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How a BJT Transistor works (base current version)

How a BJT Transistor works (base current version)

cabraham

Feb 12, 2015
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This is my understanding. With no electric field (Voltage) the average motion (current) of electrons within the junctions cancel out to be zero. When an electric field is applied across the junction the additional motion of the electrons in between their collision times (mean path time) is termed the drift current. The diffusion current is when the concentration of carriers on one side of the junction cause the attraction of the opposite carrier.

When an electric field (Voltage) is applied to the base of a transistor with respect to the emitter this electric field removes electrons from the base region. When enough have been removed, electrons from the emitter all rush in to try and fill these holes. The electrons from the emitter are not labelled as such so they now where they are going, they all feel this attraction to fill these holes they are all bouncing all over the place. When they all rush in to fill these holes most of them are attracted by the collector and swept into the collector region (diffusion) and only a few fill these holes in the base region (Base Current). Without this electric field none of this would happen.

Thanks
Adam
I've said that numerous times to Ratch, he doesn't like it. What you just described is DRIFT. Drift is charge motion due to E field being present, exerting force on carriers. Your error is as follows. We never "apply a voltage to the base-emitter" at all. Such is doom for the bjt. An external source, current source, or voltage source plus resistor is used. Positive current enters the base, and leaves the emitter. Or if you wish, electrons exit the base lead and enter emitter lead.

The electrons rushing from the emitter towards base are drifting with the field. They are not "trying to fill base holes", it is something that occasionally happens along the way. They are attracted to collector due to b-c E field and become collector current. The emitter CURRENT became collector current. The Vbe is a result of carriers recombining in both emitter and base regions forming a depletion zone and potential barrier.

This barrier is NOT controlling the emitter current nor collector current. My sims demonstrate positively that with a bjt in an amp stage, or in the raw, the emitter current controls the collector current, and Vbe trails Ie & Ic. After Ie/Ic settle, Vbe catches up. No control here, this is a figment of the imagination.

But look at my FET sims. It is obvious that Id is clearly shadowing NOT Id, the drain current, but rather Vgs, the gate-source voltage. Please note the following.

All science is based upon testing, measuring, and observing, over widely varying conditions. Every lab test, scope observation, simulation I've ever done affirms these sim results. The critics keep saying that what is going on is such and such, yet they cannot demonstrate how to see it. How do they know? Why would sims suggest otherwise?

Claude
 

Ratch

Mar 10, 2013
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"Charge transport is due to drift, where an electric field imparts force on charges, and diffusion, where density gradients determine charge motion."
No, drift is an insignificant contribution to current in the forward direction. It is limited by its thermal equilibrium value. See first file attachment, p.49.

"Never said otherwise."
You asked how diffusion could be the cause of charge motion.

"Sorry, that does not compute. A forward biased diode has its charge carriers subjected to an E field from an external source. That is drift, plain and simple. Where do you see any reputable text saying otherwise?"
The drift current is insignificant as shown in the attached file on page 49.

Drift is the reason the carriers cross the junction in the first place.
No, drift and diffusion work independently of each other.

They diffuse as well per the density gradient. The barrier voltage does not prevent drift.

Whoever said it did?

If the external source providing the E field has sufficient potential the drift continues right through the junction barrier.

No, the applied voltage Va lowers the barrier voltage allowing more diffusion to take place. Drift has nothing to do with it.

Take a 12 volt supply and a 10 kohm resistor in series with a p-n diode. When first energized, current exits the battery, through the resistor, into the diode. The original barrier voltage is just 25.7 mV. But carriers cross junction, and increase depletion zone charge and barrier potential. A 0.70 volt barrier is reached and the current is simply (12.0 - 0.70)/10kohm = 1.13 mA.

Where do you get those figures from?

They both are active. W/o drift, how can charge carriers move in a uniform direction?

By diffusion.

But external fields influence what happens on interior. Otherwise how do charges move? They need external stimulus. Really Ratch, where do you get this stuff?
I never said that external voltages do not control current. I disagree with you on how they do it.

I never said diffusion does not matter, quite the contrary. It is one important mechanism in understanding semicon behavior, but it is not the only thing going on. Drift and diffusion both must be included in any serious discussion on this subject.

Yes, but diffusion is the primary source of current in the forward direction. Drift is not. Drift current is limited by its thermal equilibrium. Diffusion can increase until the diode burns out.

Again, I have always stated that diffusion is important, but so is drift. You need both to fully understand the topic.

I do understand both, especially their relationship with respect to forward and reverse bias.

But remember that the Si atoms greatly outnumber the donor/acceptor atoms.
What difference does that make?

Take the n type material. The donor atoms, pentavalent, like arsenic, As, shed the 5th electron with little energy needed, and the key is that NO HOLE is left behind. The As atom bonds with 3 Si atoms in a tetrahedron and 4 electrons are covalent. Hence all 4 bonds are complete and a free electron does not see a hole here.

That is wrong. In the P-type material, the As needs another electron to form a complete bond with its 4th silicon atom. That is why it readily accepts a electron from the N-type material.

But remember that thermal vibrations due indeed knock electrons out of Si valence bands as well. It takes more energy to liberate a host atom's electron (Si), vs. a donor atom electron (As). If the 2 atom types had equal density, then the majority of electrons available for conduction would originate from donor atoms. But please remember that the host Si atoms greatly outnumber donor (As) atoms.

What difference does that make? The charges made available by thermal activity are insignificant when compared to diffusion in the forward direction.

Electrons liberated from the Si host atoms are also part of the conduction process.
Yes, but they don't count for much in the drift process.


The doping of the Si lattice with donor atoms provides an excess of electrons not the case in intrinsic Si.

The donor atoms provide the excess of electrons, not the silicon.

The electrons liberated from host Si leaves behind a hole,

The electrons leaving the donor atom leave behind a stationary ion which can be termed a hole.

which a free electron may combine with. Not so with As donor atoms. The donors create an excess of electrons. But current conduction consists of electrons from donors, and host, and external source. All participate in conduction.

Acceptor atoms also take in electrons and create a current.

I said thermal vibrations generate electron-hole pairs in "Si", maybe "Si lattice" is a better choice of words. The thermal energy associated w/ lattice vibrations liberates electrons from both donor and host atoms, but much less energy is needed to liberate electrons from donors. But host atoms are the overwhelming majority.

Doesn't matter. Silicon atoms are so tightly bound that their contribution to drift current is insignificant. Diffusion rules in forward bias.

No need to be hostile.
No need to be close minded.

With no junction voltage, how can drift happen? No external E field, so how do charges drift?
With no applied voltage, the drift current equals the diffusion current for a net current of zero.

Diffusion is one of two TYPES OF charge motion. Something has to give rise to charge motion, i.e. an E field. Diffusion happens at room temp with a device sitting unconnected on the table. A little diffusion, and no drift.

They both happen at room temperature, but their equality cancels each other.

Both can exist independently of each other. But w/o an external E field, how can you have drift?
Study the internal fields of a diode.

Just how is it "known to be otherwise"? In all my course work, BE, MS, and Ph.D., drift in forward diodes is very significant. I'll post quotes later.

In the forward direction, drift is limited by the thermal equilibrium. Diffusion is not. See the attachments with this posting.

So do I, and many others. But taking courses is the best way. On our own we can pick up misconceptions and bad theories that would be quickly corrected in a class and with homework. Self study can get you started, but it leaves large gaps in your data base.

It would be interesting to see your course material.

I will, promise.
Please do.

Ratch


Sorry for the rotation of the text. Be sure to rotate the page to see the complete text. Look at the last paragraph of "Va greater than zero". It directly states that drift current is limited and diffusion current predominates.
 

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Arouse1973

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I've said that numerous times to Ratch, he doesn't like it. What you just described is DRIFT. Drift is charge motion due to E field being present, exerting force on carriers. Your error is as follows. We never "apply a voltage to the base-emitter" at all. Such is doom for the bjt. An external source, current source, or voltage source plus resistor is used. Positive current enters the base, and leaves the emitter. Or if you wish, electrons exit the base lead and enter emitter lead.

The electrons rushing from the emitter towards base are drifting with the field. They are not "trying to fill base holes", it is something that occasionally happens along the way. They are attracted to collector due to b-c E field and become collector current. The emitter CURRENT became collector current. The Vbe is a result of carriers recombining in both emitter and base regions forming a depletion zone and potential barrier.

This barrier is NOT controlling the emitter current nor collector current. My sims demonstrate positively that with a bjt in an amp stage, or in the raw, the emitter current controls the collector current, and Vbe trails Ie & Ic. After Ie/Ic settle, Vbe catches up. No control here, this is a figment of the imagination.

But look at my FET sims. It is obvious that Id is clearly shadowing NOT Id, the drain current, but rather Vgs, the gate-source voltage. Please note the following.

All science is based upon testing, measuring, and observing, over widely varying conditions. Every lab test, scope observation, simulation I've ever done affirms these sim results. The critics keep saying that what is going on is such and such, yet they cannot demonstrate how to see it. How do they know? Why would sims suggest otherwise?

Claude

All the electrons in the emitter feel an extra attractive force on them from the base region now having electrons removed from it. These have to be replaced to obey the conservation of charge. But some of the electrons from the emitter will defiantly fill these holes, they have no choice. But which ones?? they don't know either so that's why I said they all rush in... The electrons that don't fill these holes are swept across to the collector terminal.

Adam
 

LvW

Apr 12, 2014
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Gentlemen - I was absent for 6 weeks (travelling through Bolivia/Chile) and I am a bit surprised that this funny discussion does not yet came to an end.
I am sure, we cannot convince Claude - but that´s not OUR problem.
I think, one of his fundamental misunderstandings is summarized in his sentence:

Just as Id is a f(Vd), it is also true that Vd is a f(Id). Which is the "cause vs. effect" is an endless chicken-egg vicious circle.

No - it is not a "chicken-egg question". In all the simple BJT based circuits we are discussing here it is the VOLTAGE that is the primary source only.
No currents without a driving voltage! It is simply impossible (!!) that a small current should be able to DIRECTLY control the value of a larger current.
Even if we speak of an "injected base current" - in fact, we do nothing than to use a voltage source with a large series resistor Rs , thus creating a voltage division between Rs and the B-E path.
More than that, all his simulations never can answer physical "cause-and-effect" questions. I think, this was discussed/explained sufficiently in the past.

We all know that (a) the DC collector current Ic rises as a a result of a temperature increase and (b) that a reduction of the B-E voltage of app. 2mV/K will bring Ic back to its initial value.
This value is verified by measurements as well as theoretical calculations based on carrier physics.
Question to Claude: If the base current would be the controlling quantity, shouldn`t a corresponding analysis exist showing by which amount the base current Ib has to be reduced (in mA/K)?
Do you know such a key parameter?

LvW
 

Arouse1973

Adam
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Welcome back LvW, I thought you had left us. Hope you enjoyed your trip.
Adam
 

LvW

Apr 12, 2014
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Welcome back LvW, I thought you had left us. Hope you enjoyed your trip.
Adam
Adam - thank you. Yes, the trip (through Bolivia and Chile) was - of course - interesting (and, sometimes, exhausting because of a very high trafiic noise level). My son lives in Bolivia since 3 years (Cochabamba) and he was celebrating his 30th birthday.
 

davenn

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welcome home LvW

if you would like to, please post a few choice photos of your trip in the off-topic members lounge section of the forum .... would be cool to see them :)


Dave
 

cabraham

Feb 12, 2015
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Gentlemen - I was absent for 6 weeks (travelling through Bolivia/Chile) and I am a bit surprised that this funny discussion does not yet came to an end.
I am sure, we cannot convince Claude - but that´s not OUR problem.
I think, one of his fundamental misunderstandings is summarized in his sentence:

Just as Id is a f(Vd), it is also true that Vd is a f(Id). Which is the "cause vs. effect" is an endless chicken-egg vicious circle.

No - it is not a "chicken-egg question". In all the simple BJT based circuits we are discussing here it is the VOLTAGE that is the primary source only.
No currents without a driving voltage! It is simply impossible (!!) that a small current should be able to DIRECTLY control the value of a larger current.
Even if we speak of an "injected base current" - in fact, we do nothing than to use a voltage source with a large series resistor Rs , thus creating a voltage division between Rs and the B-E path.
More than that, all his simulations never can answer physical "cause-and-effect" questions. I think, this was discussed/explained sufficiently in the past.

We all know that (a) the DC collector current Ic rises as a a result of a temperature increase and (b) that a reduction of the B-E voltage of app. 2mV/K will bring Ic back to its initial value.
This value is verified by measurements as well as theoretical calculations based on carrier physics.
Question to Claude: If the base current would be the controlling quantity, shouldn`t a corresponding analysis exist showing by which amount the base current Ib has to be reduced (in mA/K)?
Do you know such a key parameter?

LvW
Is this serious? We covered this. On this forum and others, I made it clear that emitter current is what controls collector current. Please consult a good text so you can learn that emitter current and base current are NOT the same thing. "Ib" is base current, "Ie" is emitter current. Please learn the difference. The b-e junction serves as the "input port" for the bjt, but the 2 terminals DO NOT have the same current. The "current controlled" description of the bjt refers to emitter, not base current. LvW, please believe me, I would not lie to you or anyone, Ib is NOT the same as Ie. Please learn the distinction.

I posted an excerpt from the original 1954 Ebers-Moll paper, and Ie is modeled as the control quantity. Since 1954, Ie, not Ib, has always controlled Ic. I keep re-iterating, yet you keep knocking down the base current straw man.

"No currents without a 'driving voltage'"?! I've stated ad infinitum that Vbe is NOT a "driving voltage". Vbe is a drop, not an emf. It doesn't drive anything, the power source does, i.e. microphone, antenna, CD player output, AM/FM tuner output, etc. Please note that voltage cannot exist w/o a current to displace the charges. A battery is a prime example. The electric field across the terminals is made possible by the redox reaction which propels ions against the electric field. Positive ions are forced towards the positive terminal, and likewise for negative. The current inside the battery moves against the E field.

Otherwise the E field would de-energize and the current would cease as well as the voltage.

Question to Claude: If the base current would be the controlling quantity, shouldn`t a corresponding analysis exist showing by which amount the base current Ib has to be reduced (in mA/K)?
Do you know such a key parameter?

Base current is not the controlling quantity, emitter current is. Until you learn that base and emitter currents are not the same, I can't reach you. But re temp, the Shockley diode relation, also present in Ebers-Moll relation is as follows: Id = Is*exp((Vd/Vt)-1), or likewise Vd = Vt*ln((Id/Is)+1).

The "Is" factor, reverse saturation current, or a scale current if you prefer, is a stron function of temperature. This current is spec'd in "neper-amp per degree Kelvin". As temp increases, Id increases w/ temp non-linearly, or in a power fashion. When I worked on log amps in the 90's/00's, I measured a 1N914B diode "Is" value temp coefficient at 0.12 neper/Kelvin. For every degree K (or C) the Is value increased by a factor of exp (0.12), approx.1.1275. For a 35 degree C rise, Is increased by a factor of 66.7.

The temp coefficient is semiconductors is due to increased conduction via freeing of carriers due to thermal lattice vibrations. We bias a bjt with a fixed value of dc current. Should temp rise, the silicon is more conductive since thermal energy generates more electron-hole pairs, i.e. more carriers available for conduction. But if the bjt is biased via a current source, or a voltage source plus a sufficiently large series resistor, the same current results in a lower voltage drop across the junction. This should not be hard to understand. If the bias current is 1.0 mA, and the temp is 25 C, let's say the Vbe is 0.650 volts. If temp increases to 75 C, more carriers are in the conduction band. The same 1.0 mA of bias current results in less than 0.650 volts for Vbe because less energy per charge is needed for the same current.

But we bias b-e junctions w/ fixed current, not fixed voltage. Thermal runaway takes place if we force a 0.650 volt source right across a junction. Temp increase will result in a current avalanche given in amp-neper/deg C. For a 1N914B diode junction, if Is is 10 pA at room temp of 25 C, at 75 C it is 667 pA. Junction current has a temp coefficient as well, but we seldom use it because we never bias w/ constant voltage right on a junction.

But people like me, who have designed and patented logarithmic amplifiers (United States no. 5, 670,775; 1997), refer to current temp coefficient as well as voltage. I used them both.

Regarding simulations answering cause/effect questions, I beg to differ. Sims can indicate strong relations between specific quantities as well as prove that the lagging quantity cannot cause the leading quantity. In my sims w/ diodes and inductor de-energizing, the sim clearly shows that the inductor current enters the diode first, then forward diode voltage drop develops as a result. This is when the inductor de-energizes. When the inductor is acquiring energy, diode is connected right across a constant voltage source, the input supply. Here, the diode voltage, Vd, determines the reverse current Id. It works both ways.

With the bjt, I showed that Ie clearly changes well before Vbe, and there can be no doubt that Ie is NOT controlled by Vbe. Yet Ic, the collector current responds and tracks Ie like a shadow, plateaus and settles with Ie in unison, while Vbe pokes along and eventually catches up. You claimed that Re "spoiled" the sim, skewing the relation. You asked for a sim with no resistance in base, emitter, nor collector. I did that and the sim revealed that Ie is clearly incontrol of Ic.

I also simmed with a FET, and it was beyond doubt, that Id (drain current), tracked and followed Vgs, the gate-source voltage. Although Id the drain current was way out in front od Vgs, Id t=locked on to Vgs and followed the same like a shadow. The sim was unequivocally clear that in a FET, it is Vgs that controls Id.

If my sims skew the results towards current control in the bjt case, why did they indicate voltage control in the FET case? Here is the answer:

Drum roll please ---------------------

The bjt is current controlled. The FET is voltage controlled. It's that simple.

Claude
 

(*steve*)

¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd
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The bjt is current controlled. The FET is voltage controlled. It's that simple.

That's the simple answer, but at the physical level it's almost exactly the opposite :D
 

Ratch

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Is this serious? We covered this. On this forum and others, I made it clear that emitter current is what controls collector current. Please consult a good text so you can learn that emitter current and base current are NOT the same thing. "Ib" is base current, "Ie" is emitter current. Please learn the difference. The b-e junction serves as the "input port" for the bjt, but the 2 terminals DO NOT have the same current. The "current controlled" description of the bjt refers to emitter, not base current. LvW, please believe me, I would not lie to you or anyone, Ib is NOT the same as Ie. Please learn the distinction.

I believe we all know the differences between the Ic, Ib, and Ic currents. You made the statement about Ie "controlling" Ic previously, but you never addressed my reply. In the active region of a BJT, the Ic is almost the same as Ie. The only difference is that Ic is a little smaller due to the relatively small amount of waste current (Ib) that gets lost from Ie when it crosses the base slab into the collector slab. So that begs the question, what controls Ie? I explained before many times how Vbe is what controls both Ie and Ic by lowering the barrier voltage of the emitter base junction caused by the uncovered charges from the diffusion process.

I posted an excerpt from the original 1954 Ebers-Moll paper, and Ie is modeled as the control quantity. Since 1954, Ie, not Ib, has always controlled Ic. I keep re-iterating, yet you keep knocking down the base current straw man.

So you did. However, you cannot legitimately use empirical models like Ebers-Moll to explain the internal processes of a BJT. Not unless they model the internal process themselves, like diffusion, doping concentration, ionization, mobility, etc, which they do not.

No currents without a 'driving voltage'"?! I've stated ad infinitum that Vbe is NOT a "driving voltage

It certainly is. I can change the Ic by manipulating the Vbe.

Vbe is a drop, not an emf.

Like any voltage, it is both.

It doesn't drive anything, the power source does, i.e. microphone, antenna, CD player output, AM/FM tuner output, etc.

A signal source like a microphone, antenna, etc. are not considered power sources. The internal voltage across the BJT is main power source.

Please note that voltage cannot exist w/o a current to displace the charges.

Yes, it can. A voltage can exist between two points forever without any current present.

A battery is a prime example. The electric field across the terminals is made possible by the redox reaction which propels ions against the electric field. Positive ions are forced towards the positive terminal, and likewise for negative. The current inside the battery moves against the E field.

Otherwise the E field would de-energize and the current would cease as well as the voltage.

A battery can supply current and voltage. The voltage still exists without current being present.

Question to Claude
: If the base current would be the controlling quantity, shouldn`t a corresponding analysis exist showing by which amount the base current Ib has to be reduced (in mA/K)?
Do you know such a key parameter?

Base current is not the controlling quantity, emitter current is. Until you learn that base and emitter currents are not the same, I can't reach you.

Before you can expect anyone to believe that, you have to prove and explain it. We can expect Ic and Ie to be related to each other because they are in series with each other. So a new question is what controls Ie? I explained many times that Vbe controls Ie and Ic by controlling the barrier voltage caused by the uncovered charges when diffusion takes place.

But re temp, the Shockley diode relation, also present in Ebers-Moll relation is as follows: Id = Is*exp((Vd/Vt)-1), or likewise Vd = Vt*ln((Id/Is)+1).

The "Is" factor, reverse saturation current, or a scale current if you prefer, is a stron function of temperature. This current is spec'd in "neper-amp per degree Kelvin". As temp increases, Id increases w/ temp non-linearly, or in a power fashion. When I worked on log amps in the 90's/00's, I measured a 1N914B diode "Is" value temp coefficient at 0.12 neper/Kelvin. For every degree K (or C) the Is value increased by a factor of exp (0.12), approx.1.1275. For a 35 degree C rise, Is increased by a factor of 66.7.

The temp coefficient is semiconductors is due to increased conduction via freeing of carriers due to thermal lattice vibrations. We bias a bjt with a fixed value of dc current. Should temp rise, the silicon is more conductive since thermal energy generates more electron-hole pairs, i.e. more carriers available for conduction. But if the bjt is biased via a current source, or a voltage source plus a sufficiently large series resistor, the same current results in a lower voltage drop across the junction. This should not be hard to understand. If the bias current is 1.0 mA, and the temp is 25 C, let's say the Vbe is 0.650 volts. If temp increases to 75 C, more carriers are in the conduction band. The same 1.0 mA of bias current results in less than 0.650 volts for Vbe because less energy per charge is needed for the same current.

But we bias b-e junctions w/ fixed current, not fixed voltage. Thermal runaway takes place if we force a 0.650 volt source right across a junction. Temp increase will result in a current avalanche given in amp-neper/deg C. For a 1N914B diode junction, if Is is 10 pA at room temp of 25 C, at 75 C it is 667 pA. Junction current has a temp coefficient as well, but we seldom use it because we never bias w/ constant voltage right on a junction.

But people like me, who have designed and patented logarithmic amplifiers (United States no. 5, 670,775; 1997), refer to current temp coefficient as well as voltage. I used them both.

What does the above discourse on temperature effects have to do with current or voltage control of a BJT?

Regarding simulations answering cause/effect questions, I beg to differ. Sims can indicate strong relations between specific quantities as well as prove that the lagging quantity cannot cause the leading quantity. In my sims w/ diodes and inductor de-energizing, the sim clearly shows that the inductor current enters the diode first, then forward diode voltage drop develops as a result. This is when the inductor de-energizes. When the inductor is acquiring energy, diode is connected right across a constant voltage source, the input supply. Here, the diode voltage, Vd, determines the reverse current Id. It works both ways.

Simulations using empirical models cannot discern the internal working principles of a device. An Ebers-Moll model does not take diffusion, doping concentration, ionization, mobility, etc. into considereation.

With the bjt, I showed that Ie clearly changes well before Vbe, and there can be no doubt that Ie is NOT controlled by Vbe. Yet Ic, the collector current responds and tracks Ie like a shadow, plateaus and settles with Ie in unison, while Vbe pokes along and eventually catches up. You claimed that Re "spoiled" the sim, skewing the relation. You asked for a sim with no resistance in base, emitter, nor collector. I did that and the sim revealed that Ie is clearly incontrol of Ic.

Are you saying that you applied a slow non-repetitive signal to the base-emitter and observed the Ic and Ic responding before the Vbe voltage?

I also simmed with a FET, and it was beyond doubt, that Id (drain current), tracked and followed Vgs, the gate-source voltage. Although Id the drain current was way out in front od Vgs, Id t=locked on to Vgs and followed the same like a shadow. The sim was unequivocally clear that in a FET, it is Vgs that controls Id.

Irrelevant to the question of what controls a BJT. A FET works on a different principle that does not include diffusion.

If my sims skew the results towards current control in the bjt case, why did they indicate voltage control in the FET case?

Invalid sims with respect to why a device works. No explanation given explaining the process.

Here is the answer:

Drum roll please ---------------------

Dead silence.

The bjt is current controlled. The FET is voltage controlled.

They are both voltage controlled. Explanation already given for the BJT.

It's that simple.

Simply wrong, by reason of using invalid techniques.

Ratch

P.S. The last time you posted, you averred that the drift current in a junction diode was an important constituent of the diode current. Do you still believe that now after reading the attachment I included in my answer?
 

LvW

Apr 12, 2014
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I made it clear that emitter current is what controls collector current. Please consult a good text so you can learn that emitter current and base current are NOT the same thing. "Ib" is base current, "Ie" is emitter current. Please learn the difference. The b-e junction serves as the "input port" for the bjt, but the 2 terminals DO NOT have the same current. The "current controlled" description of the bjt refers to emitter, not base current. LvW, please believe me, I would not lie to you or anyone, Ib is NOT the same as Ie. Please learn the distinction.

"No currents without a 'driving voltage'"?! I've stated ad infinitum that Vbe is NOT a "driving voltage". Vbe is a drop, not an emf. It doesn't drive anything,

Question to Claude: If the base current would be the controlling quantity, shouldn`t a corresponding analysis exist showing by which amount the base current Ib has to be reduced (in mA/K)?
Do you know such a key parameter?

Base current is not the controlling quantity, emitter current is. Until you learn that base and emitter currents are not the same, I can't reach you.
Claude

Only some short remarks from my side:
* Claude - I am really disappointed and surprised about the way you try to "discuss". Do you really think it is necessary to teach me that "emitter current and base current are NOT the same thing"? Where did I mention such silly stuff? Please, try to be serious. Otherwise, one could get the impression that you are trying to avoid answering the main questions.

* Do you realize some contradictions in your text?
(a) You say that "The b-e junction serves as the input port for the bjt", and
(b) "Base current is not the controlling quantity, emitter current is".

Therefore, I ask you to answer the following questions:
1.) What do YOU think is the role of the "input port" (because - in your view - neither the base current nor the B-E voltage is the controlling quantity) ?
2.) Everybody knows (and it was never questioned) that Ic depends on Ie - but which externally applied quantities control the emitter current Ie ?
(The common understanding of the term "control" is the following: To control the output of any device refers to an externally applied quantity that is able to change the output).
 

Arouse1973

Adam
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If Ie controls Ic then how does this circuit work?
Thanks
Adam

REV TRAN.PNG
 

cabraham

Feb 12, 2015
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If Ie controls Ic then how does this circuit work?
Thanks
Adam

View attachment 20032
The bjt is operating in the "inverse mode". Collector is acting as emitter and vice versa. So Ic controls Ie, but the 2 roles are interchanged. What we call "Ic" is effectively the emitter current "Ieinv" operating in inverse mode. Hence Ieinv controls Icinv, where Ieinv = Ic, and Icinv = Ie.

Claude
 

cabraham

Feb 12, 2015
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Only some short remarks from my side:
* Claude - I am really disappointed and surprised about the way you try to "discuss". Do you really think it is necessary to teach me that "emitter current and base current are NOT the same thing"? Where did I mention such silly stuff? Please, try to be serious. Otherwise, one could get the impression that you are trying to avoid answering the main questions.

* Do you realize some contradictions in your text?
(a) You say that "The b-e junction serves as the input port for the bjt", and
(b) "Base current is not the controlling quantity, emitter current is".

Therefore, I ask you to answer the following questions:
1.) What do YOU think is the role of the "input port" (because - in your view - neither the base current nor the B-E voltage is the controlling quantity) ?
2.) Everybody knows (and it was never questioned) that Ic depends on Ie - but which externally applied quantities control the emitter current Ie ?
(The common understanding of the term "control" is the following: To control the output of any device refers to an externally applied quantity that is able to change the output).

I stated and restated that Ie, not Ib, is what controls Ic. You keep replying with posts knocking down base current as control variable. You cannot mentally conceptualize that the "current control" model is Ie, not Ib, that controls Ic. You always attack Ib, in an attempt to refute current control, when you should focus on Ie.

1) The role of the input port is to input a signal from a source with the intent of varying the emitter current, so as to implement a corresponding variation in collector current. Also, the base region performs the necessary function of separating collector from emitter, otherwise punch through happens, and amplification does not happen. We can input Ib, base current, as the control quantity. But for linear amplification, the result is a condition called "beta dependency". So we set up the bias network for constant emitter current, and the signal generator, whether it outputs a constant current or voltage. This process could work w/o Vbe. If the p-n junction at the base and emitter pins was ideally rectifying, it could support any forward current with zero voltage drop in forward direction, and any reverse voltage with zero reverse current in reverse direction.

Charges transported through base and emitter regions are holes from b to e, electrons from e to b. The emitted electrons enter base then collector, becoming Ic. Vbe changes as a result of Ie, but does not control Ie. Ie must change first. Those against me insist that Ie is changing because of a change already taken place via Vbe. But not only do my many sims refute that, but just look at charge transport through the base & emitter regions. To change Vbe, you must transport charge through base & emitter regions, in order to build up or diminish depletion zone charge profile and E field. Vbe changes after charge density per unit time, i.e. current, has changed. My sims match what I've seen in scope plots in labs.

2) What external quantity controls Ie? You seem to think of Vbe as external, I see it as internal. Vbe has a terminal value right where the pins meet the outside world, and internal, the depletion zone barrier potential. My sims have shown that with or without resistors around the device, Ic always tracks Ie. One can apply Ie from an external source. A voltage source inputted directly to the base, with emitter connected to resistor to ground is a case where the voltage source value gets divided between Re and Vbe. Ie changes, and Vbe catches up eventually, but Ic changes in response to Ie change, before Vbe has had time to settle to new value. Clearly it is Ie, not Vbe.

Because Ie and Vbe are closely inter-related, one cannot change w/o the other changing as well. So if a bjt amp stage is measured at one set of Ie/Vbe values, then we change one of them, the other will change and Ic will change as well. The contentious question is as follows: was it the Vbe change, or the Ie change, that was responsible for Ic changing?

The only way to solve that riddle is to observe Ic, Vbe, and Ie on a good storage scope, and it becomes apparent. If Ie increases and settles to its new plateau, , but Vbe has not had time to settle yet, Ic will increase and settle to its plateau due solely to Ie increasing. When the number of emitted electrons per unit time increases, the number collected per unit time must also increase, regardless of whether or not Vbe has "caught up". Eventually Vbe will setlle to its new value consistent w/ Ebers-Moll, but the change in Ie is what determines the change in Ic.

Claude
 

LvW

Apr 12, 2014
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1) The role of the input port is to input a signal from a source with the intent of varying the emitter current, so as to implement a corresponding variation in collector current.
This answer would be helpful (without the rest) if you would be a little more precise: What means "signal"? Voltage or current?
In any case, from this I derive that now you agree that the emitter current is varied (controlled) either by a voltage signal (VBE) or by a current signal (IB). Is this interpretation correct?

2) What external quantity controls Ie? You seem to think of Vbe as external, I see it as internal. ............
...........................
Eventually Vbe will setlle to its new value consistent w/ Ebers-Moll, but the change in Ie is what determines the change in Ic.
Claude

I think, everybody who uses the term VBE means the voltage that exists between the B and E terminal. And I am sure you are aware that I am following this terminology (and others too).
Hence, it makes no sense to speak about time to "settle" VBE. In contrary, for classical gain stages it is our goal to produce a DC operating point witha "stiff" voltage VBE that "opens" the transistor (this is our goal, I know that there are some other constraints).
It makes no sense to continuously repeat that a change in IC follows a change in IE. This was never put into question. The main question is how we can vary the current IE.
And now your answer is: A "signal" between B and E (see 1, above)
Look at basic BJT gain stages - and you`ll have the answer. In all three basic configuration we fix the DC currents using a proper voltage VBE.
(It is really not necessary to argue again - as you did before - that, in practice, nobody will connect a voltage source directly between B and E. But this has nothing to do with the main question; it just has practical reasons).
 

Martaine2005

May 12, 2015
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When I first started out in electronics, I used the words Not Pointing iN for the arrow head on a NPN and the other one was a PNP.
I like that!!
But what was your PNP? Just curious sir!!
I have still got 6 pages to read yet.

Martin
 
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