How we can physically imagine transistor saturation current?

[goodphy]

Hello.

I've known that saturation current occurs when V_CE is less than V_BE, which means, the base-emitter is forward-biased and so it is for collector-base.

For NPN transistor, electrons from emitter (majority carrier of emitter) is able to go up to only base, no further travels into emitter. The electrons from collector (also majority carrier of collector) can go only up to base.

So, all currents from both ends of the transistor finally arrives at the base and it should go through base terminal. It means base current is the most strong than emitter and collector current?

I've always though base current is weakest one. Where am I wrong?

 

[Svein]

I always think of saturation as the point where the collector current no longer depends on the base current.

 

[meBigGuy]

So, all currents from both ends of the transistor finally arrives at the base and it should go through base terminal. It means base current is the most strong than emitter and collector current?

What the heck does that even mean?

 

[goodphy]

What the heck does that even mean?

In this case of NPN transistor, base-emitter and collector-base are both forward-biased, thus we can expect that there are two currents, one is from emitter to base, other is from collector to base. (We have two forward-biased junction.) Then at the base, two currents are combined and only place they can go through is base-terminal. Thus base current is sum of two currents.

Do you think my picture of saturation current right?

 

[LvW]

Thanks for giving detailed comments.
Oh..thus...basically my picture of saturation current is right?
Then...hmm...I'm very surprised that base current is the most strongest. It looks I_C = βI_B no longer holds in this regime.
In this case, why is this regime called "saturation"? What is saturated? collector current is saturated or maximized? why?

Sorry - I have deleted my post because it was not correct (sign error).
In the saturation region, the base current IB is larger than the expression IB=IC/beta..

EDIT/UPDATE:
At first, let me say that it is not easy to find a "visual" description of the effects which can be observed for saturation operation. Let me try the following explanation:

1.) For VCE=0 we have only one energy source in the circuit (VBE) and it is true that the base current IB is divided into two parts: IE and IC. In this case, IB is larger than IC. That means: The equation I have given in my first (deleted) post holds: IB=IE+IC. Note: IC is in a direction opposite to the classical one.

2.) However, as soon as VBE>VCE>0 the situation changes a bit. The reason is (a) that the base region is very thin and (b) the doping profile of the transistors. Because both pn junctions are now forward biased they are effective in a series connection and provide a rather small resistance. Therefore, even a rather small voltage VCE is able to allow a current between C and E. This is possible also because the emitted electrons from the emitter have enough moving energy to cross the very small base region and go to the collector.

3.) But - at the same time - the effect as described above under 1.) makes the collector current smaller as dictated by the formula IC=IB*beta (because there is a small current from B to C in the opposite direction).
Thus, the resulting current IC is smaller than IC=IB*beta which means: IB>IC/beta.

4.) Therefore, the saturation region is decribed by this condition: IB>IC/beta.

UPDATE 2:
Some additional considerations regarding 1:) and 2.) :
The described behaviour can also be observed with a closer look to the output characteristics IC=f(VCE) of a BJT: The various curves for different fixed values for IB resp..VBE do NOT cross the origin. For VCE=0 the current IC is slightly negative (collector part of the base current) and - as a consequence, we have IC=0 (both parts in opposite direction cancel each other) for a very small value of VCE (some tenth of mV).

 

[meBigGuy]

In this case of NPN transistor, base-emitter and collector-base are both forward-biased, thus we can expect that there are two currents, one is from emitter to base, other is from collector to base. (We have two forward-biased junction.) Then at the base, two currents are combined and only place they can go through is base-terminal. Thus base current is sum of two currents.

Do you think my picture of saturation current right?

Nothing about it is right. In NPN, there is never current from emitter to base or from collector to base.

Please look at a formula that relates base-to-emitter current to collector-to-emitter current in the linear region. What is that ratio called? When that ratio is reduced to 10 by overdriving the base, most manufacturers quote that as saturation and measure Vbe, and Vce.

Please read some articles on saturation and stop trying to make up your own models.

 

[LvW]

meBigGuy - I think, you are a bit too harsh. For my opinion, the questioner only has mixed-up the direction of the electron movements and the direction of current (which is nothing else than a definition!), see his first posting.

 

[goodphy]

Sorry - I have deleted my post because it was not correct (sign error).
In the saturation region, the base current IB is larger than the expression IB=IC/beta..

EDIT/UPDATE:
At first, let me say that it is not easy to find a "visual" description of the effects which can be observed for saturation operation. Let me try the following explanation:

1.) For VCE=0 we have only one energy source in the circuit (VBE) and it is true that the base current IB is divided into two parts: IE and IC. In this case, IB is larger than IC. That means: The equation I have given in my first (deleted) post holds: IB=IE+IC. Note: IC is in a direction opposite to the classical one.

2.) However, as soon as VBE>VCE>0 the situation changes a bit. The reason is (a) that the base region is very thin and (b) the doping profile of the transistors. Because both pn junctions are now forward biased they are effective in a series connection and provide a rather small resistance. Therefore, even a rather small voltage VCE is able to allow a current between C and E. This is possible also because the emitted electrons from the emitter have enough moving energy to cross the very small base region and go to the collector.

3.) But - at the same time - the effect as described above under 1.) makes the collector current smaller as dictated by the formula IC=IB*beta (because there is a small current from B to C in the opposite direction).
Thus, the resulting current IC is smaller than IC=IB*beta which means: IB>IC/beta.

4.) Therefore, the saturation region is decribed by this condition: IB>IC/beta.

UPDATE 2:
Some additional considerations regarding 1:) and 2.) :
The described behaviour can also be observed with a closer look to the output characteristics IC=f(VCE) of a BJT: The various curves for different fixed values for IB resp..VBE do NOT cross the origin. For VCE=0 the current IC is slightly negative (collector part of the base current) and - as a consequence, we have IC=0 (both parts in opposite direction cancel each other) for a very small value of VCE (some tenth of mV).

Thanks for giving me very detailed comments!

I feel like I'm still going around the first place I starts. According to your answer, my original picture of saturation current, where both forward-biased junctions will give "individual" current towards to base from them and base current is simply sum of them, seems right. I would like to ask you confirm this once more.

I feel that I'm still staying to think about saturation mode of the transistor very simply as two diodes of opposite directions are series connected with base terminal at the center and both are forward-biased.

 

[LvW]

So, all currents from both ends of the transistor finally arrives at the base and it should go through base terminal. It means base current is the most strong than emitter and collector current?

I feel that I'm still staying to think about saturation mode of the transistor very simply as two diodes of opposite directions are series connected with base terminal at the center and both are forward-biased.

* At first - as mentioned already by meBigGuy - your current directions are inverse. For npn transistors the base current goes from the base node to the emitter node (and also to the collector node in saturation mode). In the latter case, this base current (consisting of two parts) is the only current for the case VCE=0 only!.
* However, as I have tried to explain, for a finite value of VCE>0 (but VCE<VBE) there is another current from the collector to the emitter (like in "normal" linear mode). Hence, we have two currents through the collector node but in opposite directions. Therefore, the resulting current Ic is smaller as in linear mode - which means IC<IB*beta (identical to IB>IC/beta).
* Note that there is a collector current in the "normal" direction (from C to E) - even in case of rather small VCE voltages (VCE=0.1 or so). This is due to the very thin base region, the typical doping profile and excessive carrier density in the base region (due to VBE).
* Note also that this explanation is based on the physical fact that the BJT is a voltage-controlled device IC=f(VBE) ; (maybe you have learned something else, but a small current never can directly control a larger current).

 

[Jony130]

Maybe here you can find the answer
https://www.physicsforums.com/threads/carrier-flow-in-bjt-when-saturated.778000/#post-4894261

 

[goodphy]

Maybe here you can find the answer
https://www.physicsforums.com/threads/carrier-flow-in-bjt-when-saturated.778000/#post-4894261

Yes, right it is essentially the same question as I have here. Thanks!

 

[goodphy]

I've found good website of visualization for transistor operation. It include visual doping profile, which can help transistor beginner as me.

http://www.learnabout-electronics.org/bipolar_junction_transistors_04.php

 

[LvW]

I've found good website of visualization for transistor operation. It include visual doping profile, which can help transistor beginner as me.
http://www.learnabout-electronics.org/bipolar_junction_transistors_04.php

I hesitate a bit to "destroy" the picture you may have about the working principle of a BJT - but to tell you the truth: The explanation in the given reference is wrong!
Let me try to comment some parts of the text:

During normal operation, a potential is applied across the base/emitter junction so that the base is approximately 0.6v more positive than the emitter, this makes the base/emitter junction forward biased.

Correct. This clearly shows the dominatiing role of the voltage VBE. Without this voltage biasing nothing happens!

When the base emitter junction is forward biased, a small current will flow into the base. Therefore holes are injected into the P type material. These holes attract electrons across the forward biased base/emitter junction to combine with the holes.

Also correct. This constitutes the base current which cannot be avoided.

However, because the emitter region is very heavily doped, many more electrons cross into the base region than are able to combine with holes. This means there is a large concentration of electrons in the base region and most of these electrons are swept straight through the very thin base, and into the base/collector depletion layer. Once here, they come under the influence of the strong electric field across the base/collector junction. This field is so strong due to the potential gradient in the collector material mentioned earlier, that the electrons are swept across the depletion layer and into the collector material, and so towards the collector terminal.

Yes. Correct description.
(Question: Has anybody the feeling that the current IB is more than a "byproduct" only? If yes - for what reason?)

Varying the current flowing into the base, affects the number of electrons attracted from the emitter.
OK - let`s assume that there is an increase in base current corresponding to 100 additional wholes. Hence, 100 additional electrons can recombine (and do not enter the collector). Does anybody see any amplification?

In this way very small changes in base current cause very large changes in the current flowing from emitter to collector, so current amplification is taking place.

Does anybody see any justification or explanation for the claim that a "very small changes in base current cause very large changes in the current flowing from emitter to collector" ? It is simply an assertion without any verification. And it is wrong! I know that there are many contributions (even textbooks) claiming that the BJT would be current-controlled. Perhaps - because the authors think that the (correct) relation IB=IC/beta is identical with a cause-and-effect relation. But that is not the case.
Finally: There are many proofs (not on charged-carrier niveau but on circuit level - to be measured!) that the BJT clearly is voltage controlled - based on Shockleys famous exponential equation IE=f(VBE).

 

[goodphy]

I hesitate a bit to "destroy" the picture you may have about the working principle of a BJT - but to tell you the truth: The explanation in the given reference is wrong!
Let me try to comment some parts of the text:

During normal operation, a potential is applied across the base/emitter junction so that the base is approximately 0.6v more positive than the emitter, this makes the base/emitter junction forward biased.

Correct. This clearly shows the dominatiing role of the voltage VBE. Without this voltage biasing nothing happens!

When the base emitter junction is forward biased, a small current will flow into the base. Therefore holes are injected into the P type material. These holes attract electrons across the forward biased base/emitter junction to combine with the holes.

Also correct. This constitutes the base current which cannot be avoided.

However, because the emitter region is very heavily doped, many more electrons cross into the base region than are able to combine with holes. This means there is a large concentration of electrons in the base region and most of these electrons are swept straight through the very thin base, and into the base/collector depletion layer. Once here, they come under the influence of the strong electric field across the base/collector junction. This field is so strong due to the potential gradient in the collector material mentioned earlier, that the electrons are swept across the depletion layer and into the collector material, and so towards the collector terminal.

Yes. Correct description.
(Question: Has anybody the feeling that the current IB is more than a "byproduct" only? If yes - for what reason?)

Varying the current flowing into the base, affects the number of electrons attracted from the emitter.
OK - let`s assume that there is an increase in base current corresponding to 100 additional wholes. Hence, 100 additional electrons can recombine (and do not enter the collector). Does anybody see any amplification?

In this way very small changes in base current cause very large changes in the current flowing from emitter to collector, so current amplification is taking place.

Does anybody see any justification or explanation for the claim that a "very small changes in base current cause very large changes in the current flowing from emitter to collector" ? It is simply an assertion without any verification. And it is wrong! I know that there are many contributions (even textbooks) claiming that the BJT would be current-controlled. Perhaps - because the authors think that the (correct) relation IB=IC/beta is identical with a cause-and-effect relation. But that is not the case.
Finally: There are many proofs (not on charged-carrier niveau but on circuit level - to be measured!) that the BJT clearly is voltage controlled - based on Shockleys famous exponential equation IE=f(VBE).

I've also known that BJT is "typically" considered as current-controls. But I've not taken this seriously as current and voltage for PN junction is one-to-one relation. Let's say we ave very good current regulator which keeps current at certain level, When it is connected to silicon diode, The only way this current flows through the diode is that whole circuit behaves to build voltage across the diode about 0.6 V. Can somebody imagine current injection to the diode without voltage developed across it? Without proper voltage, the diode is nothing but insulator!

I'm just wondering voltage-control or current-control is really important manner in not only practical view but also theory.

 

[LvW]

I've also known that BJT is "typically" considered as current-controls. But I've not taken this seriously as current and voltage for PN junction is one-to-one relation. Let's say we ave very good current regulator which keeps current at certain level, When it is connected to silicon diode, The only way this current flows through the diode is that whole circuit behaves to build voltage across the diode about 0.6 V. Can somebody imagine current injection to the diode without voltage developed across it? Without proper voltage, the diode is nothing but insulator!

I'm just wondering voltage-control or current-control is really important manner in not only practical view but also theory.

Hi goodphi - the mentioned arguments are not new for me. In this context, I very often hear the term "chicken-and-egg" problem.
For my opinion it is not - because: No current without a driving voltage. That means: Applying what we call "current source" to a BJT or a diode is nothing else than to have a (large) voltage source with a very large source resistance Rs which forms a simple voltage divider between Rs and the non-linear device (BJT or diode). Remember the classical scheme how the operating point of a diode is fixed using a voltage source and a resistor (exponential curve and resistor load line in a common diagram).
In any case, the non-linear device allows a current only according to the voltage across it (equilibrium between V and I).

To me, it is really funny (surprising) that everybody starts designing a BJT stage with VBE=(0.6..0.7V) - in order to "open" the transistor (with a certain DC quiescent current). However, suddenly any change in the output current should be caused by the base current and NOT by a change in VBE?
I cannot understand how somebody who knows
(a) how a pn diode works and
(b) who is able to use his knowledge properly
can think that the pn junction within the transistor behaves completely different.

Example: Compare the gain values for two common-emitter stages (of course, the same dc bias point): One with beta=100 and the other with beta=200.
Wouldn`t you expext a larger voltage gain for the second case (if the BJT would be current-controlled)?
However, the gain is equal in both cases because it is the transconductance gm=dIc/dVbe that determines the gain value (that means: the cause is dVbe).
On the other hand, it is well known that the gain can be increased for a given design by increasing the DC quiescent current. Why?
How can this be explained using current-control? The explanation is that we just have an increased slope gm on the Ic=f(Vbe) curve.

I think, one must "stumble" over such contradictions.

Finally, yes - it is important to know the real physics behind the working principle of parts. Otherwise, no circuit inventions are possible and we can rely on 30 years old books.
(Current mirrors and diff. amplifiers do function only because of voltage control; the same applies to B. Gilberts famous translinear circuit concept).

 

[meBigGuy]

Your example of parallel transistors with different betas confuses me a bit. Are you saying that the voltage gain (Vc/Vbe) will match?

I had a very negative response on this forum when I said that BJT are transconductance devices. Did I say it wrong back then?
https://www.physicsforums.com/threa...ransistor-amplify-thread.709791/#post-4500825

Maybe you can sort out where I miss-communicated.

 

[LvW]

Your example of parallel transistors with different betas confuses me a bit. Are you saying that the voltage gain (Vc/Vbe) will match?
.

Yes - of course. The gain is Vc/Vbe=-gm*Rc. And gm is independent on beta. As you know, without Re feedback such a circuit is not practical.
However, the same applies in case of feedback: Gain Vc/Vin=-gm*Rc/(1+gm*Re).

I had a very negative response on this forum when I said that BJT are transconductance devices. Did I say it wrong back then?
https://www.physicsforums.com/threa...ransistor-amplify-thread.709791/#post-4500825
Maybe you can sort out where I miss-communicated.

I know Claude Abraham already from some other discussions. He very strongly always is claiming that the BJT would be current-controlled.
However, I`ve got the impression that he is not open for counter arguments - e.g. from leading instititions in the US (Berkeley, Stanford, MIT,..).
Did you ever hear or read a proof or justification for current-control?

 

[cabraham]

Yes - of course. The gain is Vc/Vbe=-gm*Rc. And gm is independent on beta. As you know, without Re feedback such a circuit is not practical.
However, the same applies in case of feedback: Gain Vc/Vin=-gm*Rc/(1+gm*Re).
I know Claude Abraham already from some other discussions. He very strongly always is claiming that the BJT would be current-controlled.
However, I`ve got the impression that he is not open for counter arguments - e.g. from leading instititions in the US (Berkeley, Stanford, MIT,..).
Did you ever hear or read a proof or justification for current-control?

The original 1954 Ebers-Moll paper from Bell Labs modeled the bjt as emitter current controlling collector current. The emitter current is functionally related to Vbe, the base-emitter voltage, but the circuit model describes Ic as alpha*Ie. Berkely, Stanford, and MIT to my knowledge do not claim the bjt is VC at all. In the small signal model, aka "hybrid pi" model, the small signal collector current can be computed as "gm*vbe" or as "hfe*ib". If the signal ac swing is small enough, current or voltage, the equations hold very well. But for large signal analysis, or using a bjt as a switch, this "gm*vbe" model is grossly non-linear and not used.
I posted plots on another forum where the signal generator driving the 1-stage bjt amp operates at hundreds of kilohertz up to tens of megahertz. At these speeds it is apparent that the change in Ie precedes the change in Vbe. Ic responds immediately to Ie, and Vbe catches up after Ic already settled. Clearly it is Ie in control, not Vbe. One of those plots is attached here.
I've probed many circuits in the lab, be it switching power converters, LED drivers, motor drivers, using diodes and bjt. When the frequency is in the rf range, it is all too easy to see that emitter current precedes base-emitter voltage, and collector current trackes emitter current very precisely. For a diode in a SPC, the forward voltage drop lags behind the forward current. The fallacy in your argument is based on Shockley's diode equation.
1a) Id = Is*exp((Vd/Vt)-1) is just one way to express this relation. Another form is:
1b) Vd = Vt*ln((Id/Is)+1)
Form 1a is what most are familiar with, but it cannot be overstated that the voltage value across the diode, Vd, does not "control" forward diode current Id. The diffusion capacitance formed at the p-n junction is non-linear, but it displays the same "Eli the ice man" properties of caps. A change in current will **precede** a change in voltage, always. Your position is based on the theory that although Ie determines Ic, it is Vbe that ultimately controls Ie, which is a mere assumption based on intuition. Ie does not depend directly on Vbe. Form 1b of SE should be regarded as well. Every critic of current control insists that currents are controlled by a corresponding voltage. It is assumed but cannot be proven because it isn't so.
You once asked me about the control mode of a motor. I stated that motor speed is controlled by voltage, motor torque is controlled by current. Not jumping topics, but we must be precise as to what that means. You stated, correct me if I misquote you, that a motor is "controlled by voltage", with no conditions. I will not put words into anybody's mouth, but am I correct in presuming you mean that the current in the windings controlling torque is a function of terminal voltage so that torque is ultimately controlled by voltage, as well as speed?
Here is my rebuttal, a dc motor made of superconducting windings has zero R, thus zero V drop, but non-zero current. Current is seen to be related to torque. Now study a copper wound motor with a small R in the windings. Current is still related to torque, albeit a small forward V drop occurs due to winding R. But if we construct a motor with carbon wire, 100 times the R of copper, we see that the same current produces the same torque, with a higher voltage drop. THe voltage in the 3 cases is zero, small, and large, and 3 currents are the same. The torques are the same as well.
The extra voltage needed in the higher resistance windings does not contribute to torque, only to heat. The extra voltage results in heat dissipation, it is an undesirable loss.
Just because a voltage drop is inevitable with resistive windings does not mean that V "controls" I at all. Likewise in a p-n junction, the forward drop is incurred due to charge diffusion, drift, recombination, and so forth. Vd/Vbe is a DROP, not an EMF. Drops do not "drive current". Anyway I will elaborate for those interested. Best regards to all.

Claude

 

[LvW]

My first reaction was not to reply to the above contribution because there are no new arguments. I don`t know why I cannot resist to reply.
Perhaps because I was engaged in teaching electronics for more than 25 years - and I never got tired to explain things again and again.

Claude, are you aware that your long and extensively contribution does not contain one single proof or justification for your claims?
Sometimes you repeat very well known facts about „functional relations“ (which never were put into question).
Or you speak about alternative calculations ("gm*vbe" or as "hfe*ib"), which do not proove anything.
Sometimes you speak about circuit models (which not necessarily reflect the real physical behaviour).
Or you manipulate Shockleys equation (solving for Vd) - do you really think this mathematical procedure has any physical meaning?
Ohh yes - and, finally, you cannot resist to mention „Eli the ice man“.
Not the best argument because during steady-state conditions a phase of -90 deg is equivalent to +270deg. Hence, it says nothing about cause and effect.

I have told you already (with examples) that - to justify the voltage-control picture - it is not necessary to go down to carrier physics.
We just need to analyze some properties of selected BJT circuits.

PS: If desired I can give you references from Stanford, Berkeley, MIT,...
For example: Berkeley University: ...IB is an undesirable but inevitable side effect of producing IC by forward biasing the BE junction...

 

[Jony130]

Example: Compare the gain values for two common-emitter stages (of course, the same dc bias point): One with beta=100 and the other with beta=200.
Wouldn`t you expext a larger voltage gain for the second case (if the BJT would be current-controlled)?
However, the gain is equal in both cases because it is the transconductance gm=dIc/dVbe that determines the gain value (that means: the cause is dVbe).
On the other hand, it is well known that the gain can be increased for a given design by increasing the DC quiescent current. Why?
How can this be explained using current-control? The explanation is that we just have an increased slope gm on the Ic=f(Vbe) curve.

I don't want to be picky or pedantic. But the voltage gain will depend on beat value, not by much but it will ( β/(β+1 ). But of course in general case I agree with your whole post.

 

[LvW]

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.

 

[cabraham]

My first reaction was not to reply to the above contribution because there are no new arguments. I don`t know why I cannot resist to reply.
Perhaps because I was engaged in teaching electronics for more than 25 years - and I never got tired to explain things again and again.

Claude, are you aware that your long and extensively contribution does not contain one single proof or justification for your claims?
Sometimes you repeat very well known facts about „functional relations“ (which never were put into question).
Or you speak about alternative calculations ("gm*vbe" or as "hfe*ib"), which do not proove anything.
Sometimes you speak about circuit models (which not necessarily reflect the real physical behaviour).
Or you manipulate Shockleys equation (solving for Vd) - do you really think this mathematical procedure has any physical meaning?
Actually, you are the one who relies on Shockley equation, SE, for "proof" that Vbe controls Ie. I was demonstrating that the quation simply expresses a functional relation and does not prove which, if either, controls the other.
Ohh yes - and, finally, you cannot resist to mention „Eli the ice man“.
Eli the ice man is forever your nemesis. He proves that the voltage across a capacitance can never be the cause of the current in the same capacitance. The cause MUST ALWAYS PRECEDE the effect. Irrefutable, sacrosanct, immutable, beyond debate, not open for discussion. My plots demonstrate that Ic closely tracks Ie, and Vbe cannot be the control because after Ic settles around the same time as Ie, Vbe continues to change. Clearly Vbe is not what controls Ie.
Not the best argument because during steady-state conditions a phase of -90 deg is equivalent to +270deg. Hence, it says nothing about cause and effect.
I have attached several sims of R-C and R_L-C networks under transient stimulation. Your "+270 degress" is the same as "-90 degrees" is flawed. In trigonometry, one can say that sin (x - 90) is equivalent to sin (x + 270) in degrees. But in pure math, it is assumed that the sine function is everlasting, i.e. it has no start or stop, always was always will be from minus infinity to plus infinity. But circuits are excited with truncated sines, i.e. they are zero before time 0 and then take on a sine function. The signal source is switched on at t=0, and a sine curve is impressed on the network. This is mathematically expressed as a unit step function u(t) times the sine function. If this composite function is shifted a cycle, the 2 are NOT the same. In everlasting sines, the -90 and 270 degree phase functions overlap and cannot be distinguished. But with a unit step multiplying the sines, the function u(t-90)sin(t-90) and u(t+270)*sin(t+270) are NOT equivalent. From t=-90 to t=+270 these 2 functions differ. One is zero while the other looks like a sine function. I've attached several plots.

The sources include step functions and sine functions. With RLC as well as RC networks, the capacitor current is always ahead of its voltage. The 270 degrees is pure nonsense. Examine please. The rlc-2 plot file shows 20 cycles starting at turn on. The red current trace is always ahead of the blue voltage trace.

I have told you already (with examples) that - to justify the voltage-control picture - it is not necessary to go down to carrier physics.
We just need to analyze some properties of selected BJT circuits.

PS: If desired I can give you references from Stanford, Berkeley, MIT,...
For example: Berkeley University: ...IB is an undesirable but inevitable side effect of producing IC by forward biasing the BE junction...

Every time we talk, I make it clear that Ie is what controls Ic, NOT Ib. Now you are attacking the Ib straw man again. Until you learn that Ie is NOT HE SAME as Ib, my discussions are pointless. Emitter current and base current, LvW, are NOT the same, trust me.

Claude

 

[meBigGuy]

I really have a problem with the idea that Ie *controls* Ic. Ie is simply Ib + Ic. That is a side effect (sort of) not a cause. The point that two transistors with the same gm will show the same voltaqge gain INDEPENDENT OF beta makes it clear the Vb is the primary cause and that Ib is an effect. Therefore Ie is an effect also since it is Ic + Ib.

Looking at the two transistor example, common emitter NPN (with different beta and assuming identical collector resistors), there will be identical currents through each collector (Vsupply - Vce divided by Rc) independent of Ib and Ie (since the betas are different, but not the gm). Ie will be different since Ib is different for each transistor.

So , Ie is not determining Ic. It just simply happens to be Ib + Ic. Of course any change in Ic will be seen immediately in Ie since Ie is "mostly" Ic.

Hopefully Claude can explain in two or three succinct sentences why this is not a valid view.

 

[LvW]

meBigGuy - I am completely with you.
Meanwhile, I`ve got the impression that Claude has a very "extraordinary" understanding of the term "control".
To me - and also for you, if I understand you correctly - it is absolute nonsense to claim that IE would "control" IC. Instead, IE simply is splitted into two currents, that´s all.
But he does not enlighten us which external quantity determines IE.
I think, it is a common understanding that to "control" something means to influence any internal system parameter externally.

 

[LvW]

To Claude:
I think, it makes no sense to continue the discussion with you as long as you believe it is necessary to invite me to "learn that Ie is NOT HE SAME as Ib".
Therefore, no more technical arguments from my side - just some comments to your contribution in red.
I am sorry to say but with all of your lengthy expositions you do not meet the points:
1.) I spoke about your "manipulations" of Shockleys equation (solving for Vd). Not about the validity of the original equation.
2.) You are spending 15 lines to "explain" that -90 deg and +270 deg are not equivalent under certain conditions. Do you think this has anything to do with the main question?
3.) You again and again refer to your simulations - which are based on models only. I must admit that I didn`t have one single look on these graphs. I know what such models can do and what they cannot do.
4.) Repeatedly you claim that IE would control IC. You simply are ignoring the fact that the B-E path is the controlling part.
Answering a corresponding question (at the end of a similar discussion in another forum) you wrote: "The role of the input port is to input a signal from a source with the intent of varying the emitter current". Unfortunately, you gave no answer to the subsequent question from my side "what kind of signal - voltage or current"?
5.) Quote: Berkely, Stanford, and MIT to my knowledge do not claim the bjt is VC at all.
Surprisingly, you seem to be not interested to see the documents I was referring to.

 

[meBigGuy]

Yeah, you nailed it. I don't know why I didn't see that sooner. I guess I should have asked what external thingy controls Ie. Or, is it simply magically sucked out by ground.

 

[cabraham]

I really have a problem with the idea that Ie *controls* Ic. Ie is simply Ib + Ic. That is a side effect (sort of) not a cause. The point that two transistors with the same gm will show the same voltaqge gain INDEPENDENT OF beta makes it clear the Vb is the primary cause and that Ib is an effect. Therefore Ie is an effect also since it is Ic + Ib.

Looking at the two transistor example, common emitter NPN (with different beta and assuming identical collector resistors), there will be identical currents through each collector (Vsupply - Vce divided by Rc) independent of Ib and Ie (since the betas are different, but not the gm). Ie will be different since Ib is different for each transistor.

So , Ie is not determining Ic. It just simply happens to be Ib + Ic. Of course any change in Ic will be seen immediately in Ie since Ie is "mostly" Ic.

Hopefully Claude can explain in two or three succinct sentences why this is not a valid view.

Claude can explain. Ie is indeed Ib+Ic, nobody claims otherwise. But how does Vbe develop across the b-e junction? Charges inputted from an external source enter the base and emitter regions. This is Ib and Ie. If the signal source is a microphone, inflections in the singers voice result in variations in the signal voltage and current. The mic cable carries a time varying I & V. Charges enter the b-e junction and feel a repulsive potential barrier due to the depletion layer at the b-e junction. An increased charge sensity crosses the junction and the depletion layer has been increased due to higher charge density. Vbe increases as a result of this increased current. Vbe is literally the line integral of the E field due to the charges in the depletion layer. You claim that Ib and Ie is merely an "effect" of Vbe. THis is a common myth.

In order for that to be true, a change in Vbe would **precede** a change in Ib/Ie. My sim plots seem to be dismissed on the grounds that they are only "models". But sim models are based on real world measured data. The parameters of the bjt under simulation, such as Ies, hfe, Ics, Vt, hie, Cpi, hoe, Cmu, hre, etc., were arrived at by measurements made in a lab using actual devices. A test set up in a lab would affirm the following.

Changes in Ib/Ie take place **ahead** of Vbe changes. The reason this is so is easy to visualize. In order to change Ic, what needs to be done? Ic is the number of electrons (npn ref) per time collected, hence the collector is so named. To increase collection we must - increase **emission**. That is Ie, and Ib is related as well. THe collector responds to changes in Ie, an increase in emitted electrons will inevitable result in an increase in collected electrons, as well as a rise in Vbe. As soon as Ie increases, Ic will increase very shortly afterward. The delay between Ie and Ic is the transit time through the base, very short indeed. But for Vbe to increase charges, i.e. electrons just emitted by emitter, must recombine in base, and holes from base must recombine in emitter. So to change Ic, a change must first happen in Ib and Ie. Ib is not the controller of Ie, but it is NOT an effect of Vbe either. Plots affirm the same. If you still believe that the sim models are not 100% reliable, you can always run a test in a lab with a scope, current, and voltage probes, or a current sense resistor and differential probe setup.

Regarding "gain", you only speak about voltage gain. There is a current gain as well. Two amps with identical inputs and bjt part both are set for the same fixed voltage gain. But one has a load impedance half of the other, demanding twice the current. Beta matters because a stage has a current gain always less than that of the raw bjt. Likewise the trans-conductance of the stage cannot exceed that of the raw bjt, i.e. gm.

The "gm*Rc'" you speak of only covers the voltage gain from base to output. But the signal source has an internal resistance which drops signal voltage. The input bias resistors also attenuate the signal by forming a divider with the source resistance. The best way to minimize gain loss is to use high valued bias resistors. But this can make the gain sensitive to beta variations.

A high beta bjt allows the user freedom to use an optimum resistor value for Re as well as R1 & R2 on the base side. THe gm value affects voltage gain only. But beta not only affects current gain, but voltage gain as well. A very high beta value allows the use of a smaller Re value, and/or larger R1/R2 values. The voltage gain approaches the limit "-Rc'/Re", so that smaller Re allows higher gain. Beta does indeed give the user freedom to get not only more current gain but voltage gain as well.

I've designed several products that include a single stage amp made with one bjt. The tradeoffs involved and analysis of such a stage make one really appreciate the interactive nature of beta, gm, Re, R1/R2, Rsource, etc.

Claude

 

[LvW]

Claude can explain. Ie is indeed Ib+Ic, nobody claims otherwise. But how does Vbe develop across the b-e junction? Charges inputted from an external source enter the base and emitter regions.
What kind of force causes the charges to move? Isn`t it a voltage? Why such vague expressions?
This is Ib and Ie. If the signal source is a microphone, inflections in the singers voice result in variations in the signal voltage and current. The mic cable carries a time varying I & V. Charges enter the b-e junction and feel a repulsive potential barrier due to the depletion layer at the b-e junction. An increased charge sensity crosses the junction and the depletion layer has been increased due to higher charge density. Vbe increases as a result of this increased current. Vbe is literally the line integral of the E field due to the charges in the depletion layer. You claim that Ib and Ie is merely an "effect" of Vbe. THis is a common myth.

Vbe is applied externally (resp. Vb only in case of RE feedback). That is the reason we use a bias scheme with a voltage divider as low-resistive as possible (in conjunction with other constraints: input resistance). Other bias schemes use a single resistor to the base node for "injecting" a base current. However - THIS is a myth! It is just a sloopy term that is rather common. However, it is not possible to "inject" or "pump" a current into any node. In each case, we need a voltage to drive such a current - with other words (as I have explained already earlier) we always create a voltage divider using a voltage source and a large source resistor which primarily determines the current into a lower resistive element (linear or non-linear). In our case: Base-emitter junction..

In order for that to be true, a change in Vbe would **precede** a change in Ib/Ie. My sim plots seem to be dismissed on the grounds that they are only "models". But sim models are based on real world measured data. The parameters of the bjt under simulation, such as Ies, hfe, Ics, Vt, hie, Cpi, hoe, Cmu, hre, etc., were arrived at by measurements made in a lab using actual devices. A test set up in a lab would affirm the following.
Changes in Ib/Ie take place **ahead** of Vbe changes. The reason this is so is easy to visualize. In order to change Ic, what needs to be done?

Now, everybody waits for an answer - "what needs to be done"? Let´s wait for an explanation..

Ic is the number of electrons (npn ref) per time collected, hence the collector is so named. To increase collection we must - increase **emission**. That is Ie, and Ib is related as well. THe collector responds to changes in Ie, an increase in emitted electrons will inevitable result in an increase in collected electrons, as well as a rise in Vbe. As soon as Ie increases, Ic will increase very shortly afterward. The delay between Ie and Ic is the transit time through the base, very short indeed. But for Vbe to increase charges, i.e. electrons just emitted by emitter, must recombine in base, and holes from base must recombine in emitter. So to change Ic, a change must first happen in Ib and Ie. Ib is not the controller of Ie, but it is NOT an effect of Vbe either. Plots affirm the same. If you still believe that the sim models are not 100% reliable, you can always run a test in a lab with a scope, current, and voltage probes, or a current sense resistor and differential probe setup.

And where is the answer? Which effect causes the emitter to increase emission? Ib is not the "controller" - OK.
(There are earlier contributions from you stating the opposite). But which quantity is the controller?

Regarding "gain", you only speak about voltage gain. There is a current gain as well. Two amps with identical inputs and bjt part both are set for the same fixed voltage gain. But one has a load impedance half of the other, demanding twice the current. Beta matters because a stage has a current gain always less than that of the raw bjt. Likewise the trans-conductance of the stage cannot exceed that of the raw bjt, i.e. gm.

Example unclear. Common emitter? Load impedance? Twice the current? Which current (Ic or I(load) or both) ?

The "gm*Rc'" you speak of only covers the voltage gain from base to output. But the signal source has an internal resistance which drops signal voltage. The input bias resistors also attenuate the signal by forming a divider with the source resistance. The best way to minimize gain loss is to use high valued bias resistors. But this can make the gain sensitive to beta variations.
A high beta bjt allows the user freedom to use an optimum resistor value for Re as well as R1 & R2 on the base side.

Nobody will deny that a larger beta may have advantages. Again, you misssed the point.

THe gm value affects voltage gain only. But beta not only affects current gain, but voltage gain as well. A very high beta value allows the use of a smaller Re value, and/or larger R1/R2 values. The voltage gain approaches the limit "-Rc'/Re", so that smaller Re allows higher gain. Beta does indeed give the user freedom to get not only more current gain but voltage gain as well.
I've designed several products that include a single stage amp made with one bjt. The tradeoffs involved and analysis of such a stage make one really appreciate the interactive nature of beta, gm, Re, R1/R2, Rsource, etc.

All these basic considerations are well-known but have nothing to do with the main question.

Claude

 

[cabraham]

Vbe is not applied externally. Let's examine a voltage divider. The crux of this issue is not bjt operation, but the simple relation between I & V. 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?

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? That is a question we should have explored months ago. I don't think we view the universe the same way at all, regardless of whether we are discussing the bjt. There is a disconnect between us regarding the I/V relation in any electric device.

I believe I know your answer, but it is not sporting to speak for you, so I'll let you state your position. Here is mine. 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. In the battery positive current exits the pos terminal and enters the neg terminal. This means that the battery is delivering power. Each resistor has a positive current entering its pos terminal and exiting the neg terminal, opposite to the battery. This indicates that the resistor is absorbing power, which it is.

The battery's 12 volts is actually an emf. An emf is the energy per unit charge gained. THhe voltages across the resistors are voltage drops. A drop is the energy per unit charge lost. You asked me what "force" drives the current. Here is my answer.

The battery internal redox reaction (reduction/oxidation) propels positive ions toward the pos terminal. and neg ions towards the neg terminal. In doing so an electric field is built. Since the pos terminal tends to repel pos ions, it takes work to do this. This work ends up increasing the E field energy.

When the switch closes, this E field exerts forces on the free charges in the wires and resistors. For an electric field E, the force on a charge q is simply F = qE, the Lorentz force law. So we can say that the battery E field provides the Lorentz force that ultimately drives the whole loop current. Remember how this field is created. The internal redox action generates a current internally which builds an electric field. This field can exert forces on the network charges. The terminal voltage is built up by virtue of internal current.

Not only does it take current to create the E field, but to sustain it as well. As positive charges exit the battery positive terminal, and pos charge enter the neg terminal, the battery E field decreases as does the terminal voltage. As soon as an E field moves charges it gives up its own energy. How does the battery replenish its E field? It does so by redox action generating current which replenishes the E field and voltage. Not only is current needed to establish the battery terminal voltage, but to sustain it as well.

Inside the battery, redox generates current, which in turn generates E field and voltage which is the integral of said E field. Outside the battery, E field acts on charges moving them around the loop. Doing so decreases field energy. Redox action generates more current, replenishing voltage/E field.

But once charges move around loop, when they enter resistor material, charges, i.e. electrons, collide with lattice atoms in the crystal structure. When an electron in conduction incurs a collision, it can lose enough energy to fall down into the valence band and ionize the atom. To conserve energy a photon is emitted. This photonic emission is in the long infrared band, felt as heat, which is easily felt. When a resistor is dissipating power its temperature rises, which we feel as heat.

When said electron ionizes the atom, it tends to produce a local E field which repels incoming electrons. This is a potential difference known as "voltage drop". The E field formed by many electrons falling from conduction to valence and ionizing atoms, forming a barrier is a drop, which represents loss of energy, not gain. Thus the resistors own voltage drop can NOT be the "cause" or "controller" of said resistor's own current. The "driver" of the entire loop's current is the battery internal redox action. Each resistor and the wires drops some voltage and dissipates some power. The battery produces 12 joules of energy for every coulomb of charge. Each resistor drops just under 6 joules for every coulomb, and a slight drop by the wires accounts for the 12 volts. Only the battery and its internal energy conversion redox process drives anything.

Forgive me if I put words into your mouth, but you seem to infer, I could be wrong, that the voltage drop across said resistor is the "driver" of that same resistor's own current. Is that your position? Again, my apologies if that is not the case. Best regards.

Claude Abraham

 

[meBigGuy]

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.

 

[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.