Are they based on a common basic idea? If so, what is it? Can we see it in other circuits? Is this only a specific electronic idea? If it is more general, can we see it in our life?
It is always interesting and useful for understanding to see and show the common basic idea behind apparently different electronic devices. Here I would like to show that there is such a connection between the old common-base transistor stage and the modern op-amp inverting amplifier. I have proposed this speculation in the question about the common-base transistor 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, and begin discussing it with Prof. Lutz Wangenheim.
To show the relationship between the two circuits, I have drawn the attached picture below. Let's consider the two arrangements in parallel. Here are my speculations:
* Both the circuits are negative feedback systems (followers, stabilizers) trying to keep steady voltages Ve at the emitter of the transistor and V(-) at the inverting input of the op-amp equal to the reference voltages VREF applied respectively to the base and the non-inverting input.
* They try to do that by passing currents I through the input sources VIN. If there is some (internal) resistance Re or an additional resistor R1, voltage drops appears across them that compensate the variations of the input sources.
* The currents are produced by current sources consisting of two elements (a voltage source and a resistor) in series. In the transistor stage, the voltage (Vcc) is constant while the resistance (Rt or Rce in the picture) varies.In the op-amp stage, the voltage (VOUT) varies while the resistance (R2) is constant.
* When the input voltage increases, the transistor increases its instant (chordal) collector-emitter resistance Rt to decrease the current I and the voltage drop VRe, to reach the equilibrium again. The voltage drop VRc (the true output voltage) across the collector resistor Rc decreases as well, but its complement to VCC (the collector voltage Vc = Vcc - VRc), i.e. the "output voltage", increases and the circuit is non-inverting.
* When the input voltage increases, the op-amp decreases its output voltage to decrease the current I and the voltage drop VR1, so the circuit is inverting.
* If the resistances Re and R1 are too small or zero, the transistor and the op-amp will continue trying to reach the equilibrium and will not manage.
Hi Cyril,
it will be no surprise to you that I have problems to follow your explanations.
I really must admit that - in addition to some visual similarities - I can see no commonalities between both circuts. In contrary:
* One circuit has inverting gain, the other one non-inverting
* Gain of the opamp is (nearly) independent on opamp parameters, whereas the gain of the BJT stage will always depend on BJT parameters: A=h21*Rc/(h21*Re+h11+Re).
(As a counter example: Remember that a common emitter stage with strong Re feedback has a gain of app. A=-Rc/Re without h-parameter dependence)
This is, of course, a result of the missing negative feedback (in my opinion).
That means: I cannot identify any possibility how the output voltage or current can react back to the input voltage/current at the emitter.
Therefore - again - my question to your "speculation": "Both the circuits are negative feedback systems".
Where is the feedback element? Where is the feedback loop?
Hi Cyril,
in the attached document A and C are a common base and a common gate, respectively, while in the midlle you can see the non inverting amplifier.
All can be viewed as current buffers but only in B this is accomplished with an evident feedback.
Furthermore A and B are prototype of quasi ideal current amplifier (A: hi=re, hr=0, hf=alfaF, ho=0).
In the common base you can introduce a feedback only considering ro, i.e. the Early effect. But if you consider the transport in the BJT as independet from the inverse polarization I don't see any feedback.
Simone and Lutz, a picture is worth a thousand words... Allow me, before to continue the discussion about these immortal circuit ideas, to complement what has been said by me above by two amusing drawings on the whiteboard. They suggest the idea that the main task of both the configurations (transistor and op-amp) is the same - to keep up a steady voltage (Ve or V-) in the middle point betwen the two circuit parts equal to a reference voltage Vref (i.e. to keep up a virtual ground). So, both the configurations are voltage followers with respect to the relation between this voltage and Vref.
To do this work, we should produce a current by the right circuit part, pass it through the left part (the resistor and the input source), and change it so that to reach the equilibrium. We can produce the current by connecting a resistor in series to a voltage source. Then, we can change the current by varying either the resistance or the voltage.
1. A man acting as a "common-base amplifier". In this case, we vary the resistance Rvar (Rce of the transistor) to change the current through Re and the voltage drop across it to zero the difference between the "virtual ground" and Vef.
2. A man acting as an "inverting amplifier". Now, we vary the voltage (Vout of the op-amp) to change the current through R1 and the voltage drop across it to zero the difference between the virtual ground and Vef.
Dear Cyrill,
if I can agree with you, to a certain extent, with the metaphor of the inverting amplifier realized with a OpAmp (Fig. 2), I don't understand the analogy of the current buffer (Fig. 1). In particular who is the voltage generator?
The analogy in Fig. 2 works also if you add a load, Fig. 1 don't.
An interesting discussion... and maybe the most interesting is the fact that only three of thousands electronics RG members think about the essence of transistor basic circuits...
Simone, sorry that I have forgotten to draw an arrow over the Vin voltage source to show that it is the input in both the circuits. But "input" is a relative concept here since, in the beginning, we assume Vref as an input (voltage follower) and then Vin becomes an input (in a differential arrangament, both they are inputs)...
About the load... Where do we connect it? Maybe first we should answer what is the output quantity. In my opinion, it is the current. So, in the transistor configuraion, we have either to connect a current load in the output circuit or to convert the collector current into voltage. With this purpose (current-to-voltage convertion) we connect the collector resistor (not shown in Fig. 1). We prefer to have a grounded load, so we can connect Rc between Vcc and ground (a grounded collector resistor), and get the voltage drop (negative) across it as an output... but then Vcc will be floating. That is why we use the "complement voltage" Vcc -VRc (Vc) as an output...
Fortunately, in the op-amp implementation, we have a genuine varying voltage (Vout)... so we do no need to convert the current to voltage... we simply get it as a grounded voltage output... Of course, in the case of a current load, we can connect it in the place of R2 to use the current output...
Hi Cyril,
I still have problems to see a close analogy between both circuits.
Here are my doubts:
Since you consider CURRENT as the output quantity:
* Opamp: We have two voltage sources with different sign (signal input and opamp output). As a result, there is a current through both resistors.
This current automatically adjusts itself - due to the feedback action - to a value that leads to the virtual ground in the middle between both resistors.
Changing the values of both (equal) resistors does NOT change the current.
* Transistor (common base): There is also (nearly the same) current through both resistors - however, this current is determined by the signal input only. The „midpoint“ voltage remains nearly constant (0.7 V) NOT because of feedback but only because of the steepness of the I-V characteristic.
But now: Changing the values of both (equal) resistors does change the current.
In summary: The only commonalities between both circuits are:
* Through both resistors goes (nearly) the same current, and
* Both circuits show at the „midpoint“ between the resistors (in the transistor case not really a common node) a voltage which is nearly constant - even if signal inputs and resistor values are changing.
But the physical principle behind these „commonalities“ is different:
* Opamp: Voltage controlled feedback
* Transistor: Voltage controlled current source at the „midpoint“.
_____________________________________________________
Final comment: I think, such exercises keep the brain alive.
Regards
Lutz
Dear Cyril,
I have explained myself badly.
I was referring to the output "V" generator.
I agree that the CB is a current amplifier (low input resistance, high output resistance) but I have difficult to see a CCCS as a Maxwell demon that varies a resistance connected to that output "V" generator.
sim
Furthermore
I must agree with Lutz that the two case are different, but IMHO:
1. CB : Current Controlled Current Source. See my pdf. (I think that this model is also more fitted to the physical mechanism of transport).
2.. OpAmp: a transresistance amplifier obtained with an ideal voltage amplifier with shunt-shunt feedback.
A very exciting discussion... and that is because we have the courage to express your personal opinion without worrying that we may be wrong and will "discredit" ourselves...
Lutz, sorry but I have serious objections to your claims:
1. I can not agree that in the op-amp inverting amplifier "changing the values of both (equal) resistors does NOT change the current". This could be seen formally from Vout/Vin = -R2/R1and I = (Vin - Vout)/(R1 + R2). For example, if we increase proportionally both the resistances (at a steady input voltage), the output voltage will not change... the total voltage (Vin - Vout) will not change as well... but the total resistance (R1 + R2) will increase... so the common current will decrease... Or we can see it simply in the input (left) circuit part where I = Vin/R1, i.e., the current is determined by both the input voltage and the input resistor R1. Maybe you mean to change the resistances in opposite directions? But then the output voltage will change and, as a result, the current will change as well...
2. Also, I can not agree that in "the transistor (common base)... the current through both resistors... is determined by the signal input only. IMO, as above, it is determined by both the input voltage Vin and the emitter resistor Re.
3. I can not agree as well with your assertion that the " 'midpoint' voltage remains nearly constant (0.7 V) NOT because of feedback but only because of the steepness of the I-V characteristic". This would be right if, instead the transistor, we would connected only a bare diode in the place of the base-emitter junction. IMO the " 'midpoint' voltage remains nearly constant" because of the big collector current injected through the emitter resistor... not because of the small base current... (BTW the 'midpoint' voltage is about Vref - 0.7 V)...
As a conclusion, the two resistors in both the transistor and op-amp configuration have the same functions: R1 (Re) acts as a voltage-to-curent converter; R2 (Rc) - as a current-to-voltage converter. We can see these functions in the "common emitter stage with strong Re feedback with a gain of app. A=-Rc/Re" as well...
So, both the configurations are voltage controlled current sources if we consider the current as an output quantity.
I would like also to note that the biasing voltage Vref in the op-amp inverting amplifier does not change its inverting nature.
Hi Cyril - yes you are right.
Both of my comments regarding the currents and their dependence on the R values are wrong (your points 1 and 2). I don`t know what has happened in my mind.
Point 3: I believe there is a misunderstanding. When I speak about the steep V_I characteristic, of course, I mean the Ic=f(Vbe) curve and NOT the base current.
Even when both resistors are NOT equal, we have about the same currents and the "midpoint" voltage (0.7 V) remains nearly unchanged. This was my claim - and, as you know, this is the working principle of the common base amplifier (same current through different resistors). Of course, this is the result of the transistor action which can be modeled as a "voltage controlled current source".
Lutz, thanks for sincerity... it happens so rarely (to be exact, it's not happened to me ever). I understand what the problem is - it is my specific way of thinking that does not match yours... and yet you're trying to penetrate it... for which I thank you very much again... I would like to add more thoughts on these circuits but I feel very tired and exhausted now midnight... and I can not...
Good morning, Cyril! (Hope you had a good night).
At first, let me tell you that I like your "way of thinking" and your desire to look behind the various circuits in order to find common principles which can lead to a better understanding.
Secondly, I am afraid that the main point of disagreement between us (luckily, disa greement regarding technical questions only) is still existent:
* In case of opamp, the property of the "midpoint" (virtual ground) is caused by negative feedback only,
* In case of BJT (common base), the property of the "midpoint" (nearly constant voltage) is NOT caused by feedback but instead by a voltage controlled current source with an exponential characteristic around 0.65 V.
Furthermore, I must confess that the search for an analogy between the two circuits - in this specific case - has been brought for me still no gain in knowledge.
Sincerely
Lutz
Hi Simone and Lutz, I keep thinking about those "devilish" circuits where we seemed to be caught in a vicious circle...
Simone suggests to think of the common-base amplifier as of a current-controlled current source... or, as he frequently says, a current buffer (follower)... IMO, if we control it by a current, this will be the famous cascode arrangement that deserves our special attention (a separate discussion). This arangement is also widely used in differential stages to set the collector currents by the side of the emitter (by connecting a current source in the common emitters)... another candidate for a separate discussion... But I ask myself, "Who will be interested in these old-fashioned ideas nowadays, in these modern times?"...
Lutz, I wonder why you see a negative feedback in the op-amp inverting amplifer...
* In case of opamp, the property of the "midpoint" (virtual ground) is caused by negative feedback only"
...but you do not (want to) see the negative feedback in the common-base stage:
* In case of BJT (common base), the property of the "midpoint" (nearly constant voltage) is NOT caused by feedback but instead by a voltage controlled current source with an exponential characteristic around 0.65 V."
The transistor does the same (but worse) as the op-amp - it compares the reference and the "midpoint" voltage by a serial subtracting (Vbe = Vref - Ve), "analyzes" the differences by its base-emitter junction (the voltage input) and then changes its collector current to keep it constant. The "exponential characteristic around 0.65 V" is not of such importance as well as the open-loop gain of the op-amp... just because there is a negative feedback which makes the op-amp to keep the "midpoint" voltage... Both the arangements do this "donkey work" in the same way - by pasing a curent through the input part of the circuit (the "thing" connected to the emitter and the inverting input)...
Cyril - in this context I think of an integrated circuit invented by Burr-Brown several years ago (OPA600) that was called "ideal transistor" or "diamond transistor". In fact, it was a current conveyor CCII (second generation) with two inputs - one high-resistive (inverting, analog to base) and one low-resistive (non-inverting, analog to emitter). The output was a current (high source resistance).
Moreover, when we add a current sensing circuitry and low-resistive output stage we arrive at the well-known current-feedback amplifier (CFA).
Thus: CC+output stage=CFA.
Surprisingly there is only one single IC on the market which allows both applications (because it has a current output and a voltage output): AD844.
This IC is extensivly used in novel filter/oscillator circuits as proposed in various published articles.
Dear Lutz,
Your last comment about CFA had a major impact on my creative activity. You have an incredible ability to realize problems in their depth and make connections with similar problems from other areas of circuitry. Thank you for the great opportunity!
I have ever met this weird device but I did not pay attention to it because I just did not understand what the hell was all this... But now it was enough to cast a glance at its internal circuit diagram to be amazed and admire this great circuit solution! This is a chain of ingenious ideas... a "necklace of diamonds"! Just to see how beautiful, simple and symmetrical its circuit diagram is enough to admire it!
For me it was a pleasure to notice that we have already discussed and continue to discuss most of the ideas set out in this circuit solution. Meanwhile, I noticed that attempts to explain it have not been very successful - instead of being made related to the classical tricks used widely in transistor circuits, it is presented as something new. Some assertions (e.g. an output acting as an input?!?) would look strange for people trying to understand this odd circuit (but not for us:); here are some examples:
"The -IN input has the characteristics of a buffer output which are very low impedance..."
(http://www.intersil.com/content/dam/Intersil/documents/an97/an9787.pdf)
"The noninverting input is the high-impedance input of a unity gain buffer, and the inverting input is its low-impedance output terminal. "
(http://www.analog.com/library/analogdialogue/anniversary/22.html)
"The inverting input is the common emitter node of a complementary pair of grounded base stages and behaves as a current summing node."
(http://www.analog.com/static/imported-files/data_sheets/AD844.pdf)
So, I have a suggestion to you - to ask a separate guestion about the CFA where to puzzle out its secrets by using the understanding gained from our previous discussions. I could ask this question; then I could write a detailed survey of the problems around this circuit solution and the possible reasons for originating of this clever idea.
Thus we "old war horses" will find a good use of all these old ideas in a modern device and so we will show today's young people that the new is often something old that is well wrapped in fancy packaging...
Hi Cyril - yes, I agree. It seems to be useful to discuss a bit around this interesting circuit approach - and its application areas.
Done!
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
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