BJT - a current-controlled device?

[Winterstone]

I would (and did) say, it depends on how you use it.

Are you really happy with such an answer? A device like a BJT allows for a current Ic that can be controlled externally.
And the question was NOT if a voltage or a current is injected in the controlling node.
The question was what happens physically in the base region! Which quantity influences the current flow - Vbe or Ib ?
But I don`t intend to start the discussion again. This was just to clarify the point of discussion.

 

[Ratchit]

vlad,

Vbe controls Ib , Ib controls Ic.

You got that wrong. Vbe controls both Ic and Ib. That is why Ic and Ib have a relationship to each other. In the active region of a BJT, Ib is an indicator of Ic, but does not control Ic. Vbe does that.

So voltage causes current but also current can cause voltage.

In a static situation, one can have voltage without current. I think you mean that current can result from voltage if there is a conduction path.

Ratch

 

[Winterstone]

And the question was NOT if a voltage or a current is injected in the controlling node.

In a certain sense I have to correct myself.
In practice we do not "inject" a current.
Think of a simple resistive voltage divider R1-R2. Do we inject a current into R2?
No, we apply a voltage to R2, which is reduced automatically due to the current and the corresponding voltage drop across R1.
And the same happens when we "inject" a current into the base node.
In reality, we apply a voltage to the B-E junction that automatically is reduced due to the voltage drop across the large Rb value that mostly determines the current value.
As a consequence, a voltage of app. 0.7 volts is developed across the B-E junction.

Edit: By the way - that is exactly the method that is applied (voltage divider rules) for calculation of the desired value for the base resistor Rb.

 

[Claude Abraham]

Are you really happy with such an answer? A device like a BJT allows for a current Ic that can be controlled externally.
And the question was NOT if a voltage or a current is injected in the controlling node.
The question was what happens physically in the base region! Which quantity influences the current flow - Vbe or Ib ?
But I don`t intend to start the discussion again. This was just to clarify the point of discussion.

Neither Vbe nor Ib, rather Ie is what controls Ic. After dozens of posts where I have demonstrated Ebers-Moll original 1954 paper w/ schematic detailing Ie controlling Ic, the Ib straw man is still being erected for the sake of being knocked down. I have shown a power converter where inductor current is established via FET turning on while diode is reverse biased. When FET turns off free wheeling diode carries inductor current. Id is already established, Vd is a result of Id and temp. Id controls Vd in forward direction, vice-versa in reverse direction.

Eli the Ice man put a period on the crackpot heresy that somehow Vd/Vbe "controls" Id/Ic. Ie changes before Vbe, and Ic follows the Ie change quickly, whereas Vbe catches up slightly later. I have stated repeaedly that this is so well known debating is crazy.

Every bjt OEM classifies the device as "current controlled" at the external level, black box, not considering internal charge storage or geometry. The next closer detailed view involves "charge control", a model I use in the switching mode, as well as FET operation. When a bjt is operating at a speed where device limitations due to internal stored charge cannot be ignored, the charge control model works well. At the sub-atomic view, only QM can provide answers, beyond the scope of this discussion, and over my head as well.

What happens physically in the base region is well known and what still remains to be clarified? If something I stated needs clarified, or if you dispute a fact I've entered, I will clarify. What do you think happens in the base region? Second, why cannot one inject a current? Of course in a non-zero impedance any current is accompanied by a corresponding voltage. I will answer any question. Point out any discrepancies that your explanation has with mine. I will do my best to examine and state what is going on in said base region.

This issue is old as the hills. Our world is filled with power sources designed for constant voltage. It is very easy to slip into the mental trap of viewing V as the independent variable, with I being dependent on V and Z. Sometimes, that is how it works. But we emphasize the current control nature of bjt because unlike FET or vacuum tube, a CVS right across the b-e junction is untenable. We need to use a resistance to control Ie or Ib (not used a lot due to beta variation), and Ic is alpha*Ie, a very stable parameter.

The fact that a bjt is current controlled does not mean we exclude Vbe from consideration. Likewise a FET is modeled as voltage controlled, but gate current must be examined. Too large a gate current switches the FET on/off very quickly, and ringing, overshoot, and emi can be a problem. Too low a gate current turns the device on/off too slowly, and large dissipation occurs, potentially failure.

We never control FET gate current Ig directly, but we control Vgs letting Ig be incidental. The CVS will charge up the g-s junction by outputting gate current. When charged to the CVS value, current decays to zero. Vgs is directly controlled, Ig is incidental, but important. A CCS across g-s results in a ramp of voltage increasing until punch through happens.

So a bjt relies on both Vbe and Ib/Ie for operation. We just have to be sure we drive Ie (or Ib) direcly letting Vbe be determined indirectly. It isn't too complicated. I & V cannot act independently. One changing cannot happen w/o the other changing as well. It just happens that I changes first due to capacitive nature of diffusion layer. When Ie changes, Ic follows immediately, Vbe eventually.

Vbe simply cannot be controlling Ic.

 

[Winterstone]

Neither Vbe nor Ib, rather Ie is what controls Ic.

Hello Claude,

this assertion from you I could read already several times in this forum. I cannot remember one single book or any other serious text , which can support this (in my eyes: somewhat "surprising") opinion.
Up to now - you are the only person I know who is stating that Ic=alpha*Ie describes the control properties of a BJT. Of course, I can be in error and my opinion is of less importance.
Therefore, I promise you - I will change my mind if you answer the following question (and it is not the first time, I did ask you already at least twice, but without any response):

How do you explain the stabilizing effect without using the voltage Vbe, which an emitter resistor Re has on unwanted and unexpected changes resp. variations of the dc quiescent collector current Ic?

With regards
W.

 

[Claude Abraham]

Hello Claude,

this assertion from you I could read already several times in this forum. I cannot remember one single book or any other serious text , which can support this (in my eyes: somewhat "surprising") opinion.
Up to now - you are the only person I know who is stating that Ic=alpha*Ie describes the control properties of a BJT. Of course, I can be in error and my opinion is of less importance.
Therefore, I promise you - I will change my mind if you answer the following question (and it is not the first time, I did ask you already at least twice, but without any response):

How do you explain the stabilizing effect without using the voltage Vbe, which an emitter resistor Re has on unwanted and unexpected changes resp. variations of the dc quiescent collector current Ic?

With regards
W.

The stabilizing effect is due to the fact that the signal source driving the base is constant voltage. Being a current controlled device, thermally unstable w/ voltage control, a bjt stage driven from a CVS must have a resistor either in the base or emitter lead. This resistor Re determines the bias current, or quiescent point of the bjt stage. Let's say that Vin (power supply bias source) is 5.0 volts, with Re = 1.0 kohm, rbb' = 20 ohm, w/ beta = 100. Vbe does not directly control Ic, but its 0.7V drop factors in so that Ie = (5.0 -0.7)/((20/101)+1.0k = 4.3 mA. I've already stated that most physical power/signal sources are intentionally built for CVS operation, not CCS. Inserting the proper value of R or a constant current source/sink alleviates this when a bjt is employed.

Temperature increases, effectively raising bets, raising Ies, raising Vt, lowering Vbe. The resistor Re is a series type of feedback. A temp rise would lower Vbe to 0.65V, so that Ie is now tending towards 4.35 mA, up from 4.30 mA. But, the increase in Ie immediately results in a larger voltage drop across Re. Since Vin is fixed, a larger Vre means that the drop across rbb' and Vbe must decrease. But did Ie decrease because of the Vbe decrease? I say it cannot, since Vbe takes time to respond.

When temp changes, more carriers are made available through electron-hole pair generation, ehpg. More holes are injected base to emitter and more electrons injected emitter to base. When holes enter the emitter, since there are now more of them, collisions with bulk emitter resistance as well as recombination in edge of emitter depletion zone result in a build up of charges above that previously encountered. As a result new holes entering the emitter incur a larger barrier due to increased charge accumulation. Now turn to Re, the increased carrier motion results in more collisions with Re lattice ions. The charge accumulation at the Re terminal connecting the emitter has increased.

This results in an increased E field and barrier to new incoming holes. So this barrier repels emitter holes as they transit towards Re resulting in a buildup of additional holes in emitter. Likewise injection of holes from base towards emitter must decrease due to charge buildup in emitter and Re. Again, the build up is a result of increased Ie, eventually Vbe decreases. Vbe is a result of interaction between the external power source E field forward biasing the b-e junction, minus the barrier E field due to accumulation of additional charges.

Before Vbe decreases, Ie first increased. Again, the external source, or ambient temp, or other condition, brought about the change in Ie, then Vbe followed as expected since Eli the Ice man predicts in a capacitance that I leads V.

Vbe decreasing did not force Ie to decrease. Rather inevitable build up of charges due to increased conduction due to increased carrier concentration brought on by increased thermal generation of electron-hole pairs, is what forced Vbe to decrease.
Vbe is involved, but not in the manner you think. If I'm wrong correct me because only you know what you think, I do not wish to put wors in your mouth. You seem to suggest, my apologies if otherwise, that any change in Ie due to temp or other action, is corrected by Vbe, i.e. that Vbe responds to the change and forces Ie by controlling it. I have shown that Vbe is incidental, it responds to the change, but is not the agent forcing the change.

Why does temp change conduction? Because more ehpg takes place at higher temp. So what does this mean? Increased conduction current in emitter and base immediately increases the charge accumulation in the emitter region as well as Re. This results in an increased voltage drop on Re as well as emitter region bulk resistance. But the base lead is connected to a CVS. So Vbe is the difference between Vb and Ve. Ve is moved up towards the base due to increased Ie as well as increased barrier formed by additional charge carriers, brought into existence via ehpg.

Vbe is not what forced Ie to reverse and correct itself. Again, Vbe adjusts after other parameters have settled. This notion that changes in Ib and Ie are merely responses to changes in Vbe cannot withstand even a mild scrutiny.

You asked me to explain how the Re provides correction to changes w/o using Vbe. But Ohm's law and Shockley state that such is impossible. In explaining how Re corrects drift due to temp, I must discuss Ib, Vbe, and Ie, all 3 of them. They all interact. But to answer your question, Vbe is NOT the agent forcing Ie to correct itself. Charge buildup does that.

If Re is too small in value, the charge build up is too small to counter-act the increased carrier conduction from ehpg. Re must be a sufficiently large value to be effective. Large Re insures more lattice collisions in Re bulk material, forcing a big buildup of charge, which forces Ve to rise towards Vb, forcing Vbe to decrease. Vbe decreasing is a result of Ie and Re as well as ehpg. Vbe change takes place after Ie and Re and the barrier have changed.

Or look at it this way, Vbe = Vb-Ve. Since Vb is fixed by the source, raising Ve lowers Vbe. But raising Ve is done by ehpg, Re collisions, Ie change, resulting in Ie lowering due to higher barrier in Re and emitter.

Ie/Vbe/Ib are so tied together, it is impossible to say that this one is the driver, or active agent, while these 2 are controlled by the other. They interact. I don't know how else to put it.

This is negative feedback, SERIES applied. It could be implemented in a parallel method as well. If the output of an amp stage is fed back parallel with the input source at the base terminal, the output change injects a changed current component into the base. Of course Vbe will catch up, it must. But Ie/Ic changed per the new current, Vbe lags behind catching up.

The best answer to your question is a phototransistor example. An emitter resistor Re connects to the bjt whose stimulus is incident light. A light beam excited the PT, Ie is established,and Vbe will be determined by device characteristics. If the light increases, the emitter current Ie increases and there is no feedback to correct it. The base and emitter float up to a new value. Here there is no CVS attached to the base forcing Vb to a fixed value. Vbe will adjust to a new value, but both Vb and Ve will rise. Vbe actually INcreases here instead of decrease.

Did this help? I will clarify, feel free to ask for specific details.

 

[Winterstone]

To Claude Abraham:

Yes - that`s what I have expected. Not a short and clear answer but a very, very long excursus about
"electron-hole pairs, ehpg, bulk emitter resistance, Re lattice ions, charge accumulation, charge build-up, external power E-field, barrier E-field, lattice collisions in Re bulk material, Re collisions " (what`s that?),....

For my opinion, this obscures the truth, which - in fact - is not too complicated.

Do you remember that Einstein was able to explain and visualize much of his insights and findings using a train or an elevator?
Thus, a good teacher should be able to concentrate on the core of the question and to give a suitable answer in a comprehensible form.
And in this case, it is possible in one sentence only:

An Ic increase due to a higher temperature causes a corresponding increased voltage drop across Re - leading to a Vbe reduction that reduces Ic again to a value which is only slightly above the initial Ic value (but considerably lower than in case of Re=0).

This is the classical explanation of a current-controlled voltage feedback effect as can be found in all textbooks as well as all other contributions (university and other scientific organizations).
Of course, also textbooks can be (and are in fact sometimes) not error-free. However, the probability that your theory tells the truth is not very high when there is no single scientific source that can support you.

By the way - I am surprised that your claim (very often contained in your former postings) that Ie would control Ic due to Ic=alpha*Ie was NOT used in the course of your last explanation (not even mentioned again). Up to now, this was your main argument against Ic=f(Vbe).

Question: If you deny that Vbe controls Ic, how do you explain that Vbe goes down 2 mV per degree C for a constant Ic value? (dVbe/dIc=-2mV/deg). As you know, this value is calculated based on transistor physics and confirmed by measurements!

For my opinion, your contribution contains a lot of questionable and contradictory statements.
Some examples:

*Vbe does not directly control Ic, but its 0.7V drop factors in...
(comment: ???)

*A temp rise would lower Vbe to 0.65V, so that Ie is now tending towards 4.35 mA, up from 4.30 mA.
*But did Ie decrease because of the Vbe decrease?
(comment: Ie increase or decrease?)

*Vbe takes time to respond.
*Before Vbe decreases, Ie first increased
*Ie/Vbe/Ib are so tied together, it is impossible to say that this one is the driver, or active agent, while these 2 are controlled by the other.
(Comment: „time to respond“,“ before...first“, and nevertheless no „driver“ or „active agent“?)

Without any doubt, a BJT with an emitter resistor Re is device with feedback and, thus, can be modeled in form of a classical control loop.
That means: It has an output (Ic=alpha*Ie) that can be controlled by another quantity (Vbe), which is developed by a comparison (difference) between the input quantity (Vb) and the feedback signal, which is Ve=Ie*Re. That`s the whole story. And this is the way - the only one (!) - that is used to design the Re-stabilized circuit (calculation of Re as well as the base divider resistors).

More than that, of course it is possible for each such control loop to show which quantity causes a change of another quantity - even if they are „tied together“. Otherwise, it wouldn`t be possible to design the loop components.

Final remark:
To me, it is not important if I can convince you or not. However, I doubt that many newcomers, beginners or other people who wants to understand the operation of a BJT and the role of stabilizing elements could be confused by your contributions, which are in contrast to all published explanations.

Regards, Winterstone

PS: It would be interesting for me to learn if there are forum members who can follow and support Claude´s theory.

 

[Nigel Goodwin]

The whole thing is a complete waste of time - it's dead simple:

A BJT takes current in it's base - so is current operated.

An FET doesn't take current in it's gate - so is voltage operated.

Easy as that, anything else is complicating things to try hide the truth.

But what does it matter anyway?.

 

[MrAl]

Hello,


Winterstone what do you want to stabilize?


If the DC gain is A and you change the Vbe by x volts, then the output voltage changes by A*x volts. So with a DC gain of 10 if the output voltage is 5.8v and Vbe goes down by 0.1v then the output goes down to 4.8v approximately. That's when biasing with a low impedance voltage divider and using an emitter resistor too.

 

[Claude Abraham]

Winterstone you missed my point. I'm not as good as Einstein, never claimed to be, so I cannot use a train car or elevator to illustrate this concept. But Einstein's rigorous analysis used tensor theory and high level math as well. His simple illustrations were useful for lay people and beginners in science, but he did not rely entirely on simplicity. You claim that Vbe decreases due to Re voltage drop increasing, therefore Ie/Ic decrease as a result of Vbe decreasing. But how does it happen, can you explain the sequence? Let's start with an Re value sufficient to be effective regarding degeneration.

A bjt is biased to a q point, Ie/Ic, and Vce. We wish that temp changes have minimal impact on q point stability, so we have Re, base tied to a CVS of low internal resistance, so how does the temp change affect the stage? Again, let's say that temp increases. What happens inside the bjt silicon?

A semiconductor becomes more conductive at elevated temp due to more carriers being made available for conduction. That is why for a constant I, Vd (Vbe) goes down 2mV/C. With more carriers available, more drift is obtained for a given external E field. Inside bjt Si, ehpg increases. The base and emitter regions both incur increased carrier population density, but more so with the emitter as it is doped much heavier than the base. So for a constant E field, hole injection b to e goes up mildly, while electron injection e to b goes up more. The natural tendency is for Ie and Ib to both increase, Ie more so than Ib since beta is larger at higher temp. This would result, if not for Re, in Vbe increasing as well due to increased current density. But w/o Re, the input voltage source at the base would require a base resistor to prevent thermal instability.

Anyway, bjt Si has enhanced conductivity, as soon as emitter carrier flow increases, Re incurs more collisions and higher voltage drop. The boundary region between Re and emitter develops a larger charge density, which reduces Ie. Here is an important point. An increase in Ie results in a very small increase in Vbe since Vbe is a logarithmic function of Ie. If Ie goes up by 1%, Vbe will eventually increase to way less than 1%, the logarithmic nature of p-n junctions being what it is.

But resistor Re has a linear I-V relation, a 1% increase in Re results in a 1% increase in voltage drop across Re. Thus we have a stronger local E field in the Re-emitter boundary region. This increased local E field opposes that from the CVS attached to the base. This results in decreases Ib/Ie. As Ie decreases, Vbe will follow.

But if you wish to view this in simpler terms, suppose we have again, Vb = 5.00V, Re = 1.00 kohm, Vbe is originally at 0.70V. A temp increase will result in a slight increase in Ie, and since Vbe has a negative temperature coefficient, it will drop. The rise in Ie would tend to increase Vbe, but Re limits the increase in Ie to a slight amount. Vb at 5.00V, Vbe dropping slightly, results in Ie increasing slightly.

What is happening here is that Vb >> Vbe, so that temp induced changes in Vbe have minimal impact on Ie. In simplest terms, Ie = (Vb - Vbe)/Re. You claim that Ie gets reduced due to Vbe reduction. That is impossible. Vbe can be reduced only after Ie has already gone down. When carriers cross the junction and form the depletion region, there is a finite lifetime before carriers recombine. Electrons from emitter enter a hole in the base forming ionization and Vbe is the integral of the E field over a region. Eventually an electron is expelled via the base lead to maintain charge neutrality. If the flow rate of electrons is reduced, then the recombination takes place at a lower rate, as does ionization. But carrier lifetime has not gone down. With lower incoming electron flow, the local E field is reduced, so Vbe being the integral of E, also goes down.

You and Ratchit keep asserting that Ie/Ic going down (or up) is a result of Vbe going down (or up). That would mean that Vbe is what controls Ie. Clearly Vbe is the integral of the local E field in the depletion region due to carrier recombination/ionization. That E field and consequently Vbe can change only after an external agent, be it a mic, a light beam, temp change (thermal energy) perturbs the system. But Ie must first change in order to perturb the carrier flow rate through the base region. Once this rate changes, the depletion zone charge density changes as does Vbe.

Vbe is clearly NOT in control of Ie nor Ic. Re provides stabilization, due to Ohm's law. You claim that Ie going up forces Vbe to go down per Re, therefore Ic goes down, but you haven't accounted for carrier distribution. There is a lot going on. If you wish to focus solely on Shockley's equation, please remember that Ie = Ies*exp((Vbe/Vt)-1), where Ies is a very strong function of temp.

If the base were driven by a CCS (constant current source), Re would not stabilize like it does with base tied to CVS. If temp increases, beta increases, but with Ib being fixed, Ic is the new elevated beta value times fixed Ib. Ic will go up a little since beta is higher at high temp.

I will clarify if needed. BR.

 

[Winterstone]

Winterstone what do you want to stabilize?

If the DC gain is A and you change the Vbe by x volts, then the output voltage changes by A*x volts. So with a DC gain of 10 if the output voltage is 5.8v and Vbe goes down by 0.1v then the output goes down to 4.8v approximately. That's when biasing with a low impedance voltage divider and using an emitter resistor too.

MrAl - I must confess, that I am confused.
Do you really ask what I want to stabilize with Re?
There must be something else behind your question.
Is it really necessary to mention the task of Re, which is stabilization of the design value for Ic against unwanted temperature and tolerance influence?
Was this your question?

W.

 

[Winterstone]

The whole thing is a complete waste of time - it's dead simple:

A BJT takes current in it's base - so is current operated.

An FET doesn't take current in it's gate - so is voltage operated.

Easy as that, anything else is complicating things to try hide the truth.

But what does it matter anyway?.

Primarily, we are discussing the role of Re in stabilizing the design value for Ic. Perhaps you didn`t realize.
I think, each newcomer/beginner cannot understand how a BJT - if "current operated" - can be stabilized with a feedback voltage.

Could you explain?

W.

 

[ericgibbs]

hi,

This Thread is going nowhere and is just being used as a platform to create further arguments as each new post appears.

As neither side is not prepared to change their position, common sense says the Thread should be Closed, as most likely agreement will never be reached.

E

 

[Claude Abraham]

Primarily, we are discussing the role of Re in stabilizing the design value for Ic. Perhaps you didn`t realize.
I think, each newcomer/beginner cannot understand how a BJT - if "current operated" - can be stabilized with a feedback voltage.

Could you explain?

W.

The answer lies in the miracle of science known as a (drum roll) resistor! A current through it is translated into a voltage. A voltage across it is translated into a current. With CC devices, such as bjt, LED, feedback can be series or parallel. Ditto for VC devices (FET, IGBT, tube). A voltage feedback op amp with a high Z inverting input feeds back the output voltage, sometimes through a resistive voltage divider. The fraction of output at the -ve input is determined by fb resistance ratios. A current in each resistor and the R values determine the divider ratio.

A feedback can be series as well, where the output and input voltages combine. But the voltage on Re is related to its current. Also remember that the base is driven by a voltage source This topology is effective only if Vb >> Vbe. The the CVS attached to the base is a mere 0.80v, while Vbe is at 0.70V, then Re stabilizing effect is greatly diminished. Ditto for a FET.

Real world applications demand that we be able to use FET or bjt, in conjunction with both V & I sources, series or parallel. A good example is a summing amp built w/ a FET op amp input stage and 2 summing resistors. The FETs at the op amp input is a voltage controlled device. The summing action at the input is a result of the 2 resistor currents summing. Negative feedback from output sums with input source(s) in the form of current, although VC device, with V feedback.

The answer to your question is how a "current operated bjt" can be stabilized with fb voltage is that said fb voltage is simply emitter current times Re. The fb voltage is a replica of Ie. In a resistor, I & V are in phase and linearly related. Thus any change in Ie is quickly translated into a series voltage. Vbe follows but what stabilizes Ie is as follows.

If Vb is 0.80V, then slight changes in temp have an adverse effect that Re is not very effective at neutralizing. Ie = (0.80-0.70)/1.00k = 0.10 mA at start. If temp forces bjt into heavier conduction while dropping Vbe to 0.65V, then Ie = (0.80-0,65)/1.00k = 0.15 mA. The Ie value is raised by 50%, not very good.

But if Vb is 10V, then a drop from 0.70V to 0.65 due to temp increase results in a change in Ie from 9.3 mA, up to 9.35 mA, less than 1%. The key here is to understand that the base of the bjt is held at a fixed voltage by the CVS. Then Vbe drops around 0.70V, and the balance of Vb loop voltage is across Re. If Vb >> Vbe, and Re is large enough, Vbe variations due to speciman, and temp, are diminished because nearly all of Vb source voltage is dropped across Re.

With 10V value of Vb, Vbe can vary from 0.30V, up to 1.00V, with Ie varying from 9.0 to 9.7 mA, less than 10%. The reason voltage series feedback works with a current controlled device lies in the fact that said CC device is biased with a voltage source and a resistor. Voltage changes are translated into current.

A bjt can be driven from a voltage source as long as resistance is properly employed. By placing a resistance is series with base or emitter in series with voltage source, we are controlling bjt current per V source and R value. By forcing nearly all voltage to appear across R, we control the current.

 

[Winterstone]

Hello Claude A.

Before answering - at least, I will try it - your posts#50 and #54. I like to present a feedback model in form of a closed control loop.
(I hope Eric will not close the thread before posting; creating "further arguments" is not the worst thing, is it? Is an "agreement" necessary to keep the thread open? )

I think, we all agree that the resistor Re provides feedback. Thus, why not create a corresponding control loop that can visualize what´s going on?
I have prepared a very rough (somewhat simplified) drawing with ic~ie, which shows two cases:

A) Control loop based on a voltage control model of the BJT with the corresponding loop gain LG(A)
B) Control loop based on current control model of the BJT - also with the corresponding loop gain LG(B).

Please see the attached pdf document.

Comment to case B): For a current control model I have found no other method than to build the difference of two currents through the base region. Other methods (if they exist) are welcome.
In this context I have used a base current ib2 driven by the emitter voltage Ve. The value of ib2 is determined by the input resistance seen at the emitter node, which is re=1/g. Thus: ib2=Ve/re=Ve*g

Result: When we compare the loop gain LG(A)=-g*Re with

*the known expressions for the amplifier input resistance at the base node rin=rbe*(1+g*Re), and
*the known expression for gain (with feedback) G=-g*Rc/(1+g*Re)

it can be seen that only the voltage control model (A) confirms the theoretical expressions (which are confirmed by measurements).
(Correction: Please read (1+g*Re) in the last line instead of (1-g*Re) )

In contrast, the loop gain for the current control case is larger by a factor beta, which is NOT in accordance with measurements.
Did I make a severe error?

Winterstone

In the foregoing, I have assumed that you know how the loop gain modifies the input resistance and the gain
*rin is increased by a factor (1-loop gain)
*gain is reduced by a factor (1-loop gain).

 

[Claude Abraham]

Hello Claude A.

Before answering - at least, I will try it - your posts#50 and #54. I like to present a feedback model in form of a closed control loop.
(I hope Eric will not close the thread before posting; creating "further arguments" is not the worst thing, is it? Is an "agreement" necessary to keep the thread open? )

I think, we all agree that the resistor Re provides feedback. Thus, why not create a corresponding control loop that can visualize what´s going on?
I have prepared a very rough (somewhat simplified) drawing with ic~ie, which shows two cases:

A) Control loop based on a voltage control model of the BJT with the corresponding loop gain LG(A)
B) Control loop based on current control model of the BJT - also with the corresponding loop gain LG(B).

Please see the attached pdf document.

Comment to case B): For a current control model I have found no other method than to build the difference of two currents through the base region. Other methods (if they exist) are welcome.
In this context I have used a base current ib2 driven by the emitter voltage Ve. The value of ib2 is determined by the input resistance seen at the emitter node, which is re=1/g. Thus: ib2=Ve/re=Ve*g

Result: When we compare the loop gain LG(A)=-g*Re with

*the known expressions for the amplifier input resistance at the base node rin=rbe*(1+g*Re), and
*the known expression for gain (with feedback) G=-g*Rc/(1+g*Re)

it can be seen that only the voltage control model (A) confirms the theoretical expressions (which are confirmed by measurements).
(Correction: Please read (1+g*Re) in the last line instead of (1-g*Re) )

In contrast, the loop gain for the current control case is larger by a factor beta, which is NOT in accordance with measurements.
Did I make a severe error?

Winterstone


In the foregoing, I have assumed that you know how the loop gain modifies the input resistance and the gain
*rin is increased by a factor (1-loop gain)
*gain is reduced by a factor (1-loop gain).

Your paper makes no sense. Is "g" supposed to be "gm"? G at the stage level is what I called "Avs" in my 4 page paper. I explained all of this. The small signal collector current ic can be expressed as hfe*ib, or gm*vbe. But what is vbe? To compute, we must consider Re', as well as r_pi. How do we get r_pi? We compute it as r_pi = hfe/gm. So it is impossible to omit hfe from any stage analysis/calculations, nor alpha. My 4 page paper spelled this out. If need be I can reattach it. I've answered every question you asked. This sketch makes no sense at all. I computed the gain in my 4 page paper. Feel free to redirect any question re the paper, or anything else. I will respond. I am trying to be helpful to you and everyone else hear. If I come across as difficult, I assure you it is unintentional.

Also this must be emphasized. Current control is the raw device model. When a bjt is installed in a bias network and subjected to an input signal, the stage has 2 "gains" to compute. One is current gain. The current in the collector supplying the load, or a following stage, is generally greater than input current. So beta, or "hfe", is a measure of the current gain of the raw device. Since a portion of the input signal generator never reaches the base, but is diverted into the bias resistors, the base current is less than the input signal current. Thus stage current gain is always less than the raw device current gain. Hfe is an upper limit on stage current gain.

The second gain under discussion is stage voltage gain. The bjt outputs a constant current source which is terminated in collector resistance Rc. A bjt has an input current ib and voltage vbe. The transconductance of the whole stage can be called "Gm", and that of the raw device as "gm". A part of the input signal voltage is dropped across source resistance, and emitter degeneration resistance Re. Not all of vin reaches vbe. Thus Gm < gm, i.e. the bjt "gm" value is the upper limit for the stage value of transconductance "Gm".

When the input signal is a voltage source, and we wish to compute stage voltage gain into a load resistance, we compute the equivalent circuit, Re', r_pi, Ie, Ic, gm, etc. To obtain stage voltage gain, not only do we need to know gm, but we need r_pi to determine vbe (or "v_pi"). That requires knowing hfe, since r_pi = hfe/gm. My 4 page paper is reattached next post, it covers this in detail.

Also, your equations for "loop gain" make me ask the question "are you referring to parallel feedback, where output is fed back to input through 1 or more resistors forming a loop"? "Loop gain" usually refers to an amp with global feedback, where output is fed back to input via resistive divider. If "A" is open loop gain, and "B" is feedback factor (fraction of output returned to input), then loop gain = T = A/(1+AB). I don't think that model covers this case which is series feedback. What is "open loop gain" here, and what is "feedback factor". Please explain. Thanks.

 

[Claude Abraham]

4 page calculation sheet

The attached details how stage gain is computed based on network part values and bjt device parameters. Questions will be answered if need be.

 

[Claude Abraham]

hi,

This Thread is going nowhere and is just being used as a platform to create further arguments as each new post appears.

As neither side is not prepared to change their position, common sense says the Thread should be Closed, as most likely agreement will never be reached.

E

I understand how frustrating this can be, but those involved have been polite and I believe there is more to learn from discussion. As long as the discourse is civil, I would ask that the thread continue so we can thoroughly dissect each question. Thanks.

 

[Claude Abraham]

Hello Claude A.

Before answering - at least, I will try it - your posts#50 and #54. I like to present a feedback model in form of a closed control loop.
(I hope Eric will not close the thread before posting; creating "further arguments" is not the worst thing, is it? Is an "agreement" necessary to keep the thread open? )

I think, we all agree that the resistor Re provides feedback. Thus, why not create a corresponding control loop that can visualize what´s going on?
I have prepared a very rough (somewhat simplified) drawing with ic~ie, which shows two cases:

A) Control loop based on a voltage control model of the BJT with the corresponding loop gain LG(A)
B) Control loop based on current control model of the BJT - also with the corresponding loop gain LG(B).

Please see the attached pdf document.

Comment to case B): For a current control model I have found no other method than to build the difference of two currents through the base region. Other methods (if they exist) are welcome.
In this context I have used a base current ib2 driven by the emitter voltage Ve. The value of ib2 is determined by the input resistance seen at the emitter node, which is re=1/g. Thus: ib2=Ve/re=Ve*g

Result: When we compare the loop gain LG(A)=-g*Re with

*the known expressions for the amplifier input resistance at the base node rin=rbe*(1+g*Re), and
*the known expression for gain (with feedback) G=-g*Rc/(1+g*Re)

it can be seen that only the voltage control model (A) confirms the theoretical expressions (which are confirmed by measurements).
(Correction: Please read (1+g*Re) in the last line instead of (1-g*Re) )

In contrast, the loop gain for the current control case is larger by a factor beta, which is NOT in accordance with measurements.
Did I make a severe error?

Winterstone

In the foregoing, I have assumed that you know how the loop gain modifies the input resistance and the gain
*rin is increased by a factor (1-loop gain)
*gain is reduced by a factor (1-loop gain).

Your errors are as follows. In "current model" you show ib as the sum of "ib1" and "ib2". But there is only one ib provided by voltage source Vb. So ib times beta gives ic, which is correct. Then ic feeds back through Re producing "Ve". You then multiply "Ve" by "g' (gm?) to obtain ib. I don't understand, "Ve" is just the voltage on the emitter measured against ground. Ve is not Vbe. Multiplying Ve by "g' makes no sense. Also why is ib separated into 2 components. My paper covers the stage voltage gain with Re feedback down to very fine detail. I ask that all examine it then reply. BR.

 

[Claude Abraham]

current gain and transconductance block diagrams withRe feedback

The block diagram using hfe and gm with Re feedback was brought up by Winterstone, so I modeled them both and attached in pdf. It confirms the interactive nature of this issue. The transconductance times vbe equals ic which is hfe times ib. In small signal mode of operation, the b-e junction is simplified to a resistor r_pi. Thus a resistor is a simple Ohm law linear relation.

Also, r_pi determines vbe since Re forms a voltage divider. To compute vbe from gm, we need r_pi and hfe. One cannot model a bjt device using gm only ignoring hfe (beta). The 2 models are consistent with all known measured data.

For the record, a bjt amp stage amplifies a small input signal producing an output whose current and voltage are both greater than input. For an output current swing, some base current swing is needed, hence hfe denotes current gain. Also, some vbe swing is needed since transconductance is not infinite. This is gm. The degenerative negative feedback action by Re must be accounted for in both current gain and transconductance calculations.

My block diagrams show no discrepancy between well known bjt parameters, stage modeling, and measured results. If I've erred I'll accept correction, please review and comment. BR.