# Transistor theory question involving the base current

Summary:
I know that the base to emitter current is what makes the transistor essentially turn on but why doesn't that deplete the second barrier and only the first
Help

DaveE
Gold Member
I'm not sure I understand your question. For an NPN BJT in normal (linear) operation, the BE junction is forward biased and the CB junction is reverse biased with an associated depletion region. The base current doesn't deplete anything, it allows electrons to enter the depletion region and generate more current across the CB junction.

One (not very precise) way of thinking about it is some of the electrons that flow from the emitter end up in the depletion region, because the base region is thin, and are swept to the collector by the strong e-field in the depletion region.

https://www.allaboutcircuits.com/te...nction transistor shown,and a base in between.

Do not overestimate the role of the base current.
(In this context, one of the worlds best known D&D engineers - the late Barrie Gilbert - has used the term "defect" and "nuisance" for the base current).
Without any doubt , there is always a base curent (which can be accounted for during the design of BJT-circuits), however, it is only the base-emitter voltage Vbe which determines the collector current (Shockleys famous equation).
This can be (and was) demonstrated and prooved.
(There is not a single proof or explanation for current control)

tech99
Do not overestimate the role of the base current.
(In this context, one of the worlds best known D&D engineers - the late Barrie Gilbert - has used the term "defect" and "nuisance" for the base current).
Without any doubt , there is always a base curent (which can be accounted for during the design of BJT-circuits), however, it is only the base-emitter voltage Vbe which determines the collector current (Shockleys famous equation).
This can be (and was) demonstrated and prooved.
(There is not a single proof or explanation for current control)
Barrie Gilbert had little formal tech education beyond high school. He went far with what little he knew but his explanations are sheer nonsense.
Nuisance - can be applied to Vbe as well. Germanium bjt had Vbe of 0.2-0.3 volts. Silicon goes 0.6-0.7. This limits the swing of an emitter follower driving a load.
"Defect"? A p type base & n type emitter are forward biased by an external source. Emitter electrons move towards base & collector. Base holes move towards emitter.
These base holes comprise nearly all the base current at low frequency. A forward biased p-n junction exhibits forward current. No defect here.
Gilbert said many times that base current should be zero, but how can a forward biased diode have zero current. Ideally Vbe should be zero since a forward diode ideally has no voltage drop.
Current control model was published in the Ebers-Moll 1954 paper. Emitter current controls collector current per
Ic = alpha*Ie.
Shockley equation relates p-n junction current & voltage, Id = Is*(exp(Vd/Vt)+1)
Combining gives Ic = Ies*(exp(Vbe/Vt)+1).
Ie & Vbe are related via Schockley. But Ic = alpha*Ie, is the equation for transistor action.
As electrons cross the base into collector, while holes cross from base to emitter, depletion zones are firmed resulting in barrier potential. This barrier is nearly equal to Vbe, & is determined by the current. A lab test affirms that Ie changes ahead of Vbe.

Hi Claude....You will remember some of our earlier discussions on this matter.
I agree that - as you wrote - emitter current controls collector current per Ic = alpha*Ie.
And therefore (as you have stated): Ic = Ies*(exp(Vbe/Vt)+1).

Final conclusion: Vbe determines (controls) Ic.
The base current Ib plays no major role - as far as the voltage gain of amplifier stages is concerned. The existence of Ib just reduces the input resistance of BJT based circuits.
(As THIS was the original question).
Do you agree?

(By the way: According to my knowledge, Barrie Gilbert has "never said many times that base current should be zero"). His position was that Ib does not "control" (determine) Ic in a cause-effect-sense. And therefore, this current would be more or less parasitic.)

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Hi Claude....You will remember some of our earlier discussions on this matter.
I agree that - as you wrote - emitter current controls collector current per Ic = alpha*Ie.
And therefore (as you have stated): Ic = Ies*(exp(Vbe/Vt)+1).

Final conclusion: Vbe determines (controls) Ic.
The base current Ib plays no major role - as far as the voltage gain of amplifier stages is concerned. The existence of Ib just reduces the input resistance of BJT based circuits.
(As THIS was the original question).
Do you agree?

(By the way: According to my knowledge, Barrie Gilbert has "never said many times that base current should be zero"). His position was that Ib does not "control" (determine) Ic in a cause-effect-sense. And therefore, this current would be more or less parasitic.)
Parasitic? That means it does nothing. Base current can be useful.
Emitter follower - load is on emitter side. Load current equals Ie. But Ie = Ib + Ic. So Ic & I are powering the load. Nothing parasitic here.
But the input source is Vg. An emitter follower ideally buffers a voltage source by presenting high input Z & low output Z.
The output, Ve (emitter voltage wrt ground) is one forward drop below Vg, ie Ve = Vg - Vbe. In this anecdotal case, Vbe does not power the load, & Vbe is a fraction of the input source Vg, that does not reach the output load.
In this case, Ib powers the load, Vbe is a loss, or parasite.
Common emitter - the load is on the collector side, powered by Ic & Vcc, the rail supply. Hence one can view Ib as a parasite, since it doesn't power the load. But the same can be said for Vbe. Both can be called parasitic.

Common base - same as common emitter.

Ib can be viewed as parasitic as it does not power the load in 2 topologies, CE & CB. In CC (EF) topology, Ib powers load.
Vbe never drives a load. It is always a fraction of input voltage lost due to forward drop of b-e junction.

In addition, even when Ib does not power load, it's existence is an inevitable result of base region doping during fabrication.
A low density of acceptor ions diffused into base results in high beta, low Ib. But tradeoff is low Vce breakdown voltage & high collector base leakage current. These are undesirable.
A bjt with low Vce breakdown rating will *punch through* if Vcemax is exceeded. The Ib value is nice & low, as we wish. But collector to emitter punch through can be catastrophic.
So kit's increase the doping density of acceptor ions, ie boron, in the base region during fab. The result is a bjt with lower collector to base leakage current, higher Vcemax, & less prone to punch through. The tradeoff is higher Ib. Not something we want, but increased Ib is worth it to avoid punch through, no debate there.
Vbe is not useful in & of itself, but it is a byproduct of fabrication to avoid bigger problems. Early bjt was made with germanium. This material has lower forward voltage drop, but higher reverse leakage current.
The Ge parts collector to base reverse leakage was bad at temps as low as 100 C. When you were turned off, they were not fully off.
1959 - Silicon arrives. Si parts have much lower reverse leakage current. Ge diodes & bjt had high reverse leakage, but low forward drops.
So diodes & bjt have low reverse leakage with higher forward drops. Tradeoff just like the Ib case.
Why use Si when Ge offers lower drop? The crossover distortion in class B amps is a problem. Instead of 0.7 volt drop, lower is better.
We use Si material & accept 0.7 volts, not because more Vbe is good, but because less reverse leakage is great. Vbe, like Ib, is a quantity we wish to minimize.

The "current driven" model if the bjt refers to the EMITTER current, NOT BASE current.
Vbe & Ib are not that useful, except in emitter follower where Ib powers load.

Questions welcome.

Anyway, the ebers-moll model published in 1954 shows bjt as Ic = alpha*Ie, not
beta*Ib. The i-v diode files are a simulation. Signal generator voltage is stepped into resistor & diode. The diode current is a quick staircase waveform. The diode voltage is slower, the current hits full value as the voltage hardly budged. Diode voltage is slow but settles to final value Ling after current settled. Next jump repeats the process.
Shockley: 2 forms of equation
1) Id = Is*(exp(Vd/Vt)-1)
2) Vd = Vt*ln((Id/Is)+1)

Many blindly assume that the voltage determines or "controls" the current. But the simulation shows that voltage on a p-n junction changes as a result of a current change. The current goes from existing value to the new value quickly. Voltage slowly catches up.
Diode current is NOT "controlled" by diode voltage.

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DaveE
Gold Member
Is it voltage or current controlled? Isn't that like asking is the heat in a resistor caused by it's voltage or current? You have both; you always have both. If you have a model that you like and that works, use that one.

I also think talking about "gain" depends greatly on the application. In some circuits voltage gain makes sense, in others current gain is easier to use.

sandy stone
Baluncore
Many blindly assume that the voltage determines or "controls" the current. But the simulation shows that voltage on a p-n junction changes as a result of a current change. The current goes from existing value to the new value quickly. Voltage slowly catches up.
At last we have narrowed it down to two possible answers to the question of which came first, the chicken or the egg. We can simulate a chicken laying an egg, or we can simulate an egg hatching into a chicken. A simulation can show anything we want, depending on the order and direction of computation.

At last we have narrowed it down to two possible answers to the question of which came first, the chicken or the egg. We can simulate a chicken laying an egg, or we can simulate an egg hatching into a chicken. A simulation can show anything we want, depending on the order and direction of computation.
Yes - I fully agree. Simulation is mathematics, nothing else. Our simulation programs can never reveal if the effect A is the cause or the result of the effect B. These simulations are based on relations only (forward/backwards).
However, concerning the question voltage/current, I think it is clear: Current is always the result of a voltage. No current (in our electronic devices) without a driving E-field (caused by a voltage).

Parasitic? That means it does nothing. Base current can be useful...............
Vbe is a loss, or parasite.................
Hence one can view Ib as a parasite, since it doesn't power the load. But the same can be said for Vbe. Both can be called parasitic.
OK - perhaps I have used a term ("parasitic") which is not correct....or it must be defined before.
To me - "parasitic" means: A property which is not intended and which has an unwanted influence on the desired circuit behaviour (like the parasitic capacitance of a circuit node).
In this sense - the voltage Vbe is certainly NOT a parasitic one because it is a voltage which is externally applied (and not the result of an unwanted effect).

Many blindly assume that the voltage determines or "controls" the current. But the simulation shows that voltage on a p-n junction changes as a result of a current change. The current goes from existing value to the new value quickly. Voltage slowly catches up.
Diode current is NOT "controlled" by diode voltage.
I do not intend to place a comment to this surprising claim.
Just one remark: Are we allowed to "blindly" trust simulation results ?
I doubt if such a simulation can correctly reproduce the time relationships between physical processes (cause and effect), unless they are intentionally included in the models.

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At last we have narrowed it down to two possible answers to the question of which came first, the chicken or the egg. We can simulate a chicken laying an egg, or we can simulate an egg hatching into a chicken. A simulation can show anything we want, depending on the order and direction of computation.
Very good point! The l-V relation pretty much is chickens & eggs. I've said that for years.

OK - perhaps I have used a term ("parasitic") which is not correct....or it must be defined before.
To me - "parasitic" means: A property which is not intended and which has an unwanted influence on the desired circuit behaviour (like the parasitic capacitance of a circuit node).
In this sense - the voltage Vbe is certainly NOT a parasitic one because it is a voltage which is externally applied (and not the result of an unwanted effect).
But Ib is externally applied as well. The external source powers resistors & the b-e junction. Vbe is a drop incurred by charges moving through the n & p material. You claim that Vbe is externally applied, which is only true if b-e is shorted directly across a voltage source, which is certain doom for the bjt.

But Ib is externally applied as well. The external source powers resistors & the b-e junction. Vbe is a drop incurred by charges moving through the n & p material. You claim that Vbe is externally applied, which is only true if b-e is shorted directly across a voltage source, which is certain doom for the bjt.
"...charges moving..."
May I simply ask - which force causes the charges to move?

"...charges moving..."
May I simply ask - which force causes the charges to move?
The signal generator. Could be an antenna, phone cartridge, microphone, etc.
For example, Sue sings into a mic, charges are moved by her vocal energy. If mic is dynamic, her air pressure from her voice moves the mic diaphragm & charges are moved. This wave travels through the mic cable & reaches the bjt in the mic preamp. The charges move through base & emitter regions, crossing junction & forming depletion zone barrier.
My Sim showed I-V relation with diode. Charges move through anode & cathode, resulting in forward voltage drop.

Sue the singer causes charges to move right at the mic diaphragm. The charges travel through the cable reaching bjt b-e junction. The depletion barrier potential has not changed yet, since charges have not crossed barrier.
Ib & Ie have changed due to Sue singing. These charges cross the barrier changing depletion barrier. The barrier potential cannot change without charge moving through base & emitter regions.

Vbe very nearly equals built in potential Vbi. Sue vocal pressure moves charges from mic to b-e junction. Vbe changes due to Sue, as does Ib & Ie. But Vbi changes as a consequence if charges crossing junction & recombinant. Then Vbi changes, & incoming charges feel an opposition due to the local E field in the depletion region.

Sue caused all this to happen. Ib & Ie are not caused by Vbe. They are all caused by Sue. Ib & Ie change BEFORE barrier potential has the chance to change.
The idea that Ib/Ie change as a *consequence* of Vbe changing is pure prejudice. Shockley equation is steady state, not transient.

After the depletion zone changes due to Sue, & things settle, the Ib-Vbe-Ie relation per Shockley is on effect. Momentarily during transients, the I-V relation does not follow Shockley.

Thank you for the long and very detailed answer.
However, with all respect - I did not want to learn something about Sue ....I only wanted to clarify which electronic force causes the charges to move.
To my opinion, it is an E-field within a conducting material. Perhaps I am wrong and it is the "vocal energy of Sue" ? A hard discussion....
(Sorry...but I could not resist to be ironic).

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And how are E fields set up inside conducting material?
By transporting charges, of course. If Sue stays silent, how can the bjt have E fields created inside its b-e region?

E fields are the result of the external mic, after being energized by Sue's vocal energy. Charges are transported the cable to the bjt. When charges cross b-e junction, electrons from emitter cross base region, where they then enter collector. Emitter electrons mostly survive the trip thru base, a few get recombined with base holes. The great number that reach the collector have high mobility. The collector is n type so electrons are majority carriers inside collector & emitter as well, being n type.
But the few electrons that recombine in base with holes form one edge of the b-e depletion zone. The electrons are MINORITY carriers when they are inside the p type base region.

Holes from base move towards emitter, but once they enter emitter, n type region, the holes are a MINORITY carrier, & recombine quickly near the edge of the emitter. This is the other end of the depletion zone.

Thus an E field is formed by these minority carriers. This E field integrated over the depletion zone width forms the Vbe. Before Sue stimulated the mic, Vbe was at its quiescent value determined by bias network.

in order to stimulate bjt charges from outside world enter b-e region. These charges in motion are Ie & Ib. When the charges reach the edges of emitter & base regions Ie & Ib have been perturbed from their quiescent value.
A moment later after charges crossing depletion zone & recombinant, Vbe gets perturbed.

Ib, Ie, & Vbe all change as a result of mic action sending charge carriers towards bjt.
Ie & Ib change immediately when new charges enter b-e regions. Vbe changes a bit later due to junction crossing & recombination.
I'm at a loss to make it clearer.

A bjt is not easy to understand. The 3 dimensional charge motion, diffusion, recombination, etc., is not trivial. Even brilliant minds need time & effort to understand it.

Even when I was age 30, I was not fully aware of all this. It took me years to find out all of this. Even a very smart person will struggle at first with this stuff I reviewed.

berkeman
Baluncore
Looking at it differently, the base drive energy arrives along a bundle of Poynting vectors, that reach the BE junction as an EM wave guided between the two conductor (probably asymmetric) transmission line. Let us assume that the impedance of the line is matched to the BE zone of the transistor.

As the electric field appears across the junction, the magnetic field guiding current will flow over and begin to diffuse into the BE junction that terminates the line.
What is it about the BE junction? Do the charge carriers react to the transverse electric field, to the magnetic field, or to the skin surface current?
Do PNP and NPN behave differently? Is one electric and the other magnetic?

berkeman
And how are E fields set up inside conducting material?
By transporting charges, of course.
OK - when it is not the E-field (caused by the applied voltage) which is responsible for the movement of charges (because you say that the E-field is "set up" by the moved charges), I must ask again ( as in my post#15):
Which force causes the charges to move?

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OK - when it is not the E-field (caused by the applied voltage) which is responsible for the movement of charges (because you say that the E-field is "set up" by the moved charges), I must ask again ( as in my post#15):
Which force causes the charges to move?
Electric force does that. Singer Sue outputs acoustic energy that mic transduces into E field. This sets charges in motion. Charges move through mic cable & reach bjt b-e junction. Ib, Vbe, & Ie get changed from these new charges showing up. Any elaboration still needed.

Electric force does that. Singer Sue outputs acoustic energy that mic transduces into E field.
OK - from the beginning, this was my conviction, of course.
But in post#14 you wrote: "Vbe is a drop incurred by charges moving through the n & p material."
To me, this is a kind of contradiction.
Either Vbe is a drop - caused by moved charges; or it provides the electric force (E-field) which causes the charges to move.

anorlunda
Staff Emeritus
Either Vbe is a drop - caused by moved charges; or it provides the electric force (E-field) which causes the charges to move.
I think you are trying too hard to apply cause and effect, first comes X then Y. The relationship V=I/R is circular. (Even when R is nonlinear and non-constant, it's still circular.)

In a circular relationship, there is no first and last. Voltage causes the current and current causes voltage. Resistance (constant or not constant) merely describes the relationship between V and I. You can turn that around by saying V and I define R, so we can't even say that R comes first.

Once again, it is circular. Therefore, there is no either or as you said.

I think you are trying too hard to apply cause and effect, first comes X then Y. Voltage causes the current and current causes voltage.
Sorry, I cannot agree to that.
Current causes voltage?

Lets assume we look at a simple resistor.
When the current through this resistor can produce the voltage across this device - which force drives resp. allows the current ? Current is identical to movement of charges - without an E-field there can be no continuous flow of charges.
(I speak about the common understanding of the quantity "current" within an electrical circuit resp. within parts like resistors or semiconductors; I do not refer to movement of charges caused by a chemical process or by diffusion/drift effects.)

Of course, during calculations and analysis of electrical circuits we can ASSUME that according to V=I*R the current I could "produce" the voltage V - from the math point of view this works.

However, asking which comes first (as a cause of the effect) it is always the voltage which comes "first". No current without a driving E-field within a conducting material.

anorlunda
Staff Emeritus
However, asking which comes first (as a cause of the effect) it is always the voltage which comes "first". No current without a driving E-field within a conducting material.
You said that before. It's not true. Consider an inductor L with an initial current I0. Now we switch the inductor to disconnect it from the charging circuit, and put in in parallel with a resistance R or a Diode D. From the R's view, a current suddenly appeared which causes a voltage to appear across the R or D device.

The inductor is just a current source. There are many other kinds of current sources. Consider a simple solar cell. Depending on the operating point, it can act like a current source (I versus V nearly horizontal) or like a voltage source (I versus V nearly vertical).