"The fact that ic/ie/ib settle to final plateau values while vbe continues to rise is proof positive that vbe is NOT the mechanism that "controls" ie/ib/ic."
"It is ie that controls ic, not ib."
Hi Claude, some time ago we already had a discussion about the same subject - and I do not want to repeat all my arguments again.
And I am happy to agree with you that "it is ie that controlsic, not ib". However, the question rises "which quantity is the cause of ie"? I couldn`t find an answer in your contribution.
Perhaps you have seen some of my arguments in favour of voltage control in this thread - unfortunately, there was not a real and serious discussion.
In contrary, one forum member was so free to consider my attempts for describing my position as "rubbish" - to me it is really funny, it seems that for some people this question is something like a religion that cannot be discussed.
In addition to my earlier presented arguments, I can mention the Early effect which clearly prooves that it is the electrical field within the base region (prop. to the voltage Vbe) that influences Ic. Finally, I like to mention that this sight is clearly supported by papers from Stanford Univ., Berkeley Univ and also the MIT. References can be provided.
LvW
The CAUSE of ie is the external signal source whose signal is inputted to the bjt amp stage, i.e. a microphone, tuner, CD player, antenna, etc. Let's say that the bjt amp in question is the 1st stage of a PA system, and Sue is singing into a mic plugged into this PA 1st stage.
The mic signal is a steady value, then Sue inflects her voice and increases the mic output. The mic diaphragm vibrates with greater amplitude. For a condenser tye of mic, charge in the electret element undergoes an increase in displacement resulting in an increased current out of the mic. Likewise the charge separation increases, which increases the mic output voltage.
But the bjt b-e junction does not respond instantly. As these increased charges propagate through the mic cable it takes finite time, ic, the bjt collector current has not increased yet. When this increased charge density arrives at the bjt b-e junction, what happens?
The holes enter the base, electrons enter the emitter. From that point they easily drift through their respective regions since holes are very mobile in the base, being the majority carrier, likewise for electrons in the emitter. The small signal b-e voltage vbe has not yet responded to this change.
When the electrons have completed their journey through the n-type emitter region (bjt device is npn), they encounter the p-type base. In this region the electrons are minority carriers and if the base were thick, most electrons recombine with base atoms and ionize them. But a bjt is made with an intentionally narrow base region so that electrons from the emitter survive the trip through the base and reach the collector with a very high percentage, 99.98% or so.
The increased density of electrons entering the base results in this increased number entering collector, increasing collector current ic. THe plots attached in my last post illustrate this. However not all electrons entering emitter make it to the collector.
The increased density of holes entering the base results in more holes recombining at the edge of the emitter region near the base, ionizing the b-e depletion region to a greater extent due to increased hole density. As a result a slight increase in the number of emitter electrons that recombine in this hole-enhanced edge of emitter region takes place.
Likewise an increase in electrons transiting through the base increases the number that recombine in the base region, ionizing that edge of the b-e depletion zone. This increase in depletion zone ionization at both edges increases the local E field and barrier potential.
The b-e voltage vbe, increases in response to increased emitter current ie density. Hoever let us remember that a very small fraction of emitter electrons get left behind at the emitter edge near base, and in the base itself. So the collector current increases immediately after emitter current increases. Since electron emission increases, so does collection, with the delay being the transit time from emitter through base to collector.
But how does vbe get charged up to the new increased value? As electrons ionize both edges of the depletion zone, the barrier potential builds up with increased charges transiting through. But for a bjt whose beta value is over 100, quite common, less that 1 electron per 100 contributes to this increased ionization and voltage increase. Thus many electrons must transit through emitter and base in order to eventually build up the depletion zone ion density to its new increased value.
The plots illustrate this. Notice how collector, base, and emitter current reach their plateaus long before vbe does the same. The increase in ie is propagated very quickly to ic, but the increase in vbe takes more time. The depletion layer gets increased by electrons/holes that recombine and that is a small fraction of the total.
It is clear that Sue the singer is what "controls ie", and clearly NOT vbe. The ic has settled on its plateau while vbe is still catching up. Then vbe keeps increasing to finally and eventually reach its plateau whereas ic has been constant for a long time.
There is no doubt that ic is controlled by ie, and ie is controlled by the signal acting as stimulus, which is Sue's microphone. Vbe cannot be the cause of ic because a cause must ALWAYS PRECEDE its effect. This is not the case. Ic stops changing while vbe continues to change. That is a no brainer.
I will clarify if necessary, but you have asked me that question before which I've answered. If you have issues w/ my treatise, please elaborate. Best regards to all.
Claude
