How does a transistor amplify current or voltage?

[Ratchit]

Claude,

But how does Vbe relate to being "the rudder". Vbe exists because p-n jcns are not ideal. An ideal revtifying jcn would be open in reverse, short in forward direction. If a semiconductor w/ zero bandgap energy existed, charges entering the emitter would transit through the base onward to the collector. The no. of e- collected is controlled by the no. of e- emitted. After Ib/Ie are changed by Sue, Ic increases as a result, but a little later, Vbe increases. But Vbe increasing is not the control mechanisn. In order to supply more electrons for the emitter to emit, work must be done. When Sue cranks it up, it is the increased current/voltage from the mic resposible for Ie/Ib increasing. As a consequence, Vbe increases after the fact, but Ic has already begun to increase w/o Vbe increasing yet.

Vbe exists to lower the barrier voltage caused by the uncovered charges, which in turn are caused by the diffusion of electrons from the n-material and holes from the p-material. An ideal diode cannot exist as a bipolar junction because of this barrier voltage. I already acknowledged that energy is involved from a outside source in creating and maintaining Vbe. But this does not detract from the fact that it is the Vbe voltage that will control Ib and Ic, because Vbe is what lowers the barrier voltage. It also ties in with the equations from Sedra and Smith. And any claim about what occurs first does not determine what is the control

The intermediate variable at the jcn is Ie. Although Ib & Vbe change as well, only Ie contributes to Ic. Ib & Vbe are necessary because perfect semiconductors cannot be produced. The Ie is not controlled by the Vbe. The eqns you keep referencing are functional relations, not causal.

Start w/ the n-p-n as a pair of back to back diodes. The upper diode is the b-c jcn, described by Shockley's diode eqn
Ic = Ics*exp((Vbc/Vt)-1). Since Vbc is negative, a very small Ic exists, leakage of the reverse biased p-n b-c jcn.

The lower p-n jcn, b-e is forward biased. So we have
Ie = Ies*exp(Vbe/Vt)-1), a large current since Vbe is positive.

If the base region was very wide, say 1.0 mm, the above relations are what we get. Ie = Ib, a large value, w/ Ic being a small leakage.

But if the base region is made ultra thin, say 1.0 um, we find that Ic measures almost as large as Ie, w/ Ib being quite small, Ib = Ie-Ic. Instead of 2 mere series diodes, we have transistor action. The b-c jcn is reverse biased, but Ic is almost Ie. The Shockley eqn needs another term. The additional current is Ic = alpha*Ie. So we get

Ic = alpha*Ies*exp((Vbe/Vt)-1) + Ics*exp((Vbc/Vt)-1).

But sonce the 2nd term is much smaller than the 1st, we round off & use Ic = alpha*Ies*exp((Vbe/Vt)-1). Also, alpha is very near to unity, so some texts omit alpha, w/ the resulting eror being 1 to 2%.

I don't follow your reasoning, and why you insist on postulating about perfect semiconductors. Bipolar transistors will never be perfect due to their barrier voltage. The collector exists to isolate the output from the input. Without the collector, the output voltage would intefere with the input signal. Yes, the base if made thin so as to not capture any more base current that necessary, and any it does catch is wasted.

The Vbe eqn, which I've called eqn 2), is derived from Shockley's diode eqn, plus the eqn 3), the transistor action eqn. Ie & Vbe do have a functional relationship per eqn 2). But Vbe is directly related to Ib & Ie, but indirectly related to Ic. Ie has a direct relation w/ Ic per eqn 3).

In order for Ic to change, all that is needed is a change in Ie. In the process, Ib & Vbe change as well, but that is not what changes Ic. Ic starts increasing prior to Vbe increasing, & Vbe continues to increase after Ic has settled

Sedra and Smith say that Ic is exponentially related to Vbe. The reason Vbe exists is to lower the barrier voltage. That is the basis for regarding it as a VC device with respect to causality.

No brainer at all. Transistor action is Ic = alpha*Ie. Any test w/ instruments will affirm. A good software simulator might show this, but the models have to be really accurate. Low end packages won't work.

That relationship is true, but it is not the causing factor.

Ratch

 

[BrownOut]

Vbe controls Ic and thereby Ie as stated in Sedra and Smith.

It is not, the Vbe at the junction controls Ic as stated in Sedra and Smith.

Sedra and Smith have done so.

Your explanations do not encompass the equations given in Sedra and Smith.

Here is what Sedra and Smith says in the section "Basic Semiconductor Concepts" :

The electrons carrying the current I in the external circuit from the P material to the n material cause holes to be extracted from the n region and electrons to be extracted from the p region. The results in majoriy carriers being supplied to both sides fo the juntion by external circuit: holes to the p material and electrons to the n material. These majority carriers will neutralize some fo the uncovered bound charge, causing less charge to be stored in the depletion layer. Thus the depletion layer narrows and the depletion barrier volteage reduces. ... thus the diffusion current increases until equilibruim is achieved with ID=IS=I, the externally supplied forward current.

Then why did you name five people what agree with you?

I didn't name anyone. I only said there is consensus, with at least 5 people agreeing.

 

[Ratchit]

MrAl,

I was not saying that causality is impossible, i was saying that to prove it we would have to go back to the big bang to do it.

For a transistor, no.

Sure there are extra charges in a transistor that have to be cleared before the signal begins to control the output. Sure there is a delay so that things do not happen simultaneously. But all that is irrelevant. That is not what is determining causality. Vbe is determining Ic according to the equations in Sedra and Smith, and the time order is not a factor. Neither is storage time and internal capacitance.

The base emitter voltage of one transistor can be 0.6v and the other transistor 0.65v but if the current gains are the same 10ma will produce the same collector current in both transistors.

The equations in Sedra and Smith show that different BJT's will have different Ic's for the same Vbe. The doping concentration and other factors may not be the same for each BJT.

... and if we view that potential as external (ie we entered the picture at some late (t,x,y,z)) we might claim voltage control.

Remember, time is not a factor in casuality. The effect of the cause is.

Ratch

 

[Ratchit]

Brownout,

Here is what Sedra and Smith says in the section "Basic Semiconductor Concepts" :

The electrons carrying the current I in the external circuit from the P material to the n material cause holes to be extracted from the n region and electrons to be extracted from the p region. The results in majoriy carriers being supplied to both sides fo the juntion by external circuit: holes to the p material and electrons to the n material. These majority carriers will neutralize some fo the uncovered bound charge, causing less charge to be stored in the depletion layer. Thus the depletion layer narrows and the depletion barrier volteage reduces. ... thus the diffusion current increases until equilibruim is achieved with ID=IS=I, the externally supplied forward current.

I have no problem with that statement. It is not in conflict with what I said. Vbe lowers the barrier voltage and causes some of the N and P charges to neutralize each other, thereby causing the barrier to become thinner and allow the voltage from the collector to sweep the charges through the base and into the collector output circuit. The fact that the current, voltage and energy is supplied externally does not alter the causality. Invariably, a small amount of negative charges get swept into the base, but they are a result of Vbe.

My Smith & Sedra is Microelectronic Circuits, third edition, p. 196. Under section "Operation of the NPN transistor in the active mode", subsection "The collector current", there is the equation Ic = Is*e^Vbe/Vt , where Is and Vt are considered constants. There is a similar equation for Ib. That tells me that Vbe is controlling the transistor, making it a VC device from a causality viewpoint.

I didn't name anyone. I only said there is consensus, with at least 5 people agreeing.

Fair enough, so you did. I am sure there are a lot more, for what it's worth.

Ratch

 

[BrownOut]

I have no problem with that statement. It is not in conflict with what I said. Vbe lowers the barrier voltage and causes some of the N and P charges to neutralize each other, thereby causing the barrier to become thinner and allow the voltage from the collector to sweep the charges through the base and into the collector output circuit. The fact that the current, voltage and energy is supplied externally does not alter the causality. Invariably, a small amount of negative charges get swept into the base, but they are a result of Vbe.

The statement says that current flows into the junction and nutralizes the charge. It says nothing about vbe. Note the bolded words, all about current control.

The electrons carrying the current I in the external circuit from the P material to the n material cause holes to be extracted from the n region and electrons to be extracted from the p region. This results in majoriy carriers being supplied to both sides fo the juntion by external circuit: holes to the p material and electrons to the n material. These majority carriers will neutralize some fo the uncovered bound charge, causing less charge to be stored in the depletion layer. Thus the depletion layer narrows and the depletion barrier volteage reduces. ... thus the diffusion current increases until equilibruim is achieved with ID=IS=I, the externally supplied forward current.

My Smith & Sedra is Microelectronic Circuits, third edition, p. 196. Under section "Operation of the NPN transistor in the active mode", subsection "The collector current", there is the equation Ic = Is*e^Vbe/Vt , where Is and Vt are considered constants. There is a similar equation for Ib. That tells me that Vbe is controlling the transistor, making it a VC device from a causality viewpoint.

That equation only relates current and terminal voltage, how and where control takes place is in the physical section that I quoted, and it talks only of current control.

 

[Ratchit]

Brownout,

The statement says that current flows into the junction and nutralizes the charge. It says nothing about vbe. Note the bolded words, all about current control.

The electrons carrying the current I in the external circuit from the P material to the n material cause holes to be extracted from the n region and electrons to be extracted from the p region. This results in majoriy carriers being supplied to both sides fo the juntion by external circuit: holes to the p material and electrons to the n material. These majority carriers will neutralize some fo the uncovered bound charge, causing less charge to be stored in the depletion layer. Thus the depletion layer narrows and the depletion barrier volteage reduces. ... thus the diffusion current increases until equilibruim is achieved with ID=IS=I, the externally supplied forward current.

No, not about current control. It is about the effects of current released or held back by Vbe. The Sedra equations show that Vbe controls Ic.

That equation only relates current and terminal voltage, how and where control takes place is in the physical section that I quoted, and it talks only of current control.

That terminal voltage goes across the emitter base junction and controls the barrier voltage. The barrier voltage controls the emitter current. Most of the emitter current goes to the collector, but a small part is wasted in the base. That is the chain of causation, voltage controls current. But for reasons stated many times, the design and calculation of currents in BJT's should be done by functionally assuming that Ib controls Ic.

Ratch

 

[Claude Abraham]

This Vbe is causal is a religion/cult. To say that Vbe lowers the barrier voltage is circular reasoning. Would you please define "barrier voltage". How does it differ from Vbe? Seriously you make no sense. What is the eqn for barrier voltage? What is the mechanism? How does Vbe play a role?

The barrier is due to charge build up. The Vbe is due to minority carrier concentration in the base region. This is the barrier. The voltage which opposes this barrier, not lower it, is that of the mic output, as well as the current. How can "Vbe lower the barrier voltage". You're implying that Vbe causes the barrier voltage to lower, then Ib/Ie changes. But how do you affirm w/ measuring instruments. I can probe Vbe, but how do I probe "barrier voltage"? Define barrier voltage in terms of E field & path.

You throw out vague terms & concepts not found in textbooks then offer that as proof. The Ebers-Moll eqn, E-M, does not make the bjt VC. Drs. E & M in their 1954 paper, explicitly illustrated the bjt model w/ current controlled current sources. The exponential eqn you keep re-iterating is E-M. The authors of said eqn explicitly declared the bjt as CC.

Ib/Ie change ahead of Vbe, so there goes your "Vbe lowers the barrier voltage so that current can change". What part of Ib/Ie change before Vbe do you not understand? Causes can not lag their efects. That alone demolished your theory. I cannot believe that you believe your own position. You are simply unwilling to believe that you have it wrong. Your arguments are beyond anything I've ever seen. You said you're retired. What EE field did you work in? Analog, digital, software, power, field apps, manufacturing, quality, testing, optical, emc, ? Any patents or publications?

Ideal p-n junctions would need no forward voltage, & leak no reverse current. Emitter electrons would reach the collector w/o Vbe. Real world devices have band gap energy requirements making some forward drop, Vbe, inevitable. So in the real world, Vbe cannot be neglected, it is a stark reality. Because of Vbe, a bjt can only offer a finite value of gm. Thus the gm of the stage is less than the gm of the bjt. Ditto for current gain. Although Ib & Vbe do not contribute to Ic, they are indispensible quantities for real world non-ideal devices, & must be accounted for. So the E-M eqn , my ref eqn 2), gives us voltage gain limits, & eqn 1) gives us that for current gain.

Again, all 3, Ib/Vbe/Ie, are necessary, but Ie is what ultimately determines Ic final value. But Ie wouldn't exist w/o Ib/Vbe, so we have a vicious circle. Causality is nothing but trouble, & those who invoke it do so only to be trolls. It is impossible to settle a causality argument w/o going back to Genesis or the big bang, depending on what you believe.

 

[BrownOut]

No, not about current control. It is about the effects of current released or held back by Vbe. The Sedra equations show that Vbe controls Ic.

It only refers to current. Says absolutely nothing about vbe. Lets' try again:

The electrons carrying the current I in the external circuit from the P material to the n material cause holes to be extracted from the n region and electrons to be extracted from the p region

Blah blah, nothing about vbe, all about current. Current nutralizes the uncovered charge, and that lowers the barrier voltage:

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

There's you causal effect, and it's due to current. So says Sedra and Smith.

That terminal voltage goes across the emitter base junction and controls the barrier voltage.

Once again from Sedra and Smith:

These majority carriers will neutralize some fo the uncovered bound charge

Thus the depletion layer narrows and the depletion barrier volteage reduces

This shows that current is directly responsible for reducing the barrier voltage. This is directly quoted from Sedra and Smith ( except for some unavoidable typos) It proves current control over the barrier voltage and thus vbe. If you continue to write otherwise, it doens't change what Sedra and Smith directly says about it.

That is the chain of causation, voltage controls current

Not according you your favorite reference, Sedra and Smith:

The electrons carrying the current I in the external circuit from the P material to the n material cause holes to be extracted from the n region and electrons to be extracted from the p region.

See the big word? CAUSE, as in causal. There is your proof of causal.

 

[Ratchit]

Brownout,

It only refers to current. Says absolutely nothing about vbe. Lets' try again:

The electrons carrying the current I in the external circuit from the P material to the n material cause holes to be extracted from the n region and electrons to be extracted from the p region

Blah blah, nothing about vbe, all about current. Current nutralizes the uncovered charge, and that lowers the barrier voltage:

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

There's you causal effect, and it's due to current. So says Sedra and Smith.

I maintain they are talking about the effects of the current, and they do not say that current controls anything. Of course charge will flow in and reduce the barrier width when the barrier voltage is lowered. It will happen because of the application of Vbe. And it would happen even if there were no base current. That is why I and others say that the base current is an unfortunate side effect and not the true cause of the Ie. Let me say that again. If it were possible to open the base lead and somehow induce a voltage across the emitter base segment, you would still be able to control Ie with Vbe alone. The majority carriers would still be able to combine and pass through the base segment to the collector to complete the transistor action without any base current being present. Only Vbe would control Ic. I will have more to say about cause and effect below.

Once again from Sedra and Smith:

These majority carriers will neutralize some fo the uncovered bound charge

Thus the depletion layer narrows and the depletion barrier volteage reduces

This shows that current is directly responsible for reducing the barrier voltage. This is directly quoted from Sedra and Smith ( except for some unavoidable typos) It proves current control over the barrier voltage and thus vbe. If you continue to write otherwise, it doens't change what Sedra and Smith directly says about it.

Again, current control is your interpretation. I call it a current effect. And the charge is able to flow because Vbe has changed. The effect of the charge flow made possible by Vbe is to reduce the barrier width. That does not contradict the portion of Sedra that I quoted with the equations, which do say something about Vbe control.

Not according you your favorite reference, Sedra and Smith:

The electrons carrying the current I in the external circuit from the P material to the n material cause holes to be extracted from the n region and electrons to be extracted from the p region.

See the big word? CAUSE, as in causal. There is your proof of causal.

Not really. Cause is used widely to denote more than one thing. In the case of Sedra, it is used to denote cause of effect. I maintain that, for the reasons given above, Vbe is the cause of control. Two different things.

Ratch

 

[BrownOut]

I maintain they are talking about the effects of the current, and they do not say that current controls anything. Of course charge will flow in and reduce the barrier width when the barrier voltage is lowered. It will happen because of the application of Vbe.

It's plain and simple, current flows into the uncovered charges and reduces the barrier voltage. S&S says this very clearly.

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.

The reduction of the barrier voltage is reduces as a result of current. That's exactly what Sedra and Smith says,and it's the same as we've said all along.

And it would happen even if there were no base current.

That's false, and not supported by Sedra and Smith

Again, current control is your interpretation.

Nope. It's clearly shown in Sedra and Smith that current reduces the barrier voltage. The barrier voltage, and hence vbe,is controlled by the current.

The effect of the charge flow made possible by Vbe is to reduce the barrier width.

Sedra and Smith disagrees:

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

The change in barrier voltage results from the current. Current changes vbe, not the other way round, according to Sedra and Smith.

Not really. Cause is used widely to denote more than one thing. In the case of Sedra, it is used to denote cause of effect. I maintain that, for the reasons given above, Vbe is the cause of control. Two different things.

HA! Yes, it's cause and effect. The cause is current, and the effect is vbe. That's what Sedra and Smith says. Causal is causal, there are no other meanings.

 

[MikeMl]

How many angels can dance on the head of a pin?
**broken link removed**

 

[Reloadron]

How many angels can dance on the head of a pin?
**broken link removed**

Now that was really a good one.

Ron

 

[misterT]

**broken link removed**
You mean like this?

 

[MrAl]

Brownout,

I maintain they are talking about the effects of the current, and they do not say that current controls anything. Of course charge will flow in and reduce the barrier width when the barrier voltage is lowered. It will happen because of the application of Vbe. And it would happen even if there were no base current. That is why I and others say that the base current is an unfortunate side effect and not the true cause of the Ie. Let me say that again. If it were possible to open the base lead and somehow induce a voltage across the emitter base segment, you would still be able to control Ie with Vbe alone. The majority carriers would still be able to combine and pass through the base segment to the collector to complete the transistor action without any base current being present. Only Vbe would control Ic. I will have more to say about cause and effect below.

Again, current control is your interpretation. I call it a current effect. And the charge is able to flow because Vbe has changed. The effect of the charge flow made possible by Vbe is to reduce the barrier width. That does not contradict the portion of Sedra that I quoted with the equations, which do say something about Vbe control.

Not really. Cause is used widely to denote more than one thing. In the case of Sedra, it is used to denote cause of effect. I maintain that, for the reasons given above, Vbe is the cause of control. Two different things.

Ratch

Hello again,

Ok, it appears that you want to disregard the capacitance altogether. I dont feel comfortable doing this because capacitance is a part of nature, just like inductance, so i have to take the nature way of thinking about it. I was more involved with control systems so it's not like me to ignore something like capacitance, because that alters the response unquestionably, and you can not control something unless you know its H(s) and its G(s), and ultimately the capacitance will either be in H(s) or in G(s) and there's no getting around that. It's not possible to control something without knowing both H(s) and G(s) so i can not ignore capacitance.
It's like taking a system like:
H(s)=G/(s+1)
and trying to say that you can develop a feedback system for that system by only considering H(s) (of the above system) to be:
H(s)=G/1
which just can't happen for the most part unless you get lucky. If it did work out by chance, the response would not be what we calculated it to be for all time.

In any case, it appears that the model you like best is the one that you have ignored some parameters in favor of some other parameters, or rather some internal parts.

I probably shouldnt have gotten into this discussion in the first place
because I guess i dont take any extreme view on either form of control anymore because i mostly look at the transistor from the external viewpoint anyway, which means the transistor appears as some transfer function which of course has to include time. If i am going to use a transistor in a circuit then i would use the model most appropriate to the application, and ignore any theoretical views that there is a more perfect way of looking at it.
BTW if the transistor did not require a time factor in there then we would have perfect 100 percent switching and our switching regulators sure would be nice.

The number of angels that can stand on the head of a pin is equal to the area of the pin head divided by the footprint area of each angel rounded down to the nearest whole number, unless they each stand on one foot and then it's half the area of the footprint:
N=Ap/Fpa
where
N is the number of angeles
Ap is the area of the head of the pin, and
Fpa is the footprint of each angel

Since angels are spiritual, they dont have any footprint area, so Fpa tends to zero, so N becomes:
N=lim[Fpa --> 0]Ap/Fpa (read as the limit of Ap/Fpa as Fpa tends to zero)
which of course equals infinity, so the number is infinite.

 

[Ratchit]

Claude,

This Vbe is causal is a religion/cult. To say that Vbe lowers the barrier voltage is circular reasoning. Would you please define "barrier voltage". How does it differ from Vbe? Seriously you make no sense. What is the eqn for barrier voltage? What is the mechanism? How does Vbe play a role?

Not a cult, but a hard core group. Why is is circular reasoning? Barrier voltage, also called "built-in potential", also called "contact difference of potential", is the voltage caused by the uncovered charges when electrons from N-type material diffuse into P-type material, and holes from P-type material diffuse into N-type material. The barrier voltage is caused by the two types of semiconductors losing their majority carriers at their contact boundary due to diffusion, thereby creating a depletion region and leaving charged ions of opposite polarity behind. These charged ions cause a barrier voltage to appear. For what it's worth, the equation is Vbi=Vt*ln((Na*Nd)/ni^2), where Vbi=built-in voltage, Vt=0.026 volts, Na=concentration of acceptor ions, Nd=concentration of donor ions, ni^2 =n*p, where n=electron concentration and p=hole concentration. Vbe is the external applied voltage. Vbe opposes the barrier voltage, allowing more majority charge carriers to diffuse into the base segment where they are swept into the collector circuit. All this is "staple goods" found in any good semiconductor physics text. Some of the charges don't make it to the collector, and are snared by the base segment, and make up the base current. But this base current does not control Ic/Ic. Vbe does.

The barrier is due to charge build up. The Vbe is due to minority carrier concentration in the base region. This is the barrier. The voltage which opposes this barrier, not lower it, is that of the mic output, as well as the current. How can "Vbe lower the barrier voltage". You're implying that Vbe causes the barrier voltage to lower, then Ib/Ie changes. But how do you affirm w/ measuring instruments. I can probe Vbe, but how do I probe "barrier voltage"? Define barrier voltage in terms of E field & path

Yes, the barrier is due to uncovered charge buildup. No, Vbe is the applied voltage, not the barrier voltage. An opposing voltage means a net lowering. The mic input causes Vbe to change. Yes, that is what I am saying. You cannot probe the depletion region because it is only 0.5 microns, which is of the order of the wavelength of visible light. I doubt any voltmeter can measure it anyway without changing the reading. Barrier voltage depends principally on the doping concentrations of the two semiconductor types, but V=-∫ ε*dx .

You throw out vague terms & concepts not found in textbooks then offer that as proof. The Ebers-Moll eqn, E-M, does not make the bjt VC. Drs. E & M in their 1954 paper, explicitly illustrated the bjt model w/ current controlled current sources. The exponential eqn you keep re-iterating is E-M. The authors of said eqn explicitly declared the bjt as CC.

What vague terms? Remember what I said about models being good at what but not at how and why?

Ib/Ie change ahead of Vbe, so there goes your "Vbe lowers the barrier voltage so that current can change". What part of Ib/Ie change before Vbe do you not understand? Causes can not lag their efects. That alone demolished your theory. I cannot believe that you believe your own position. You are simply unwilling to believe that you have it wrong. Your arguments are beyond anything I've ever seen. You said you're retired. What EE field did you work in? Analog, digital, software, power, field apps, manufacturing, quality, testing, optical, emc, ? Any patents or publications?

I thougt I made it clear before that lag, lead, or concurrency is not a factor in determining control. Making something happen is. What difference does it make what I did? The prof has much better qualifications than I do, including may publications and patents, and you do not believe him. I stand on my arguments.

Ideal p-n junctions would need no forward voltage, & leak no reverse current. Emitter electrons would reach the collector w/o Vbe. Real world devices have band gap energy requirements making some forward drop, Vbe, inevitable. So in the real world, Vbe cannot be neglected, it is a stark reality. Because of Vbe, a bjt can only offer a finite value of gm. Thus the gm of the stage is less than the gm of the bjt. Ditto for current gain. Although Ib & Vbe do not contribute to Ic, they are indispensible quantities for real world non-ideal devices, & must be accounted for. So the E-M eqn , my ref eqn 2), gives us voltage gain limits, & eqn 1) gives us that for current gain.

The Shockley ideal diode equation assumes the current is only caused by diffusion and thermal recombination-generation (r-g). It also assumes no r-g in the depletion region. So a ideal diode will have a forward voltage drop, and a reverse saturation current even if it does not have any leakage resistance or reverse voltage breakdown. If you can find a diode with no forward voltage drop or reverse current, it cannot possibly a bipolar device due to its physics.

Again, all 3, Ib/Vbe/Ie, are necessary, but Ie is what ultimately determines Ic final value. But Ie wouldn't exist w/o Ib/Vbe, so we have a vicious circle. Causality is nothing but trouble, & those who invoke it do so only to be trolls. It is impossible to settle a causality argument w/o going back to Genesis or the big bang, depending on what you believe.

I don't believe that for a second. One can establish the control element in a device by what it does with respect to controlling the output.

Ratch

 

[Ratchit]

Brownout,

It's plain and simple, current flows into the uncovered charges and reduces the barrier voltage. S&S says this very clearly.

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.

The reduction of the barrier voltage is reduces as a result of current. That's exactly what Sedra and Smith says,and it's the same as we've said all along.

That's true as far as it goes, but without Vbe changing, the majority carriers cannot move. In other words, after the majority carriers have diffused as far as they can into each others segment, everything is locked up and nothing can move anymore. Only when Vbe lowers the barrier voltage can further diffusion take place.

That's false, and not supported by Sedra and Smith

The S & S equations support it.

Nope. It's clearly shown in Sedra and Smith that current reduces the barrier voltage. The barrier voltage, and hence vbe,is controlled by the current.

Nope, the current reduces the width of the depletion region, but it cannot do it until Vbe opposes the barrier voltage. Otherwise, how would Ie be locked up in the first place?

Sedra and Smith disagrees:

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

The change in barrier voltage results from the current. Current changes vbe, not the other way round, according to Sedra and Smith.

Read what you quoted from S & S very carefully. They say that the increased charge flow reduces the barrier voltage. I agree with that. But S & S does not say that the current controls Vbe. They cannot, because Vbe is externally controlled. So Vbe causes the majority carriers to increase, by whatever method you like. And the barrier width has to reduce or the carriers could not increase. But it was Vbe that caused the carriers to change.

HA! Yes, it's cause and effect. The cause is current, and the effect is vbe. That's what Sedra and Smith says. Causal is causal, there are no other meanings.

Vbe is externally applied, the barrier depletion voltage is internally generated. So Vbe in the casual control.

Ratch

 

[Ratchit]

MrAl,

Ok, it appears that you want to disregard the capacitance altogether. I dont feel comfortable doing this because capacitance is a part of nature, just like inductance, so i have to take the nature way of thinking about it. I was more involved with control systems so it's not like me to ignore something like capacitance, because that alters the response unquestionably, and you can not control something unless you know its H(s) and its G(s), and ultimately the capacitance will either be in H(s) or in G(s) and there's no getting around that. It's not possible to control something without knowing both H(s) and G(s) so i can not ignore capacitance.
It's like taking a system like:
H(s)=G/(s+1)
and trying to say that you can develop a feedback system for that system by only considering H(s) (of the above system) to be:
H(s)=G/1
which just can't happen for the most part unless you get lucky. If it did work out by chance, the response would not be what we calculated it to be for all time.

In any case, it appears that the model you like best is the one that you have ignored some parameters in favor of some other parameters, or rather some internal parts.

I ignored capacitance, AC, and transients because I wanted to keep the discussion simple. Already, you are submitting some transfer and feedback terms that have no relevance to this discussion. Remember, we are discussion whether a BJT is, causually speaking, a current controlled or voltage controlled device.

I probably shouldnt have gotten into this discussion in the first place
because I guess i dont take any extreme view on either form of control anymore because i mostly look at the transistor from the external viewpoint anyway, which means the transistor appears as some transfer function which of course has to include time. If i am going to use a transistor in a circuit then i would use the model most appropriate to the application, and ignore any theoretical views that there is a more perfect way of looking at it.
BTW if the transistor did not require a time factor in there then we would have perfect 100 percent switching and our switching regulators sure would be nice.

All that is fine and good, but it does not clairify the discussion.

Ratch

 

[BrownOut]

That's true as far as it goes, but without Vbe changing, the majority carriers cannot move. In other words, after the majority carriers have diffused as far as they can into each others segment, everything is locked up and nothing can move anymore. Only when Vbe lowers the barrier voltage can further diffusion take place.

According to S&S, it is current that lowers vbe. Current moves under the influence of the external circuit, not under the infulence of vbe. vbe changes as a result of the input current:

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

Do you get it yet? The sequence of events is the current nertralizes some of the charge, and barrier (vbe) changes as a result. There it is in plain, simple english. Anyone should be able to understand that.

The S & S equations support it.

S&S does not support your false ideas. Even a basic text like that would not support anything so blatanaly untrue.

They say that the increased charge flow reduces the barrier voltage. I agree with that. But S & S does not say that the current controls Vbe. They cannot, because Vbe is externally controlled. So Vbe causes the majority carriers to increase, by whatever method you like. And the barrier width has to reduce or the carriers could not increase. But it was Vbe that caused the carriers to change.

They say barrier voltage decreases as a result, and current causes the decrease. vbe is internal, not externally controlled. vbe is the voltage measured across the PN junction of the base and emitter. It is the result of the change of the barrier potential. That which controls the barrier potential controls vbe. There is no equation or any text that supports vbe as an external source. It's totally internal to the bjt, and it occures at the PN junction, where the bjt is controlled from. Every single semiconductor physics text says this. If you actually read any semiconductor physics text rather than going on about things you don't know about, you would understand that.

Vbe is externally applied, the barrier depletion voltage is internally generated. So Vbe in the casual control.

That's so pathetically absurd. Every time you make such an utterly false statement, you further discredit yourself. vbe is the junction potential, and nothing else.

 

[MrAl]

MrAl,



I ignored capacitance, AC, and transients because I wanted to keep the discussion simple. Already, you are submitting some transfer and feedback terms that have no relevance to this discussion. Remember, we are discussion whether a BJT is, causually speaking, a current controlled or voltage controlled device.



All that is fine and good, but it does not clairify the discussion.

Ratch

Hi again,


You mean does not clarify 'your' discussion?

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.

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? Is it because the field has to be there first before the charge can even think about moving? Is this similar to I=V/R or V=I*R, where V has to be there first?

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

 

[Ratchit]

Brownout,

According to S&S, it is current that lowers vbe. Current moves under the influence of the external circuit, not under the infulence of vbe. vbe changes as a result of the input current:

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

Do you get it yet? The sequence of events is the current nertralizes some of the charge, and barrier (vbe) changes as a result. There it is in plain, simple english. Anyone should be able to understand that.

Your statements are wrong, and are not backup by S & S. Here is why. When a N-type and a P-type are put together to form a metallurgical junction (transistor speak for close contact), a diffusion voltage will form and current will exist until enough uncovered charges are present to form a back voltage called the built-in voltage (Vbi). This cancels out the diffusion voltage. Per Claude's request, I presented the formula for the intrisic built-in voltage at as Vbi=Vt*ln((Na*Nd)/ni^2). As you can see, Vbi depends only on the doping concentrations and the temperature. No external voltage or current influences this value. This voltage cannot be determined by any measurement at the Vbe terminal, because the diffusion voltage is cancelled by the built-in voltage and the diffusion current is cancelled by the reverse thermal current. So the metallurgical junction is invisible to any passive measurements at the base-emitter terminals. We know that is true, because if voltage or current were present, the diode would be supplying energy, which of course it does not. So here we have a metallurgical b-e junction with the terminals Vbe=0, and a Vbi and currents constantly existing within the metallurigical junction. In other words, the b-e terminals are shielded from the activity of the junction.

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.

S&S does not support your false ideas. Even a basic text like that would not support anything so blatanaly untrue.

Nothing I said is contradicted by S & S.

They say barrier voltage decreases as a result, and current causes the decrease. vbe is internal, not externally controlled. vbe is the voltage measured across the PN junction of the base and emitter. It is the result of the change of the barrier potential. That which controls the barrier potential controls vbe. There is no equation or any text that supports vbe as an external source. It's totally internal to the bjt, and it occures at the PN junction, where the bjt is controlled from. Every single semiconductor physics text says this. If you actually read any semiconductor physics text rather than going on about things you don't know about, you would understand that.

The barrier voltage decreases because Vbe opposes the always present Vbi. Vbi is determined only by the doping concentrations and a temperature sensitive constant, not Vbe. That is what the semiconductor texts say. Vbe is applied externally and influences the metallurgical junction internally. You are confusing the Vbi with Vbe, they are different. I have read semiconductor texts and do understand what is hapenning.

That's so pathetically absurd. Every time you make such an utterly false statement, you further discredit yourself. vbe is the junction potential, and nothing else.

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

Ratch