Transistor theory question involving the base current

[eq1]

It is not a solution to bjt, but solution to diode.

You're right. I was being sloppy with my terminology there. The BJT math you're using is a solution to a very particular set of equations but that solution is not names the Shockley Equation.

The SE describes the math
A bjt b-e junction is a diode & thus described by SE. But SE gives the relation between Ie & Vbe, or Ib & Vbe.
Transistor action is as follows:
Ic = alpha*Ie. This is the TAE, transistor action equation.
Thus when we combine SE with TAE, we get
Ic = alpha*Ies*(exp(Vbe/Vt)-1).
...
A lab test will affirm that p-n junction current changes ahead of the junction voltage.

You're loosing me in this argument.

I thought you were trying to prove Ic is not determined by Vbe. I can't understand how the simulation of a diode, or the measurement of a diode in lab, shows that when a diode has no collector.

If BE in a transistor is just a diode what is the explanation for the current gain in the collector? If the answer is the transistor action equation given, where does alpha come from? Is alpha a parameter that can be manipulated by the designer of the BJT? How do they do it?

As a side note, I think you'll also notice the diffusion constant of the base is a member of the Ies constant given in the equation for Ic, why is that there?

Seriously. Read Dr. Hu's book. I think you'll get a lot from it.

 

[cabraham]

You're right. I was being sloppy with my terminology there. The BJT math you're using is a solution to a very particular set of equations but that solution is not names the Shockley Equation.
You're loosing me in this argument.

I thought you were trying to prove Ic is not determined by Vbe. I can't understand how the simulation of a diode, or the measurement of a diode in lab, shows that when a diode has no collector.

If BE in a transistor is just a diode what is the explanation for the current gain in the collector? If the answer is the transistor action equation given, where does alpha come from? Is alpha a parameter that can be manipulated by the designer of the BJT? How do they do it?

As a side note, I think you'll also notice the diffusion constant of the base is a member of the Ies constant given in the equation for Ic, why is that there?

Seriously. Read Dr. Hu's book. I think you'll get a lot from it.

Fair question re where does alpha come from. Here is a fair answer.
Alpha is Ic/Ie. When the b-e junction is forward biased, Ib & Ie are nonzero, as well as Vbe. It is desirable to make Ic as close to Ie as possible, minimizing Ib.
Alpha = beta/(1+beta). Infinite beta means alpha = 1, & Ib = 0, so Ic = Ie.
To make beta large & make alpha approach a value of 1, the base region is doped lightly while the emitter region gets heavy doping. This way, under forward bias, the base region hole density is light resulting in a small number of holes moving towards emitter. The emitter doping is heavy resulting in large number of electrons moving towards base & collector.
This is why current gain beta is large. The base region has small volume with low density of acceptor ions, while emitter region has larger volume with heavy density of donor ions.
This is done to insure that Ie >> Ib.

 

[cabraham]

This attached pdf demonstrated the relation between Ie, Ib, Ic, & Vbe. The Vbe lags behind Ib, Ie, as well as Ic. Vbe cannot be the controlling quantity.

 

[Tom.G]

The Vbe lags behind Ib, Ie, as well as Ic.

Not according to the plot you posted. The corners of the various traces perfectly line up at the same instant. The slopes do vary from drift time, parasistics, etc.

This can easily be seen at the 12ns mark where ALL the signals begin their transition at the same instant. Magnify the plot to about 1600% or more and align the Vbe corner with the edge of the window, now scroll vertically. You will see the other trace corners also align at the edge of the window.

Try scaling either the plot traces or the supplies/component values so that all traces have the same amplitude, that should highlight any delays that exist.

Cheers,
Tom

 

[eq1]

This is why current gain beta is large. The base region has small volume with low density of acceptor ions, while emitter region has larger volume with heavy density of donor ions.
This is done to insure that Ie >> Ib.

Is the geometry of the collector region not important? It doesn't seem to factor into the explanation.

Notice the structure on slide 3. (Jaeger's Microelectronics is also a great text for this) Given your explanation, what is the n+ buried layer for? (you might also notice the emitter volume is shown as smaller for this particular device)
https://engineering.purdue.edu/~ee255d3/files/MCD4thEdChap05_[Compatibility_Mode].pdf

In any case, the equation for beta has two other very important components besides doping. The width of the base and emitter depletion region as well as the diffusion constants. See equation 5.3.11 from the link below. How do those factor into the explanation?

https://ecee.colorado.edu/~bart/book/book/chapter5/ch5_3.htm#5_3_1

Have you ever see a figure like 5.3.1? (link below) Since your claim is Ic is not determined by Vbe how do you account the distance wb-xp,bc? Is that distance not a function of Vbe?

https://ecee.colorado.edu/~bart/book/book/chapter5/ch5_3.htm#fig5_3_1

 

[eq1]

Earlier we saw:

Ic is NOT "determined" by Vbe.

But the "related thread thing" at the bottom just showed me this:
https://www.physicsforums.com/threads/transistor-base-collector-current-mystery.169710/post-1329920

cabraham: In a nutshell, Ic is strongly determined by the forward bias on the b-e junction, and weakly influenced by Vce.

Am I getting trolled here or something...

 

[cabraham]

Not according to the plot you posted. The corners of the various traces perfectly line up at the same instant. The slopes do vary from drift time, parasistics, etc.

This can easily be seen at the 12ns mark where ALL the signals begin their transition at the same instant. Magnify the plot to about 1600% or more and align the Vbe corner with the edge of the window, now scroll vertically. You will see the other trace corners also align at the edge of the window.

Try scaling either the plot traces or the supplies/component values so that all traces have the same amplitude, that should highlight any delays that exist.

Cheers,
Tom

Thanks for examining my plots. Of course the trace align at the edge of window, we agree on that. But notice that Vne rises, eventually settling. But the currents, Ib/Ie/Ic rise faster, overshoot very slightly, then decrease, settling to final value.
Ic responds to Ie rising by rising, Ie overshoot its final value, so does Ic, as it must. If emitter emits more electrons, then collector will collect more electrons,, inevitably always..
Ie, & Ic, both settle to final values, as will Vbe, but Ic overshoot then decreases while Vbe is monotpnically rising. If Ic was controlled by Vbe, then Ic would monotlnically rise then settle to its final value, which it does not. A short time span exists where Ic is decreasing while Vbe still increased. Tjis is because electron collection in collecror is controlled by electron emission from emitter.
The emitter emission of electrons increased, overshot, then decreased & settled. The collector collection if electrons increased, overshot, then decreased & settled. Just like emitter emission did.
But Vne monolithically rose. After overshoot by Ie & Ic, Vbe was increasing while Ie was decreasing. Ic followed Ie, not Vbe. Ic was decreasing while Ie was decreasing & Vbe still increasing.
Ic responds to changes in Ie, while Vbe is responding to changes in Ib as well, but with extra delay.
I realize these are subtle points. It takes much examination to reveal these responses.
Thanks for your interest. Feel free to ask further questions.