How does a transistor amplify current or voltage?

[tvtech]

Guys please leave it alone. You are scaring potential new members away...

Even I am scared of this thread. It has gotten out of hand for Newbies...never mind regulars here. There is absolutely nothing constructive here.

It has become a war of wisdom. And that helps nobody. A simple question has become a complicated thread involving 24 Pages.

Kindly stop your crap guys. Agree to disagree. And get over it??

Cheers

 

[ericgibbs]

Guys please leave it alone. You are scaring potential new members away...

Even I am scared of this thread. It has gotten out of hand for Newbies...never mind regulars here. There is absolutely nothing constructive here.

It has become a war of wisdom. And that helps nobody. A simple question has become a complicated thread involving 24 Pages.

Kindly stop your crap guys. Agree to disagree. And get over it??

Cheers

hi TT,
I agree, I thought the same way as you in my post #30.

I think most of have the experience to know when its time to get off the merry go round.

 

[MrAl]

Hello again,


Yes i think it is time to go too

 

[BrownOut]

Vbe is changed externally. I can put a voltage source in the b-e terminals, and lock the voltage so the transistor cannot change it. Your equation above shows that vbarrier has to change to accomodate Vbe, because Vbi does not change at constant temperature once the transistor is manufactured.

Nodoby ever said you could't force vbe to equal an external value. That's a trival application of basic electronics, and proves nothing.

You cannot say Vbe does not change anything, it cancels some of the barrier voltage. Here is a quote from the textbook Semiconductor Circuit Analysis, by Phillip Cutler, p 19. I only deleted references to a diagram which shows a diode in series with a variable voltage power supply.

"At first I slowly increases, because because the applied voltage V has not sufficiently reduced the inherent potential barrier. As we increase V further, say to a few tenth of a volt, we find that I rapidly increases because V has finally become large enough to reducxt significantly the internal barrier potential and cause the recombination current to become quite large."

vbe is the junction voltage, and nothing more. It is made up of barrier voltage and contact voltage. vbe changes as barrier voltage is changed. To say a voltage changes itself makes no sense.

"For such a diode, the height of the potential barrier at the junction will be lowered by the applied forward voltage V. The equilibrium initially established between the forces tending to produce diffusion of majority carriers and the restraining influence of the potential-energy barrier at the junction will be disturbed."

So this text too, believes that Vbe cancels some of the barrier voltage.

Sedra and Smith says otherwise.

By the way, the reason S & S always talk about current is because they are analyzing the BJT with a current source. But it is still the Vbe that determines what the junction will be to support the current source.

It doesn't matter how the bjt is biased. The physics of the junction doesn't change.

Read my statement again. I said it opposes the energizing voltage. That is what a back voltage does.

I know what your statement said. I said a voltage can't change another voltage, and you've not proven otherwise. There is no "back voltage" in capacitors. There is only the voltage it is charged to.

Wrong twice. Look again at the quotes and especially the link. The link says the junction voltage is Vbi-Vbe, so Vbe is not the junction voltage.

No, I'm correct. The links doesn't prove otherwise. vbe is the junction voltage. If you want to prove that vbe is not the same as junction voltage, then you must show another voltage source. I don't think that equation appears in the link, as it would be quite impossible. When you cherry pick the potential equation, Φ = Φ1 - vbe, you conveniently neglect that Φ1 produces no voltage, as it only cancles contact potential, which results in ZERO voltage both at the junction and at the contacts with no external current supplied. So, the equation becomes Φ = ZERO - vbe = -vbe, for any measurable voltage.

As I said before S & S analyze the BJT with a constant current source, but that does not change Vbe controlling the forward current.

And as I said, it doesn't change the junction physics. It's the same if the bias is voltage or current, vbe is changed by neutralized change in the junction. A voltage cannot change another voltage. This chapter by Sedra and Smith is the definitive explanation of how junction voltage, and thus current, is controlled in bjt's.

the equation for forward current should be Id-Is=I, which means diffusion current minus thermal current equals the forward current.

That's correct. But the important thing to keep in mind is the process that give rise to this current starts when charge is injected into the junction, that lowers the barrier voltage, and thus vbe. Also, the typo notwithstanding, thermal current is small, and not affected by input voltage or current, so the statement by Sedra and Smith that current through the junction is controlled by injected current is accurate.

The problem with some of these beginners texts is that the cartoons they show has vbe forced to some external voltage. That's OK for showing how bjt's work at a functional level, but does not show what is going in under the hood. And so, for example, when they show equations like the following:

Φ = Φ1 - va, which is technically correct, all they are showing is that is it possible to force the junction voltage to be equal to an external voltage, per kirchoff's law. Unfortunately, however, to understand how vbe is changes, one needs junction physics theory such as in Sedra and Smith. One also needs to keep in mind basic laws of electronics physics. For example, the mesurement of potential between any two points includes voltage sources between those points. No voltage outside the measurement points can affect the measurement. One cannot show a single circuit where this principle is violated. Thus, although the junction voltage may be forced to be equal to an externally connected voltage ( which is never done in practice ) the external voltage can not change the junction voltage. It can only be changed by charge injected into the junction, per Sedra and Smith. And this, vbe is the same as junction voltage, as there are no other sources.

.

 

[Claude Abraham]

Guys please leave it alone. You are scaring potential new members away...

Even I am scared of this thread. It has gotten out of hand for Newbies...never mind regulars here. There is absolutely nothing constructive here.

It has become a war of wisdom. And that helps nobody. A simple question has become a complicated thread involving 24 Pages.

Kindly stop your crap guys. Agree to disagree. And get over it??

Cheers

But this isn't a question of "you say tomato, I say tomato". The entire OEM semiconductor communtity says bjt is CC. Every fact I've presented supports the OEMs. The critics keep stating the contrary but only give circular reasoning. Their arguments would suggest that all electrical devices in the universe are "voltage controlled". So why would OEMs & the entire electronics community bother classifying some devices as CC & others as VC? This "voltage is the cause of current" concept is the root of all of the debates in electronics. Science does not support this causal viewpoint, yet its supporters constantly invoke it as if it were law.

If it takes 50 or 75 pages, there is no harm. The problem w/ locking such a thread is that whenever a bjt question is asked, or current feedback op amp, or any current mode device is pondered, the critics take the thread off on a tangent insisting that the device in question is really voltage mode. Then tens of pages of debate ensue, & the moderator lockes the thread thinking it is the right thing to do. But locking the thread limits & discourages discussion about any current mode device.

The critics are so hostile to any device being defined as current mode, that they will engage & argue for as long as it takes to lock out the thread. So we can have threads about FETs w/o any issue at all because we agree they are VC, likewise for voltage feedback op amps. But if the critics cannot redefine everything as VC, they will ruin & disrupt any thread involving devices classified as CC.

If locking the thread gives the moderators a sense of satisfaction, so be it. The OEMs universally describe bjt devices as CC. The critics are exploiting the universal characteristic that all electrical devices possess, namely that I cannot exist w/o V. But it also holds that V cannot exist w/o I. They don't like to deal with that side of it. Since it is impossible to establish either as the "cause" of the other, I just say they mutually co-exist. We externally drive the bjt w/ current, hence CC. At the internal level, it is impossible to say what causes what.

Internally its always been charge control & quantum mechanics. The CC model is never intended to cover internal phenomena. Why can't we leave it at that? Peace to all.

 

[BrownOut]

tvtech, eric, et al, you can use the "thread tools" drop down menue at the top of every page of the thread to "unsubscribe" to the thread, and you won't even be notified when a post has been made. It is not required of any member to read every thread. I sure don't.

 

[ericgibbs]

tvtech, eric, et al, you can use the "thread tools" drop down menue at the top of every page of the thread to "unsubscribe" to the thread, and you won't even be notified when a post has been made. It is not required of any member to read every thread. I sure don't.

hi D,
What the CC camp have been saying is the I way I was taught, I have used it for longer than I care to remember and its never let me down.

It seems a shame that newbies and wannabes are being misguided by some posters who will not accept the facts presented to them.

Regards
Eric

 

[tvtech]

hi D,
What the CC camp have been saying is the I way I was taught, I have used it for longer than I care to remember and its never let me down.

It seems a shame that newbies and wannabes are being misguided by some posters who will not accept the facts presented to them.

Regards
Eric

I am with the CC camp too. The other camp are trying to re - invent the wheel.

And we all know that a wheel works best when it is round. Not round with square edges here and there. Simply round.

Cheers

 

[Claude Abraham]

Also, FWIW, analyzing the device at dc ( 0 freq) serves little to no purpose. This dc argument is along the lines of "frozen time". If we have a bjt device in the steady state condition, or dc, then it becomes difficult if not impossible to ascertain what controls what.

Ie/Vbe/Ib/Ic/Vbc/Vce are established & settled. How can anyone say just from measuring these staedy parameters, that Ic is controlled specifically by this variable? And forget about causality because all variables mutually co-exist. One cannot decide which causes which.

But if the b-e terminals are excited with a current source pulse generator, or voltage pulse plus resistor, then we can observe the voltage & current waveforms at the device terminals. This is the only way to ascertain which is the control variable, & which are incidental. Any such test will affirm that as soon as the pulse generator current increases, Ie is already increased, bit Vbi is still its former value.

Ie enters the emitter terminal, then e- transit through emitter into base then collector. As the e- transit through the base a small percentage get recombined, adding to the already existing minority charge storage region. The result is that the Vbi eventually increases. But Ie increases immediately because Ie is in series w/ the pulse generator. Ic begins increasing as soon as Ie increases & emitter e- transit from emitter to collector. The control mechanism is that Ie increasing directly resulted in Ic increasing.

But, I will remind us, that ultimately Vbe will increase as a consequence. The E-M eqns tell us that increasing Ie and/or Ic inevitably increases Vbe per

Ie = Ies*exp((Vbe/Vt)-1), which can be expresed as Vbe = Vt*ln((Ie/Ies)+1).

Now we can examine the dc case for good measure. An Ie/Vbe/Ib/Ic is established steady state. A current source is inputted to the emitter terminal. Just to use numbers, Isource = Ie = 101 uA, Vbe = 0.600 V, Ib = 1.0 uA, & Ic = 100 uA. Now let the generator driving the b-e terminals increase its output from 101 uA to 201 uA.

Ie immediately increases, but Vbi, the built in potential has not had time to respond. Since the output current increased, the generator Vout has increased as well. Vout/Iout is still Zo the t-line impedance. The increased emiiter current Ie proagates through the emitter, into the base. At this point recombination results in Vbi commencing to rise. Nearly all e- transit through the base into the collector & Ic increases. Eventually equilibrium is attained. Ie = 210 uA, Vbe = 0.618 V, Ib = 2.0 uA, & Ic = 200 uA.

As Ie continues, Vbi increases as minority charges increase around the depletion zone. But eventually the density * Vbi will level off. So in order to sustain this new value of increases current, Vbe & Vbi must increase, which they do shortly afterward. This is well known. The exponential/logarithmic relation between Ie & Vbe tells us that only a small change in Vbe is needed due to a large change in Ie.

But the sequence of events makes it clear that Ie already increases before the internal depletion width & Vbi is even aware what is going on. Also, the terminal voltage Vbe (not Vbi), increases as a result of the increased current. But Vbe has a capacitance associated with the depletion zone charges. A cap's voltage, i.e. Vbe, cannot increases until charging current is established. Thus Vbe, external terminal potential, cannot "cause" Ie or Ic to change. Vbe is incidental/consequential. Ditto for Vbi.

Again, eventually Vbe/Vbi must increase to the new value, slightly greater since the current increased. I kept hearing the critics say "lowering the barrier voltage" which is unclear. In order to increase Ie/Ib/Ic, the Vbi & Vbe inevitably increase. Vbi never gets lowered. Of course the Vbe-Vbi difference increases. There is a resistance re for the emitter region. Increased current times re gives increased difference voltage Vbe-Vbi. But as we know, Vbe & Vbi are both consequential.

If I haven't been clear as to why I believe in the OEM viewpoint of CC, I'll elaborate. From this point on, I hope we can focus on device behavior w/o the causality issue. Cheers.

 

[MrAl]

tvtech, eric, et al, you can use the "thread tools" drop down menue at the top of every page of the thread to "unsubscribe" to the thread, and you won't even be notified when a post has been made. It is not required of any member to read every thread. I sure don't.

Hi BrownOut,


Very good point. It's a more complicated discussion then most others, that's all.

Also, I cant help but remember a saying i made up back in the early 90's if i may repeat here:

"There are some things that are perfectly clear that render other things perfectly useless, and there are other things that are perfectly useless that render other things that are perfectly clear, useless".

 

[BrownOut]

Hi BrownOut,


"There are some things that are perfectly clear that render other things perfectly useless, and there are other things that are perfectly useless that render other things that are perfectly clear, useless".

I'll remember that!

 

[Ratchit]

To the Ineffable All,

I agree with Brownout about one thing. No one is forcing anyone to read this thread. This is not a mandatory assignment or duty.

Brownout,

Φ = Φ1 - va, which is technically correct, all they are showing is that is it possible to force the junction voltage to be equal to an external voltage, per kirchoff's law. Unfortunately, however, to understand how vbe is changes, one needs junction physics theory such as in Sedra and Smith. One also needs to keep in mind basic laws of electronics physics. For example, the mesurement of potential between any two points includes voltage sources between those points. No voltage outside the measurement points can affect the measurement. One cannot show a single circuit where this principle is violated. Thus, although the junction voltage may be forced to be equal to an externally connected voltage ( which is never done in practice ) the external voltage can not change the junction voltage. It can only be changed by charge injected into the junction, per Sedra and Smith. And this, vbe is the same as junction voltage, as there are no other sources.

I am showing a attachment which shows that the junction voltage (Vj) is not, repeat not the junction voltage of Vbe as you state above. This page is from the book The PN Junction Diode, by Gerold W. Neudeck of Purdue University. Notice Fig 2.6a where the diode is shorted, and still there is a voltage of Vbi across the diode. What is more, Neudeck proves that is true by using loop equations. And when a forward voltage is applied, the junction voltage becomes Vbi-Va as shown in Fig. 2.6b. This agrees with the link I showed in my previous post. If you cannot accept that, then nothing will convince you what the value of Vj is.

Ratch

 

[Ratchit]

Claude,

Instead of replying to every paragraph, this time I will just touch on the highlights.

If you do not like my saying steady state DC, I will change my verbiage to the word equilibrium, which the transisitor texts use frequently.

In your discussion, you mention that Vbi changes with the minority charges or whatever. All the textbooks I have say that Vbi is only dependent on the doping, and a constant which varies slightly with temperature. I can send you an attachment from Neudeck's book if you want. Anyway, you have to show me that Vbi changes dynamically like you say. I understand it to be set permanently during transistor manufacture by the doping concentration.

Ratch

 

[BrownOut]

I am showing a attachment which shows that the junction voltage (Vj) is not, repeat not the junction voltage of Vbe as you state above. This page is from the book The PN Junction Diode, by Gerold W. Neudeck of Purdue University. Notice Fig 2.6a where the diode is shorted, and still there is a voltage of Vbi across the diode. What is more, Neudeck proves that is true by using loop equations. And when a forward voltage is applied, the junction voltage becomes Vbi-Va as shown in Fig. 2.6b. This agrees with the link I showed in my previous post. If you cannot accept that, then nothing will convince you what the value of Vj is.

]

Do you see in the illustration where the diode is shorted, and it shows I=0? If there is a junction voltage across the diode, then why would the current be zero? It's a basic law of electronics that for nonzero voltage and nonzero conductivity, there is non zero current, so whats going on here? Why is the diode voltage zero while inside the diode ther is a voltage? Things that make you go Hmmmmm.....?

Very simple, as I've said all along, your cartoon ingnores that very fact that the depletion potential is nertralized by the bulk n and p excess carrier potential. Thus, for the diode, the total junction voltage is zero, when the leads are shorted. The reason you are confused about what the junction voltage really is comes from the vocabulary, where your reference calls junction voltage what Sedra and Smith calls barrier potential. So to say "junction" voltage is vbi - va, in your reference, is equivilent to what Sedra and Smith would call voltage in the depletion layer being equal to vbi - va. However, you neglect that vbi is equal to excess carrier potential in the n and p regions, so in fact, for the total PN junction, the voltage is zero for the no external source case. Or in other words, for the bulk PN junction, vgs = vbi - v(bulk) - vgs = 0 - vgs = -vgs. So, in accordance with Sedra and Smith, when current lowers the barrier voltage by nertualizing uncovered charge, it changes vgs by the same amount.

Also, notice the equation vj = vn + vp. vn and vp are the potential that cancel vj, as I've said many time. It also says vn and vn are only dependent on the material, and not external sources. Thus, since only the depletion voltage changes, and vn and vp stay the same, and since vn and vp cancel vj in the non biased case, ie

vj + vn + vp = 0, and vp, vn doesn't change, then the formula given above is true, and Sedra and Smith are validated.

 

[Ratchit]

Brownout,

Do you see in the illustration where the diode is shorted, and it shows I=0? If there is a junction voltage across the diode, then why would the current be zero? It's a basic law of electronics that for nonzero voltage and nonzero conductivity, there is non zero current, so whats going on here? Why is the diode voltage zero while inside the diode ther is a voltage? Things that make you go Hmmmmm.....?

Why do I get the feeling that you still do not get it? See further explanation below.

Very simple, as I've said all along, your cartoon ingnores that very fact that the depletion potential is nertralized by the bulk n and p excess carrier potential.

Do you always call well drawn illustrations in textbooks "cartoons"? The neutralization is not important with respect to the point that Vbe is directly increasing or decreasing the effects of Vbi, thereby permitting more or less charge to flow.

Also, notice the equation vj = vn + vp. vn and vp are the potential that cancel vj, as I've said many time. It also says vn and vn are only dependent on the material, and not external sources. Thus, since only the depletion voltage changes, and vn and vp stay the same, and since vn and vp cancel vj in the non biased case, ie

vj + vn + vp = 0, and vp, vn doesn't change, then the formula given above is true, and Sedra and Smith are validated.

No, Vj = Vn+Vp only at equilibrium.

Let's look at the attachment again. Neudeck is talking about the contact potential of the ohmic contacts of the metal-semiconductor. He is not talking about Vp and Vn being the contact potiental of the PN junction. If fact, metal-semiconductor contacts are used to produce Schottky barrier diodes. Anyway and always, Vn+Vp=Vbi, thereby cancelling, out the voltage at terminals b-e at equilibrium. At Vbe = 0, the junction voltage is Vbi. So changing the Vbe raises or lowers the junction voltage according to Vj = Vbi - Vbe.

Ratch

 

[Claude Abraham]

Claude,

Instead of replying to every paragraph, this time I will just touch on the highlights.

If you do not like my saying steady state DC, I will change my verbiage to the word equilibrium, which the transisitor texts use frequently.

In your discussion, you mention that Vbi changes with the minority charges or whatever. All the textbooks I have say that Vbi is only dependent on the doping, and a constant which varies slightly with temperature. I can send you an attachment from Neudeck's book if you want. Anyway, you have to show me that Vbi changes dynamically like you say. I understand it to be set permanently during transistor manufacture by the doping concentration.

Ratch

For all practical purposes, Vbi can be regarded as constant. In Sze, it is computed as the space integral of the local E field. This E field value changes w/ charge accumulation slightly, but can be approximated as constant. I gave some values where Ie is 101 uA, with Vbe as 0.600 V. When Ie increased to 201 uA, Vbe only increases to 0.618 V. A doubling of current resulted in only a 3% increases in Vbe. Vbi is internal & changes even less. So it's basically correct to treat Vbi as a constant.

The point I was making was to show the sequence of events. If one examines only equilibrium conditions, which variable changes first is unknown. In my example, steady state or equilibrium values Ie/Ib/Vbe were 101uA/1.0ua/600mV. The new equilibrium values went to 201uA/2.0uA/618mV. So which variable "controlled" the change? One can argue that "the increase in Ie resulted in the increase in Vbe/Ic". But another can say "it was the increase in Vbe that changed the other variable values".

In equilibrium conditions 1 & 2, we expected a rise in Ib/Ie to include a rise in Vbe, & just as importantly, vice-versa. But which changed first, & which is the control quantity, or in your words "which is 'causal'?" Only by examining & measuring transient behavior can we begin to answer this question. But I remind you, that just because Ie changes 1st, does not make Ie the "cause" of Vbe. The fact that Ie/Ib change ahead of Vbe only means that Vbe is not causal.

But, Vbe is an essential part of bjt operation. The band gap energy requirement of the semicon material determines that a voltage drop is needed for a given current in equilibrium conditions. The Ie/Ib quantities are indispensible as well. All 3 equations relating Ic to Ib/Vbe/Ie provide useful info detailing the bjt behavior. Calling a bjt CC is not implying that Vbe is unimportant, because it most certainly is important.

To summarize, we have universal concensus that CC is a good bjt model at the external level. But this external condition is all I ever claimed that the CC model can cover. For internal viewing of the bjt device, I've already discarded the CC view in favor of QC (charge control). You've insisted that VC is the internal behavior of the bjt. But you can only support your position by neglecting time. Even in equilibrium your proclamation of VC is purely arbitrary. If we view the bjt variables after the transients have settled, in eq condition, which variable is causal & which is consequential cannot be determined. To insist that it is Vbe is nothing but dogma. To determine which variables are leading & lagging involves time, & only transient behavior will enlighten us.

CC works externally - you agree - I agree.

CC does not adequately cover internal behavior - you agree - I agree.

The difference in our views is that the OEMs & semicon physicists use QC & QM to describe internal behavior. You insist that the voltage eqn (eqn 2)) covers bjt internal operation. But when it comes to time delays, storage, rise & fall times, stored charge, etc., neither CC nor VC gives the answer. Only QC does. So when I need to take a closer look at the bjt, I scrap CC & adopt QC.

 

[BrownOut]

Why do I get the feeling that you still do not get it?

Get what? I know what I'm talking about

Do you always call well drawn illustrations in textbooks "cartoons"?

I couldn't care less how well they are drawn, but rather how accurately they describe reality.

The neutralization is not important with respect to the point that Vbe is directly increasing or decreasing the effects of Vbi, thereby permitting more or less charge to flow

.

The neutralization is very important.

No, Vj = Vn+Vp only at equilibrium.

EDIT 8/9 2:46pm - Yes, as I said, in the non-biased case.

Let's look at the attachment again. Neudeck is talking about the contact potential of the ohmic contacts of the metal-semiconductor. He is not talking about Vp and Vn being the contact potiental of the PN junction. If fact, metal-semiconductor contacts are used to produce Schottky barrier diodes.

vp and vn are important to the operatoin of the PN junction, and can't be dismissed.

Anyway and always, Vn+Vp=Vbi, thereby cancelling, out the voltage at terminals b-e at equilibrium.

vn and vp cancel vbi always, equilibrium or not.

At Vbe = 0, the junction voltage is Vbi. So changing the Vbe raises or lowers the junction voltage according to Vj = Vbi - Vbe.

vj = vbi - vbe is valid across the depletion layer only. Across the entire junction, vj is vbe. That's why you never see vbi in any of the formulas for current.

 

[Ratchit]

Brownout,

I couldn't care less how well they are drawn, but rather how accurately they describe reality.

Then perhaps you could explain where the diagrams depart from reality.

No, Vj = Vn+Vp only at equilibrium.

EDIT 8/9 2:46pm - Yes, as I said, in the non-biased case.

That is not what you said before, but OK, now we agree that Vj=Vn+Vp=Vbi at equilibrium

vn and vp cancel vbi always, equilibrium or not.

Not true, as explained below.

vj = vbi - vbe is valid across the depletion layer only. Across the entire junction, vj is vbe.

The depletion layer encompasses the entire junction. Across the whole diode, the voltage is zero at equilibrium, even though the junction voltage is Vbi. So there are three contact voltages in play here, the PN contact voltage (Vbi) and the two ohmic voltages at the metal-semiconductor interface. At equilibrium, their algebraic sum is zero. Upon forward bias, Vj = Vbi-Va, so the junction voltage is lowered and more charge can flow.

That's why you never see vbi in any of the formulas for current.

The current value is given by the diode equation references the voltage across the whole diode and takes into consideration the temp, type of semiconductor, and a fudge factor. The voltage across the junction is internal and cannot be measured directly.

Ratch

 

[Ratchit]

Claude,

For all practical purposes, Vbi can be regarded as constant. In Sze, it is computed as the space integral of the local E field. This E field value changes w/ charge accumulation slightly, but can be approximated as constant. I gave some values where Ie is 101 uA, with Vbe as 0.600 V. When Ie increased to 201 uA, Vbe only increases to 0.618 V. A doubling of current resulted in only a 3% increases in Vbe. Vbi is internal & changes even less. So it's basically correct to treat Vbi as a constant.

OK, we agree on Vbi. Small changes in Vbe make large changes in Ie/Ic due to the exponential relationship.

The point I was making was to show the sequence of events. If one examines only equilibrium conditions, which variable changes first is unknown. In my example, steady state or equilibrium values Ie/Ib/Vbe were 101uA/1.0ua/600mV. The new equilibrium values went to 201uA/2.0uA/618mV. So which variable "controlled" the change? One can argue that "the increase in Ie resulted in the increase in Vbe/Ic". But another can say "it was the increase in Vbe that changed the other variable values".

I like to think of the voltage being the cause, because, as I was trying to explain to Brownout, the Vbe lowers the junction voltage, thereby allowing the diffusion forces to push more charges across the junction. The b-e terminals are usually driven by a current supply, but the b-e diode already has plenty of charges that cannot go anywhere due to the barrier voltage. So Vbe will accomodate to whatever is necessary to make the current exist. I consider this a backdoor method to adjust the nonlinear Vbe to the correct value, and the causal reason for voltage control.

In equilibrium conditions 1 & 2, we expected a rise in Ib/Ie to include a rise in Vbe, & just as importantly, vice-versa. But which changed first, & which is the control quantity, or in your words "which is 'causal'?" Only by examining & measuring transient behavior can we begin to answer this question. But I remind you, that just because Ie changes 1st, does not make Ie the "cause" of Vbe. The fact that Ie/Ib change ahead of Vbe only means that Vbe is not causal.

But, Vbe is an essential part of bjt operation. The band gap energy requirement of the semicon material determines that a voltage drop is needed for a given current in equilibrium conditions. The Ie/Ib quantities are indispensible as well. All 3 equations relating Ic to Ib/Vbe/Ie provide useful info detailing the bjt behavior. Calling a bjt CC is not implying that Vbe is unimportant, because it most certainly is important.

I explained my reasons why I believe Vbe is causal. It directly affects the internal and separate junction voltage, and it does not matter if it happens before or after something else. I also agree that Ie/Ic are important, because they are the reason for the bJT's existence. You will disagree that Ib is necessary to bring Vbe to the junction, but is not the control. However, it is an indicator of output current.

To summarize, we have universal concensus that CC is a good bjt model at the external level. But this external condition is all I ever claimed that the CC model can cover. For internal viewing of the bjt device, I've already discarded the CC view in favor of QC (charge control). You've insisted that VC is the internal behavior of the bjt. But you can only support your position by neglecting time. Even in equilibrium your proclamation of VC is purely arbitrary. If we view the bjt variables after the transients have settled, in eq condition, which variable is causal & which is consequential cannot be determined. To insist that it is Vbe is nothing but dogma. To determine which variables are leading & lagging involves time, & only transient behavior will enlighten us.

Yes, charge is one way to look at it, but to move charge requires voltage or diffusion. Remember what the prof said about a BJT resembling a FET at low signal levels before secondary effects kick in? Does anyone have a quibble about a FET or a tube being a voltage control device?

CC works externally - you agree - I agree.

CC does not adequately cover internal behavior - you agree - I agree.

Yes, in fact, a lot of transistor texts do not get involved in whether a BJT is CC or CV.

The difference in our views is that the OEMs & semicon physicists use QC & QM to describe internal behavior. You insist that the voltage eqn (eqn 2)) covers bjt internal operation. But when it comes to time delays, storage, rise & fall times, stored charge, etc., neither CC nor VC gives the answer. Only QC does. So when I need to take a closer look at the bjt, I scrap CC & adopt QC.

Whatever works best, as long as you treat FET's and tubes the same way.

Ratch

 

[BrownOut]

Then perhaps you could explain where the diagrams depart from reality.

They have various weaknesses. Most are too simple to be dependable.

That is not what you said before, but OK, now we agree that Vj=Vn+Vp=Vbi at equilibrium

It's exactly what I said before. I've re-read what I wrote, evidently you haven't.

vn and vp cancel vbi always, equilibrium or not.
Not true, as explained below.

My statement is correct, as explained below.

The depletion layer encompasses the entire junction. Across the whole diode, the voltage is zero at equilibrium, even though the junction voltage is Vbi. So there are three contact voltages in play here, the PN contact voltage (Vbi) and the two ohmic voltages at the metal-semiconductor interface. At equilibrium, their algebraic sum is zero. Upon forward bias, Vj = Vbi-Va, so the junction voltage is lowered and more charge can flow.

A PN junction is more than just a depletion region. What goes on in the bulk silicon is important too. vbi is cancelled by vn and vp during equalibrium and during forward bias. Nothing you've written has shown otherwise. Va changes and total depletion voltage changes, but vbi is still cancled, as it is nothing more then the intrinsic voltage, and does not change. In other words, the voltage in the depleation region is reduced to vbi - va, under the condition of forced applied voltage. vbi is a constant, however, and does not change. Hence, the voltage across the diode is, from the perspective of the steady state voltage, equal to -va.

The current value is given by the diode equation references the voltage across the whole diode and takes into consideration the temp, type of semiconductor, and a fudge factor. The voltage across the junction is internal and cannot be measured directly.

The reason is because the change of barrier potential is identically equal to the voltage across the whole diode, the junction voltage ( as defined by Sedra and Smith ). This is bacause of the resaons given above.