How a BJT Transistor works (base current version)

[cabraham]

I studied up a little about batteries from some chemistry textbooks. It appears that only the cations and anions travel through the electrolyte, but no electrons. Electron transfer is caused by reduction at the cathode and oxidation at the anode, but the electrons only travel along the conduction path of the external circuit. The textbook explanation is confirmed by this link. http://www.qrg.northwestern.edu/projects/vss/docs/power/2-how-do-batteries-work.html It would appear that if the electrons could or did move through the electrolyte, the battery would short out internally.

Ratch

I have a sometimes sloppy habit of referring to current in terms of "holes & electrons. I use these terms in a loose sense, i.e. colloquially. In a battery the negatively charged ions, i.e. anions, move towards the negative terminal inside the battery. Likewise the positively charged cations move towards the positive terminal. This is true only inside the battery. Anyway, like I said, holes and electrons are terms I use loosely, but in electrolyte the carriers are not literally holes/electrons but +/- ions. I was speaking colloquially, not literally. But nonetheless I agree about the anions/cations.

Claude

 

[LvW]

Claude - as I have mentioned already: We should agree not to agree.
I do not intend to go into details of your long contribution (post#100).
Either you are explaining physical details not under discussion or you are misinterpreting some effects - exclusively under your point of view: Ie controls Ic.

Several times you have claimed that Ie would control Ic - but how is Ie controlled? Can you answer this?
Do you even deny that the B-E junction is the controlling part of a BJT?
Here are some excerpts from your posts (all in italics) - just claims without justification.:

* What controls current is the external network, which is usually designed to fix the emitter current at a set dc operating point,

* A bjt must be current driven. An externally applied current into the emitter is best. An externally applied current to the base results in beta dependency.

* Instead, let us apply an external **current source** to the emitter terminal and force Ie to a fixed value, letting Vbe settle to whatever value Ies, and temp dictate. Ie is the value of the CCS, so Vbe=Vt*exp((Ie/Ies)+1)

I have never seen this relationship. Vbe rises exponentially with Ie?

* I've stated repeatedly that the 1954 Ebers-Moll paper stated Ie as the controlling current for the Ic current source, they include a picture I will attach later.

Not true. The Ebers-Moll equations show that Ic resp. Ie depend on Vbe with an exponential relationship.
When will you attach your Ebers-Moll interpretation?
Several times you were asked to provide a diagram which can demonstrate that „an externally applied current into the emitter is best“.
Such a visual demonstration could help to understand your position.

 

[cabraham]

Claude - as I have mentioned already: We should agree not to agree.
I do not intend to go into details of your long contribution (post#100).
Either you are explaining physical details not under discussion or you are misinterpreting some effects - exclusively under your point of view: Ie controls Ic.

Several times you have claimed that Ie would control Ic - but how is Ie controlled? Can you answer this?
Do you even deny that the B-E junction is the controlling part of a BJT?
Here are some excerpts from your posts (all in italics) - just claims without justification.:

* What controls current is the external network, which is usually designed to fix the emitter current at a set dc operating point,

* A bjt must be current driven. An externally applied current into the emitter is best. An externally applied current to the base results in beta dependency.

* Instead, let us apply an external **current source** to the emitter terminal and force Ie to a fixed value, letting Vbe settle to whatever value Ies, and temp dictate. Ie is the value of the CCS, so Vbe=Vt*exp((Ie/Ies)+1)

I have never seen this relationship. Vbe rises exponentially with Ie?

* I've stated repeatedly that the 1954 Ebers-Moll paper stated Ie as the controlling current for the Ic current source, they include a picture I will attach later.

Not true. The Ebers-Moll equations show that Ic resp. Ie depend on Vbe with an exponential relationship.
When will you attach your Ebers-Moll interpretation?
Several times you were asked to provide a diagram which can demonstrate that „an externally applied current into the emitter is best“.
Such a visual demonstration could help to understand your position.

Figure 5 from the original Ebers-Moll 1954 paper is attached. The collector current is modeled as a current source controlled by alpha*Ie, which is what I have been stating since day one. Since 1954, it has been standard practice to model a bjt Ic as controlled by alpha*Ie. In a small signal mode of operation, the b-e junction is modeled as r_pi, and the relations ic=beta*ib and ic-gm*vbe are equivalent, since r_pi=beta/gm. For large signal operation "gm" is not constant and hybrid pi model is not used. Anyway, whenever I mention Ebers and Moll modeling the bjt as current controlled, people question it, but that model is in their paper. I can elaborate later.

Claude

 

[LvW]

Claude - thank you for providing the excerpt from the Ebers-Moll paper.
Without reading (and understanding) the whole text - my first impression is the following:
1.) This part discusses common-base configuration only.
2.) The current Ie contains an exponential term with the emitter potential (phi,e) within the exponenet, correct?
3.) Because of the common base configuration this potential (phi,e) is identical to the E-B voltage, correct?
4) Result: There is an exponential relationship between the voltage Vbe and the emitter current, correct?
________________
Question: Is my interpretation correct?

 

[cabraham]

Claude - thank you for providing the excerpt from the Ebers-Moll paper.
Without reading (and understanding) the whole text - my first impression is the following:
1.) This part discusses common-base configuration only.

The relation Ic = alpha*Ie is true for all 3 configurations. Also, the following 2 relations Ic = alpha*Ies*exp((Vbe/Vt)-1), and Ic = beta*Ib, are always true for all 3 configurations.

2.) The current Ie contains an exponential term with the emitter potential (phi,e) within the exponenet, correct? Yes.
3.) Because of the common base configuration this potential (phi,e) is identical to the E-B voltage, correct? Yes.
4) Result: There is an exponential relationship between the voltage Vbe and the emitter current, correct?
That has never been contested, Ie is exponentially related to Vbe, and Vbe is logarithmically related to Ie. This is a functional relationship, but valid only under steady state conditions, i.e. after transients have settled. Just as one can write Ie = Ies*exp((Vbe/Vt)-1), it is equally valid to state the following: Vbe = Vt*ln((Ie/Ies)+1). DOes the value of Vbe determine the value of Ie, or the other way around? In steady state that question cannot be answered. We set a q point, 1.0 mA for Ie, 0.647 V for Vbe, then we perturb using an external source. After settling the new values are Ie = 1.1 mA, and Vbe = 0.648 V. Which one changed first, and which followed? Without examining the behavior during the perturbation, there is no answer, unless we have a priori knowledge of device behavior.

The plots I attached show a perturbation, and it is clear that Ie settles first, then Vbe follows. The critics of current control insist that although Ie controls Ic, Vbe ultimately is what controls Ie, but that is pure assertion, no evidence to support it. Notice in my plots that Ic ascends exactly when Ie begins its ascent, and Vbe of course increases with Ie/Ic. But when Ie settles to its plateau, so does Ic, byt Vbe takes extra time to settle. In my power supply plots, diode current Id has already begun to descend while Vd diode voltage is still ascending towards its final plateau. This clearly demonstrates that there is more going on here than meets the eye.

To say that Vbe controls Ie requires that you show how. Equations do not help because just as y = mx +b (algebra linear equation), it is equally true that x = (y-b)/m. Which variable is independent vs. dependent cannot be ascertained by merely inspecting an equation, not a single valued function anyway. Please tell us hoe a change in Ie is done by a change in Vbe first. Without transporting charge through the b-e junction (Ie/Ib), how does Vbe change first? How does that change in Vbe then motivate Ie to change? My plots say otherwise. The reason I am invoking transient response is that static observations tell us nothing re which variable is independent. I will add more later, a plot that takes this further. I thank all persons, whether they agree or not with me, for having the interest, and for conducting themselves well. BR.

Claude
________________
Question: Is my interpretation correct?

 

[LvW]

Claude - here is my conclusion: You are the only person who does not use a Vbe voltage of app. 0.65 volts for "opening" the BJT to have an emitter current of some milliamps.
I wonder how you are designing BJT gain stages. I suggest to stop the exchange of opinions (assertions) now.
LvW

 

[Ratch]

Claude,

You are still ignoring the physics of the BJT. The Vbe decreases the back-voltage caused by the uncovered charges due to diffusion, and allows more Ic charge flow driven by diffusion to take place until a new equilibrium is established. That proves Vbe is the control of Ic in a BJT in the active region, and no number of equations, plots and models can prove otherwise.

Ratch

 

[LvW]

Hi Ratch,
it was just two days ago that something came into my mind:
There is a handwritten paper from Claude (a pdf copy must be somewhere on my hard disc) which contains a comprehensive calculation of a classical BJT amplifier with RE-stabilization.
Given are all resistor values, the ratio B=Ic/Ib and the supply voltage - nothing else. The task was to calculate the bias point.
Claude has done a very good job and he has found the correct solution.
Why is this worth mentioning? Because he has started the calculation assuming a constant voltage VBE=0.7 volts.
No surprise, because that is the only way to solve the task.
However, I ask myself: Why on earth is he continously claiming that VBE would be the RESULT of an injected current Ie ?

 

[Ratch]

Hi Ratch,
it was just two days ago that something came into my mind:
There is a handwritten paper from Claude (a pdf copy must be somewhere on my hard disc) which contains a comprehensive calculation of a classical BJT amplifier with RE-stabilization.
Given are all resistor values, the ratio B=Ic/Ib and the supply voltage - nothing else. The task was to calculate the bias point.
Claude has done a very good job and he has found the correct solution.
Why is this worth mentioning? Because he has started the calculation assuming a constant voltage VBE=0.7 volts.
No surprise, because that is the only way to solve the task.
However, I ask myself: Why on earth is he continously claiming that VBE would be the RESULT of an injected current Ie ?

The collector current Ic will be present at any small Vbe voltage above zero, but it becomes significant at around 0.6 to 0.8 volts due to the exponential relationship of Vbe to Ic. Many folks use 0.7 volts as a approximation of what the Vbe is in the active region of the transistor.

Claude continually disregards the physics of the BJT and uses circuits with external components instead of the device itself to "prove" that Ic is not dependent on Vbe. I think by "injected", he means the diffusion of charge carriers from the emitter into the base of the transistor where they are whisked away to the collector circuit to form Ic. He can't seem to realize that the diffusion is controlled by Vbe and therefore Vbe is controlling Ic, and Vbe is not the result of Ic.

Ratch

 

[cabraham]

Hi Ratch,
it was just two days ago that something came into my mind:
There is a handwritten paper from Claude (a pdf copy must be somewhere on my hard disc) which contains a comprehensive calculation of a classical BJT amplifier with RE-stabilization.
Given are all resistor values, the ratio B=Ic/Ib and the supply voltage - nothing else. The task was to calculate the bias point.
Claude has done a very good job and he has found the correct solution.
Why is this worth mentioning? Because he has started the calculation assuming a constant voltage VBE=0.7 volts.
No surprise, because that is the only way to solve the task.
However, I ask myself: Why on earth is he continously claiming that VBE would be the RESULT of an injected current Ie ?

Because Vbe is the result of Ie! That is why I claim it, because it is so. Sir Oliver Heaviside's transmission line theory lays many perplexing questions to rest if properly applied. A switch is turned on and a power supply is up and running, let's call it Vcc, which supplies the bias for the bjt stage. A b-e junction is in series with a resistor. A Vcc supply powers the resistor and p-n junction. What happens?

Actually, let's reduce this to extreme simplicity, a battery, resistor, and diode. The switch is thrown, and the battery voltage is 12 volts, resistor is 1.0 kohm, and the diode is a garden variety 1N914, or any other you wish to use. Diode is oriented in forward direction. I don't need a pic. you can draw one, or just mentally visualize this case. So the switch is thrown, and the resistor-diode series combination now has 12 volts across it. What determines the current? The battery has very low internal resistance, consider it an ideal constant voltage source. The truth is that the diode has near zero volts across it, around 26 mV, due to thermal energy via lattice vibrations.

Consider the connecting wires a t-line, so the 12 volts travels through the t-line towards the R-D series network. The current is initially Vbatt/Zo, where Zo is the t-line characteristic impedance, which we can call 120 ohm for illustration. So I, the current, is 12.0V/120 ohm = 0.100A, or 100 mA. At the initial transit of charge from the battery, there is no feedback indicating that a 1.0 kohm R and a diode D lie ahead. This R-D network is not seen by the battery. The 12 volts and 100 mA travel in unison along the t-line and reach the R-D network at a speed just below that of light, maybe 50 to 80% of c.

But what happens when the charges reach the R-D network? We have an impedance transition from 120 ohm to 1,000 ohm plus the resistance of the diode. A reflection occurs, where the voltage reflects with a positive coefficient, current has negative. Some charge enters the R-D. When charges transit through the R, collisions between charge carriers (electrons, holes) and lattice ions occur. Ionization occurs, and electrons are dropped from conduction to the valence energy band. In accordance with conservation of energy law, CEL, photons are radiated per Planck's law E = hf. A resistor conducting current radiates heat, as we know.

So what happens when charges enter the crystal lattice structure of the diode? They have energy already, imparted by the battery's redox (reduction/oxidation) chemical reaction. The positive charges enter the p side (forward biased), and electrons enter n side. The depletion zone is as it was before the switch initiated the charge motion, Vt = 26 mV. The charges move straight through their respective n and p silicon regions. Why would they not? In p material, holes easily conduct, as they have high mobility being majority carriers. Likewise for electrons in the n region. So please note that charges conduct through the diode with ease in the forward direction without anything being done to the depletion zone. A hole entering a p region encounters little resistance since a hole has a statistically smaller chance of recombination vs. an electron, and vice-versa in n region.

Once the holes have transited through the p region, and electrons through n region, they cross the depletion zone and ionize with atoms in the local region, enhancing the depletion zone. The increased hole density from the battery has crossed into the n region and ionized local n region atoms, adding to the small ionization already present due to thermal energy. Likewise, in the p region, this new addition of electrons has increased the ionization. In p region, electrons have low mobility being minority carriers, and don't get far before recombination.

A stronger E field occurs in this new enhanced depletion zone, and the line integral of this larger E field is a larger barrier voltage. The diode forward voltage drop increased due to its increased forward current. The change in Id gave rise to a change in Vd. So now what happens?

New charges crossing the n and p regions will encounter this new increased barrier potential, this E field will oppose the E field from the battery. Charges in the vicinity feel the battery E field propelling them towards the diode junction, along with the barrier potential oriented oppositely, pushing against their motion. Thus the current begins to decrease due to the Vd build up. After several dozen or so t-line reflections, the equilibrium condition will be attained. The current will be determined by the net voltage on the R, which is 12.0-0.70 volts = 11.30 volts. So I = 11.30V/1.0 kohm = 11.3 mA.

The relation between V & I in the resistor R is too easy as R = V/I, I =V/R, and V = IR. But the diode behaves per Shockley, i.e. Vd = Vt*ln((Id/Is)+1), or Id = Is*exp((Vd/Vt)-1). One could attempt to solve the 2 equations, but the result is a transcendental equation, unsolvable. Math tables have been computed for similar type functions, one being a Lambert function. If temperature is very constant, Lambert functions can solve this equation. But for any significant power, the diode parameter "Is", the scaling or saturation current, has a strong temperature coefficient. Small changes in temp produce large changes in Ies.

But V is logarithmic wrt I in the diode, meaning that Vd does not change much as I varies slightly. At room temp, we know a priori that when the dust settles, Vd is not far from 0.60 to 0.70 volts. If we start at Vd = 0.70V, then Id = 11.3 mA. If Vd = 0.60V, then Vd = 11.4 mA, a change of just 1%. Otherwise we would have to iteratively compute via Ohm's law and Shockley to get the result.

Spice type simulators do just that. If I assume 0.70V, get 11.3 mA, then plug the network into Spice, it may compute, Vd = 0.677 V, and Id = 11.323 mA. My error would be -0.2%, not bad.

Although Vbe in the bjt amp stage commences at 26 mV, then climbs up to near 0.70 volts after current has been passed through the junction, I know from past observation, that at the current levels in question, the Vbe will lie in the range of 0.60 to 0.70 volts. So I was leveraging this repeatable property of b-e junctions to estimate the forward drop at 0.70 volts.

Once all transients settle, Vbe will be pretty close to that value. I also allow a tolerance, computing the network with 0.60 volts Vbe, and obtaining results based on that value. I vary Vbe from 0.50 to 0.80 volts, vary beta from worst case min at min temp, up to worst case max at max temp, then compute. I then observe the spread in gain values. A well designed amp stage is robust enough to maintain consistent performance despite Vbe and beta variations.

Do I need to elaborate? BR.

Claude

 

[cabraham]

Claude,

You are still ignoring the physics of the BJT. The Vbe decreases the back-voltage caused by the uncovered charges due to diffusion, and allows more Ic charge flow driven by diffusion to take place until a new equilibrium is established. That proves Vbe is the control of Ic in a BJT in the active region, and no number of equations, plots and models can prove otherwise.

Ratch

Uncovered charges? Please define that term. Back-voltage, I presume you mean barrier potential, which I covered thoroughly. An increase in Ie/Ic results in increase in Vbe, which results in a larger barrier potential or what you term "back-voltage". Please examine the charge transit in the silicon. New carriers entering the base and emitter move easily through said regions as they are majority carriers. But when they approach the depletion zone's barrier potential they feel a repulsive force. You are assuming that in order to get the charges past this barrier, the Vbe at the terminals must increase. But the increase in Ib/Ie is due to an increase in I/V at an external source. How do we increase Vbe? If the signal source driving the bjt incurs an increase in I/V how does this change trigger the sequence of events.

So let's use a simple dipole antenna as the source, its output feeds a bjt amp stage. The fm radio signal increases, the antenna is 300 ohm, so an increase in I & V occur due to increased electromagnetic energy in space. This increased I and V transit through the wires and into the b-e junction. The base and emitter current have already incurred the change, increased above its original value. But the barrier potential has not yet changed. We have a new increased value of Ie, Ib, and terminal Vbe, but the barrier Vbe is still at ther old value. It is the barrier Vbe that appears in the Shockley equation. The exponential relation is between the barrier potential and current. Once the increased emitter current and base current cross the junction, the barrier potential does increase in response, per Shockley. Vbe is logarithmic wrt Ie/Ib, so a 5% or 10% change in Ie/Ib, incurs a smaller change in Vbe. Then due to this increased barrier, the current is slightly reduced. The difference between the antenna output voltage and Vbe is slightly less due to Vbe increasing which happened due to Ie/Ib increasing.

The physics is all there. BR.

Claude

 

[Ratch]

Because Vbe is the result of Ie! That is why I claim it, because it is so. Sir Oliver Heaviside's transmission line theory lays many perplexing questions to rest if properly applied. A switch is turned on and a power supply is up and running, let's call it Vcc, which supplies the bias for the bjt stage. A b-e junction is in series with a resistor. A Vcc supply powers the resistor and p-n junction. What happens?

The bias circuit may or may not be be powered by the Vcc supply. It could be powered by its own power supply. That point is irrelevant. If you insert a external resistor in series with a PN junction like the base-emitter slabs, then you are looking at a circuit, and not just the junction itself.

Actually, let's reduce this to extreme simplicity, a battery, resistor, and diode. The switch is thrown, and the battery voltage is 12 volts, resistor is 1.0 kohm, and the diode is a garden variety 1N914, or any other you wish to use. Diode is oriented in forward direction. I don't need a pic. you can draw one, or just mentally visualize this case. So the switch is thrown, and the resistor-diode series combination now has 12 volts across it. What determines the current? The battery has very low internal resistance, consider it an ideal constant voltage source. The truth is that the diode has near zero volts across it, around 26 mV, due to thermal energy via lattice vibrations.

Again, you are analyzing a circuit which includes a PN junction. What you should really do is connect a voltage across the PN junction with no resistor and plot the current from 0 volts to around 0.8 volts. Observe its exponential response due to diffusion. You cannot drive it with current because a current generator is a voltage generator with a high external resistance, and thereby does not represent a PN device only.

Consider the connecting wires a t-line, so the 12 volts travels through the t-line towards the R-D series network. The current is initially Vbatt/Zo, where Zo is the t-line characteristic impedance, which we can call 120 ohm for illustration. So I, the current, is 12.0V/120 ohm = 0.100A, or 100 mA. At the initial transit of charge from the battery, there is no feedback indicating that a 1.0 kohm R and a diode D lie ahead. This R-D network is not seen by the battery. The 12 volts and 100 mA travel in unison along the t-line and reach the R-D network at a speed just below that of light, maybe 50 to 80% of c.

Why should anyone treat that setup as a transmission line and get involved with internal capacitances and possible wire inductances only applicable at high frequencies?

But what happens when the charges reach the R-D network? We have an impedance transition from 120 ohm to 1,000 ohm plus the resistance of the diode. A reflection occurs, where the voltage reflects with a positive coefficient, current has negative. Some charge enters the R-D. When charges transit through the R, collisions between charge carriers (electrons, holes) and lattice ions occur. Ionization occurs, and electrons are dropped from conduction to the valence energy band. In accordance with conservation of energy law, CEL, photons are radiated per Planck's law E = hf. A resistor conducting current radiates heat, as we know.

Same answer as above.

So what happens when charges enter the crystal lattice structure of the diode? They have energy already, imparted by the battery's redox (reduction/oxidation) chemical reaction. The positive charges enter the p side (forward biased), and electrons enter n side. The depletion zone is as it was before the switch initiated the charge motion, Vt = 26 mV. The charges move straight through their respective n and p silicon regions. Why would they not? In p material, holes easily conduct, as they have high mobility being majority carriers. Likewise for electrons in the n region. So please note that charges conduct through the diode with ease in the forward direction without anything being done to the depletion zone. A hole entering a p region encounters little resistance since a hole has a statistically smaller chance of recombination vs. an electron, and vice-versa in n region.

Only electrons enter and leave the diode. There are no holes or positive charges in the wires or the battery. If there were, they would be annihilated immediately due to the overwhelming presence of electrons in the wire. Holes are only present in the semiconductor due to the doping of the P-type semiconductor.

Once the holes have transited through the p region, and electrons through n region, they cross the depletion zone and ionize with atoms in the local region, enhancing the depletion zone. The increased hole density from the battery has crossed into the n region and ionized local n region atoms, adding to the small ionization already present due to thermal energy. Likewise, in the p region, this new addition of electrons has increased the ionization. In p region, electrons have low mobility being minority carriers, and don't get far before recombination.

Like I said before, no holes come from the battery, only electrons.

A stronger E field occurs in this new enhanced depletion zone, and the line integral of this larger E field is a larger barrier voltage. The diode forward voltage drop increased due to its increased forward current. The change in Id gave rise to a change in Vd. So now what happens?

The Id is caused by diffusion, not an E-field like it is in a resistor. The voltage across the diode counteracts the back-voltage caused by the uncovered charges. This is explained repeatedly in textbooks.

New charges crossing the n and p regions will encounter this new increased barrier potential, this E field will oppose the E field from the battery. Charges in the vicinity feel the battery E field propelling them towards the diode junction, along with the barrier potential oriented oppositely, pushing against their motion. Thus the current begins to decrease due to the Vd build up. After several dozen or so t-line reflections, the equilibrium condition will be attained. The current will be determined by the net voltage on the R, which is 12.0-0.70 volts = 11.30 volts. So I = 11.30V/1.0 kohm = 11.3 mA.

The relation between V & I in the resistor R is too easy as R = V/I, I =V/R, and V = IR. But the diode behaves per Shockley, i.e. Vd = Vt*ln((Id/Is)+1), or Id = Is*exp((Vd/Vt)-1). One could attempt to solve the 2 equations, but the result is a transcendental equation, unsolvable. Math tables have been computed for similar type functions, one being a Lambert function. If temperature is very constant, Lambert functions can solve this equation. But for any significant power, the diode parameter "Is", the scaling or saturation current, has a strong temperature coefficient. Small changes in temp produce large changes in Ies.

The Shockley equation is an exponential equation due to diffusion. Why are you bringing in transmission line concepts?

But V is logarithmic wrt I in the diode, meaning that Vd does not change much as I varies slightly. At room temp, we know a priori that when the dust settles, Vd is not far from 0.60 to 0.70 volts. If we start at Vd = 0.70V, then Id = 11.3 mA. If Vd = 0.60V, then Vd = 11.4 mA, a change of just 1%. Otherwise we would have to iteratively compute via Ohm's law and Shockley to get the result.

Spice type simulators do just that. If I assume 0.70V, get 11.3 mA, then plug the network into Spice, it may compute, Vd = 0.677 V, and Id = 11.323 mA. My error would be -0.2%, not bad.

Although Vbe in the bjt amp stage commences at 26 mV, then climbs up to near 0.70 volts after current has been passed through the junction, I know from past observation, that at the current levels in question, the Vbe will lie in the range of 0.60 to 0.70 volts. So I was leveraging this repeatable property of b-e junctions to estimate the forward drop at 0.70 volts.

Once all transients settle, Vbe will be pretty close to that value. I also allow a tolerance, computing the network with 0.60 volts Vbe, and obtaining results based on that value. I vary Vbe from 0.50 to 0.80 volts, vary beta from worst case min at min temp, up to worst case max at max temp, then compute. I then observe the spread in gain values. A well designed amp stage is robust enough to maintain consistent performance despite Vbe and beta variations.

You have not once used the word "diffusion" to describe what happens. What does your elaborate explanation of how to compute the exact Vbe have to do with how a BJT works?

Do I need to elaborate? BR.

Claude

Yes.

Ratch

 

[Ratch]

Uncovered charges? Please define that term.

Look at page 27 of this link. http://users.ece.gatech.edu/mleach/ece3040/notes/chap01.pdf

Back-voltage, I presume you mean barrier potential, which I covered thoroughly. An increase in Ie/Ic results in increase in Vbe, which results in a larger barrier potential or what you term "back-voltage".

No it doesn't. The Vbe reduces the back-voltage and allows a greater value of Ic to exist due to diffusion.

Please examine the charge transit in the silicon. New carriers entering the base and emitter move easily through said regions as they are majority carriers.

When an electron moves from the N-slab into the P-slab, it is no longer a majority carrier. Similarly for a hole moving from the P-slab into the N-slab.

But when they approach the depletion zone's barrier potential they feel a repulsive force.

Yes, that repulsion is the caused by the barrier voltage.

You are assuming that in order to get the charges past this barrier, the Vbe at the terminals must increase.

Yes, to increase the number of charges, Vbe has to lower the barrier voltage so more charges can diffuse and establish a new equilibrium. So Vbe controls the diffusion, and thereby the current crossing the PN boundary.

But the increase in Ib/Ie is due to an increase in I/V at an external source. How do we increase Vbe? If the signal source driving the bjt incurs an increase in I/V how does this change trigger the sequence of events.

A BJT is a transconductance device, so only voltage controls its output. The signal source changes Vbe which in turn changes its diffusion and thereby its collector current.

So let's use a simple dipole antenna as the source, its output feeds a bjt amp stage. The fm radio signal increases, the antenna is 300 ohm, so an increase in I & V occur due to increased electromagnetic energy in space. This increased I and V transit through the wires and into the b-e junction. The base and emitter current have already incurred the change, increased above its original value. But the barrier potential has not yet changed. We have a new increased value of Ie, Ib, and terminal Vbe, but the barrier Vbe is still at ther old value. It is the barrier Vbe that appears in the Shockley equation. The exponential relation is between the barrier potential and current. Once the increased emitter current and base current cross the junction, the barrier potential does increase in response, per Shockley. Vbe is logarithmic wrt Ie/Ib, so a 5% or 10% change in Ie/Ib, incurs a smaller change in Vbe. Then due to this increased barrier, the current is slightly reduced. The difference between the antenna output voltage and Vbe is slightly less due to Vbe increasing which happened due to Ie/Ib increasing.

If you are talking about Ib causing a change in Ic, then you are extrapolating the behavior of the BJT from a circuit containing external components. That is an invalid way on determining what is controlling the BJT.

The physics is all there. BR.

Claude

Yes, it is.

Ratch

 

[cabraham]

The bias circuit may or may not be be powered by the Vcc supply. It could be powered by its own power supply. That point is irrelevant. If you insert a external resistor in series with a PN junction like the base-emitter slabs, then you are looking at a circuit, and not just the junction itself.
Diodes/bjt are used in circuits, and the internal physics is best examined under real world circuit conditions. The resistor is all important since a battery across a naked diode is certain doom for said diode.


Again, you are analyzing a circuit which includes a PN junction. What you should really do is connect a voltage across the PN junction with no resistor and plot the current from 0 volts to around 0.8 volts. Observe its exponential response due to diffusion. You cannot drive it with current because a current generator is a voltage generator with a high external resistance, and thereby does not represent a PN device only.
0.8 volts!!! Back in the mid 70's I tried doing just that. The result was incinerated diodes. With a current source I could drive a few hundred mA through the diode and the forward drop exceeded 0.80 volts. But when I tried to drive the diode in the raw from a voltage source sans resistor, I could not get anywhere near 0.80 volts. Around 0.60 volts or so, the device smoked. Now let us discuss current generators.

Who told you that a CCS (constant current source) is just a constant voltage source (CVS) with a high resistance. Who told you that Ratch? Seriously, this is a real problem. The Norton-Thevenin equivalence principle states that a CCS with a high external shunt resistance can be replaced by a CVS with a high external series resistance. Electrically they are equivalent, but not physically/thermally. We must take an example and it will become clear to anyone willing to learn.

A CCS has a 99 ohm external shunt resistance, and sources 1.0 amp. A load of 1.00 ohm is placed across the CCS. It is obvious from circuit theory that the load current is 0.990 amp, and the external resistance current is 0.010 amp, due to current division. The 1.00 ohm load resistor has a voltage of 0.990 amp * 1.00 ohm = 0.990 volt. Applying Thevenin-Norton we can replace the 1.0 amp CCS w/ 99 ohm shunt resistance with a CVS of 99 volts and 99 ohm series resistance.

The same 1.00 ohm load resistor is connected. The current is 99 volts divides by the 99 ohm external resistance plus the 1.00 ohm load, or 99V/100 ohm = 0.990 amp, same as before. This demonstrates that these 2 networks are electrically equivalent, an external load placed across either pair of terminals produces identical results. But are the 2 physically equal?

The answer can be attained by computing the power dissipation. With the Norton network, CCS, the 99 ohm external shunt resistance dissipated 9.90 mW, while the 1.00 ohm load dissipated 980.1 mW. So 99% of the total power is in the load, 1% lost in the shunt resistance. Now examine the Thevenin network, the 99 volt CVS plus 99 ohm series external resistance.

The power in the 1.00 ohm load is still 980.1 mW. But the power dissipated in the 99 ohm series external resistance is - 97.03 watts! The efficiency is now just 1%, as 99% of the power is lost in the source resistance. I assure you Ratch, that a constant current network, is NOT merely a high voltage in series with a high resistance. Consider a bjt collector driving the base of a 2nd bjt. Say the 1st bjt is a pnp whose collector sources current into the 2nd bjt base, an npn. This configuration is a CCS driving a base. If the 1st bjt is powered from a 3.3V rail, and the 2nd bjt has its emitter at ground, there is no large voltage plus large series external resistance. What we have is a current source plain and simple. Current sources are just as "real" and just as "artificial" as voltage sources, no more no less.

Regarding the bjt exponential response, that is due to carrier lifetime and energy levels. A slab of undoped silicon exhibits diffusion but not exponential I-V properties. How do you say that the exponential relation is due solely to diffusion? Makes no sense. Did you take semiconductor physics at uni? We all know that I vs. V is exponential. I never disputed that, I have reaffirmed it ad infinitum. Please look it up, diffusion is NOT the mechanism behind the I-V exponential function.



Why should anyone treat that setup as a transmission line and get involved with internal capacitances and possible wire inductances only applicable at high frequencies?



Same answer as above.



Only electrons enter and leave the diode. There are no holes or positive charges in the wires or the battery. If there were, they would be annihilated immediately due to the overwhelming presence of electrons in the wire. Holes are only present in the semiconductor due to the doping of the P-type semiconductor.

I already covered this in another reply to you. Battery electrolytes have charges motion in the form of cations and anions. Metals are electron carriers. Semiconductors carry holes and electrons. To keep things neat, I colloquially use "holes" to denote any positive current, and electrons for negative. When I say that holes enter the base, it means positive charge enters, or negative charges (electrons) leave. Equivalent it is. Maybe I am sloppy, but you're making way too much out of positive current convention.

Like I said before, no holes come from the battery, only electrons.

Old news.

The Id is caused by diffusion, not an E-field like it is in a resistor. The voltage across the diode allows counteracts the back-voltage caused by the uncovered charges. This is explained repeatedly in textbooks.

? When charges enter base, they difuse in accordance with Fick's differential equation. But diffusion does not explain the drift action, which is what happens under external bias. Under thermal; action, with the bjt unconnected to a source, electron-hole pairs are being continuously generated by lattice vibrations. They diffuse through the silicon and cross the junction forming the barrier of 26 mV. Diffusion certainly is there, but the question was how I computed currents based on a priori knowledge of Vbe of roughly 0.65 volts. That is a math and sequence of events problem, and diffusion is not really helpful to explain that.

The Shockley equation is an exponential equation due to diffusion. Why are you bringing in transmission line concepts?



You have not once used the word "diffusion" to describe what happens. What does your elaborate explanation of how to compute the exact Vbe have to do with how a BJT works?

I was replying to LvW, that is all.

Yes.

Ratch

 

[cabraham]

I uploaded the file with the common emitter bjt amp stage referenced in an earlier post so that members can refer to it. Questions/comments are welcome.

Claude

 

[cabraham]

Look at page 27 of this link. http://users.ece.gatech.edu/mleach/ece3040/notes/chap01.pdf

I looked through it, and they did mention the term w/o really defining it. Please explain "uncovered" vs. "covered" charges.

No it doesn't. The Vbe reduces the back-voltage and allows a greater value of Ic to exist due to diffusion.



When an electron moves from the N-slab into the P-slab, it is no longer a majority carrier. Similarly for a hole moving from the P-slab into the N-slab.

But I said when hole moves through p region it is majority, as are electrons in n region. When they cross the junction they become minorities. I never said otherwise, straw man. My point was that increased current density can enter base and emitter regions BEFORE the barrier has had time to change.

Yes, that repulsion is the caused by the barrier voltage.



Yes, to increase the number of charges, Vbe has to lower the barrier voltage so more charges can diffuse and establish a new equilibrium. So Vbe controls the diffusion, and thereby the current crossing the PN boundary.

But the barrier voltage never gets lowered, rather the source voltage of the generator driving the network increases, which results in an increase in barrier. If Ie is 1.2 mA, Vbe = 0.65 V, and the antenna output jumps from 1.2 mV up to 1.3 mV, the increase in current is immediate. In order to increase Ie from 1.2 to 1.3 mA, we do not lower the barrier, nor increase terminal Vbe. How would you increase terminal Vbe without first increasing current Ie? How can one remotely lower the barrier potential. The barrier is 2 walls of charge called a depletion zone. The only way to change the value is to transport charge through the region. Your theory is that in order to increase current, you must first increase the vcltage to allow it. Plots from my real world circuit sims say otherwise.

A BJT is a transconductance device, so only voltage controls its output. The signal source changes Vbe which in turn changes its diffusion and thereby its collector current.

A bjt possesses transconductance, but also current gain, as does a FET. It is impossible to change Vbe first ahead of Ie. You view the base terminal as a low resistance constant voltage source and the barrier as an opposing source. The current is formed in between the two. So to change that current, one must change the difference between Vbe terminal and Vbe barrier potential. But the base terminal is NEVER connected to a low-Z CVS. It would be destroyed. The E field which provides drift/diffusion originates in an external source, not counting thermal energy, and charges from outside the device transit through the regions/junction. You are mistaken by asserting that the emitter current is an island and it is driven by the local region potential. Vbe is formed by charges generated btth locally, and outside the device.

Take a wall outlet, 120 volts. The load is a heater, 11.9 ohms, and the power cord has 0.10 ohms of resistance. Current is 120V/(11.9+0.1 ohm) = 10 amps. There is 119 volts across the heater, and 1.0 volts across the power cord. The utility company is driving all here, the cord and heater. I'll bet you will insist that the 10 amps in the power cord is being "driven/controlled" by the 1.0 volt drop. Is that your view? If yes, then we may as well not discuss bjt at all, since our view of the relation between I & V is oil and water. If you believe that the 1.0 volt drop in the power cord is controlling the 10 amps, then I don't know what else to say. But I will not put words in your mouth, I will await a response. BR.

Claude


If you are talking about Ib causing a change in Ic, then you are extrapolating the behavior of the BJT from a circuit containing external components. That is an invalid way on determining what is controlling the BJT.

It is the only valid way. all bjt's are used w/ external components, and that is the behavior that matters. You insist that I force a voltage source upon Vbe and plot Ie/Ic to show the exponential function. But what does that prove? Something we've all long known. If I force a current in Ie, then plot Vbe as a log function, that does not give us new insight. We need to know what happens sequence wise when a bjt is in a "free" operating mode, not forced by a CVS or CCS, but used with an actual amp stage. How does a change in the antenna voltage or current affect the bjt parameters? Forcing the device across a pure source I or V, while said device is naked proves nothing. What use is a naked bjt, or FET for that matter?

Yes, it is.

Ratch

 

[Ratch]

Claude,

"Diodes/bjt are used in circuits, and the internal physics is best examined under real world circuit conditions. The resistor is all important since a battery across a naked diode is certain doom for said diode."

"0.8 volts!!! Back in the mid 70's I tried doing just that. The result was incinerated diodes. With a current source I could drive a few hundred mA through the diode and the forward drop exceeded 0.80 volts. But when I tried to drive the diode in the raw from a voltage source sans resistor, I could not get anywhere near 0.80 volts. Around 0.60 volts or so, the device smoked. Now let us discuss current generators."

No, diodes and other components are best examined without any external components. Whoever said one should put so much voltage across a diode that it destructs? The voltage should be increased only to whatever gives the rated current is.

"Who told you that a CCS (constant current source) is just a constant voltage source (CVS) with a high resistance. Who told you that Ratch? Seriously, this is a real problem. The Norton-Thevenin equivalence principle states that a CCS with a high external shunt resistance can be replaced by a CVS with a high external series resistance. Electrically they are equivalent, but not physically/thermally. We must take an example and it will become clear to anyone willing to learn."

Like you say, any linear Thevenin circuit can be replaced by an equivalent Norton circuit. If they are electrically equal, who cares what their thermal characteristics are if they are external to the component? In any case, you are attaching external components to the component under study and thereby forming a expanded circuit.

"A CCS has a 99 ohm external shunt resistance, and sources 1.0 amp. A load of 1.00 ohm is placed across the CCS. It is obvious from circuit theory that the load current is 0.990 amp, and the external resistance current is 0.010 amp, due to current division. The 1.00 ohm load resistor has a voltage of 0.990 amp * 1.00 ohm = 0.990 volt. Applying Thevenin-Norton we can replace the 1.0 amp CCS w/ 99 ohm shunt resistance with a CVS of 99 volts and 99 ohm series resistance. The same 1.00 ohm load resistor is connected. The current is 99 volts divides by the 99 ohm external resistance plus the 1.00 ohm load, or 99V/100 ohm = 0.990 amp, same as before. This demonstrates that these 2 networks are electrically equivalent, an external load placed across either pair of terminals produces identical results. But are the 2 physically equal? The answer can be attained by computing the power dissipation. With the Norton network, CCS, the 99 ohm external shunt resistance dissipated 9.90 mW, while the 1.00 ohm load dissipated 980.1 mW. So 99% of the total power is in the load, 1% lost in the shunt resistance. Now examine the Thevenin network, the 99 volt CVS plus 99 ohm series external resistance. The power in the 1.00 ohm load is still 980.1 mW. But the power dissipated in the 99 ohm series external resistance is - 97.03 watts! The efficiency is now just 1%, as 99% of the power is lost in the source resistance. I assure you Ratch, that a constant current network, is NOT merely a high voltage in series with a high resistance. Consider a bjt collector driving the base of a 2nd bjt. Say the 1st bjt is a pnp whose collector sources current into the 2nd bjt base, an npn. This configuration is a CCS driving a base. If the 1st bjt is powered from a 3.3V rail, and the 2nd bjt has its emitter at ground, there is no large voltage plus large series external resistance. What we have is a current source plain and simple. Current sources are just as "real" and just as "artificial" as voltage sources, no more no less."

You are not telling me or anyone else who is familiar with Thevenin or Norton anything new. We are concerned only with the electrical characteristics of the component, are we not? Anytime you introduce a current source into a circuit, you are also inserting a lot of resistance also.

"Regarding the bjt exponential response, that is due to carrier lifetime and energy levels. A slab of undoped silicon exhibits diffusion but not exponential I-V properties. How do you say that the exponential relation is due solely to diffusion? Makes no sense. Did you take semiconductor physics at uni? We all know that I vs. V is exponential. I never disputed that, I have reaffirmed it ad infinitum. Please look it up, diffusion is NOT the mechanism behind the I-V exponential function.
"
Undoped silicon does not exhibit diffusion. Diffusion occurs when P-type and N-type material are put in contact with each other. And yes, diffusion is why we use Schockley's exponential equation instead of the I=V/R as in a resistor.

"I already covered this in another reply to you. Battery electrolytes have charges motion in the form of cations and anions. Metals are electron carriers. Semiconductors carry holes and electrons. To keep things neat, I colloquially use "holes" to denote any positive current, and electrons for negative. When I say that holes enter the base, it means positive charge enters, or negative charges (electrons) leave. Equivalent it is. Maybe I am sloppy, but you're making way too much out of positive current convention."

Yes, you were sloppy in your description. Especially in averring that electrons move through the electrolyte in one of your previous responses. I am happy you explained yourself better.

"? When charges enter base, they difuse in accordance with Fick's differential equation. But diffusion does not explain the drift action, which is what happens under external bias. Under thermal; action, with the bjt unconnected to a source, electron-hole pairs are being continuously generated by lattice vibrations. They diffuse through the silicon and cross the junction forming the barrier of 26 mV. Diffusion certainly is there, but the question was how I computed currents based on a priori knowledge of Vbe of roughly 0.65 volts. That is a math and sequence of events problem, and diffusion is not really helpful to explain that."

Diffusion is the primary cause of current in a diffusion diode and BJT in the active region. The drift and recombination-generation (R-G) current is miniscule in the forward direction. I mentioned this before in another thread in another forum.

Ratch

 

[Ratch]

Claude,

"I looked through it, and they did mention the term w/o really defining it. Please explain "uncovered" vs. "covered" charges."

My apology. The page number should be 17, not 27.

"But I said when hole moves through p region it is majority, as are electrons in n region. When they cross the junction they become minorities. I never said otherwise, straw man. My point was that increased current density can enter base and emitter regions BEFORE the barrier has had time to change."

Not really. When a charge carrier moves through the junction it immediately leaves an ion or "uncovered charge" behind, which increases the barrier voltage. That happens immediately. It is stating the obvious to say that holes are in the majority in P-type material and electrons in N-type material.

"But the barrier voltage never gets lowered, rather the source voltage of the generator driving the network increases, which results in an increase in barrier. If Ie is 1.2 mA, Vbe = 0.65 V, and the antenna output jumps from 1.2 mV up to 1.3 mV, the increase in current is immediate. In order to increase Ie from 1.2 to 1.3 mA, we do not lower the barrier, nor increase terminal Vbe. How would you increase terminal Vbe without first increasing current Ie? How can one remotely lower the barrier potential. The barrier is 2 walls of charge called a depletion zone. The only way to change the value is to transport charge through the region. Your theory is that in order to increase current, you must first increase the vcltage to allow it. Plots from my real world circuit sims say otherwise."

Vbe combines with the barrier voltage and produces a resultant voltage which controls the diffusion. That is not my theory. It is what is taught in transistor physics books. If the resultant voltage is lowered, the current will increase and produce more ions or uncovered charges until a new equilibrium is reached. The depletion region occurs because there are few charges in this area, not because of two charge walls. One must increase Vbe to reduce the barrier voltage so that more charges can diffuse. More current is not going to happen until the barrier voltage is lowered. I think it is up to you to explain why your sims say why current increases without an increase in Vbe.

"A bjt possesses transconductance, but also current gain, as does a FET. It is impossible to change Vbe first ahead of Ie. You view the base terminal as a low resistance constant voltage source and the barrier as an opposing source. The current is formed in between the two. So to change that current, one must change the difference between Vbe terminal and Vbe barrier potential. But the base terminal is NEVER connected to a low-Z CVS. It would be destroyed. The E field which provides drift/diffusion originates in an external source, not counting thermal energy, and charges from outside the device transit through the regions/junction. You are mistaken by asserting that the emitter current is an island and it is driven by the local region potential. Vbe is formed by charges generated btth locally, and outside the device."

A BJT, FET, and tube are all transconductance devices. They can be made into current amplifiers by additional circuitry, but are not current amplifiers by themselves. The current in the base-emitter is waste current that "leaks" through the emitter-collector circuit. Vbe is comes from the base voltage source and not from any internal source within the BJT.

"Take a wall outlet, 120 volts. The load is a heater, 11.9 ohms, and the power cord has 0.10 ohms of resistance. Current is 120V/(11.9+0.1 ohm) = 10 amps. There is 119 volts across the heater, and 1.0 volts across the power cord. The utility company is driving all here, the cord and heater. I'll bet you will insist that the 10 amps in the power cord is being "driven/controlled" by the 1.0 volt drop. Is that your view? If yes, then we may as well not discuss bjt at all, since our view of the relation between I & V is oil and water. If you believe that the 1.0 volt drop in the power cord is controlling the 10 amps, then I don't know what else to say. But I will not put words in your mouth, I will await a response."

Wait no longer. Of course I will not say that the voltage drop across one component of a series circuit is going to control the total current in the circuit. A series circuit by itself is not a transconductance device like a transistor is. But the small voltage applied to the base of the BJT transistor does control a lot of current in the collector.

"It is the only valid way. all bjt's are used w/ external components, and that is the behavior that matters. You insist that I force a voltage source upon Vbe and plot Ie/Ic to show the exponential function. But what does that prove? Something we've all long known. If I force a current in Ie, then plot Vbe as a log function, that does not give us new insight. We need to know what happens sequence wise when a bjt is in a "free" operating mode, not forced by a CVS or CCS, but used with an actual amp stage. How does a change in the antenna voltage or current affect the bjt parameters? Forcing the device across a pure source I or V, while said device is naked proves nothing. What use is a naked bjt, or FET for that matter?"

You cannot put a BJT into a circuit and say its intrinsic behavior is what the circuit does. If you want to see what the intrinsic behavior is, then you have to measure it by itself and formulate an explanation based on its physics. For instance, take a general purpose op-amp. It is a dual voltage amplifier with a high amplification factor. By itself it is useless. However, its parameters are not measured in an inverting or noninverting circuit with feedback and whatever else. Same with any other active component.

Ratch

 

[LvW]

Gentlemen - I must admit that I am not a specialist in charged carrier physics. Hence, I cannot comment on most of your last statements.
I was educated as an engineer - therefore, I think as an engineer and I have worked as an engineer for many years.
For this reason, I primarily rely on measurements and observations - followed by calculations, evaluations and interpretations..
And I am still waiting for a single example (based on measurements, observations) which could clearly demonstrate if the BJT is current-controlled.

On the other hand, in my various posts I have mentioned many (at least 10) phenomena which can be explained only assuming that the current Ic is controlled/determined by the voltage Vbe (transconductance model). See experiment description below.

Regarding Claude`s statement (applying a voltage Vbe would destroy the device): Did you never record the characteristic of a pn diode or a BJT ?
(This can be done with an xy recorder in seconds, without any thermal problems). I did it - resp. my students did it - in lab experiments rather often..

In this context, here is the description of a very simple but instructive experiment:
1.) Grounded emitter E and grounded collector C : Connect the base terminal B to a varying dc voltage (0.1...0.6 volts) and record the input characteristic Ib=f(Vbe).
Note that only Vbe can be the cause of the currents Ib, Ic and Ie.
2.) Keep the B-E voltage at at a fixed level and measure Ib (within the allowed current range, uncritical with respect to temperature) ;
3.) Now - connect C to a voltage source and slowly increase the C-E voltage from zero up to some volts. During this procedure record the base current Ib and the collector current Ic.
4.) You will notice the following: For rising voltages Vce the current Ib will decrease below the initial value and - at the same time - the current Ic will increase, thus fulfilling the known relation Ie=Ic+Ib.
5.) This measurement/simulation can be repeated for another initial voltage Vbe (see 2))
6.) Evaluation: Which electric quantity determines the final currents Ib and Ic for voltages above Vce,sat (normal BJT operation) ? Claude, what is your answer?

 

[Arouse1973]

I am still waiting for Claude to explain why this anomaly is actually happening. Please excuse me if I have missed it in the 100 or so posts. And I hope he can explain why a similar simulation using only a voltage controlled current source with BJT parasitic and associated components as in his original simulation of the BJT produce the same results. No transistor, emitter current, base current or collector current and no p-n junction.
Adam