (Vin with R1) is virtually shorted and it sources a current I = Vin/R1 = Vout/R2.
Thus these two seemengly different circuits (with different names and different active elements) are based on the same concept - "neutralizing a resistance by an antiresistance". The result is the same (AC) virtual ground at the emitter and at the inverting input.
Your simulations are very useful but for the purposes of understanding they seem too detailed. Visualizing quantities in digital form with all these fast-changing numbers is confusing. Also, the graphs separated from the circuit diagram are not so convenient for me. I solve these problems by visualizing the quantities on the circuit diagram as in the service diagrams of TV receivers in the good old days. However, I show the values, usually in three successive pictures (for 0, +Vin, -Vin) only at certain typical points in time (without a sweep)... like a "snapshot" of the circuit operation. For the purposes of understanding, I use analog indicators - voltage vars and current loops with proportional height/thickness.
I agree with your answer but let me say what I agree with. Regarding the DC mode, this circuit is a common-emitter stage with negative feedback ("emitter degeneration" implemented by Re). A constant (DC) input voltage is produced by the R1- R2 voltage divider…. "stiffen" by the decoupling capacitor Cb and applied to the base so that the output collector voltage is somewhere in the middle of the supply voltage.
Another (AC) input voltage is applied through the capacitor Ce to the emitter… and this is the true input voltage. Since it "wiggles" around ground, it is "lifted" by the charged coupling capacitor Ce to the value of the initial emitter voltage. The input source has some internal resistance Rs that, when roughly explaining the circuit, can be ignored. Regarding AC variations, the AC input voltage source is connected through Rs to the emitter.
Due to the negative feedback, the transistor keeps its emitter voltage almost constant, the input voltage variations are applied across Rs. Since (almost) the same current flows through the collector resistor Rc, they create proportional voltage variations across Rc. Hence, K = VRc/Ve = Rc/Re.
I think this way of visualizing the voltages can be realized in the simulators in order to roughly illustrate the circuit operation... OK, If you have nothing against, I will visualize the CB stage operation by these qualitative means (voltage bars and current loops) for the three typical cases - zero, positive and negative Vi.
I would like to give you a practical advice. You have explained this circuit phenomenon as a mathematician. Try to explain it as a technician and future engineer. As they say, find a "physical" explanation for the problem... Simple electrical and non-electrical analogies can help you see the idea in all this. This is how technicians understand circuits - through intuition and imagination. Formulas only give the quantitative relationship between electrical quantities... they do not show ideas... and this is what is necessary for understanding.
To truly understand this circuit, you need to imagine where all these currents flow and what voltage drops they create for three typical cases of the input voltage - zero, positive and negative. I would illustrate this with current loops and voltage bars imposed on the circuit diagram...
I have done it in my last edit of my answer. It would be useful for you to try to explain to yourself (before I do this) why currents flow like this and why voltage drops are like this...
I think it should be explained how "the emitter current Ie is turned into a roughly equal collector current Ic". Also, I think it should be explained how the "increasing vin (signal) will cause an increase in Ie"... because, in my opinion, in both cases it cannot be done directly...
Also, I can't understand what is the need, in the first circuit diagrams, for the base to be truly grounded and the input and power source - AC grounded?
In my opinion, the input source forces the transistor to adjust its collector and emitter current equal to the input current through the emitter voltage and through the negative feedback mechanism. It cannot directly set the emitter current.
Still let's follow the causality: Iin -> Ib -> Ic -> Ie... not directly Iin -> Ie.
I can't agree with your opinion about the role of the causality. For the purposes of understanding, explaining and Inventing circuits, we need it. We deal with circuit building blocks that have inputs and outputs. We change the input quantity (cause ) and, as a result, the output quantity (effect) changes. We know this from the experience we have from life. In this way, seeing causal relationships in unrecognised phenomena, we understand them.
... Please, unlock my answer since I want to replace Fig. 2 with a more informative version where the charging currents are shown with dotted lines.
... What do you mind if I improve one of my figures? And don't I deserve at least an answer to my request? I have done something useful for your site that costs me effort and time but you create the false impression that it costs nothing to you. You allow yourself a very rude administration to me, while someone else is enough to just push the button and you immediately fulfill his whims. Could you give any reasonable explanation for this behavior?

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