Where is the "dynamic load"?

-

ResearchGate: Where is the "dynamic load"?

My question

There are great ideas in circuitry that make us marvel and admire the incredible human imagination of their inventors... Such a great idea is the so-called "dynamic load" in amplifier stages...

Unfortunately, as a rule, this brilliant circuit idea is presented in a formal and sterile way in textbooks and company manuals... and, as a result, it remains misunderstood.

Seeing this paradox, I dedicated years of my life to figure out what it really is... and I still keep thinking about it...

Five years ago, I put this idea for a discussion in the RG forum with a little "provocative" question. I expected that well-known and experienced circuit designers would be involved in the discussion... and by joint efforts we would reveal the "philosophy" of this technique... but, contrary to my expectations, I did not get any reaction.

One year later, in the hot discussion about the exotic CFA, I showed the common in seemingly different circuit topologies of various dynamic load amplifiers (copied in the first my comments below).

I had lost interest in this discussion when, one day ago, on the eve of the New Year, a short insertion of Josef Punčochář to the question below made me begin thinking again about this idea (obviously it has been in my mind all these years).

Thus a new idea arose in my head - that in fact, in these circuit configurations, the "dynamic load" is something conditional... and we can not say exactly what is the dynamic load and what - the amplifying stage... because, in practice, they are the same thing...

My answers

January 1, 2018 (a copy from "What is the truth about CFA?") - four years ago.

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

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

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

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

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

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

"But... what kind is the active load in the CMOS amplifying stage - NMOS or PMOS?"

January 2, 2018. You are absolutely right with your clear explanations... but they explain well the simple dynamic load acting as a "constant current source" (see the picture Dynamic load.jpg in my first comment above).
In this question, I propose to discuss the idea of the more complex dynamic load acting as an "opposite-controlled current source"... usually implemented with the help of a current mirror (see the picture Differential dynamic load.jpg in my fourth comment above)...
I have graphically illustrated both techniques in the attached pictures with two kinds of "cutters".

......

A great discussion... although with only three participants... for the shame of great circuit designers:)
Josef, I agree with you that the famous Baron Munchausen's idea - bootstrapping, is a good example of a dynamic load... but again this is a simple dynamic load.
In your circuit, you have realized a constant current source by an ohmic resistor (R1) and a following voltage source (the emitter follower T2) whose voltage is applied to the upper R1's terminal through a "shifting" capacitor (CB). This combination acts as a dynamic load with infinite (differential!) resistance since the voltage is dynamic... and this creates the illusion that the ohmic resistance has become infinite... Actually, it is the same finite ohmic resistance (R1); only the whole network resistance has become infinite...
The trick of the "complex dynamic load" is more sophisticated - not only to keep a constant current but even to change it in the opposite direction.

......

January 2, 2018. Let me first add a little more wisdom to your philosophy about "differential, balanced and symmetrical circuits". Yes, they are especially elegant... but those of them, in which the output signals of their two constituent parts are not just subtracted (compensated) but they interact... are extremely interesting. Amplifying stages with "differential dynamic load" (let me use your terminology) are just such circuits.

In these circuits, two identical current sources are connected, contrary to "good" circuit rules, in series (like in differential pairs where two voltage sources are connected, again contrary to "good" circuit rules, in parallel). Then, the two current sources are driven by the input voltage signal(s) in opposite directions. As a result, they oppose to each other and the voltage of the common node vigorously changes.
The CMOS inverter is a typical example of this configuration. Imagine the two transistors are exactly the same. Initially, we apply "zero" input voltage (exactly equal to half of the supply voltage) and then begin to wiggle it around the "zero".
In this configuration, the constituing transistors are both (common-source) amplifying stages and dynamic loads... and each of them can be considered both as an amplifier and dynamic load of the other...
But now, probably as a result of your prophetic thoughts, a new idea attacked me and did not give me peace of mind in those hours after midnight when "normal" people are sweetly sleeping... That is what it is...
If the simple dynamic load (constant current source) has infinite 
differential resistance what resistance should have the more sophisticated "differential load" (contrary-driven current source)?
https://en.wikibooks.org/wiki/Circuit_Idea/Negative_Differential_Resistance#"Increasing"_negative_differential_resistance
Will it not have "negative resistance"? Because the negative resistance is what is beyond the infinite resistance…

......

Regardless of the dynamic load type - simple (constant current source) or complex (opposite-controlled current source), the configuration is the same - two current sources in series. More precisely speaking, they are current-stabilizing non-linear elements. When we introduce a ground, the one of them becomes a source (pulling up) and the other - sink (pulling down).
The CMOS pair is the "perfect simplicity" since both elements (PMOS and NMOS transistors) are automatically controlled in different directions by a common input voltage. As if from this voltage two complementary one another (to the supply voltage) input voltages are created... and each of them controls its own current source. So there are no unnecessary elements here...
But in the case of a differential pair with a current mirror dynamic load, there are two separate input voltages each of them controlling its own current source. Unfortunately, both input voltage sources are grounded... so we cannot put the one current source on the other. Then?
The next ingenious idea is to "clone" one of the current sources (T1) to a new location (T4) - above the other current source (T2), by means of a current mirror (T3 - T4). So, the combination T1 + T3 + T4 serves as a composed pull-up current source while T2 - as before, as a pull-down current sink. Again there are two equally acting opposing current sources (transconductance amplifiers).
But what is the amplifier and what - the dynamic load here?

......

I am not sure if this new point of view will change the course of history (because the CMOS circuits are the basis of modern computers)... but at least it can serve as an illustration of a non-standard thinking in circuitry:)
Really, "people rarely use the CMOS inverter as a linear amplifier"... but still it is an amplifier (in the middle linear region).
By the way, I have corrected the links above so that they point to the WB story about differential negative resistance, not to the story about true negative resistance... since there are no internal sources in transistors... and they behave as the simpler differential (dynamic) resistors.
Now about bridge circuits (they are my favorites too)...
Your idea to make a bridge amplifier is wonderful... but I am afraid it exists and is even widely used; for example, I have seen it in car audio systems. But do not despair, yet you have done a "subjective new invention" which, from the point of view of the inventive creation, is no less important than a real invention:)
Your idea is closely related to our discussion about the presence of negative resistance in these circuits. I think the negative resistance can exist in the bridge amplifier... and it is another kind (S-shaped) of true negative resistance... It is also a manifestation of the Miller theorem. Two years ago, I even dedicated a question about this phenomenon in the bridge amplifier:
https://www.researchgate.net/post/Why_do_bridged_amplifier_outputs_see_half_the_load_impedance2

......

IMO мost of electronic devices have a voltage output; hence, the main task in circuitry is controlling voltage. We do it through the well-known 19-century technique - by means of the so-called "voltage divider".
For this purpose, we simply "stretch" a network of two resistive elements (electrically-controlled resistors) in series between the supply rails: one - "pulling up", and other - "pulling down" the common point between them. Then we begin controlling the resistance of the one, the other or both (in opposite directions). In the latter case, the 2-input (differential) signal is converted into a single-ended output signal.
In our dynamic load configurations, both elements are implemented by transistors acting as voltage-controlled current-stabilizing resistors (transconductance amplifying elements). So, their output quantity is current... but the final output quantity is voltage - the voltage of the common middle point. Thus two problems are solved - extremely high gain and single-ended output.
The output is single-ended since initially we have two pull-down elements... and we have "cloned" the one of them above the other. If we want to keep the differential output, we have to "clone" each of them above itself... i.e., to build our favorite bridge amplifier…

......

You have written a very interesting insertion that seriosly made me think about the output node... It reminds me the hot discussion about the current-feedback amplifier where you have developed the same idea about the emitter summing node...
IMO you have posed a question about the output equilibrium - although the output voltage vigorously change, it has to find the quiescient point somewhere between the supply rails (to stay in active mode).
I think this is achieved if:
the current sources are not ideal (as shown in the attached graphical representation, their IV curves are not absolutely horizontal)... or at least one of them is not ideal... and the output can be unloaded
the output is loaded by the second stage that sinks/sources the excess current (your suggestion)
Otherwise (ideal current sources and no load connected), the output voltage will reach the supply rails (saturate).
Now I will explain the first case. If the two elements (transistors) are not ideal (have finite differential resistance), then when the one of them increases its momentary CE resistance, the other decreases its CE resistance... The total resistance and accordingly, the common current, stays constant. As a result, the common output voltage changes while the common output current does not change... like when moving the slider of a potentiometer.

January 4, 2018. I thank you for such interesting and in-depth analyzes in your comments. They made me think deeply and try to fully understand the mechanism of this kind of dynamic load. Throughout the day I have been trying to imagine, as an animation in my mind, the operation of this fully symmetric 2-element network... and finally I succeeded...
But at one point, a vague remembrance began to emerge in my mind - that I had already asked such a question. It turned out that it was 4 years ago:
https://www.researchgate.net/post/What_is_current_steering_and_how_is_it_implemented_Is_there_a_dual_voltage_steering_If_so_what_is_it_and_how_is_it_implemented2
It was so interesting, informative and exciting that I started to read it... and that is what I am doing right now... But let answer some of your questions above.
Abdelhalim, I started thinking about your idea of summarizing the most accurate and appropriate comments. I think this can be done periodically by summarizing the answers to the present and then enriching this summary with new thoughts derived from further comments. Perhaps this should be done by the author of the question? And perhaps the name of the author should be mentioned in order to preserve copyright... or not? I will try it soon.

January 5, 2018. Before continuing, I would like to point out that our discussion is a good example of unselfish cooperation for the sake of an ideal goal - revealing the absolute truth about great circuit solutions. And note that all this takes place in a friendly atmosphere of mutual respect and support that favors the emergence of new ideas.
I thank everyone involved in this noble undertaking.

Now I will focus your attention entirely to the influence of "the second stage connected to this point" (Lutz) and the "current balance challenge at the output node" (Abdelhalim). In other words, I will try to reveal the "absolute truth" about the dynamic load mechanism... only how do I stop the time to do all this:)?
Regarding the Lutz's remark, I think that he has nothing to fear from the influence of the next stage since:
Although the two transistors (amplifier and dynamic load), constituting the dynamic voltage divider, have individually very high differential resistances, the output resistance of the whole dynamic divider is (can be) very small as the constant DC current through this network is (can be) very big.
So, the dynamic voltage divider is composed by high-resistive elements but it has low output resistance (behaves as a voltage-type output) what we want. Am I right?

......

I have the feeling that, in the end, we will be able to completely unravel this intricate circuit solution.
Still, I wonder, should not it have already been done... to be explained somewhere, and not to do it now, almost half a century after it was invented?
I will graphicall illustrate the circuit operation by a series of drawings... but now I аm in a hurry to throw some initial observations.

I agree with you that in the initial position the circuit must be in a balanced symmetric state (for this purpose it might have been better for the current source to be connected to -VEE instead to ground): V1 = V2, IL = IR, VOUT = VCC/2.

January 6, 2018. This picture drawn in an old-fashioned manner... from the times when there were not computers (but there were thinking people:)... is a gift for you!
......
So the output cannot be unloaded?

......

January 7, 2018. I do not know why, but your comments are very stimulating to my creative thinking... and not only with their content... but somehow with their benevolent style of expression. They do not deter me but sooner make me continue in the same direction with even greater enthusiasm. From this point of view, you are the perfect contributors for such a creative discussion:)
Nicholai, I am very fond of your "philosophical" reflections on the structure and function of cuircuit building blocks, particularly about the differential pair above... How much I love such wise thoughts... as though they are "native pearls" extracted from the information slag... In fact, they are what explains circuits... they are the know-how of understanding...
You said, "Consider the simple differential stage with resistors at the collectors"... and then you removed one of them... since you know the function of these resistors - they act as current-to-voltage converters. And if we do not need a voltage output, we do not need a resistor too. So, the "perl" here is: The function of the collector resistor is to convert the collector current into voltage. By comparison, in a conventional textbook, they would only say that "there is a resistor RC included in the collector" .. as if it was given by God...
"But if the input voltage is purely differential, then this "ground" doesn't have to be a reference potential for the actual input signal. It may play a role for the input biasing, but that's a secondary consideration" reminds me the input circuit of your CMOS bridge amplifier... and my question asked five years ago:
https://www.researchgate.net/post/Can_we_connect_a_floating_voltage_source_between_the_two_inputs_of_a_differential_amplifier_If_so_at_which_conditions

......

Do you really believe that the output quantity of dynamic pairs (of "contradicting" transistors) is a small difference current... and they cannot operate without a second stage (load)?

......

If you ask why, in the differential pair, a second identical stage is added to the single transistor amplifier, 99.999...9% of people (including most of circuit designers) will respond in this way - "Because the circuit is so made", "Because the circuit is such"... or something similar... And only 0.000...1% of them will begin to think "why"...
And because we, here in this forum, pretend that we are from the latter:), let's find at least another reason to include such a "helper" (second symmetrical stage). I described it a few years ago in the WP and WB stories below:
https://en.wikipedia.org/wiki/Talk:Emitter-coupled_logic#Revealing_the_truth_about_ECL_circuits
https://en.wikibooks.org/wiki/Circuit_Idea/Revealing_the_Truth_about_ECL_Circuits#...from_the_side_of_the_emitter
... but briefly here is the idea:
To improve the circuit, we decide to bias the single amplifying transistor stage from the side of emitter (by including an emitter current source). But the stage becomes uncontrollable from the side of base (since the emitter voltage follows the input base voltage). Then an ingenious idea comes to us - to "harden" the "soft" emitter by connecting the Nikolai's "helper" (another identical stage) to it.
Thus, at common-mode signals, the emitter is "soft", "easily movable"... and there is no amplification; in the case of differential signal, it becomes "hard", "immovable" - there is high amplification.

......

This new "amplifier" does not amplify:) since the emitter voltage follows the base voltage... and, when varying the input base voltage the emitter-base voltage does not change... the collector current and, accordingly, the collector voltage - also... and you do nothing.
Now return the "helper" back to its place and, of course, apply a steady input voltage to the base (thus making an emitter-coupled amplifier). Vary the input base voltage - now the emitter-base voltage will change... the collector current and the voltage will vigorously change…

January 8, 2018. Let's see another dynamic load circuit in the attached figure...

Imagine we change VIN and VREF simultaneously but in opposite directions ("differentially").
What is the output - voltage or current? Can we left it unloaded (do we need a second stage)?

......

Or, if you prefer, let's see the internal structure of the more complex current-feedback amplifier (CFA)...
What kind is the output of the dynamic load pair (the common point between the collectors of Q4 and Q6) - voltage or current? Is there an output difference current? Can we left the output unloaded (without connecting it to the buffer +1)?

......

Here is my answer (about the CFA above)...
(the transistor output characteristics are approximated by straight lines)

January 9, 2018. As you can see in the graphical solution above, the dynamic voltage divider is not loaded... the current through it stays constant... there is no output current... but the output voltage vigorously changes…

......

The tiny moment here is that they are not ideal:)

......

... "we try to make the sources as ideal as possible"... but no more than this:)
The more ideal they will be, the greater the voltage gain will be... and finally, when they become ideal, it will become infinite... and the output will saturate...
The circuit can be used with a fully voltage output without any output (load) current drawn... To produce voltage, it does not need any buffer stage…

......

Now let me try to explain in an even simpler way the role of the two current mirrors in the CFA configuration. Here is my short story...
In the complementary emitter follower, the emitters of transistors Q1 and Q2 are joined, and their collectors are connected to the supply rails. To realize the dynamic load configuration, we need just the opposite - the collectors to be joined, and the emitters connected to the supply rails... That is, we want to turn the transistors 180 degrees.
For this purpose, we decide to "clone" Q1 and Q2 by the help of two current mirrors, and then connect their rotated counterparts Q4 and Q6 as we want. Thus, there are two complementary stages - original (Q1 and Q2) and copy (Q4 and Q6).
Conclusions:
In the circuit of a current-feedback amplifier, both transistors are cloned and rotated since they are not connected as we need
In the circuit of a long-tailed pair, only one of transistors is cloned and rotated since the other is connected as we need
In the circuit of a CMOS amplifier, transistors are neither cloned nor rotated since they are connected as we need.

January 10, 2018. It is very interesting for me to see how you are trying to understand and explain this bizarre circuit solution (CFA). You are clever and well-educated scholars and educators... and maybe because of that you do it in such a formal, linear and logical way... but this is not sufficient for a true understanding of this sophisticated circuit...
I do not consider myself to be either smart or very erudite... but there is something in my way of thinking that allows me to imagine the circuit operation at the lowest levels of understanding... and to grasp the underlying ideas behind circuits... sometimes even better than their creators. I do not know exactly how to call it - "technical sense", "visual thinking", "analogical thinking", "generalizing thinking" or something else... but I know it is a kind of some innate special arrangement of the brain that is very difficult to acquire…

......

I will use this mental technique to show again that there is no need for the output of the "dynamic voltage divider" (Q4 and Q6) in the current-feedback amplifier to be loaded with a next stage... that consumes the difference between collector currents in the input complementary emitter follower (Q1 and Q2)... since there is no such difference... there is no output current... only the output voltage vigorously changes...
The first key point of understanding the operation of this part of circuit is to realize that the current sources (Q4 and Q6) are not perfect (ideal)... and, in addition, they are not loaded with a zero resistance load. This means that the output currents (IQ4 and IQ6) of the current mirrors are not exactly (only) determined by the input base-emitter voltages (VBE4 and VBE6)... but they also depend on the load resistance connected in the collector (of the other transistor). If they were loaded with zero resistance loads (e. g., the Q4 collector is connected to ground and Q6 collector - to VCC), their output currents would be exactly determined by their input base-emitter voltages (VBE4 and VBE6). So, it is quite possible that different (not equal) base-emitter voltages to be applied to Q4 and Q6... but their collector currents IQ4 and IQ6 to be equal. Let's see how it is possible...
The second key point of understanding the circuit operation is to think in terms of static (instant, chordal) collector-emitter resistances (RCE4 and RCE6) instead of currents flowing through them. This means to think of the two collector-emitter junctions (CE4 and CE6) as of two partial resistances (RCE4 and RCE6) of a potentiometer as shown in the attached picture. In the graphical representation, static resistances are represented by lines starting from the origin of the coordinate system.
Operation. Initially, the input base-emitter voltages are equal (VBE4 = VBE6). The static collector-emitter resistances are equal as well (RCE4 = RCE6)... the potentiometer slider in the analogy is in the middle.
When the input base-emitter voltages (VBE4 and VBE6) change differentially - e.g., the magnitute of VBE4 increases while of VBE6 decreases, RCE4 decreases but simultaneously RCE6 increases... like the two partial resistances of the potentiometer when moving the slider to right. But the total resistance RCE4 + RCE6 remains constant... so the common current flowing through the network remains constant as well.... and the output voltage VA vigorously changes.
In the graphical representation, RCE4 IV curve rotates clockwise... and RCE6 IV curve simultaneously rotates in the same direction... so the operating point A vigorously moves to right along the horizontal blue line.
When VBE4 and VBE6 change differentially but in the opposite direction, the processes are reversed…

......

Final remarks. I understand that this situation - the input base-emitter voltages of the two transistors are different but their collector currents are identical, is quite confusing... but it can be easily explained by the help of considerations above...
The reason of this behavior is that an opposite variable (dynamic) load is included in the collector circuit of each of transistors - and that is the other transistor. It changes its static resistance... thus changing the voltage across the amplifying transistor... and accordingly, the current through it (due to the Early effect)... so the current does not change.
In other words, the collector current of the transistor depends on two variables - the base-emitter voltage and the voltage across the load... and if we simultaneously change them in the same direction (both increase/decrease), the current will stay unchanged…

January 12, 2018. I have finally realized what I do to imagine circuit operation at the lowest level of understanding - I replace the more abstract electronic circuits with equivalent simple electrical circuits where I perform the functions of actively operating semiconductor devices - "doers".
This way of thinking - a kind of "empathic technique", allows me to slow the operation of devices down to step-by-step mode of investigation... so I completely understand what they actually do (feel:).
Such electrical substituents of electronic devices are manually-controlled voltage and current sources, variable resistors (rheostats), potentiometers... For example, by using a humble rheostat, I have revealed what differential resistors actually do in the story below:
https://en.wikibooks.org/wiki/Circuit_Idea/Negative_Differential_Resistance
With the same success we can use more sophisticated electrical substituents, e. g., bridge circuits...I will demonstrate it in my next comment…

......

A side but useful effect of this discussion for me was that I finally imagined how the mysterious current feedback amplifier (CFA), particularly its non-inverting amplifier implementation, works. I have already suggested this explanation in the hot discussion below...
https://www.researchgate.net/post/What_is_the_truth_about_the_exotic_current_feedback_amplifier_Is_it_something_new_or_just_well_known_old_Is_it_really_a_current_feedback_device/1
... but now I have finally realized it by the help of the concept "balanced bridge circuit":
https://www.researchgate.net/post/What_is_the_idea_of_a_balanced_bridge_circuit_Why_it_is_named_in_this_way_What_is_the_bridge_in_the_bridge_circuit
Saying "bridge circuit", I mean the most generalized bridge idea - two grounded voltage sources joined with a meter (the "bridge")...

......

Nice associations... Of course, we should first mention Maxwell's demons:) But let's continue with the bridge analogy...
In the past, looking at the first balanced bridge circuit, I was very surprised that a low-resistance ammeter (galvanometer) was used instead of the high-resistance voltmeter. Quite later I realized the simple truth that, when the bridge is balanced, both voltage across and current through the meter are zero... so we can use both voltmeter and ammeter...
Apparently they used a galvanometer to get the maximum sensitivity and accuracy of the measurement. Then there were not voltage amplifiers, and voltmeters were made by a galvanometer and resistor in series... so that in any case voltmeters were more insensitive than the single galvanometer. The only problem was when the two voltage sources and galvanometer were perfect - then the current would be significant.
We can reproduce this famous 19's arrangement today, as shown in the attached picture. Here a constant (or slowly varying) perfect voltage source V(-) is connected from the left side of the ammeter, and an imperfect voltage source - from the right side. The latter is implemented by the varying voltage source VOUT and the voltage divider Rf-Rg. Imagine someone controls the left voltage source and we control the right source.
1. Assume that initially the two voltages are equal - V(-) = VRg = VOUT*Rg/(Rg + Rf). In this situation, two voltage sources - a perfect and imperfect with equal voltages are connected one to another. Thus they are subtracted from each other (compensated)... and the result is zero - there is no voltage across and no current through the galvanometer... the bridge circuit is balanced…

......

2. Now imagine the voltage V(-) decreases. In this situation, a perfect and imperfect source with different voltages are connected one to another.... as though the output of the voltage divider is shorted by the V(-) source... and the latter determines the common voltage. VOUT does not change... and it is not determined by the divider ratio Rg/(Rg + Rf).
A current I begins entering the galvanometer from the right and the meter's arrow turns to the right... the bridge circuit is imbalanced.

......

3. We see that there is imbalance and begin decreasing VOUT until we restore the balance - the current becomes (almost) zero again. VOUT is again determined by the divider ratio - VOUT = VRg*(Rg + Rf)/Rg.
Thus, if the (input) voltage V(-) slowly varies, and we accurately keep the circuit balanced, the (output) voltage VOUT will proportionally change - VOUT = VRg*(Rg + Rf)/Rg... and this arrangement will act as a "non-inverting amplifier with negative feedback" having a gain of (Rg + Rf)/Rg.

......

If we decide to realize this arrangement today, more likely we will use an electronic current-to-voltage converter as a current-sensing element instead the "floating" ammeter.
And since the converter is usually grounded, we can short the bridge output by a piece of wire... and move the "ammeter" somewhere in the left part of the bridge circuit, e. g. between the left voltage source and ground... and even inside the source - a key point of understanding CFA...
If the left (input) voltage source V(-) consists of two parts - pull-up and pull-down (a usual case in electronics), we can insert two current-to-voltage converters in each of parts... and subtract their output quantities - another key point of understanding CFA...
After these remarks, we are ready to present the operation of the real current-feedback amplifier from this viewpoint…

......

The input voltage V(+) is buffered by the complementary emitter follower (Q1 and Q2) and its "copy" V(-) appears at its output (the joined emitters). This part of the circuit serves as the left voltage source in the bridge analogy above.
By the voltage divider Rf-Rg, a part of the output voltage VOUT is produced and applied to the emitter follower's output. This part of the circuit serves as the right voltage source in the bridge analogy.
Thus the two voltage "sources" are short connected by a piece of wire and the current-sensing element is hidden inside the left voltage source. It is composed of two current-sensing elements - "pull up" connected to VCC (Q3) and "pull down" connected to ground (Q5). Their current outputs (Q4 and Q6 collectors) are contrary connected to convert the current difference into voltage VOUT with high gain.
Conclusion. The difference between the conventional "voltage-feedback amplifier" and this so-called "current-feedback amplifier" is in the type of the output quantity after comparing (subtracting) the two voltages (the "static error"): in the former, it is a voltage; in the latter, it is a current.
In both non-inverting configurations, the negative feedback is implemented by the same "voltage type" network - a voltage divider... and accordingly, the voltage gain is represented by the same expression.

January 13, 2018. This afternoon I decided to review your work... but my strength and desire quickly ran out after the first few pages... What I thought while I was reading, was "Thank God that I am no longer a student... and in particular, in this highly sophisticated course:)"
I really want to ask you... and my request is honestly to answer me, "Do you strictly teach this material in that way?" Because if that is the case, I guarantee you that poor students will not be able to understand much of this...
I sincerely hope this is not the case... and, in addition to this material, you explain the matter in this way that we use when discussing here - simply, clearly and frankly... because these young people really need it…

......

Let me now continue by adding some thoughts to your sentence above, "I to V converter is just input impedance of the buffer - this is transimpedance."
The simplest I-to-V converter is simply a resistor. It is a quite strange converter since it has only two terminals (one port)... and we wonder what is the input port and what - the output port (I solve this problem by drawing the diagram appropriately as in the attached picture). This element possesses impedance determined by the ratio of voltage across and current through the same resistor.
If we connect a voltage follower (buffer) to the resistor, another "voltage copy" appears at the buffer's output (I agree that the input impedance of the buffer can serve as such a resistor). Now we can talk about another kind of "impedance"... called "transimpedance"... since it is determined by the current through the resistor and "voltage copy" at the output... although it is quite strange to make such a ratio between these two quantities located in different places.

......

Here is another viewpoint... We can think of this part of circuit as of a complementary emitter follower with built in the output current-sensing element. It senses the output current that appears when a load is connected to the follower's output.
Saying "load" I include such a strange load as a voltage source... since it is still a source... another input source... In this way, we have applied an input voltage to the buffer input (the CFA non-inverting input)... and another input voltage to the buffer output (the CFA inverting input). The second voltage source acts as a "load"... and a "load current" begins flowing into/out the output. Of course, we can connect an "ordinary load" - e.g., a variable resistor, and to change the "load current" by varying the resistance.
Now a few words about the "current-sensing element". To measure the load current, it can be connected in series to the output... as a "pass element". But then it would be bipolar and floating which is difficult to realize. If we insert it between the n-p-n transistor and ground (to ground it), it will sense only the entering current; if we insert it between the p-n-p transistor and VCC, it will sense only the exiting current. So, to sense both entering and exiting curents, it is implemented by two separate current-sensing elements (input parts of current mirrors) which are connected accordingly to ground and VCC. Then their output currents are subtracted and converted to voltage by the dynamic load network.
At the end, I will emphasize again that the output quantity of the dynamic load stage is voltage rather than current... and therefore there is no need for another conversion from current to voltage.

January 16, 2018. I was just thinking for a while to devote myself to a more routine editorial work to complete the first summary... and you once again aroused my desire for discovery with your intriguing words. I still wonder how you manage to influence my thinking in this "magical" way:)
Thus, in the late afternoon, I decided to make a "creative winter walk" during which I could try to find out what is this "general and fundamental principle behind all these circuits" since, as you said, "we are getting closer to it"... I was walking through the park lanes, thinking aloud... while my phone was accurately recording my thoughts...
Now I am listenning to the recording at home, extracting the most valuable thoughts... and describing them in this comment... This is my usual technology through which I create my writings on the web…

......

"What would be that great idea?", I was asking myself... (what has always impressed me, is that great designers - "gods" in circuitry, are not so interested in great ideas... but are satisfied and feel happy with concrete circuit realizations... while we, "mortals", are looking for the great idea in every circuit...)
I have long know the most general (not-only-electrical) idea here is that we provoke a conflict between two systems (beings, devices, etc.) by making them struggle to change the same quantity in different directions. Doing that, they strain and their efforts (the output quantity) are a measure of the extent of our "provocation" (the input quantity):
http://www.circuit-fantasia.com/my_work/conferences/cs_2005/paper2.htm
As I have summarized above, in the electrical domain, the two opposing systems are constant sources connected through variable resistors to the common conflict point. They change their impacts on the common point by varying their resistances. Thus, depending on the configuration (serial or parallel, voltage or current divider) either the partial voltages across resistors vary at constant current or the partial currents through resistors vary at constant voltage.
In the first case, the voltages on the serial connected resistors are redistributed while the voltage sum remains constant - there is a voltage redistribution. In the second case the currents through the parallel connected resistors are redistributed while the current sum remains constant - there is a current redistribution ("current steering").
--------------------------------------
All this was before my walk... But during it, at one point I realized that the more general (primary) electrical idea than the voltage and current redistribution here is the resistance redistribution... since in both cases, to change voltages or currents, resistances are redistributed while the resistance sum remains constant... Depending on the configuration, the partial resistances vary either at constant current or voltage.
Instead of "redistribution", we can also use a figurative word from the field of cinematography and audio engineering - "crossfading"... the partial resistances are crossfaded…

January 17, 2018. You are more than inventor... you are scientist "trying to think in even more abstract terms "... and you can think scientifically by looking for scientific theories to justify naked facts... I am only a "humble inventor"... but both we are "philosophers with imagination":) So let's add more "philosophy" to this subject...
Since there are both negative and positive feedback in the hair-dryer motor&heater system, perhaps we can think of it as of a kind of negative impedance converter (NIC)... and v.v.:)
How nice you said it - "in a circuit, there's always something more abstract than voltage or current"! I think circuit designers have to write this thought in big letters... and stick it on the wall in their offices... so that every morning, when they come to work, re-read it... so they will never forget it:)

......

Do your students really understand your "cascode structure"... and the „dynamic – dynamic“ load T7 and T8 inside it... since I have problems with understanding:)? Maybe I have to go through your course:)

......

I suggest that we continuously develop the summary in the following way:
That one of us who wants to change something in it, let copy and edit it, and then put it at the end of discussion. If one who has made the previous version has no objections, to delete it in its old place... Or, if you prefer, not to delete it?
Thus the summary will really be our collective work... similar to a Wikipedia page...
I will try it below…

January 31, 2018. I have added to the summary the observations below about the output properties of the dynamic-load stage. To see them, click the [+] at the bottom of summary.
-----------------------------------------------------------------------
Output properties of the dynamic-load stage:
The dynamic voltage divider "produces" voltage; so, the open (unloaded) circuit provides ideal load conditions for this voltage type output. Significant current flows through both parts but, in the ideal case, there is no exiting/entering load current.
The output resistance is relatively low since it is formed by two complementary ("crossfaded") partial resistances in parallel - when one of them increases, the other decreases and the total output resistance decreases as well (maximum in the middle and minimal at the ends). Each of collectors "sees" the extremely high output resistance of the other collector, but the load "sees" the low output resistance of the dynamic-load pair.
The output voltage of the middle point is single-ended. It is a result of two oppositely (differentially) changing input voltages. So, the dynamic-load stage converts a differential input voltage into a single-ended output voltage.

Summary

General idea - provoking a conflict between two systems (beings, devices, etc.) by making them "struggle" to change the same quantity in different directions. They "strain" and their efforts (the output quantity) are a measure of the magnitude of the "provocation" (the input quantity).

Electrical idea - coupling "incorrectly" two homogeneous electrical sources - both current or voltage type, and controlling their input quantities in opposite directions (by changing the input quantity of the one, the other or both). As a result, they interact - each of them tries to change (to impose its value of) the common output quantity (current/voltage), and the other common quantity (voltage/current) vigorously changes. The latter is used as an output quantity of the whole couple: if current sources are connected, the output quantity is voltage; if voltage sources are connected, it is current.

Implementation. Instead sources, two kinds of controlled non-linear elements are used in practice:

To produce output voltage, two controlled current-stabilizing elements are connected in series and supplied by a common constant voltage source thus forming a dynamic voltage divider. If the circuit is grounded, the upper element can be considered as pulling up and the lower - as pulling down (examples - the CMOS amplifier, current-feedback amplifier).

To produce output current, two controlled voltage-stabilizing elements are connected in parallel and supplied by a common constant current source thus forming a dynamic current divider (examples - the differential pair, emitter-coupled amplifier).

In both cases, to change voltages or currents, both partial resistances are changed (in contrast to the ordinary resistor-load stage, where only one of them is changed). So they are redistributed ("crossfaded") while the total resistance remains constant. Depending on the configuration, the partial resistances vary either at constant current or voltage. So, the general electrical idea of dynamic load is resistance redistribution.

Name. The current discussion is about the first version of dynamic divider - voltage type, where two transistors acting as controlled current-stabilizing elements (transconductance amplifiers) are driven by two opposite input voltages (differential or one of them is an inversed copy of the other). It can be called "differential dynamic load" in contrast to the simpler "single dynamic load", where only one of transistors is driven while the other acts as a constant current source. In the future, we will call this "differential dynamic load" simply "dynamic load".

Topologies. The two transistors of this kind of dynamic load pair are oriented with their collectors (drains) to each other to act as opposing current "sources":

  • In the CMOS stage, only two transistors are connected with their sources to Vdd and Vss (ground) and their drains are joined; so the dynamic voltage divider is composed by both "original" transistors. This pair is controlled only with one voltage referenced to ground; its complementary voltage referenced to VCC acts as a second voltage.
  • In the differential pair, where the collectors of both "original" transistors are turned up, the one of transistors is "cloned" and turned down, by the help of a current mirror, to stand against the other collector. So the dynamic voltage divider is composed by one "original" and one "cloned" transistor. This pair is controlled with two input voltages referenced to ground (differential input).
  • In the input stage of CFA, two "original" transistors are turned with their emitters to each other. By the help of two current mirrors, both transistors are "cloned" and rotated back to stand with their collectors against each other; so the dynamic voltage divider is composed by both "cloned" transistors.

So, to implement the dynamic load idea:

  • in the simplest CMOS stage, there is no need of a current mirror
  • in the ordinary differential pair, there is a need of only one current mirror
  • in the sophisticated CFA, there is a need of two current mirrors.

Output properties of the dynamic-load stage:

  • The dynamic voltage divider "produces" voltage; so, the open (unloaded) circuit provides ideal load conditions for this voltage type output. Significant current flows through both parts but, in the ideal case, there is no exiting/entering load current.
  • The output resistance is relatively low since it is formed by two complementary ("crossfaded") partial resistances in parallel - when one of them increases, the other decreases and the total output resistance decreases as well (maximum in the middle and minimal at the ends). Each of collectors "sees" the extremely high output resistance of the other collector, but the load "sees" the low output resistance of the dynamic-load pair.
  • The output single-ended voltage is a result of two oppositely (differentially) changing input voltages; so, the dynamic-load stage converts a differential input voltage into a single-ended output voltage.

Comments

Post a Comment