How we can physically imagine transistor saturation current?

[LvW]

Claude - I only can hope that you never will face the situation for explaining to students how a transistor works.
Sorry - but I`ve got the impression that you cannot identify the main part of a certain problem (you cannot see the wood for the trees).
LvW

 

[cabraham]

Claude - I only can hope that you never will face the situation for explaining to students how a transistor works.
Sorry - but I`ve got the impression that you cannot identify the main part of a certain problem (you cannot see the wood for the trees).
LvW

Been there done that. My students praised my teaching methods. Please answer me this one LvW. What courses have you taken to acquire bjt skills?

I took 2 quarters of electronic circuit theory, 1 quarter of rf (amps, tuned networks, modulation, detection, etc.), one quarter of integrated electronics (op amp, comparator, logic gates, architecture), 1 year of circuit theory, 1 year of e&m fields, 1 year of digital logic, 2 quarters of energy conversion, 1 quarter of modern physics (kinetic theory of matter, relativity, quantum mechanics), 1 quarter of solid state physics from physics dept., 1 year of controls, and that was undergrad.

In grad school 1 quarter of semiconductor physics, 1 semester of advanced semiconductor physics, 1 semester of device fabrication, 1 semester of sensors and their physics, 1 semester of very large scale CMOS devices, 1 quarter of advanced controls, 1 quarter of microprocessors, 1 semester of power, 1 semester of signals & systems, to name some.

You never prove anything. I hoped my simple resistive divider would clarify things, please respond to that example as to why you disagree. Transistor operation is NOT what the conflict involves. What is not being agreed upon is the basic nature of voltage & current and the relation between the two. Using your logic, every current is controlled by a voltage, so therefore there is no such thing as a current controlled device. If that was true, we would never classify things as VC or CC, because VC is always understood. A FET is a FET, a bjt is just that, a motor is a motor, etc. Semiconductor OEMs call bjt parts CC, they would never do so is what you say was true.

Also, why is a motor VC? Classic uni teachings treat torque as CC and speed as VC. I know your answer but you should say it, not me.

Claude

 

[cabraham]

I see. The current is mysteriously injected into the transistors base terminal without any change in base terminal voltage. The transistor just senses it is supposed to draw more current and "sucks" it out of the input circuit in a way that keeps Vbe constant. Sorry --- not buying it.

It's interesting the way increasing the voltage on an RC circuit shows current increase at the capacitor before voltage increase. Funny how the voltage lags the current. Yet it was the voltage increase that caused the change.

BTW, I won't read explanations longer than three sentences. If you can't say it in three simple succinct sentences, you are blowing smoke.

Assumption, no more. Which voltage "caused" the current to change? Certainly not the voltage across the cap terminals. This I leads V in a cap has been law since 19th century, but it wasn't until you and LvW came around that an error was uncovered?
Claude

 

[meBigGuy]

WOW --- a braggart as well. People who have to resort to their backgrounds in an attempt to build credibility are obviously lacking in earned credibility.

It is so sad that you cannot seem to answer questions in a clear manner and feel the need to resort to belittling attacks. One more and I'm done. Either you act mature, or this ends. (I really don't care since it is apparent you are not going to answer the questions) I'm sorry (not really) you can't see that the externally applied voltage to the RC caused the increase in current, and transient capacitive effects made it appear (to some) that the voltage at the capacitor didn't change. But, everyone knows it did change because I = Cdv/dt, so there had to be a dv/dt ACROSS THE CAPACITOR to cause the I. You are letting the transient capacitive and charge transfer effects cloud your judgement regarding cause and effect. And yes, I understand you could say that current forced into the cap caused the dv/dt, but it started with the voltage applied to the base terminal.

You have not responded to explain how the following happens. Try to stick to the point. :
"The current is mysteriously injected into the transistors base terminal without any change in base terminal voltage. The transistor just senses it is supposed to draw more current and "sucks" it out of the input circuit in a way that keeps Vbe constant.

 

[LvW]

...
I took 2 quarters of electronic circuit theory, 1 quarter of rf (amps, tuned networks, modulation, detection, etc.), one quarter of integrated electronics (op amp, comparator, logic gates, architecture), 1 year of circuit theory, 1 year of e&m fields, 1 year of digital logic, 2 quarters of energy conversion, 1 quarter of modern physics (kinetic theory of matter, relativity, quantum mechanics), 1 quarter of solid state physics from physics dept., 1 year of controls, and that was undergrad.

In grad school 1 quarter of semiconductor physics, 1 semester of advanced semiconductor physics, 1 semester of device fabrication, 1 semester of sensors and their physics, 1 semester of very large scale CMOS devices, 1 quarter of advanced controls, 1 quarter of microprocessors, 1 semester of power, 1 semester of signals & systems, to name some.
...
Claude

Wow - this changes everything.
Against the described background of your academic career you now have convinced me that the BJT is, of course, a current-controlled device.

 

[meBigGuy]

You have not responded to explain how the following happens. Try to stick to the point. :
"The current is mysteriously injected into the transistors base terminal without any change in base terminal voltage. The transistor just senses it is supposed to draw more current and "sucks" it out of the input circuit in a way that keeps Vbe constant.

When you precisely and succinctly reply to this issue using your years of academic expertise, please note that we are speaking of the base terminal, not the base region. Again, how, without an increase in base terminal voltage, does the base current into the base terminal increase? Keep it short and sweet.

Remember, The less you know about a subject, the longer it takes you to explain it. (or is it "If you can‘t explain it simply, you don‘t understand it well enough")

 

[cabraham]

When you precisely and succinctly reply to this issue using your years of academic expertise, please note that we are speaking of the base terminal, not the base region. Again, how, without an increase in base terminal voltage, does the base current into the base terminal increase? Keep it short and sweet.

Remember, The less you know about a subject, the longer it takes you to explain it. (or is it "If you can‘t explain it simply, you don‘t understand it well enough")

Mebigguy, you don't decide how many sentences it takes to explain something. You're asking me to summarize circui theory in 3 sentences. Why 3, not 7? Just who decides that. As far as taking a long time to explain something, maybe I have a hostile audience that won't listen.

Base current increases due to external source driving it. Say an antenna receives a radio signal. Antenna feeds an input preamp bjt stage. Program signal increases. Antenna and cable incur an increase in both I & V. Signal travels through cable, arrives at base terminal. Base terminal is at old value of voltage. Charges enter base terminal and current is increased immediately, and the bulk silicon resistance incurs an increased voltage drop. Base-emitter voltage increases due to increased antenna signal and increased current drops a larger voltage in the Si bulk resistance. But when the increased current crosses the b-e junction, recombination happens and the depletion layer is increased due to higher charge density. The depletion layer voltage increases as well, this is depletion layer Vbe'. Terminal voltage is Vbe, they are nearly equal.

Regarding your previous remark about the R-C network, the increase in cap voltage came as a result of current charging the cap. But this current increase came about how? A battery on the left side of the open switch is converting chemical energy into electrical. The redox action propels ions against the E field. A current inside the battery transports positive ions to the pos terminal, and likewise for negative. This current gives rise to the battery terminal voltage. When the switch closes, charges move to the right towards the resistor and capacitor. Until charges arrive at cap, cap voltage is zero but current is moving from battery, through switch, through resistor and towards cap. Then the current reaches the cap. Current has been established while cap voltage is zero. As current enters cap plates, said cap voltage build up, which results in current decaying.

Cap voltage increased due to current, that current was motivated by switch closure and change in voltage, and that voltage was motivated by battery internal current, that current was motivated by chemical redox.

The bold italics is my short and sweet summary. Notice that the cap voltage is NOT the same voltage that motivated the current in the 1st place. I will clarify.
Claude

 

[cabraham]

Wow - this changes everything.
Against the described background of your academic career you now have convinced me that the BJT is, of course, a current-controlled device.

Actually, this disagreement is not about bjt devices, it is all about the fundamental relation between I & V. Please answer my question if you wish regarding the resistive voltage divider. Thanks.
Claude

 

[cabraham]

WOW --- a braggart as well. People who have to resort to their backgrounds in an attempt to build credibility are obviously lacking in earned credibility.

It is so sad that you cannot seem to answer questions in a clear manner and feel the need to resort to belittling attacks. One more and I'm done. Either you act mature, or this ends. (I really don't care since it is apparent you are not going to answer the questions) I'm sorry (not really) you can't see that the externally applied voltage to the RC caused the increase in current, and transient capacitive effects made it appear (to some) that the voltage at the capacitor didn't change. But, everyone knows it did change because I = Cdv/dt, so there had to be a dv/dt ACROSS THE CAPACITOR to cause the I. You are letting the transient capacitive and charge transfer effects cloud your judgement regarding cause and effect. And yes, I understand you could say that current forced into the cap caused the dv/dt, but it started with the voltage applied to the base terminal.
To cause I there had to be a dv/dt across the cap? That is pure dogma. The cap was uncharged, charges arrived and entered cap plates, and voltage increased as a result. The quantity "dv/dt" times C is the current I. But dv/dt = I/C, is the relation under these conditions. The rate at which voltage rises, i.e. dv/dt, is determined by I/C. You're claiming the rate of voltage change "caused" I, but that is impossible. Before cap voltage even began to change I was already existing. What makes you think you can judge cause and effect. Please take us through step by step in the R-C battery switch circuit, and give us the causes and effects.

Remember, the shorter the explanation, the --- oh never mind, elaborate if need be.
Claude

You have not responded to explain how the following happens. Try to stick to the point. :
"The current is mysteriously injected into the transistors base terminal without any change in base terminal voltage. The transistor just senses it is supposed to draw more current and "sucks" it out of the input circuit in a way that keeps Vbe constant.

 

[LvW]

Actually, this disagreement is not about bjt devices, it is all about the fundamental relation between I & V. Please answer my question if you wish regarding the resistive voltage divider. Thanks.
Claude

OK - I will do you a favor - here is my answer:

Claude: Vbe is not applied externally.

For designing a BJT amplifier stage the classical procedure is to „open“ the transistor with an externally applied voltage of app. 0.65 volts.
Please note that it is not necessary to teach me that nobody uses for this purpose a battery or something like that. Everybody knows how this accomplished (even me).

Claude: Take a 12 volt battery, with a pair of 1.0 kohm resistors wired in series forming a divider. A switch is thrown and the circuit is energized.
The steady state current is 6.0 mA, and 6.0 volts is dropped across each resistor. Is this agreeable?

Do you really expect an anwser?

Claude: So examine the second resistor, connected to ground, whose voltage is 6.0 volts, one end at 6.0V, the other end ground. With me so far?
Does this 6.0 volts "drive" the 6.0 mA current in same resistor?

Why such an absurd question? Has anybody claimed so?

Claude: The voltage drop, 6.0 volts, across the 2nd resistor cannot be "driving" the 6.0 mA current in the same resistor. If you disagree, please explain.
Using your logic, every current is controlled by a voltage, so therefore there is no such thing as a current controlled device.

You are fighting against windmills. My only statement was „no current without a driving voltage“.
I ask myself: Why are you twisting the meaning of words ?

Regarding „ELI the ICE man“ and your simulation results - based on an arbritary RC circuit - there are two possibilities:

1.) If you are referring to the steady-state sinusoidal change of electric quantities the observable phase differences cannot say anything about cause and effect resp. the question „what comes first“. I hope you do not need further explanation.
2.) If you refer to the behaviour immediately after t=0 (your first graph in post#22) the result is not very surprising: Connecting a voltage source to a cap, the current hardly can be in advance to the voltage and vice versa. What do you want to proove with this simulation?
More than that, the rule of „ELI...“ does not apply at all (the rule is based on phasor representations and requires sinusoidal signals - hence, steady-state).

Summary: All your simulations and the whole „ELI..“ story is meaningless and does not proove anything.

 

[meBigGuy]

It is really very simple

Ic = Is * (e^(Vbe/Vt) - 1) Base current is a side effect. Read Gray and Meyer or any analog design textbook.
The fact that the above holds nearly regardless of Ib is not debatable.

PERIOD. End of Game

Claude is being fooled by transient effects caused by depletion capacitances. The fact that A preceeds B or vice versa is a transient effect, not an indicator of a root cause.
In the steady state, the above holds.

 

[LvW]

Base current is a side effect. Read Gray and Meyer or any analog design textbook.

...or some other reliable sources.

Univ. of Berkeley:
Ic is determined by the rate of electron injection from the emitter into the base, i.e., determined by VBE. An undesirable but unavoidable side effect of the application of VBE is a hole current flowing from the base, mostly into the emitter. This base (input) current, Ib, is related to Ic by the common-emitter current gain,

Stanford Univ.:
Conceptual View of an NPN Bipolar Transistor (Active Mode): Device acts as a voltage controlled current source: VBE controls IC.

Mass. Inst. of Tech. (MIT):
Bipolar Junction Transistors: basic operation and modeling…… how the base-emitter voltage, VBE, controls the collector current, IC:...

Barry Gilbert:
BJT is a voltage-controlled current-source; the base current is purely incidental (it is best viewed as a „defect“)

Winfield Hill (Co-author Art of Electronics):
The physics and formulas are the key: In the case of the transistor we have the solid-state physics resulting in the rigorous Ebers-Moll formulas, with their precise prediction for collector current IC from VBE. Just because you can successfully bias a few BJT circuits with current doesn't meant they're current-controlled devices.

 

[LvW]

For finalizing the discussion, here is an excerpt from W. Shockleys patent document (Sept. 1951).

If we idealize the structure for the moment and neglect any resistances at the semiconductor contacts the comparison between this device and a vacuum tube becomes clear. In place of the grid there is the p-region, which can be charged in respect to Ne by holes. This modulates the flow of electrons from Ne into P just as the charge on the grid modulates the flow of electrons from the cathode...Thus the fact that there are two processes of conduction through the p-region permits control to take place in a way similar to that in a vacuum tube.
......

The electron current due to the difference in potential between E and B is also relatively insensitive to collector voltage.

 

[cabraham]

WOW --- a braggart as well. People who have to resort to their backgrounds in an attempt to build credibility are obviously lacking in earned credibility.

It is so sad that you cannot seem to answer questions in a clear manner and feel the need to resort to belittling attacks. One more and I'm done. Either you act mature, or this ends. (I really don't care since it is apparent you are not going to answer the questions) I'm sorry (not really) you can't see that the externally applied voltage to the RC caused the increase in current, and transient capacitive effects made it appear (to some) that the voltage at the capacitor didn't change. But, everyone knows it did change because I = Cdv/dt, so there had to be a dv/dt ACROSS THE CAPACITOR to cause the I. You are letting the transient capacitive and charge transfer effects cloud your judgement regarding cause and effect. And yes, I understand you could say that current forced into the cap caused the dv/dt, but it started with the voltage applied to the base terminal.

You have not responded to explain how the following happens. Try to stick to the point. :
"The current is mysteriously injected into the transistors base terminal without any change in base terminal voltage. The transistor just senses it is supposed to draw more current and "sucks" it out of the input circuit in a way that keeps Vbe constant.

It is really very simple

Ic = Is * (e^(Vbe/Vt) - 1) Base current is a side effect. Read Gray and Meyer or any analog design textbook.
The fact that the above holds nearly regardless of Ib is not debatable.

PERIOD. End of Game

Claude is being fooled by transient effects caused by depletion capacitances. The fact that A preceeds B or vice versa is a transient effect, not an indicator of a root cause.
In the steady state, the above holds.

Ic=alpha*Ies*exp((Vbe/Vt)-1)
Ic=beta*Ib
Ic=alpha*Ie
All 3 hold. Neither equation "proves" cause & effect.

Claude

 

[cabraham]

The attached pdf's are from MIT and Georgia Tech. MIT slides 20 through 23 model the bjt using current controlled current source, i.e. Ic=alpha*Ie. Ga. Tech describes bjt as current controlled. I don't like using Ib to control Ic as it makes circuit beta dependent, but Ie to Ic relies on alpha, very precise, consistent, and reliable.

Of course Ie is indeed strongly related to Vbe. But it is equally true that Vbe is a function of Ie. Professors from leading institutions all over the world know the Ie/Vbe relation via Shockley/Ebers-Moll laws. But trying to ascertain which is the "controller" and which is "controlled" is not possible simply by examining equations. I've searched the web and found professors from Stanford, MIT, Caltech, UCLA, Ga Tech, Purdue, my current uni Case Western Reserve where I am a doctoral candidate, and you can find a prof who states that Ie controls Ic, and that Ie is controlled by Vbe since they are related. Others regard Ie as what controls Ic, and Vbe is incidental.

But this "chicken-egg" paradox can be broken. We can settle once and for all which one is in control but only under conditions. That is why I insisted that transient events be studied. I can post slides from On Semi, Fairchild, and others who produce bjt devices stating they are current controlled and have done so on this forum and others. But every prof, or practitioner who claims otherwise comes down to this:

"Sure Ie 'controls' Ic, but what controls Ie? Their answer is Vbe, and their proof is Shockley equation, herein called "SE". But SE, (and E-M, Ebers-Moll) for that matter only relate Ie and Vbe. They do not tell us which comes first or which follows. If the b-e junction is perturbed, and Ic increases, we know that Vbe and Ie both increased as well. So which is correct:

1) The increase in Vbe produced increase in Ie, and subsequently increase in Ic. Ie changed due to Vbe changing.
2) The increase in external source current and/or voltage propagated to the b-e junction changing Ie. As a result Ic changed and Vbe as well. Ie controls Ic, Vbe is incidental.

So which is in control, which is incidental? Only transient tests can answer, which I provided. Time for lunch. Although this is important, so id food. More later.

Claude

 

[LvW]

I must admit that I do not have the time (and motivation) to continue the discussion at this point.
In my post#42 I have provided some statements from - I think - reliable sources with high reputation.

For all readers/forum members who, up to now, haven`t heard about one of these knowledge sources - Barrie Gilbert - here are some background infos which show his reputation:

Between 1970-1972 he was Group Leader at Plessey Research Laboratories.
He later joined Analog Devices Inc. and was appointed ADI Fellow in 1979. He manages the development of high-performance analog ICs at the NW Labs in Beaverton.
For work on merged logic he received the IEEE “Outstanding Achievement Award” (1970) and the IEEE Solid-State Circuits Council “Outstanding Development Award” (1986).
He was Oregon Researcher of the Year in 1990, and received the Solid-State Circuits Award (1992) for “Contributions to Nonlinear Signal Processing”.
He has written extensively about analog design and has five times received ISSCC Outstanding Paper Award.
He has been issued over 40 patents and holds an Honorary Doctorate from Oregon State University.

 

[meBigGuy]

I have a real serious issue with your assertion that Ie is "in control" when I can have two transistors with different beta showing the same gain (as described way back somewhere). Ie is different in each transistor (since Ib is different and Ie = Ib + Ic), but the gain is the same. Vbe produces the same gain independent of beta, except for the transient effects caused by depletion capacitances which you have twisted to try to prove your point.

You still haven't answered how the transistor "knows to suck" more base current if Vbe did not change (and you are not describing an RC like transient effect)

There was a "failed?" movement in the past to move to current based design methodology, which Gilbert commented on. http://cas.ee.ic.ac.uk/people/dario/files/E416/gilbert-voltagemode-currentmode.pdf.

I will go through the two papers you linked to. I don't have time at the moment.

 

[meBigGuy]

Can we say that, from a "what's happening inside the device perspective", Ib controls Ic. I'm just adding that Vbe controls Ib. Ib = Vbe/rbe

Current flow from the base lowers the threshold for Ic current flow.

We have many world class analog designers here, and every one of them said they see the bjt as a transconductance device. Not one could relate to my statement that "someone is saying that Ie controls Ic" .

You can't have an Ie with Ib, so saying Ie is in control seems weak.

 

[LvW]

meBigGuy - I know the linked paper from B. Gilbert. I think - and that`s only one aspect within his contribution - that, of course, the pair voltage-current is NOT a chicken-egg case as some people believe ("Voltage can cause current and current can cause voltage" and therefore, both would be equivalent).

 

[meBigGuy]

Yeah --- that's what Gilbert said.

So, if Claude said "yeah, one way you could look at the bjt is as a transconductance device, but here is this other incomprehesible model I like but can't support mathematically, rather only through obfuscation" that would be 1 thing. But he called "Vbe controls Ic" a Myth, and says it is totally wrong.

To my mind what he is saying is a personal theory, not supported by mainstream science and borders on the stuff that is against forum rules. I've tried hard to make sense of it, but it just isn't there.

In fact, from the first pdf Claude just posted, on slide 11, it's titled " Bipolar Junction Transistors: basic operation and modeling… … how the base-emitter voltage, vBE, controls the collector current, iC"

I'd be happy if he could just say that Vbe controls Ie.

LvW, I think your two transistor example you described earlier is wrong (different beta, same gain). If there are different betas, there will be different gain unless there is exactly balancing different rbe. (since Ib = Vbe/rbe, Vbe is same for both devices, so Ib is the same for both devices, so, since beta is different, Ic is different). the gm you pointed to is a linearization.

 

[LvW]

LvW, I think your two transistor example you described earlier is wrong (different beta, same gain). If there are different betas, there will be different gain unless there is exactly balancing different rbe. (since Ib = Vbe/rbe, Vbe is same for both devices, so Ib is the same for both devices, so, since beta is different, Ic is different). the gm you pointed to is a linearization.

No - I don`t think so.
If the DC current remains the same also the transconductance does not change (gm=Ic/VT) - and the voltage gain still is gmRc/(1+gm*RE) .
This is true for RE finite oand RE=0.
The only difference is the input resistance (smaller for larger Ic).
In contrast, a larger DC current Ic (same transistor, same beta) causes an increase in gm and voltage gain.

 

[meBigGuy]

I'll have to go back to the analog designer that told me that and understand why he said that gmRl was a useful linear approximation and that actually Ib = Vbe/Rbe. Do you agree that if Ib is the same, then the gains would be different due to different beta?

 

[Jony130]

Jony130, I am still in a learning process (who is not?) - and this is not ironic - but I really don`t understand the meaning behind ...will depend on beat value, not by much but it will ( β/(β+1 ) . Please help me improving my english knowledge.

Forget about it, I was using the wrong small-signal model and this is why I jump with this ( β/(β+1) factor. And I also forget that β/rpi = gm.
Av = Rc/re * ( β/(β + 1) = (β/(β+1)*re)*Rc = (β/rpi)*Rc and β/rpi = gm.
So for a constant gm and difference in β value the voltage gain will be the same. And the only thing that has changed is rpi.

As for the "control" what about phototransistor? Is phototransistor a voltage or current controlled device? Or may be a "light/photon" controlled device.

 

[LvW]

I'll have to go back to the analog designer that told me that and understand why he said that gmRl was a useful linear approximation and that actually Ib = Vbe/Rbe. Do you agree that if Ib is the same, then the gains would be different due to different beta?

For comparing the voltage gain values of two different transistors (different beta values) I think it is only "fair" to use the same DC quiescent current in both cases. For this purpose, the DC bias must be adopted correspondingly.
Now - in both cases we have the same transconductance gm and, hence, the same voltage gain (formula given in post#51).
(That`s what also Jony130 has stated above).
Of course, in both cases the base currents (DC as well as ac) and the dynamic input resistance h11 (y11) will be different. But this has no influence at all on the voltage gain between base and collector.

To answer yout last question: Yes, if we keep Ib constant and - at the same time - have different beta values we also have diffent Ic values and, hence, different transconductances gm (different gain). But why is gain different? Again, because the transconductance gm has changed. And gm is the slope of the Ic=f(Vbe) curve!
The current gain beta (as we have seen: not a good and appropriate term) does not appear at all in the gain formulas.

By the way: Speaking about gain, input and output resistances, we always are speaking about "linear approximations". This is always the case when we are analyzing non-linear circuits. This is not any kind of "drawback", but an unavoidable consequence of small-signal analyzes.

 

[LvW]

In fact, from the first pdf Claude just posted, on slide 11, it's titled " Bipolar Junction Transistors: basic operation and modeling… … how the base-emitter voltage, vBE, controls the collector current, iC"
I'd be happy if he could just say that Vbe controls Ie.

Yes - it seems that Claude is trying to convince us and others that the BJT is voltage controlled. Funny situation.

 

[meBigGuy]

For comparing the voltage gain values of two different transistors (different beta values) I think it is only "fair" to use the same DC quiescent current in both cases. For this purpose, the DC bias must be adopted correspondingly.

You lose me here. If I take two NPN transistors of different beta, each with 1K collector resistor to 5V, Emitter to ground, and bases connected together, their gain will be different. I think this because Vbe directly controls Ie, not Ic. That's what Ebers-Moll says.

 

[LvW]

You lose me here. If I take two NPN transistors of different beta, each with 1K collector resistor to 5V, Emitter to ground, and bases connected together, their gain will be different. I think this because Vbe directly controls Ie, not Ic. That's what Ebers-Moll says.

Big misunderstanding! May I repeat my first sentence from post#54?

For comparing the voltage gain values of two different transistors (different beta values) I think it is only "fair" to use the same DC quiescent current in both cases. For this purpose, the DC bias must be adopted correspondingly.

Hence, in my example both base nodes are, of course, NOT connected because both transistors need a different bias scheme (for different IB but equal IC and equal DC operating point). That is the reason, the input resistance will be different in both cases (however, without any influence on the voltage gain).

I think, for comparing two different -but similar - circuits we must define, of course, the operating conditions for ensuring a "fair" comparison. As another example, comparing the gain capabilities of a FET and a BJT, I think both amplifiers should have the same load resistor (RC resp. RD) and the same operating point. Otherwise a comparison is useless and without any value.
And the same applies in our case: For comparing the influence of beta on the voltage gain, both BJT units must have the same operating point - otherwise a fair comparison is not possible (remember: Even one single BJT unit has different gain values for different operating points).

 

[meBigGuy]

I misunderstood you way way back in the beginning. no problem. I get it.

Moving on:

In active mode, Ie is totally controlled by Vbe in a beautiful way that includes Ib. That just wraps this thread up completely for me. (I had it wrong in an earlier post)

Claude can post all he wants to regarding his misunderstandings of transient effects caused by internal capacitances.

 

[LvW]

Claude can post all he wants to regarding his misunderstandings of transient effects caused by internal capacitances.

Yes - he continuously complicates the discussion by introducing effects which may overshadow the basic question. Therefore, the whole "ELI the ICE man" story is useless - as far as our main subject is concerned. Or he is teaching us some commonplace truths (truisms), for example "Emitter current and base current, LvW, are NOT the same, trust me." And, of course, one could ask - why such silly questions?

 

[LvW]

Readers may remember my claims that it is not necessary to go down to molecular physics (charged carrier level) for explaining how the collector current Ic is controlled (voltage Vbe or current Ib). Yesterday, by accident, another outstanding and very simple example has come into my mind: Darlington transistor:

It is known that the combined output-to-input current ratio for the classical Darlington combination (intentionally, I do not use the misleading term "current gain") is the product of the corresponding β values of each transistor: β,d=(β,1*β2,).

Now, let`s compare the voltage gain of such a Darlinton array with a single transistor - identical to the one with ic/ib=β,2.
For a fair comparison, in both cases we again should allow the same DC operational point.

As a result, we will find that the voltage gain of the Darlington combination is only 50% of the value obtained with a single transistor - although the Darlington stage has a beta value β,d that is β,1 times larger than that of the single transitor (β,2).

The reason is obvious: One can show that for the transconductances of both devices we have gm,d=0.5*gm,2.
This clearly prooves that the BJT is voltage controlled because it is the the transconductances gm=dVbe/dIc that determines the Ic variations and, hence, the voltage gain.