How does a transistor amplify current or voltage?

[Ratchit]

MrAl,

You mean does not clarify 'your' discussion?

Our discussion. I am not participating in this thread by myself.

I talked about feedback systems because if you want to use that transistor in a feedback system then you might have to carefully consider its transfer function, and if you consider its transfer function to be a constant gain (or even slightly variable) the feedback system might oscillate.

True, but not pertinent to the topic of the discussion.

You want to talk about something more basic, but isnt that what we are doing? Capacitance is basic because it is everywhere between any two points. You seem to want to say that the basic diode (equation) is more basic to the decision to decide on control than capacitance is. I guess i dont quite understand why you would want to do that, except maybe to describe the diode alone and add the capacitance later as needed. Sort of like having a light bulb (which would also have some capacitance) you want to describe the basic behavior and add the small capacitance later as it is determined to be for that particular bulb (if it is even needed at all later).
I guess i could understand that, but then we seem to be back to where we have an equation and it can be solved for either variable I or V. You would have to show that there is some physical basis for your bias of voltage control over current control. I just dont see how you can do this given that both field and charge seem to exist at the same time. Is it explainable in terms of quantum physics perhaps?

We are not discussing capacitance. Perhaps you could start another different thread?

Perhaps a diagram would help here. You can draw a picture of what you say is happening and why it appears to be voltage controlled. You seem to be very convicted in your point of view so it may help a lot to do this and i would like very much to see this.
It doesnt always help to show some professors point of view because sometimes they have their biases too, but if you would like to show a whole bunch of scientists have that same point of view that would have more weight here. I once posted an argument backed by many professors and many people still didnt believe

A diagram would explain a point of view, but not prove it. A discussion with credible references can convince someone.

Ratch

 

[BrownOut]

Now we apply an external Vbe. The metallurgical junction is not shielded from Vbe. This lowers the depletion barrier voltage across the junction by opposing the Vbi. Many sources confirm that Vbe lowers the Vbi. Vbi has not changed, that is always constant except for a temperature constant. When the depletion barrier voltage decreases, then the junction width decreases and more charge flows etc. So it is the external Vbe which causes the current to change. Now, as the prof suggested, you could Theveninize the voltage into a current source, and claim it be current controlled, but that is pretty lame, because the Shockley equation uses Vbe as its independent variable.

Except Sedra and Smith directl contradicts this:

This results in majoriy carriers being supplied

These majority carriers will neutralize some fo the uncovered bound charge

Thus the depletion layer narrows and the depletion barrier volteage reduces

According to Sedra and Smith, the junction voltage, and thus vbe, is changed as a result of current. External voltage cannot change the barrier voltage. It only changes when charge neutrailzes uncovered charge, per Sedra and Smith. If you have any other theory that proves otherwise, then reference it. In the absense of a reliable reference, then Sedra and Smith prevails.

The barrier voltage decreases because Vbe opposes the always present Vbi.

One voltage cannot decrease another voltage. The barrier voltage decreases when charge in the depletion layer is neutalized. per Sedra and Smith.

What is absurd? Please explain. The junction potential is Vbi-Vbe.

The junction poetential is vbe. The only voltage that exists at the junction is vbe and is equal to contact voltage minus the barrier potential, which is decreased by current. vbe does not exist outside the B-E junction.

 

[Ratchit]

Brownout,

Except Sedra and Smith directl contradicts this:

This results in majoriy carriers being supplied

Without a doubt, majority carriers do get supplied.

These majority carriers will neutralize some fo the uncovered bound charge

Certainly, no problem there. And that will reduce the depletion region.

Thus the depletion layer narrows and the depletion barrier volteage reduces

Yep.

According to Sedra and Smith, the junction voltage, and thus vbe, is changed as a result of current. External voltage cannot change the barrier voltage. It only changes when charge neutrailzes uncovered charge, per Sedra and Smith. If you have any other theory that proves otherwise, then reference it. In the absense of a reliable reference, then Sedra and Smith prevails.

No, S & S says the depletion barrier voltage reduces, They do not say Vbe changes due to what happens in a transistor. Vbe is the applied external voltage independent of the transistor. The depletion barrier voltage is Vbi-Vbe.

One voltage cannot decrease another voltage. The barrier voltage decreases when charge in the depletion layer is neutalized. per Sedra and Smith.

Why of course it can. The back voltage of an energizing capacitor is an example of a opposing voltage. S&S are correct, it will decrease to the value of Vbi-Vbe. Only an external source controls Vbe.

The junction poetential is vbe. The only voltage that exists at the junction is vbe and is equal to contact voltage minus the barrier potential, which is decreased by current. vbe does not exist outside the B-E junction.

No, the junction potential voltage is Vbi-Vbe. Vbe exists at the b-e terminals, not at the junction. Why of course Vbe exists outside of the junction. It exists at the b-e terminals.

Ratch

 

[BrownOut]

No, S & S says the depletion barrier voltage reduces, They do not say Vbe changes due to what happens in a transistor. Vbe is the applied external voltage independent of the transistor. The depletion barrier voltage is Vbi-Vbe.

vbe is not externally applied voltage. vbe is the voltage at the B-E junction. When the depletion voltage changes, then vbe changes proportionally.

Why of course it can. The back voltage of an energizing capacitor is an example of a opposing voltage.

One voltage cannot change another voltage. There is not "back voltage" on an energizing capacitor. You're thinking of inductance, and even then,it doesn't change any other voltage.

No, the junction potential voltage is Vbi-Vbe. Vbe exists at the b-e terminals, not at the junction. Why of course Vbe exists outside of the junction. It exists at the b-e terminals.

Wrong! There are no other voltage sources between the B-E terminals, and so by kirchff's law, the terminal voltage is the junction voltage. If you can a reference or a fundemental physical law that says otherwise, then let's see it. Repeating a falsehood won't make it correct.

 

[MrAl]

MrAl,



Our discussion. I am not participating in this thread by myself.



True, but not pertinent to the topic of the discussion.



We are not discussing capacitance. Perhaps you could start another different thread?



A diagram would explain a point of view, but not prove it. A discussion with credible references can convince someone.

Ratch

Hi again,

I'm just asking you to do a small diagram so i can see what exactly it is that you are talking about here. When it comes to highly technical issues a diagram almost always helps. You've taken the time to reply all these times so i would think it wouldnt take too long to do a quick diagram showing the internal transistor and what it is that you are saying is happening. You can draw what changes with the voltage, what changes with the current, etc., and that would give us a picture to work with. Doing this is very common with subjects like these. We dont just keep talking about the same thing, we make little diagrams and it shows what we are talking about.

If you dont see the need for some reason then i ask you as a personal favor perhaps to do a small but informative diagram. I'd like very much to understand your point of view and im certainly open to that point of view but i have to see it first. You can perhaps show the field and charges or whatever you think is pertinent to the discussion. Keep in mind i am only casually disagreeing right now and i am open to suggestion here, so a diagram could be all that i need to better understand your point(s).

Thanks much.

 

[tvtech]

Never mind.

 

[BrownOut]

FFS guys. I reckon I will leave this Forum. Fight with your clever views or interpretations or whatever. Sure, there are very clever people here......

I hope you mean you're leaving the thread, and not the forum. Aside form this thread, there are lots of interesting discussions going on with the forum.

 

[tvtech]

I hope you mean you're leaving the thread, and not the forum. Aside form this thread, there are lots of interesting discussions going on with the forum.

I seriously beg you Guys to leave this thread alone. Common. There are way more interesting things to do than this???

Cheers

 

[Ratchit]

MrAl,

I'm just asking you to do a small diagram so i can see what exactly it is that you are talking about here. When it comes to highly technical issues a diagram almost always helps. You've taken the time to reply all these times so i would think it wouldnt take too long to do a quick diagram showing the internal transistor and what it is that you are saying is happening. You can draw what changes with the voltage, what changes with the current, etc., and that would give us a picture to work with. Doing this is very common with subjects like these. We dont just keep talking about the same thing, we make little diagrams and it shows what we are talking about.

If you dont see the need for some reason then i ask you as a personal favor perhaps to do a small but informative diagram. I'd like very much to understand your point of view and im certainly open to that point of view but i have to see it first. You can perhaps show the field and charges or whatever you think is pertinent to the discussion. Keep in mind i am only casually disagreeing right now and i am open to suggestion here, so a diagram could be all that i need to better understand your point(s).

Any text book has all the diagrams you could desire, but here is a link that shows what I am talking about. **broken link removed**

Notice section 4.2.3 where the built-in voltage is calculated, and section 4.2.4 where the barrier voltage is shown to be Vbi-Vbe. Also some flatband diagrams are shown

Ratch

 

[Ratchit]

Brownout,

vbe is not externally applied voltage. vbe is the voltage at the B-E junction. When the depletion voltage changes, then vbe changes proportionally.

Certainly, Vbe is the externally applied voltage. The b-e terminals are just about as external as it gets. That is where the e-b joins the outside world and Vbe is applied. The barrier depletion voltage is Vbi-Vbe. That is a subtractive relationship, not a proportional one.

One voltage cannot change another voltage.

Correct, no voltage changes the built-in voltage Vbi. But Vbi is part of the barrier depletion voltage along with Vbe. So Vbe can change the barrier depletion voltage by subtracting itself from Vbi.

There is not "back voltage" on an energizing capacitor.

When a capacitor energizes with a voltage source, the separation of charges causes a voltage that opposes the energizing voltage. That is the back voltage.

You're thinking of inductance, and even then,it doesn't change any other voltage.

When a coil is energized by a voltage source, the magnetic induction causes a voltage that opposes the energizing voltage. That is the back voltage.

Wrong! There are no other voltage sources between the B-E terminals, and so by kirchff's law, the terminal voltage is the junction voltage. If you can a reference or a fundemental physical law that says otherwise, then let's see it. Repeating a falsehood won't make it correct.

Yes, there are two sources. The built-in voltage and the diffusion voltage. I showed you that the Vbi is determined only by the doping concentrations and a temperature sensitive constant. It cannot be changed externally without destroying the BJT, except a little by temperature change. And Vbi is constantly present across the barrier depletion region since the instant the BJT was manufactured. At rest, it is the equilibrium of Vbi with the diffusion voltage, and that makes it invisible to the b-e terminal. K's law is not violated.

Ratch

 

[MrAl]

MrAl,



Any text book has all the diagrams you could desire, but here is a link that shows what I am talking about. **broken link removed**

Notice section 4.2.3 where the built-in voltage is calculated, and section 4.2.4 where the barrier voltage is shown to be Vbi-Vbe. Also some flatband diagrams are shown

Ratch

Hi again,


I was hoping you would draw something up to help explain your point, but you instead posted a link, which is still nice i guess, and i'll take a better look at this later or tomorrow, but i would like to see your own way of describing it, and i'd even be willing to help draw it if need be.
After taking a quick look (better look later or tomorrow) i found the following in 4.2.2:

START_QUOTE
We call this region the depletion region, extending from x = -xp to x = xn. The charge due to the ionized donors and acceptors causes an electric field, which in turn causes a drift of carriers in the opposite direction.
END_QUOTE

They are stating that the charge causes an electric field, which in turn etc., etc., but that's interesting isnt it?

I'll take a better look asap. In the mean time if you feel like drawing something up yourself or you want to describe to me what to draw i will draw it. If you feel more comfortable describing this privately, PM me as needed.

Thanks again.

 

[Ratchit]

MrAl,

I was hoping you would draw something up to help explain your point, but you instead posted a link, which is still nice i guess, and i'll take a better look at this later or tomorrow, but i would like to see your own way of describing it, and i'd even be willing to help draw it if need be.
After taking a quick look (better look later or tomorrow) i found the following in 4.2.2:

Why is it better if I draw it than the link which does a much better job?

START_QUOTE
We call this region the depletion region, extending from x = -xp to x = xn. The charge due to the ionized donors and acceptors causes an electric field, which in turn causes a drift of carriers in the opposite direction.
END_QUOTE

They are stating that the charge causes an electric field, which in turn etc., etc., but that's interesting isnt it?

I guess so. It is basic physics.

I'll take a better look asap. In the mean time if you feel like drawing something up yourself or you want to describe to me what to draw i will draw it. If you feel more comfortable describing this privately, PM me as needed.

I think I will stick to discussing it only.

Ratch

 

[BrownOut]

Certainly, Vbe is the externally applied voltage. The b-e terminals are just about as external as it gets. That is where the e-b joins the outside world and Vbe is applied. The barrier depletion voltage is Vbi-Vbe. That is a subtractive relationship, not a proportional one.

The junction voltage is the same as at the B and E terminals. The terminals are directly connected to the junction, and no other voltages besides the junction voltage is present. vbe is the differenced between barrier voltage and contact voltage, and as S&S points out, is changed by current. vbe in internal, not external. m In other words:

vbe = vbi - vbarrier ( which is changed by emitter current, per Sedra and Smith )

Correct, no voltage changes the built-in voltage Vbi. But Vbi is part of the barrier depletion voltage along with Vbe. So Vbe can change the barrier depletion voltage by subtracting itself from Vbi.

Wrong. vbe doesn't change anything, but is the change in the barrier voltage. No external voltage can be measured between the B and E terminals. There is only junction voltage, which changes with barrier voltage. There are no other sources. Junction voltage is the contact voltage minus barrier voltage. I've asked for a reference that shows othewise, you have none.

When a capacitor energizes with a voltage source, the separation of charges causes a voltage that opposes the energizing voltage. That is the back voltage.

NO, it's just the capacitor voltage, and it doesn't change any other voltage. There is no back voltage.

When a coil is energized by a voltage source, the magnetic induction causes a voltage that opposes the energizing voltage. That is the back voltage.

Which changes no other voltage.

Yes, there are two sources. The built-in voltage and the diffusion voltage

Together, they make up the junction voltage, which is the same as vbe. There are no other sources. vbe is measured directly the junction and is the junction only.

And Vbi is constantly present across the barrier depletion region since the instant the BJT was manufactured. At rest, it is the equilibrium of Vbi with the diffusion voltage, and that makes it invisible to the b-e terminal. K's law is not violated

K's law is not violated because vbe is the same as junction voltage. There are no other sources. If Vbi never changes, and barrier voltage changes due to current, the resultant change is reflected in vbe. So, once again, junction voltage IS vbe. Nothing has shown otherwise. Thus, according to Sedra and Smith, emitter current changes junction voltage, and vbe. S&S also states

thus the diffusion current increases until equilibruim is achieved with ID=IS=I, the externally supplied forward current.

ID = I, which becomes collector current in bjt's.

 

[Claude Abraham]

The fallacy in Ratchit's position is the following "when Vbe is applied, it opposes....". You haven't learned that there is no such thing as "applying Vbe". How does one increase Vbe at the terminals? Only one way exists, being the injection of charge into the mic cable, which transit through said cable as current & voltage. The separation of charges in the 2 conductors & the motion of said charges occur together. Thus voltage & current are simulataneous & inclusive. No surprise, since the characteristic impedance, Zo, of the cable is resistive.

Then the moving charges reach the b-e terminals. Ie & Ib are now established. As the charges diffuse through the base & emitter regios, Vbe', the internal b-e voltage has not had time to increase. The charges cross the depletion zone, a few don't make it, & Vbe' starts to change. But Ic has already changed. Thus Ic changes while Vbe' is commencing its increase.

Ratchit, you seem to think, that we can "apply Vbe" w/o Ie/Ib at the same time. You make a prior assumption as an axiom, that Vbe can exist on its own. Nobody can "apply Vbe" & then expect currents to follow as a result. Did you ever study transmission line theory? Are you familiar with characteristic impedance, reflections, V & I reflection coefficients, proagation speed, etc. You accept as a given that one can simply "apply Vbe" to the terminals. Now let us discuss Vbi.

Vbi, the built in potential, is due to thermal energy. At any temp above 0 K, the lattice has energy due to vibrations. Quantized packets of said energy is termed "phoNons". Due to collisions with the lattice, energy from phonon activity results in valence band e- gaining energy to reach conduction band. They transit through the base & emitter regions (for the p type base, holes are the majority carrier), & cross the depletion zone. On the other side, they become minority carriers. These are separated charges, with E fields & energy. This is Vbi.

Now we wish to bias the bjt externally. If we wish to overcome Vbi, we need work. Sue sings generating mic I & V. Charges are given energy & transit through the mic cable, & I = V/Zo. They arrive at the bjt b-e terminals. Vbe has not changed 1 iota, yet Ie & Ib are now increased. The energized charges transit through the b-e jcn. Then Vbe will commence to increase, while Ic has already changed.

Thus the sequence of events is very clear. Sue's energy is transduced into electrical I & V by the mic, resulting in charge motion towards the mic, I & V. Both Ib & Ie arrive at the b-e terminals before Vbe has even begun to change.

The "opposition" to Vbi, is due to the increased I & V from the mic element, due to Sue. Ie/Ib arrive at the b-e terminals, Ic increases as the minority carriers in the depletion zone begin increasing as well as Vbe. Vbe does not control Ic at all, not by a country mile. You keep asserting "when Vbe is *applied*", etc. If you've ever studied t-lines, you would know that we can never apply a current alone, or a voltage alone, because the t-line has finite characteristic impedance.

The phrase "apply a current", or "apply a voltage" is colloquial. Most real world independent power/signal generators are built for constant voltage. So "apply a voltage" is synonomous with "energize using a constant voltage source". The current will be determined by the t-line Zo.

"Apply a voltage" is meaningless in the literal sense. Whether you realize it or not, the instant Vbe (external or internal) begins to change, the currents Ib/Ie have already been changing. It is not Vbe that is responsible for Ic changing.

 

[MrAl]

MrAl,



Why is it better if I draw it than the link which does a much better job?



I guess so. It is basic physics.



I think I will stick to discussing it only.

Ratch

Hello again,


Well ok then, but you can expect me to change my point of view if you dont answer my questions or at least draw a diagram and label the points of interest and what is happening in your own words.
Drawing a diagram is very common practice when people are trying to communicate in a technical way. An example is when talking about a circuit, it is much quicker and more apparent when we have a circuit schematic to look over. Think of what most textbooks would look like if they didnt have diagrams, figures, charts, tables, etc.

 

[tvtech]

Isn't that like creating something from nothing?

Hi jac4b. Welcome to ETO.

You should have a simple answer to your question shortly.

Cheers

 

[Claude Abraham]

Not "something from nothing". A bjt can provide both current gain & power gain together. But it needs a power source. Without said power source, the bjt is merely a pair of back to back diodes. Then, power gain is limited to unity.

The input power at the b-e terminals is much smaller than the output power. The power gain is provided by the power supply biasing the bjt device. Conservation of energy always holds.

 

[Ratchit]

Brownout,

The junction voltage is the same as at the B and E terminals. The terminals are directly connected to the junction, and no other voltages besides the junction voltage is present.

The junction voltage is Vbi-Vbe, as I will show below. See also below for what happens when Vbe forward biases the diode.

vbe is the differenced between barrier voltage and contact voltage, and as S&S points out, is changed by current. vbe in internal, not external. m In other words:

vbe = vbi - vbarrier ( which is changed by emitter current, per Sedra and Smith )

Vbe is changed externally. I can put a voltage source in the b-e terminals, and lock the voltage so the transistor cannot change it. Your equation above shows that vbarrier has to change to accomodate Vbe, because Vbi does not change at constant temperature once the transistor is manufactured.

Wrong. vbe doesn't change anything, but is the change in the barrier voltage. No external voltage can be measured between the B and E terminals. There is only junction voltage, which changes with barrier voltage. There are no other sources. Junction voltage is the contact voltage minus barrier voltage. I've asked for a reference that shows othewise, you have none.

You cannot say Vbe does not change anything, it cancels some of the barrier voltage. Here is a quote from the textbook Semiconductor Circuit Analysis, by Phillip Cutler, p 19. I only deleted references to a diagram which shows a diode in series with a variable voltage power supply.

"At first I slowly increases, because because the applied voltage V has not sufficiently reduced the inherent potential barrier. As we increase V further, say to a few tenth of a volt, we find that I rapidly increases because V has finally become large enough to reducxt significantly the internal barrier potential and cause the recombination current to become quite large."

As you can see, this text believes the applied voltage reduces the barrier voltage, not the other way around.

Next is a quote from a textbook Integrated Electronics: Analog and Digital Circuits and Systems, by Millman and Taub, p53.

"For such a diode, the height of the potential barrier at the junction will be lowered by the applied forward voltage V. The equilibrium initially established between the forces tending to produce diffusion of majority carriers and the restraining influence of the potential-energy barrier at the junction will be disturbed."

So this text too, believes that Vbe cancels some of the barrier voltage.

And finally look at the link I referenced in my message to MrAl, repeated here for your convenience. **broken link removed** . The last sentence is:

"The total potential across the semiconductor equals the built-in potential minus the applied voltage, or:**broken link removed**
"
The equation means junction voltage = Vbi-Vbe

By the way, the reason S & S always talk about current is because they are analyzing the BJT with a current source. But it is still the Vbe that determines what the junction will be to support the current source.

NO, it's just the capacitor voltage, and it doesn't change any other voltage. There is no back voltage.

Read my statement again. I said it opposes the energizing voltage. That is what a back voltage does.

Which changes no other voltage.

Same answer as above.

Together, they make up the junction voltage, which is the same as vbe. There are no other sources. vbe is measured directly the junction and is the junction only.

K's law is not violated because vbe is the same as junction voltage. There are no other sources. If Vbi never changes, and barrier voltage changes due to current, the resultant change is reflected in vbe. So, once again, junction voltage IS vbe. Nothing has shown otherwise.

Wrong twice. Look again at the quotes and especially the link. The link says the junction voltage is Vbi-Vbe, so Vbe is not the junction voltage.

Thus, according to Sedra and Smith, emitter current changes junction voltage, and vbe. S&S also states

thus the diffusion current increases until equilibruim is achieved with ID=IS=I, the externally supplied forward current.

ID = I, which becomes collector current in bjt's.

As I said before S & S analyze the BJT with a constant current source, but that does not change Vbe controlling the forward current. I should not have to proofread your post, and I am sure it is just a typo. But the equation for forward current should be Id-Is=I, which means diffusion current minus thermal current equals the forward current. The equation would be correct for reverse current, however.

Ratch

 

[Ratchit]

Claude,

The fallacy in Ratchit's position is the following "when Vbe is applied, it opposes....". You haven't learned that there is no such thing as "applying Vbe". How does one increase Vbe at the terminals? Only one way exists, being the injection of charge into the mic cable, which transit through said cable as current & voltage. The separation of charges in the 2 conductors & the motion of said charges occur together. Thus voltage & current are simulataneous & inclusive. No surprise, since the characteristic impedance, Zo, of the cable is resistive.

I can easily put a voltage source on the b-e terminals. Wouldn't that be applying Vbe? What are you getting into distributed impedances and cables?

Ratchit, you seem to think, that we can "apply Vbe" w/o Ie/Ib at the same time. You make a prior assumption as an axiom, that Vbe can exist on its own. Nobody can "apply Vbe" & then expect currents to follow as a result. Did you ever study transmission line theory? Are you familiar with characteristic impedance, reflections, V & I reflection coefficients, proagation speed, etc. You accept as a given that one can simply "apply Vbe" to the terminals. Now let us discuss Vbi.

I think I made my position on clear before, that whether there is a lag or lead with Ie/Ib, it does not matter when determining what controls what. Why are you thowing out things that are best measured with a time domain reflectometer (TDR). What we are discussing is not that involved.

Vbi, the built in potential, is due to thermal energy. At any temp above 0 K, the lattice has energy due to vibrations. Quantized packets of said energy is termed "phoNons". Due to collisions with the lattice, energy from phonon activity results in valence band e- gaining energy to reach conduction band. They transit through the base & emitter regions (for the p type base, holes are the majority carrier), & cross the depletion zone. On the other side, they become minority carriers. These are separated charges, with E fields & energy. This is Vbi.

Yes, the uncovered charges caused by diffusion.

Now we wish to bias the bjt externally. If we wish to overcome Vbi, we need work. Sue sings generating mic I & V. Charges are given energy & transit through the mic cable, & I = V/Zo. They arrive at the bjt b-e terminals. Vbe has not changed 1 iota, yet Ie & Ib are now increased. The energized charges transit through the b-e jcn. Then Vbe will commence to increase, while Ic has already changed.

It takes energy to lower the built-in voltage Vbi because some charge carriers pass through the base terminal. As I said before, Vbe still controls the current, because I can set Vbe with a voltage source, and the current will have to comply according to the Shockley diode equation.

Thus the sequence of events is very clear. Sue's energy is transduced into electrical I & V by the mic, resulting in charge motion towards the mic, I & V. Both Ib & Ie arrive at the b-e terminals before Vbe has even begun to change.

The "opposition" to Vbi, is due to the increased I & V from the mic element, due to Sue. Ie/Ib arrive at the b-e terminals, Ic increases as the minority carriers in the depletion zone begin increasing as well as Vbe. Vbe does not control Ic at all, not by a country mile. You keep asserting "when Vbe is *applied*", etc. If you've ever studied t-lines, you would know that we can never apply a current alone, or a voltage alone, because the t-line has finite characteristic impedance.

Yes, Vbe does control the current by opposing the Vbi. That is what every good textbook I read say happens. Whether it takes energy or current or voltage come before or after is not relevant. The quantity of Vbe determines the quantity of Ie/Ic.

The phrase "apply a current", or "apply a voltage" is colloquial. Most real world independent power/signal generators are built for constant voltage. So "apply a voltage" is synonomous with "energize using a constant voltage source". The current will be determined by the t-line Zo.

Colloquial or not, everyone knows what it means. Distributed impedances are not necessary for steady state dc.

"Apply a voltage" is meaningless in the literal sense. Whether you realize it or not, the instant Vbe (external or internal) begins to change, the currents Ib/Ie have already been changing. It is not Vbe that is responsible for Ic changing.

But Vbe will determine what their steady state will be.

Ratch

 

[Ratchit]

MrAl,

Well ok then, but you can expect me to change my point of view if you dont answer my questions or at least draw a diagram and label the points of interest and what is happening in your own words.
Drawing a diagram is very common practice when people are trying to communicate in a technical way. An example is when talking about a circuit, it is much quicker and more apparent when we have a circuit schematic to look over. Think of what most textbooks would look like if they didnt have diagrams, figures, charts, tables, etc.

You should change your viewpoint or not on the merits of my arguments, not on whether I draw things. It is a good thing I am not a textbook writer.

Ratch