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.