Transistor theory question involving the base current

[cabraham]

Believe me, you don't want to dig down to the fundamental physics of electricity which is QED. But if you're talking circuit analysis, which I believe we are, then the relationship is circular. Circuit analysis is compatible with physics as long as certain assumptions are valid. You can call it either simplified physics or computational tricks that are not physics; the difference is semantic.

I won't engage in a meaningless semantic debate. If you insist that every current source is really a voltage source go ahead. But your either-or is still false.

This thread kind went off topic but the answer to the original question is: charge doesn't move from C to B because that would be uphill in energy when the NPN is in active mode. [1] Why is the collector at the lowest energy? Because that's how the NPN is constructed. We dope the collector that way on purpose.

I know textbooks like to draw BJTs like back to back diodes but a BJT is not a back to back diode. (I suppose it is that, but it is also more.) If you want proof try the following experiment. Wire two diodes back to back on a bread board, see if you get a transistor. When you understand why that didn't work you'll understand BJTs.

Personally, I like to think of the BJT as voltage controlled. V(B,E) modules the height of the barrier from E-B. When the barrier is low enough charge from E can flow into B and then it's all downhill from there. Personally, I am not a fan of the water analogies but it's kinda like the lowering of a flood gate in a dam, once the gate gets below the water line, things start to move. In a dam the force moving the water is gravity. In a BJT its diffusion moving the charge. And it kinda shows how the amplification works, i.e. a small movement of the gate height leads to a large change in the amount of water in the river below. And I suppose it can even have some implications on this discussion. i.e. for the dam, which controls the movement of water? The flood gate or gravity? And of course, for the case of the dam, the answer is both. But the analogy is getting a bit stretched.

My $0.02 anyway.

[1] https://en.wikipedia.org/wiki/File:Bjt_forward_active_bands.svg

Vbe does NOT "modulate height of barrier". Also how can diffusion "move the charges"? Diffusion is the tendency for charges to spread out until all regions have equal concentration.
The barrier in a bjt without being energized is the thermal voltage, roughly 25.9 mV, usually rounded off to 26 mV.
What modulates this barrier is changes in current. Vbe changes after current has changed. Say the bjt is biased at Ic = 1.0 mA with Vbe = 0.65 V. What changes barrier value?
At 1.0 mA, charges cross junction & recombine. Vbe stays at 0.65 V since recombination rate & transport rate are in equilibrium.
The external source, Sue the singer, starts singing, which moves charges from mic, through cable, to bjt. This increase in charge results in more recombination in the depletion zone, raising the barrier slightly. Equilibrium occurs when the barrier increases to balance recombination with increased charge density.
The current increased due to Sue energizing more charges. The extra charges transported through b-e junction resulted in more charges accumulating in the depletion zone, most reach the collector resulting in increased Ic. Some recombine, increasing the recombined charges in depletion zone, increasing Vbe.
Look at a bjt data sheet & a graph is usually given for Vbe as a function of Ic or Ie.
Ie & Ib changed as a result of Sue. Vbe changes later due to increased recombination in base-emitter depletion zone.
Vbe does not force Ie to change. It does not change barrier value. Vbe changes as a result of barrier being subjected to a change in charge transport.
Semiconductor physics texts will affirm this.

 

[eq1]

Also how can diffusion "move the charges"? Diffusion is the tendency for charges to spread out until all regions have equal concentration.

In your mind, what is the difference between "move" and "spread out"?

You definitely seem curious about the topic but I think you have quite a lot of misunderstandings. I can highly recommend Chenming Hu's (TSMC Distinguished Professor Emeritus and 2020 IEEE Medal of Honor winner) Semiconductor Textbook, which should also qualify as a physics textbook. It was written with the practitioner in mind so it might be more up your alley than the more conventional physics textbooks.

I think there are many sections of it you would find interesting but I'll point out this particular section of chapter 8 as it seems to be on topic.

https://www.chu.berkeley.edu/modern...-integrated-circuits-chenming-calvin-hu-2010/

"Figure 8–1b shows that when the base–emitter junction is forward biased, electrons are injected into the more lightly doped base. They diffuse across the base to the reverse-biased base–collector junction (edge of the depletion layer) and get swept into the collector. This produces a collector current, IC. IC is independent of VCB as long as VCB is a reverse bias (or a small forward bias, as explained in Section 8.6). Rather, IC is determined by the rate of electron injection from the emitter into the base, i.e., determined by VBE. You may recall from the PN diode theory that the rate of injection is proportional to e^(qVBE ⁄ kT) . These facts are obvious in Fig. 8–1c."

 

[cabraham]

Ic is NOT "determined" by Vbe. Ic changes as a result of Ie changing. Vbe catches up eventually.
Many references, including uni texts, often use the Shockley equation
Ic = alpha*Ies*exp((Vbe/Vt)-1) as "proof" that currents Ie or Ic are "controlled" or determined by Vbe. But Shockley's equation is just as easily expressed as follows:
Vbe = Vt*ln((Ie/Ies)+1).
You & some text authors insist that changes in Ie/Ic are a *result* of changes in Vbe. A lab bench & scope test are needed.
I submitted a simulation showing diode voltage as changing after diode current has changed. LvW insisted that this Sim is only as accurate as the model used. Maybe lab set up is nevessary.

 

[eq1]

But Shockley's equation is just as easily expressed as follows:
Vbe = Vt*ln((Ie/Ies)+1).

You’re wielding Shockley’s equation like it’s an axiom but it’s not. It’s a solution to a very particular system known as a BJT. But, where did Shockley derive it from? And what experimental evidence supports those equations? And what do they have to say about this topic? I would write it down but Dr. Hu has done an excellent job for me and I doubt I can improve upon his work. In addition Berkeley has made his work available for free! I suggest you take advantage of it.

Edit: I also just remembered Sedra & Smith has a pretty decent derivation of the Shockley equation (and MOSFET and PN junction for that matter) just using dimensional analysis which might be more approachable. It's definitely not free but it's been a standard microelectronics text for decades so I think most libraries, pretty much anywhere in the world, should be able to obtain it. The problem is that derivation will only work for a 1d transistor, basically one line in the cross section. But it illustrates the point nicely and succinctly.

 

[cabraham]

You’re wielding Shockley’s equation like it’s an axiom but it’s not. It’s a solution to a very particular system known as a BJT. But, where did Shockley derive it from? And what experimental evidence supports those equations? And what do they have to say about this topic? I would write it down but Dr. Hu has done an excellent job for me and I doubt I can improve upon his work. In addition Berkeley has made his work available for free! I suggest you take advantage of it.

Edit: I also just remembered Sedra & Smith has a pretty decent derivation of the Shockley equation (and MOSFET and PN junction for that matter) just using dimensional analysis which might be more approachable. It's definitely not free but it's been a standard microelectronics text for decades so I think most libraries, pretty much anywhere in the world, should be able to obtain it. The problem is that derivation will only work for a 1d transistor, basically one line in the cross section. But it illustrates the point nicely and succinctly.

I realize SE (Shockley equation) is NOT an axiom. In semiconductor physics class,cwe derived it as homework.
It is not a solution to bjt, but solution to diode. The SE describes the math relation between diode voltage & current in forward direction, or to limited extent in reverse before Zener breakdown occurs.
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).
Read my earlier post where a p-n junction diode was connected to a staircase generator & resistor, & observe the I & V diode waveforms. It is clear that current settles before voltage.
The current in a diode determines the voltage, not the other way around.
As soon as the generator increases its voltage its current increases. The charges move towards the diode. When they arrive, anode & cathode current both increase, but it takes time for Vd, forward voltage drop of diode, to settle to its final value.
LvW insisted that simulator models are suspect, & that Sim results are only as reliable as the models, I've witnessed such waveforms in labs on a scope.
A lab test will affirm that p-n junction current changes ahead of the junction voltage.

 

[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.