Hi Josef, Hi Cyril,

I think we have a short circuit between questions ;-)

ERIK

I have attached below pictures (not so well drawn as the picture above) of the three typical situations in the AC amp operation:

1. Zero input voltage - the capacitive "batteries" C1/C2 charge to the bias voltages.

2. Positive input change - the capacitive "batteries" convey the positive/negative voltage changes and slightly discharge.

3. Negative input change - the capacitive "batteries" convey the negative/positive voltage changes and slightly charge.

Hi Cyril. It is interesting how you think of the same issue differently and how you formulate the question and give alternatives for answers that we probably may vote for one of them. Good imagination surely helps understanding/ remembering, e.g. your idea of the capacitor serving as a bridge with some slow (large) vehicles (DC) blocked while the more vibrant vehicles (cars and motorcycles) can pass easily (high frequencies). nice idea.

At the very beginning of your questions you seemed to object the traditional way of teaching the capacitor at amplifier input as DC blocker and high frequency (signal) short. For me, I do believe that this is a nice way as well provided that the students had been subjected to RLC circuit analysis and probably done some RC vector analysis and done some experimental measurements on RC impedance variation with frequency so that the concept of a blocking capacitor for signal frequencies has practical meaning and makes sense when coming to amplifier design and analysis. He should at this level be able to select a proper capacitor value to serve approximately as short circuit for signal frequencies or otherwise they should account for it in input impedance calculations.

As you like to think of the issue differently (like me sometimes) I can add one more idea to yours above. We may think of the capacitor as "Abis" on the way of the signal. High frequency components (signal) can vibrate and therefore cross the (jump over) the abis while low vibrators or static components can't pass it. Thanks. @AlDmour.

Thank you, Ismat... you are precise and penetrative as usual... so that I have no chance to hide my "a bit provocative" intentions:-)

Really, there is something confusing in the fact that I ask questions that I can answer... and really I begin answering them below the question... But it turns out that this is a successful strategy since my initial answers serve as a good basis for generating further answers. I figured out it here, in ResearchGate, a few months ago and my intention is to enlarge it with the help of enthusiastic people loving great circuit ideas...

Obviously, the traditional two ways of teaching capacitive coupling techniques ("DC blocker" and "AC shortener" concepts) are good as they are still dominating:-( But I cherish the hope that, by the help of this discussion (that is immediately listed by Google), I will manage to impose at least one of the four ways of presenting the capacitive coupling suggested above... or, at worst, this will strike at prevailing ideas:-)

Now a little more seriously... As you have also noted, the high-level explanations in the frequency domain are more suitable for circuit design; they are more quantitative than the explanations in the time domain. But as my purpose is always to reveal the basic ideas behind circuits, I prefer using the more qualitative low-level explanations in the time domain.

Thank you for the impressive "vibration" analogies; in addition to "shock-absorber" analogies, they help understanding. I only have found difficult to understand what "abis" means...

The attached Flash movie is a gift for you:-)

Regards, Cyril

Hi Cyril,

I have some problems in understanding the following sentence ("voltage sources" in parallel?)-

Can you help me?

Lutz

"The BJT emitter-grounded amplifier is a special case of this arrangement since its input (the base-emitter junction) behaves as a constant-voltage nonlinear element (a voltage stabilizer). Now we have two (almost) perfect voltage sources connected in parallel and this combination is supplied by the biasing current source (the base resistor and the power supply in the attached picture below)."

Lutz, you are no less penetrating than Ismat:-) I was absolutely sure you will react to this speculation.

Really, precisely speaking, there is only one source - from the left side. The right voltage "source" actually is not a source but a constant-voltage nonlinear resistor (I mean the base-emitter junction) or a voltage stabilizer (Vbe0). So, we have a voltage source (Vin + Vbias) connected in parallel to a voltage stabilizer. The result is almost the same (a kind of a current steering) as in the case of two voltage sources (or two voltage stabilizers) connected in parallel. There is only a difference in the current "picture"...

BTW I have begun thinking about feedback loops and will return soon to this interesting subject (negative feedback).

Regards, Cyril

Sorry Cyril, we commonly use "Abis" in communication engineering (GSM terminology) to mean the interface between base station (BTS) and the base station controller (BSC). Generally, it means a deep slot ( also it means: abyss, chasm ... from Babylon translator). For me, the capacitor looks like an Abis (or abyss) Thanks.@AlDmour.

And to be consistent in my "provocative" style-:), here is my favorite speculation:

If we connect two constant-voltage elements in parallel (voltage source||voltage source, voltage source||constant-voltage nonlinear element or constant-voltage nonlinear element|constant-voltage nonlinear element), a conflict appears and the current vigorously diverts (is steered) between them. Here we have the second situation.

Regards, Cyril

Thank you, Ismat. I like your analogy.

Regards, Cyril

Cyril - thanks for clarification (because I, normally, have problems in visualizing two voltage sources in parallel!). However, does it help to see the B-E junction as a "stabilizer" rather than as a non-linear load? I think, it is not the primary task of the B-E junction to stabilize any voltage. It acts as a load with a voltage across it that controls the Current Ic.

Regards

Lutz

Lutz, you have problems with parallel-connected voltage sources but school textbooks have not - as I can remember from my school years, they claimed that batteries can be connected both in series and parallel. It is not so difficult to imagine what the result would be in the last case...

I have named the base-emitter junction "stabilizer" just for simplicity (insted the longer "constant-voltage nonlinear resistor"). I agree with you that "to stabilize any voltage is not its primary task" but this (voltage-stable) kind of nonlinearity as though "amplifies" the current (by the current-steering phenomenon). So, here I am primarily interested how the transistor input is seen from the side of the input voltage source than what the main fuction of the junction is...

It is interesting for me to see what are your "problems with visualizing two voltage sources in parallel". I hope you do not mean ideal voltage sources; we need some (low) internal source resistance. It is important also if these sources can allow passing a "counter-current" through them (entering the positive source terminal). IMO it is extremely interesting to consider where currents flow in the three possible arrangements described above...

Regards, Cyril

Cyril,

I will not further stress the question of two voltage sources in parallel (ideal or not) - perhaps not too important in practice.

However - I do not consider the B-E junction as a "constant-voltage nonlinear resistor", because I think this is an over-simplification, which does not help at all.

Or does it help to understand things better? I doubt.

Why not treat this path what it is - a load in form of a pn junction with a non-linear V-I characteristic. And you know - in no case it is a "source" (constant or not).

May be I see the whole problem as described by you from the engineering point of view only?

The reason is perhaps the following: In another electronic forum I just have "fighted" against some guys who claim the BJT would be current-controlled because they think the simple equation Ic=beta*Ib would tell something about the physical truth. And for explaining the physical effects that take place in controlling Ic , of course, I need the base-emitter voltage as a RESULT of external biasing.

Do you understand?

Regards

Lutz

Lutz, will we ever get to a consensus about how to look at elements and circuits when understanding them?

If we talk about the diode (p-n or B-E junction), you know my approach - to simplify the natural characteristic depending on the specific application. When I consider: diode switching circuits, I use the major (over) simplification completely ignoring Vf (case 1 in the attached picture below, the IV curve is on the ordinate); voltage stabilizers - the IV curve is a vertical line shifted by Vf from the origin (2); voltage stabilizers with some low internal resistance (3) and finally, logarithmic converters - the natural IV curve (4).

When explaining current steering technique, 3 is the most suitable approximation. In this case, the exact shape of the IV curve is not so important. The only important is the two elements to have relatively low dynamic resistance so that to change vigorously their instant resistances and, as a result, the current flowing through them...

Here is an example of a tricky circuit (a 3-LED tuning voltage indicator) based on the current steering technique. I invented it in 1983 independently from the German inventor Wolfgan Grote (Industrielle Elektronik-Shaltungen) and widely used it in practice. I can't imagine a simpler tuning LED indicator... Can you imagine how it operates (where currents flow)? It is a good mental exercise:-)

A little suggestion: LED0 was green-colored (2.5V), LED2 and LED3 were red-colored (1.8V)

Regards, Cyril

In respect of your "I will not further stress the question of two voltage sources in parallel (ideal or not) - perhaps not too important in practice", I would only remember that in the diferential (long-tailed) pair there are exactly two voltage sources (emitter followers) in parallel that "fight" each other...

"Lutz, will we ever get to a consensus about how to look at elements and circuits when understanding them?"

Hi Cyril, may be, that we get no consensus in some cases. So what?

I think that is quite normal that different methods are used to explain things - and that these methods sometimes are based on a different degree of simplification.

This raises the question: Why simplification at all?

My answer: Of course, simplification is necessary and cannot be avoided - however, - for my opinion - not with the aim: As simple as possible.

Better approach: As simple as necessary.

And this automatically leads to the diode characteristic and your approach No. 3:

"When explaining current steering technique, 3 is the most suitable approximation. In this case, the exact shape of the IV curve is not so important".

For my opinion, this is an "over-simplification" because of the following reason:

It is very important to know the V-I characteristic of the pn junction (exponential law) because the slope of this curve leads to the most important transistor parameter that determines the gain properties: Transconductance gm=Ic/Vt.

Based on your case 3 the transconductance would be constant - which inhibits a application oriented design of a BJT amplifier.

Thus, I see really no reason to use a simplified diode characteristic as given with case 3.

Do you see my point?

Regards

Lutz

Cyril, regarding your differential amplifier example I must confess, that I cannot follow you. Please help me.

I neither see two voltage sources in parallel nor a "dramatic conflict" (as mentioned in the figure). Which voltage sources?

What I see is the following: Two different currents meet in a common node and the sum of both currents is flowing through a resistor. The sum of these currents creates a corresponding voltage drop across the resistor. That`s all. Did I overlook something?

"...Two different currents meet in a common node and the sum of both currents is flowing through a resistor. The sum of these currents creates a corresponding voltage drop across the resistor..." Lutz, I share this explanation too but I would say in addition also:

The two currents are produced by two voltage sources (VCVS or, particularly, emitter followers). I hope you will agree with me that in this arrangement (a long-tailed pair), the outputs of two emitter followers are joined while their inputs are driven in opposite directions. Generally speaking, these emitter followers (including the power supply) act as voltage sources driving a common load (the emitter resistor or current source). They keep up their output voltages by means of a series negative feedback so that each of them tries to keep up its output voltage equal to its input voltage.

But their input voltages change in opposite directions; so their output voltages try to change in the same opposite directions too. As a result, a kind of a "conflict" appears in the common emitter point (like everyday conflicts between people trying to change something common to opposite states). We can say that the two NFB followers disturb each other and each of them reacts to the else's disturbance by vigorously changing its own current (the instant collector-emitter resistance). As a result, the one current vigorously increases while the other - decreases; as though, the current diverts (is steered) from the one to the other leg...

Here is an electromechanical example of this phenomenon - join mechanically the actuators of two servo systems, drive them with opposite input signals and use the consumed currents (the reactions) as outputs...

Good explanation by Prof.Cyril.

P.S.

Hi Cyril, I cannot resist to place some comments to your last contribution, which I can only partly agree upon.

[Quote: The two currents are produced by two voltage sources (VCVS or, particularly, emitter followers).]

I think, the transistor alone cannot be seen as a voltage source - only together with an emitter resistance Re and the corresponding feedback effect (not an IDEAL voltage source, but with a low source resistance Re||1/gm). That means, not the currents are produced by a VCVS, but in contrary: The currents - and the associated current-controlled feedback effect - cause an emitter voltage that is provided with a relatively low source resistance (Re||1/gm).

I think, this view is supported by the diff. amplifier small-signal models, which contain controlled current sources rather than voltage sources.

[Quote: Generally speaking, these emitter followers (including the power supply) act as voltage sources driving a common load (the emitter resistor or current source)]

This sounds strange to me, because the „common load“ is IDENTICAL to the resistor that forms the emitter follower.

[Quote: We can say that the two NFB followers disturb each other and each of them reacts to the else's disturbance by vigorously changing its own current (the instant collector-emitter resistance).]

I think, the coll.-emitter resistance remains nearly uneffected and cannot serve as a reason for a current change.

However, the model provided by you can give a good explanation:

If - as a first step - we imagine two identical emitter followers (but different voltages at the base nodes) we have two different emitter voltages Ve1 and Ve1. Now, I can agree to you that there is something like a „conflict“ if we connect both emitters with each other. However, the conflict is not a dramatic one because we have non-ideal „voltage sources“ (source resistance app. 1/gm).

As a model - we can imagine two equal resistors R1=R2 in series between two ideal dc sources V1 and V2. Then there will be a current between V1 and V2 and the node between R1 and R2 will have the voltage (V1+V2)/2.

Thus, after both emitters are connected, the common emitter node has a voltage Ve between the initial values Ve1 and Ve2. Thus, of course the collector currents will be affected correspondingly.

Insofar - I agree to your picture that both stages "disturb" each other.

Final remark: Voltage sources in parallel are, of course, no problem if they are non-ideal. Such circuits - perhaps together with current sources and other resistors - are typical network examples to be calculated by students.

Regards

Lutz

Edit: In another view the diff. amplifier can be seen as an emitter follower driving a load consisting of a common base configuration. This, immediately, leads us to the known tanh characteristic.

In my experience, seeing the coupling capacitor as an "elastic membrane" for electric charge will draw a very close parallel with a mechanical equivalent. For the dc component the membrane is exponentially extended (=charges). For the ac component, it follows the "vibrations" of the source. This is the way I first understood capacitors operating in a circuit when I was at high-school. This can be also verified by comparing an electric circuit(containing capacitors) and its mechanical dual system (Ogata).

Vasile, thank you for the kind words! All we, involved in teaching and especially in basic circuit courses, need them. To reveal the truth behind the fundamental circuit ideas is a thankless job. We profit nothing by that; sooner we come in for a lot of trouble when proposing new interpretations of well-grounded doctrines. Let's see it in our situation...

Both Ray and you have managed to explain what coupling capacitors do only in 2-3 sentences. This "high-level" explanation (lying mainly in the frequency domain) is simple and clear, and it is suitable for "conventional" educational purposes (99.9999999...999% of situations:-) But what do we do when (0.0000000....01% probability in our modern time:-) a curious, profoundly thinking student (some future Widlar) would like to know what really happens inside - what particular voltages are and where currents flow? Then we have no chance to get out of the confusing situation and we should go down to lower-level explanations in the time domain that really explain the circuit operation. If I fell into such a situation, I would tell the story below about coupling capacitors to my hypothetical curious student...

------

THE PROBLEM. With respect to the input part, BJT (e.g., n-p-n) is a "nasty" amplifying element - it expects the input voltage is positive and varies around 0.6 V. Unfortunately, the AC input voltage varies around 0 V. So, we have to "lift" (bias) it with these 0.6 V (Vbias). Contrary, in the output part, the load expects the output (collector) voltage varies around the ground but it varies around 1/2Vcc (the quiescent voltage); so, we have to drop off it with 1/2Vcc.

THE REMEDY. The simple way to do it, is to add such a steady voltage in series to the varying input voltage (according to KVL). This means simply to connect a voltage source in series to the input one. But this source is floating (we suppose the input source is initially grounded); so it has to be a kind of a battery... and it has to be a rechargeable battery... Now it is easy to guess that a charged capacitor C1 can do this work if only we find a way to keep it charged at 0.6 V. For this purpose, we use a galvanic input voltage source and connect a charging resistor Rb to the power supply.

Now we have to consider the three typical situations in the circuit operation (I have copied below the three pictures from above for your convenience).

1. CHARGING THE "BATTERIES" (ZERO INPUT VOLTAGE).

For clarity, first assume in the beginning, when we switch on the power supply, the input voltage is zero and the left C1 terminal is grounded through the DC zero source resistance. The capacitor C1 and the base-emitter junction are connected in parallel.

The capacitor is initially empty and diverts all the biasing current flowing through the loop E > Rb > C1> ground > E. The base-emitter junction is shunted by the capacitor; it is cutoff and does not affect the charging process.

When the voltage across the capacitor reaches 0.6 V, the base-emitter junction begins diverting a part of the biasing current and finally diverts all the current. The capacitor does not play any role since there is no current flowing through it; the biasing current is determined completely by Rb. Thus, at the end of this initial stage, the input capacitor C1 is "automatically" charged to the needed bias voltage 0.6 V.

Similarly, the output capacitor C2 is "automatically" charged through the galvanic load to the quiescent voltage 1/2Vcc (it is interesting to see how) .

As Vasile has noted, until this initial stage, when the capacitors charge, the amplifier does not work properly and the load can be even damaged if the capacitors have too big capacitances.

We can present this initial stage in an impressive way to our students, if we made them think of the capacitors as of slowly stretching shock-absorbers (the red voltage bars Uc1 and Uc2 give some impression of such an representation).

2. DISCHARGING THE "BATTERIES" (POSITIVE INPUT VOLTAGE).

When the input voltage rises above the ground (becomes positive), the total voltage of the left composed voltage source (Vin + Vc) exceeds the base-emitter voltage (0.6 V). So this source adds an additional current to the initial bias current (set by Rb) through the base-emitter junction thus discharging. But because the capacitance is large enough, the voltage across the capacitor decreases slightly, and the input capacitor conveys the increasing input voltage variations to the transistor base (they are "lifted" with 0.6 V as the transistor "wants").

In a similar way, the output capacitor descends the collector variations with 1/2Vcc thus conveying them to the load as it "wants".

In the impressive mechanical analogy, the shock-absorbers convey the fast input/output movements up.

3. RECHARGING THE "BATTERIES" (NEGATIVE INPUT VOLTAGE).

When the input voltage descends below the ground (becomes negative), the total voltage of the left composed voltage source (Vin + Vc) becomes less than the base-emitter voltage (0.6 V); so the composed source begins diverting a part off the biasing current thus recharging. Again, because the capacitance is large enough, the voltage across the capacitor increases slightly (it restores the charge), and the input capacitor conveys the decreasing input voltage variations to the transistor base ("lifted" with 0.6 V as the transistor "wants").

In a similar way, the output capacitor conveys the increasing collector voltage variations to the load as it "wants".

In the mechanical analogy, the shock-absorbers convey the fast input/output movements down.

As a conclusion, during the positive half-wave, the capacitors discharge but during the next negative half-wave, they restore the initial charge (voltage). Thus they act as "rechargable bias batteries" with almost steady bias voltages across them...

I have told this story to you, my colleagues, in such a figurative way to show how colorfully can be this world presented to our students... I really love you too!

Regards, Cyril

Lutz, with the question below, I have started a series of questions dedicated to the current steering technique and its implementations in differential amplifiers, op-amps and other widespread electronic devices. We can continue our discussion there.

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_implemented?

But let me say a few final words about your last comment. IMO we have to distinguish between the common mode and the differential mode in the operation of the long-tailed pair.

In the common mode (e.g., with the two bases connected to a common input source), there is no doubt that we have exactly two emitter followers working in parallel with respect to the load. There is no doubt that the common emitter resistor is the common load (exactly, "the „common load“ is IDENTICAL to the resistor that forms the emitter follower" and there is nothing strange in this fact) for the two emitter followers (if you prefer, split it into two separate resistors connected in parallel). There is a negative feedback (emitter degeneration) as well.

So, the two transistors act as "cooperating" voltage sources. Of course, in respect to the collector resistors, the two transistors act as current sources (voltage-to-current converters) and the collector resistors are their loads. You can think of this arrangement as of two common-emitter amplifying stages with a common resistor in the emitters. So, really "the two currents are produced by two voltage sources (VCVS or, particularly, emitter followers)".

Now about the operation. You can explain it in various ways - that currents produce voltage or voltages produce currents, but the truest "low-level" explanation is only one: the transistors CHANGE THEIR INSTANT COLLECTOR-EMITTER RESISTANCES, accordingly, their collector (emitter) currents and partial voltages with the only purpose to keep their base-emitter voltages about 0.6 V. So, I can't understand your assertion, "the coll.-emitter resistance remains nearly uneffected and cannot serve as a reason for a current change". If the instant resistance between the collector and emitter stays "unaffected", who is this magical way in which a transistor can change the current? Remember:

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

I will support my explanations with the possibly simplest but extremely useful for understanding equivalent electrical circuit (an electrical analogy of an electronic circuit:-) Think of this arrangement (the long-tailed pair) as a circuit of three resistors - two variable "pull-up" resistors (T1 and T2, temporarily neglect the collector resistors) and one steady pull-down resistor (Re). Or, you can think of this combination as of a kind of a "double voltage divider" with two separate upper legs and one common lower leg. Or, even better, imagine a mechanical system of three springs - two pull-up and one pull-down.

In the common mode, at positive input quantity, the two pull-up elements become stronger and begin simultaneously pulling the common joint upward whlie the pull-down element slightly resist them. As a result, they lift the common point.

Now about the differential mode...

There are always some reactance associated between junctions of devices which are used for amplifiers. For proper signal transfer and amplification this reactance had to be compensated. The bypass or coupling capacitors helps for frequency response to get as desired.

Hi Erik,

do you like matured questions? :-))

Josef

What do you mean, Josef?

Hi Josef, Hi Cyril,

I think we have a short circuit between questions ;-)

ERIK

Hi Cyril,

Only that there had long been quiet.

And now the move again. Number of "wievs" is greit.

Sincierly

Josef

Erik, maybe a "bridge between questions":)

Josef, a really interesting fact (32455 views)... So much attention to a humble capacitor? How can we explain it?

You just make good questions. :-)

Josef

Well, today's "semiconductor youngsters" usually do not know what is it the proper voltage and soothing glow cathodes. Don't touch of anode! :-))

Hi, as per you mention on the statement above, DC blocking capacitor. In amplifier such  BJT FET etc there is act as the road block when AC signal flow through their input is not to disturb the quiescent point and operation point (Q-point) of the circuit. Other word said that decoupling the AC and DC.....Anyone kindly fix me, if i'm wrong.

I too have a same doubt.