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
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.
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."
"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.
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.
Read my statement again. I said it opposes the energizing voltage. That is what a back voltage does.
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.
As I said before S & S analyze the BJT with a constant current source, but that does not change Vbe controlling the forward current.
the equation for forward current should be Id-Is=I, which means diffusion current minus thermal current equals the forward current.
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
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, 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,
"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".
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.
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.
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.
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
Why do I get the feeling that you still do not get it?
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
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.
I couldn't care less how well they are drawn, but rather how accurately they describe reality.
No, Vj = Vn+Vp only at equilibrium.
EDIT 8/9 2:46pm - Yes, as I said, in the non-biased case.
vn and vp cancel vbi always, equilibrium or not.
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.
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.
Then perhaps you could explain where the diagrams depart from reality.
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.
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.
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.
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