# Understanding Transistor

Discussion in 'General Electronics Chat' started by sdmuashr, Mar 24, 2012.

1. ### WinterstoneBanned

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Hi crutschow,

I believe, we are relatively close together.
This is because I think, YOU are using the equation Ic=beta*Ib as a possibility to interpret it as an indication for current-control (simplified view for engineering tasks).
In contrast, my opinion is that this is purely a mathematical relation between two currents, which is used in the course of BJT design only in the form: Ib=Ic/beta (see my examples below).
Thus, I see absolutely no necessity for another interpretation of the BJT principle - even during design and parts calculation.

Quotation Crutschow: I was just trying to say that the view of an engineer is how a device acts as a black box when used in a design whereas a teacher emphasizes the device's inner workings (as it should be).

I must confess I cannot identify two different views.
We speak of a common emitter stage and its gain, right?
I am sure that both of us use exact the same formula in case of Re feedback) : G=g*Rc/(1+g*Re).
Where g is the transcundactance g=dIc/dVbe.
And, of course, we – as engineers - don’t use 2-port parameters for determining the transconductance but the following formula, which results from Shockleys equation : g=Ic/Vt.
Where is the problem? Where are different views?
By the way: Shockleys equation Ic=f(Vbe/Vt) clearly indicates voltage-control of Ic.

Quotation Crutschow: I believe you should confront the student/engineer with both views. The voltage-mode initially for analysis since that is the physics of the device. Then, when teaching circuit design, cover the current-mode model also, since that is a much easier way to view the transistor for many BJT design synthesis calculations.

Don’t you think we can use the relation Ib=Ic/beta as it is, just treating it as a mathematical formula without the necessity to define a „current-mode“? At which stage of the design do we need this relation?
My answer: Only as an indication for the order of magnitude of Ib – as an input for designing the bias network. We do not need the exact Ib value and – what’s important – we do not need to know if Ib has a controlling function or not. We need Ib only as a current, which equals a small percentage of Ic. More than that: We treat it as a parasitic parameter, which cannot be avoided. That's all.
In short: Why do you favor the „current-control“ mode for engineering practice? Where do you need it?

Quotation Crutschow: I think it would be very difficult to try to do a BJT circuit design without using the idea that it appears more like a current-mode device than a voltage-mode device for many of the design calculations.

Counter example: Design of a typical common emitter stage:
* Select/specify Vsupply, Rc, Re and Ic.
* Normally one does not know the value of beta (lets assume somewhere between 100 and 250)
* For the worst case assumption (beta=100) and Vbe=0.65 volts the resistive divider is calculated – based on the requirement that the current through the divider is (6...10) times the current Ib=Ic/100.
If the corresponding input resistance is to low a recalculation is necessary.

Question: Was it necessary to consider the BJT as current-controlled? No, I don’t think so.
I only have used the mathematical relation Ib=Ic/beta.
What is your engineering approach to design such a stage? I suppose, exactly the same.

Now – if somebody asks: Why Vbe=0.65 volts? My answer is: This value determines the current Ic.
And if somebody asks: Why factor (6...10)? My answer is: I try to make the source resistance of the provided bias voltage Vbe=0.65 as low as possible (approach of voltage-control!) – in accordance with the mentioned constraints.
My claim: Everybody who tries to design the bias network low-resistive (as low as allowed) exploits the voltage-control feature of BJT's (perhaps without knowing).

W.

Last edited: Aug 12, 2012
2. ### WinterstoneBanned

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Hi Mr RB,
I think your "simple proof" does not work.

Here comes my proof:
* It is a well-known fact that for constant bias conditions the current Ic has a large positive temperature coefficient. That means: rising temperature causes an increase of Ic.
The physical reason for this effect is known, of course.
This fact is the main reason for dc stabilizing the selected operating (quiescent) current Ic using emitter feedback.

* Therefore: Transistor manufacturers present graphs Ic=f(Vbe) showing this effect. For temperatures t2>t1 each Vbe value corresponds with another Ic value that is larger (if compared with t1)
This leads to graphical representations (curves) with 2 ore more similar looking characteristics from left to right for falling temperatures (left: t2, right: t1).
By the way: The slope of these graphs Ic=f(Vbe) gives the transconductance g, which is a key parameter to determine the voltage gain of a stage.
Remark: A transconductance dI/dV shows how an input voltage variation causes a variation in output current (that is: voltage-control).

* Now, to put this temperature effect to numbers, two curves can be chosen with EQUAL Ic values and the difference in Vbe is noted. This leads to the well-known value of approx. -2mV/deg.

Summary: Of course, you are right that "Vbe can be two different values" (for the same Ic value).
However, this is an effect caused by temperature only and it demonstrates how Vbe determines the current Ic for different temperatures. A clear indication of voltage-control.
If you think, this could be an indication of current-control you should EXPLAIN and VERIFY your claim (see above): BUT for both values of Vbe, Ib ALWAYS sets Ic. So Ic is current controlled

Regards
W.

Last edited: Aug 12, 2012
3. ### crutschowWell-Known MemberMost Helpful Member

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5. ### WinterstoneBanned

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Hi again crutschow,

OK I give up.
I realize that you are not willing to answer some of my specific questions. Just repeating statements does not bring the discussion to an end that's engineering like.
(Please, excuse the bad expressions- I hope you know what I mean).
In the following, quotations from crutschow are in italic:

I have no problems with that gain equation. But that is only used for design of amplifiers with little or no stage feedback, such as RF stages. Most low frequency designs use an emitter resistor for local feedback where the transistor gm value becomes of small importance.

Perhaps you have overlooked that my formula indeed did include the emitter resistor. Is the formula correct or not? Do you use another formula (although there is no other). No substantial answer from you.

I though I made that clear. It's easier to perform many circuit design calculations if you treat the BJT as a current-mode device, not a voltage-controlled due to the difference between the current-mode (relatively) linear input/output relationship and the non-linear voltage-control equations . If you can't understand that from my previous posts then I am at a loss to explain it further.

Perhaps I would be able to understand if you could give ONE SINGLE example for your claims. You repeatedly mention "many circuit design calculations" without giving any justification or at least one example.
Instead, you consider it as necessary to blame me for "not understanding" your position. Where are those many "design calculations"?
I have presented a typical example for my approach of gain calculation.
I have asked you if you use the same equation - and at which stage of the design you are using the current-control approach. No answer!
Show me one single formula used during the design process of a BJT amplifying stage that can support your view.

Vbe may physically determine the collector current but determining what collector current is generated for a Vbe of 0.65V requires solving a relatively complex non-linear equation, as does solving the inverse question of what Vbe do I need for a specific collector current. Using beta I can easily determine the collector current from the base current.

As far as I remember you have stressed several times the engineering approach (on your side) and the more academical view (on my side).
May I ask you: Do you really start your design with the base current in order to "easily determine the collector current from the base current".
Is this your understanding of a classical engineering approach? I cannot believe.
I cannot understand your "arguments" as they are far from reality.
Didn't you read my example how to design a BJT amplifying stage? Was it necessary "solving a relatively complex non-linear equation" as you claim?
Why don't you give any comment to my example?

Your voltage-control viewpoint has led you to the wrong conclusion. The normal bias network is designed to be as high resistance as possible, limited only by the constraints of the input bias current. A low-resistive circuit takes more power and can reduce the input signal voltage.

Why do you continuously ignore my statements? Didn't I mention that the resistance niveau of the bias network has to be fixed with respect to power requirements as well as input resistance of the whole amplifier?.
What is the purpose of such superfluous remarks?
I am very sorry to say that a discussion ignoring the statements from the other side is useless as it is in fact no "discussion".

Finally, one technical information for you: Each textbook tells you that the voltage feedback caused by the emitter resistor Re requires a nearly fixed base voltage (independent on Ic and Ib variations due to temperature effects). This means: The bias network should come as close as possible to a voltage source. Only in this case you can exploit the full feedback signal.
I hope you are able to agree that this means: Bias network with resistors as low as possible - with respect to power and input resistance constraints (as I have mentioned before).
I really don`t know how many designers would agree with your surprising claim that the bias network has to be "as high resistance as possible".
__________________-

I am very sorry about the form and the contents of this (my last !!) contribution within this thread. But it was not only my decision.
I am always a bit angry if a discussion seems to be not possible on a fair and objective basis.

Winterstone

Last edited: Aug 12, 2012
6. ### crutschowWell-Known MemberMost Helpful Member

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W.

Well I give up also. We obviously are not on the same wavelength.

7. ### Mr RBWell-Known Member

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I showed a case where Vbe can be two different values, but Ib:Ic ratio remains the same.

If you can show me a case where Ib can be two different values AND Vbe:Ic ratio remains the same, then you finally have an argument!

Everything you and Ratchit have said comes down to a chicken and egg argument. I'm not interested in "what comes first".

There are two ways in electronics to control the transistor, either by applying voltage Vbe to the base to set the Ic, or by applying current Ib into the base to set Ic. The better choice to control the transistor is to use Ib to control it, so the bipolar transitor is "current controlled". In contrast, the better choice with a FET is to control it with gate voltage, so the FET is best described as "voltage controlled".

8. ### RatchitWell-Known Member

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Mr RB,

No you didn't. Not when the junction temperatures are the same, as Winterstone and I pointed out to you. Why do you keep making the same false statements, when the textbook Sedra and Smith contradicts you? For a particular BJT in the active region at particular temperature, there is only one value of Vbe for each Ic value. And I already explained to you why the Ib is proportional to Ic, but does not control Ic. To advance further, you have to explain why Winterstone, I and Sedra and Smith are wrong. Can you do it?

That would not prove Ib controls Ic. I already said that Ib and Ic are proportional to each other because they are each controlled by Vbe according to Sedra and Smith. That makes Ib an indicator of Ic, not a control of Ic.

Where did that phrase come from? I did not say it. There is no issue about what comes first. The issue is about what controls what. The simple principle is that Vbe controls Ic and Ib.

The two ways you list to control a BJT are one and the same. Applying a current to the base automatically applies a Vbe to the BJT, and Vbe is what makes the transistor respond. The Ib is going to indicate what the Ic will be. We are not discussing FET's.

Ratch

Last edited: Aug 12, 2012
9. ### Mr RBWell-Known Member

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That is exactly the "chicken and egg" argument. I'm not using Ib to indicate anything! Nor do I care that much what processes are occuring inside the silicon in my transistor.

I'm applying or "controlling" Ib to give the desired control of Ic. You're too concerned with the mechanism of what is happening inside the transistor, and not thinking about "controlling" the transistor.

In all cases in electronic design we are concerned with "controlling the transistor" and it is best controlled by applying the Ib to give the desired Ic. If you think it is better in electronic design to apply a Vbe in order to control Ic then I think we have to part company.

Now to go back to my "proof", I have supplied these two real-world cases;
cold transistor; Vbe 0.6v, Ib 1mA, Ic 100mA
hot transistor; Vbe 0.55v, Ib 1mA, Ic 100mA

all you have to do is show me two cases in the real-world like this;
case1; Ib XmA, Vbe 0.6v Ic 100mA
case2; Ib YmA, Vbe 0.6v Ic 100mA

Provide those two cases and you have an argument.

10. ### Claude AbrahamMember

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The physical cause of Ie is the energy conversion in the external signal source driving the network. It could be a microphone, photodiode, antenna, signal from another amp stage, etc. The cause of Ie is NOT, repeat, is NOT - Vbe! The source driving a bjt network outputs current & voltage. Let's say the signal generator is a mic. The singer, use the name Sue, outputs acoustic power from her throat/vocal chords/diaphragm. This power in converted to electric power by the mic diaphragm, current & voltage are generated. The current enters the b-e junction & encounters a potential barrier due to charges in the depletion zone. This voltage drop, signal quantity, is "vbe", & represents the energy charges lose due to this barrier. The mic generates an emf, which is the energy gained per unit charge. The b-e junction exhibits a forward drop, which is energy lost per unit charge.

In order for Ie to exist, an active source must be imparting energy to the charges, in this case the mic. The vbe forward drop is a loss of energy per charge incurred crossing the barrier. The energy lost per charge cannot be the cause of the charges moving through the junction. To move charges we need to impart energy to the charges, i.e. energy gain per charge. Vbe is a loss of energy per charge incurred crossing the junction barrier. There is no battery or generator inside a b-e junction. How can said junction "cause" emitter current to flow.

A simple test affirms this. Use a pulse generator & scope. Initiate an emitter current puls, measure Ie, Vbe, & Ic. You will see that Ie (Ib as well) change before Vbe, and that Ic responds to theIe change, not the Vbe change. Ratchit refuse to even discuss transients, prefers to keep this discussion in the dc domain where it is impossible to disprove "voltage is causal" heresy. In general, voltage does not cause current.

Finally, re Shockley's diode equation, Id = Is*exp((Vd/Vt)-1), it is equally true that Vd = Vt*ln((Id/Is)+1). Vd, the forward diode voltage, does not control Id, not in the forward direction. A voltage source right across the junction spells certain doom for the diode. In forward bias, we almost always control diode current, & forward voltage is a function of Id & temperature. Take a buck converter, see attached sketch. When FET is on, diode D is reverse biased. Voltage is Vd, nearly equal to source voltage Vg. In this condition, Vd determines Id along with temp. But when FET turns off, D is forward biased. Inductor provides current to diode, & inductor maintains steady current. Just after FET goes off, inductor current is continuous, having same value. The forward voltage drop of D is controlled by Id & temp.

This is well known. With diodes in forward mode, we never control Id w/ Vd, rather Id controls Vd. In reverse mode we usually control Id w/ Vd. In forward mode a voltage source across a diode results in thermal runaway. Diode equation by Shockley gives functional erlation between Id & Vd, it does not imply causality at all. Vd does not control Id in forward mode at all. The only exception is a rectifier w/o a choke. When line input exceeds cap voltage, diode conducts abruptly. Diode anode is connected to ac input voltage source, & cathode is connected to cap voltage, also a voltage source. A surge current takes place wreaking havoc. These diodes have high surge current ratings for this reason. At that moment when diode conducts, Vd can force Id resulting in surge current.

11. ### RatchitWell-Known Member

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Mr RB,

The "chicken and egg" or more exactly "chicken or egg" is used to characterize the question of what happens first. That is not the question here. The questions of whether Ib or Vbe controls Ic is not about what happens first, so let's forget about that phrase. You still have not addressed the fact that according to Sedra and Smith, Ib and Ic are dependent on Vbe. Therefore Ib is an indicator of what the Ic value is. Whether you care about how transistors operate or not, they still do what they do regardless of what you think about them.

Au contraire, I am concerned and thinking about controlling the transistor. The argument is about whether Ib or Vbe controls the transistor. I aver that when you drive the transistor with a current, you also apply a definite value of Vbe which is really controlling the transistor, and also definine Ib. You say that despite what Sendra and Smith and others say, the Ib is controlling Ic. So far you have not presented any evidence of your claim that has not been easily debunked.

Now you are off on a tangent. I never said design and calculation should not use the relationship between Ib and Ic. I said that Vbe is really what controls Ib and Ic.

Invalid! Measurements taken at different temperatures.

Already have done so. Sedra and Smith showed that Ib and Ic depend only on Vbe and not on each other. Did you read that post?

Ratch

12. ### WinterstoneBanned

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Hello Mr Rb,

1.) There are two dominant temperature effects which influence the BJT operation: The emitter current Ie (rising with temp.) as well as the factor beta=Ic/Ib (also rising with temp.).
The latter effect has approx. a factor of 2 for a temperature difference 75 deg.
Thus, for different temperatures and Vbe=const the collector current Ic as welll as the Ic:Ic ratio changes (in contrast to your idealized case).

2.) Please, can you show why during design and analysis of a common emitter stage the "better choice" would be to use Ib to "control the transistor" ?
In my posts to chrutschow I have presented my way of design and calculation. It would be very helpful if you could show us a similar set of formulas which can support your view.
Do you realize that at the start of the design process the value of beta is known typically with a tolerance of 200...300% (or even more).
Assuming that you also start with specifying Ic - the current Ib also is known only with such a huge tolerance.
This is the engineering view without measurements of the BJT 4-pole parameters prior to design.

3.) I think it is not "the better choice" to control a FET with the gate voltage - it is the only way. Or do we have the choice?
It is not only "best described" as "voltage controlled" - it is in fact voltage controlled. Some wise men have prooved this.
I know that there are manufacturers who specify - surprisingly - a "current gain" even for FET's. For my opinion: Confusing - if not false (because there is no relation between input and output currents).

13. ### WinterstoneBanned

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Hello Claude,

I must confess that - up to now - I didn't read your contribution very carefully because I am not sure if it answers the problem under discussion.
Your reply is the answer to my question "what is the physical cause" of Ie. I only wanted the answer: Vbe, because you didn't mention this bias condition.
I don't know if the example with the MIC can enlighten the case. But that`s not my point.

Another problem is that we probably speak about different circuit configurations. This makes the discussion a bit problematic.
I think, the following example from you concerns the common base configuration, right? Nevertheless, it is an interesting problem.

Example (Quotation): A simple test affirms this. Use a pulse generator & scope. Initiate an emitter current puls, measure Ie, Vbe, & Ic.
You will see that Ie (Ib as well) change before Vbe, and that Ic responds to the Ie change, not the Vbe change

The main question is, indeed, if Vbe changes after Ie and Ic.
I have not measured or simulated this case, but I am bit sceptical: Does Vbe really change after Ie and Ic ?
Question: Did you perform measurements or simulations?

What do you think of my explanation: If a change in Ie is applied (externally) the emitter potential changes immediately (due to the emitter input resistance) and - as a consequence - also Vbe.
That means, we are again at the same starting point: Change in Vbe.

And - as far as the diode and Shockley's equation is concerned - I think, this touches the question: What is a "current source" resp. what means if we say "a current is applied" ?
Isn't a `technical current source nothing else than a voltage source with a large (but finite !) input resistance?

Regards
W.

Last edited: Aug 13, 2012
14. ### Claude AbrahamMember

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Emitter input resistance re is a part of the picture. But e-b diffusion capacitance is what I am referring to. A change in vbe takes place when ie charges Cdiff. All capacitors exhibit the behavior that current leads voltage. Another way to examine this is through depletion region charge density. An increase in Ie results in more e- transiting through b-e junction. The number of e- that recombine in base w/ holes injected from base region must increase. Likewise the number of e- that recombine in base increases due to increased Ie. But nearly all e- continue on into collector. The increase in Vbe is a mere consequence of increasing Ie.

A current source is not always a voltage source plus high resistance. Take an inductor with 1.0 amp. Switch this current into a 10 ohm resistance. The inductor behaves like a current source. The initial voltage is the 1.0 amp times the 10 ohm load resistance, or 10 volts. The inductor is a true current source, not a voltage source plus resistance. A test that affirms this is to short the inductor.

A voltage source plus large resistance when shorted dissipates power, whereas a true current source does not. A superconducting inductor with current dissipatesd zero power in a short. Thevenin & Norton networks are only electrically equivalent, not thermally equivalent. A 1.0 amp current source with 1.0 kohm shunt resistance is electrically equivalent to a 1.0 kvolt voltage source with a 1.0 kohm series resistance. However when shorted the current source dissipates zero power, the voltage source dissipates 1.0 kwatt. When open the current source dissipates 1.0 kW, the voltage source dissipates zero.

Many times a magnet has been suspended above a superconducting current loop indefinitely, no decay has been measured over the course of years. If the inductive current source was truly a voltage source plus high resistance the energy would decay. The 2 are not thermally equivalent. In my power supply example attached, the inductor de-energizes into the diode maintaining current source behavior. The diode voltage is literally dictated bu its current & junction temperature.

I can control V, or V can control I. In the case of the diode, it alternates with the switching cycle. I've designed power supplies for a few decades, believe me, the inductor is indeed a current source. I will not elaborate, but a photodiode terminated in a very low impedance also behaves like a current source, a true one, not voltage source plus large R.

Ie changes ahead of Vbe. Vbe changes as a consequence of Ie. But Ie changes as a consequence of singer Sue cranking up/down her volume. Ic changes because more e- were collected. But that is because more e- were emitted by the emitter. Ebers-Moll back in 1954 published their bjt paper detailing current controlled behavior. The collector current is modeled as a current source controlled by Ie per the equation Ic = alpha*Ie. Look it up, Ebers-Moll equation, based on Shockley diode equation, treats Ic as controlled by Ie. I'm not making this up. Shockley equation for the diode merely gives the math relationship between I & V, does not establish a cause/effect pecking order. The circuit where the diode is employed dictates which is independent, which is dependent. BR.

15. ### ()blivionActive Member

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For simple circuits, a good way to understand them is to play with them in a simple simulator. The falstad simulator would be perfect for this, and even has basic BJT circuits set up already. Visualization is the key.

*HERE* is a direct link to the NPN simulation. (Must have Java enabled)

And if you want more, just thumb through the circuits menu item.

Last edited: Aug 13, 2012
16. ### WinterstoneBanned

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Claude, I agree with your current source examples.
However, by mentioning a "technical source" in my former reply I had indeed a constant voltage source with a large internal resistance in mind.
Nevertheless, the main question still is (not:who killed Norma Jean): Who controls Ic, right?
According to my understanding (which may be incorrect) the critical path is the pn junction and the associated depletion region - independent on the configuration (common emitter or common base).

And my understanding of the control process is as follows:
The width of this depletion region is modulated caused by changes of the electrical field across this region. This field is caused by a voltage resp. by voltage variations across this region.
For a fixed voltage between emitter and collector the "breathing" depletion region causes a corresponding change in current through the emitter-collector path.
Because a certain percentage of this current is "lost" - that is Ib - we have Ic<Ie.
Of course, we can bring the diffusion capacitance into the game - however, I think for a fundamental understanding of the control principle this wouldn't be necessary.
Now - which external parameter is responsible for the mentioned voltage (and its variation) across the pn junction? I think, it is the voltage between base and emitter.
Now my question: What else do we need to describe the principle of Ic control?

Perhaps, this description is a bit simplified, but I think it concentrates on the main effects in order to understand the principle of Ic control. And it applies for both configurations (CE and CB).
Final question: Do you see any fundamental error in the above description?

I think tge same applies to a pn diode and the Shockley equation:
Don't you think that a voltage must be applied in order to create an electrical field that influences the depletion region?
Even if you inject a current into the diode - it is still the developed voltage that opens the diode.

W.

17. ### Claude AbrahamMember

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The b-e junction voltage variation lags the increased emitter current. It cannot be the driver. Regarding electric fields necessary to change Ie, I already stated that Sue the singer outputs acoustic power, mic diaphragm transduces acoustic power into I & V & the E field generated by mic diaphragm provides drift to emitter e-. The "applied voltage" includes an "applied current". Voltage control thinkers cannot fathom that I & V exist in unison. We use constant voltage sources everywhere, batteries, generators, bench top power supplies. We grow accustomed to viewing V as the INdependent variable, with I as DEpendent. When the mic generates power, it is outputting V & I in unison, boundary conditions like resistance & capacitance determine network behavior. Here is the all important fact you must consider. The voltage from the mic element is NOT Vbe. Of course in order to change the emitter current the mic output voltage must change. But that voltage change accompanies change in Ie, then Vbe changes as a result. Vbe is not changing Ie nor Ic. Ie does that.

The "breathing" b-e depletion region is not a result of modulating vbe at all. The breathing is a result of modulating the mic output. The increase in mic output V & I takes place before depletion region breathes. The modulated charge density transiting through the b-e junction results in a modulated collector current. Vbe is incidental it takes place as a result of the I & V variation forced by singing Sue & the mic element. That is what drives everything.

The increase in Ie, which is caused by Sue & mic, accounts for Ie change, Ib change, Vbe change, & ultimately Ic change. Vbe is merely incidental. The Ic value can change simply because Ie changes. Ic does not have to wait for a Vbe change. Increasing Ie does not require an increase in Vbe, it only requires an increase in the source driving the signal, i.e. the mic.

So we have a quiescent condition, bjt biased at Ic = 1.0 mA for easy values & computation, Vbe = 0.65 V, Ie = 1.01 mA, so that Ib = 0.01 mA. A depletion region in the b-e junction has settled to steady state value. Sue starts singing. Mic outputs I & V. Increased e- flow enters b-e junction. Ie increases. Since more e- are emitted by emitter, more e- will be collected by collector. It takes a short time for the e- to transit from emitter to base. The width of the depletion region & Vbe have not changed. A capacitance, Cdiff requires current first before the voltage changes.

So meanwhile Cdiff is charging up due to increased Ie/Ib, while Ic is increasing due to more e- being collected. Eventually enough charges accumulate in the depletion region so as to increase Vbe. When Sue sings, mic output I & V both increase. More e- transit through b-e junction, resulting in more Ic as well as more Vbe. They are both incurred by Sue imparting energy to the mic. Ib/Vbe/Ie all increase due to this action. The actual cause of Ic changing is Sue. But Ie is the variable most closely influencing Ic. Hence in 1954 Drs. Ebers & Moll published their bjt model as current controlled. Actually they modeled it as 2, not 1, but 2 CCCS (current controlled current source). In saturation, the other CCCS is active, but in cut-off or linear mode, only 1 CCCS need be considered. I will elaborate if you have questions, feel free to ask me anything. I am not averse to explaining & accepting criticism. I hope I helped. BR.

18. ### WinterstoneBanned

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However, I am not convinced that your view can explain the known effects.

Remark: BTW, Horowitz/Hill (page 80): Collector current is determined by base-to-emitter voltage.

Thanks & regards.
W.

Last edited: Aug 13, 2012
19. ### Claude AbrahamMember

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But you cannot explain why you say I'm wrong. Everything I posted is well known & confirmed. Every bjt producer calls them charge controlled at the micro level, current controlled at macro level. Because so many people like to view the electrical world in terms of voltage, there are always books which adhere to that view. Scope, pulse generator, etc. affirms the current controlled nature. I believe I have supported my position, & I ask anybody to offer a counter proof to my post.

Claude

20. ### RatchitWell-Known Member

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Claude,

Now that Winterstone has had a chance to answer your post, I feel the time is appropriate for me to challenge your assertions.

Now you have to define what "cause" means. For instance, suppose I have a circuit consisting of wires, a lamp, a battery, and a switch. Does the switch cause the lamp to light? Does the battery, the wiring, or what? If cause means control, then I think we can agree that the switch controls the lamp.

Now to continue. A BJT is a creature of diffusion. Putting a P-type semiconductor in contact with a N-type semiconductor will make a diode and cause the holes from the p-type to diffuse into the N-type slab and the electrons from the N-type to diffuse into the P-type slab. The charge diffusion from the neutral atoms causes these atoms to be "uncovered" and turn into charged ions. This will happen without any external voltage applied. The process will stop when the ions caused by the uncovered charges accumulate enough back voltage to stop the process. Applyng a forward voltage will lower the back voltage and allow the diffusion process to continue. Therefore the applied external voltage is controlling the diode current. The diffusion mechanism explains why the voltage and current are exponentially related. FETs and vacuum tubes are controlled by electrostatic fields, and do not have exponential relations. I think we can dispense with Sue, singers, and microphones. Just say that a voltage is applied.

You appear to imply that the energy for Ie, and thereby Ic, comes from the base circuit. Well it doesn't. It comes from Vcc in the collector circuit. The base controls Ic and thereby Ie through Vbe. The current and energy lost in the base circuit are a miniscule part of the collector current and energy. There is a "battery" of sorts in the b-e junction. It is the diffusion voltage and back-voltage, which is modified by the externally applied Vbe.

Again you want to complicate things by introducing storage charge and transients. I did discuss it in this thread. Look at page 5, posts 49 through 51.

I don't know what Schockley's equation and a buck converter have to do with whether a BJT is controlled by Vbe or Ib. If you apply a voltage to a junction diode, the equation tells you what the current will be. If you drive the diode with a current source, the inverse equation tells you what the voltage across the diode will be. A diode is not doomed if the voltage is kept low enough.

To summarize:

The mechanism for BJT functionality is diffusion. Any explanation that disregards that principle is doomed to failure. The charges in the emitter and base will diffuse into each other for a short time and cause a momentary current to exist just by being in contact with each other. This current will be sustained by lowering the barrier or back-voltage by applying Vbe. Most of the current from the emitter to base does not exit the base terminal, it goes instead to the collector terminal. Therefore, there is relatively little current in the base circuit and little energy expended in the base circuit. The base circuit only exists to supply the control voltage Vbe. Most of the energy for Ie and Ic comes from the collector voltage circuit. That makes a BJT a voltage controlled current source.

Ratch

21. ### WinterstoneBanned

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Claude, excuse me, but I cannot remember that I have stated you were "wrong".
The only point is I cannot follow you. In some parts your last post was only a duplication of the former post.
What about a counter proof for my explanation as given in post#94 ???

WHAT is WRONG in my explanation (Question 1)?

You certainly will agree that a technical discussion (in case of different views) should mainly consist of question and answers.
I have asked some specific questions - no answer up to now.
I have asked if your explanations apply to the common base configuration only and to show or describe a corresponding circuit - no response.

What was the reason for you to speak about the the common base configuration instead of the common emitter (which was the basis of the discussion in this thread up to now)?

Now I kindly ask you: How do you explain the control function of the common emitter configuration - based on your theory that the emitter current is the first cause of controlling Ic (Question 2).
I am sure that this would clarify some of the possible misunderstandings.

Can you please respond to both of my questions 1 and 2.

Thanks+regards
W.

Remark 1: I am sorry for numbering the main questions - but your answer to these questions is most important to me.
Remark 2: Referring to the BJT producers, you probably have read my remark, that some companies even specify a current gain for FET's. That's not a convincing argument for me.
Remark 3: If desired I can give you some more references supporting voltage control.
Remark 4: Isn't funny that more than 60 years after the BJT was invented we are still discussing how this part really works?

Last edited: Aug 13, 2012