# Collector current of an NPN common emitter circuit

Discussion in 'General Electronics Chat' started by MrAl, May 29, 2013.

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1. ### Claude AbrahamMember

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Eli the ice man demolished that theory. You can't refute Eli. Time is not negotiable. Ie/Ib change w/o Vbe, then Vbe catches up. End of debate. Vbe is a loss incurred crossing the junction. Vbe is not an active power source, like a generator or battery. It is literally the energy per unit charge LOST crossing the juction.

When a source, such as a mic, outputs a signal I./V, via a mic cable, the signal, both I & V transit through the cable and reach the b-e junction. The V & I are already present, the cable impedance Z0 = V/I. If the terminating load resistance is different from Z0, a reflection occurs, I/V return to the source where another reflection occurs, eventually settling.

Transmission line theory is generally not used for wavelengths longer than 20 times the distance spanned. But it must be acknowledged if we wish to discuss which is in control. Timing is all important. The mic outputs a signal and is totally oblivious to Ie/Vbe?Ib that lies ahead. Mic only sees Z0 the cable impedance. Ie is not a by product of Vbe, it is created when singer's acoustic energy is transferred to mic element. I & V both co-exist at that point and eventually reach base of input amp stage. If stage has large ac degeneration, the impedance of the stage input along with cable Z0 determines reflection coefficient. Eventually when settling takes place Ie still leads Vbe chronologically.

The notion that Vbe controls Ie is nothing more than an assumption which as of yet has been unable to withstand scrutiny. In addition, do you know the difference between **voltage drop* and *emf*? An emf is a ratio of energy GAINED per unit charge, vs. a drop which is energy LOST per unit charge. The potential barrier at the b-e junction, according to you, must be lowered in order for Ie/Ib to increase. Yet Shockley states that increased Ie/Ib is linked to increased Vbe.

Even before Vbe changes, I & V from the mic are established. If the I value is an increase over a moment ago, Vbe has not changed yet. I enters b-e junction and the increased carrier density eventually forces an increase in barrier potential. Clearly the mic/singer is in control of everything. The singer increases acoustic energy, increasing mic element I/V, increasing Ie/Ib, then Ic increases due to larger Ie. Vbe eventually catches up to the new increased value, but Ic increased due to increase in Ie.

Physics of bjt is quite clear. I re-iterate, Shockley states the relation between I & V, for an LED, diode, bjt, SCR, etc. V does not control I, not in the forward direction at all. My switch mode power supply illustration addresses that point. Please review.

2. ### ericgibbsWell-Known MemberMost Helpful Member

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hi All,

The Thread is 'interesting' but does anyone of the contributors to this Thread believe that either side will accept the explanation/arguments of the otherside.?

If not, I am considering Closing the Thread, unless there is any serious objection, IMHO its going nowhere.

Eric

3. ### steveBWell-Known MemberMost Helpful Member

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I think this thread might have more life if people would spend more time clearly defining what they mean by "control" when they try to promote a viewpoint. Concepts like "control from a physics point of view", or "control from a model point of view" are too vague for my liking. We are technical people here, not philosophers. So, define something clearly, and then a clear conclusion becomes possible. Otherwise, it is all just interesting "talking" without a practical use.

I'll ask a question I once asked before to show the problem. If a CEO of a company orders his VP of manufacturing to tell the production manager to tell a technician to perform a new test on a product, who is in control? Is it the CEO, or the technician? I'll bet we can't even agree on such a simple idea, and I'll bet a few people might even argue for the VP and the manager. We can't agree because each person has a different concept of what "control" is.

Biologists would have an equally hard time arguing whether a virus is a life form or not. When they do so, they are really arguing about the definition of what life is. Here we are really arguing about what "control" is and what a "physics model" really is. This is not trivial or easy to contemplate.

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5. ### RatchitWell-Known Member

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Claude,

No he did not. I don't try to refute Eli. I don't negotiable with time. I and V also change phase in a RC or RL circuit because of internal energy storage or release. Same for a BJT. You made that argument several times now. Unless you counter my arguments, it won't do any good to say that again.

You also made that point recently. I countered by saying that you cannot determine the basic physics of a BJT by an analyzing an application. At that time, I described the diffusion physics of a BJT, and how the emitter-base voltage affected the emitter/collector current. You have not made any counter arguments that address what I said. So again, repetition is not a valid answer.

Timing is not important in determining physical control of a BJT. You are obfuscating and pettifogging the issue by bringing up those subjects.

Yes, voltage drop is the loss of energy per unit charge across a load. EMF is is the force exerted on a charged particle by an electric field. (Beats me why the express that in volts.) Your definition of EMF differs from mine. Both Shockley and myself are correct. Shockley takes into consideration both the back-voltage of the uncovered charges and the externally applied Vbe. His equation gives the equilibrium value of current caused by of these two voltages.

I already stated that the phase differences are due to internal energy absorption and release, and does not change what controls what.

I agree, the physics is clear. You cannot determine what is in control by Shockley's equation. You have to delve into the physics of the device. Your SMPS is an application that cannot explain how a BJT works.

Ratch

6. ### WinterstoneBanned

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Hello Steve, I fully agree with you.
More than that, in some former posts (106 and 108) I have tried to explain my view on the difference
between "to control" (to vary a physical quantity intentionally by applying an external control signal) and "to determine".
However, as it seems - without success. I never got a comment on this.

More than that, in a related thread (BJT- a current-controlled device?) I have placed (see post#14) a question regarding the known relation Ic=alpha*Ie

"Ie controls Ic ? Thus, the question arises: How do you define „to control“?
To me, alpha is a factor that cannot be controlled externally.
"

I got no answer.

Regards
W.

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7. ### RatchitWell-Known Member

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steveB,

OK, I will do so. The control I am talking about is a physical device, specifically a real BJT, not a model of one. I want to define what input parameter will affect the output. My determination is that a voltage cross the emitter-base terminals affects the emitter/collector current. A current can be sent into the emitter-base terminals, but the base-emitter voltage will be the same as it was before at the same emitter/collector current. I can physically explain how the emitter-base voltage affects the diffusion process and thereby the emitter/collector current, but I cannot explain how a base current does so. Therefore, I conclude that the emitter-base voltage controls the emitter/collector current of a BJT.

Two different types of control. One is administrative, the other is physical.

Yes, it all depends on what the definition of life is. The cell is considered the most basic form of life, so going by that definition, a virus does not meet all the criteria to be called a fully functional life form.

Ratch

Last edited: Jun 14, 2013
8. ### Claude AbrahamMember

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Here's your answer. Alpha is the parameter which conveys how Ie controls Ic. Ie can indeed be controlled externally. Please refer to my 4 page computation sheet and it is shown how to control Ie to a pretty good consistent value using emitter degeneration. Since alpha varies so little, forcing a fixed value of Ie, results in a very consistent predictable Ic value. Vbe is to uncontrollable to control Ic. Every bias network published never applies Vbe externally because it is thermally unstable to do so.

If, however, we operate in the small signal mode, we can differentiate Shockley's equation and derive transconductance value gm = Ic/Vt (Vt = nkT/q). In small signal equivalent circuit, ic = gm*vbe = hfe*ib, as both are equivalent. The quantity "vbe" is computed from the hybrid pi network, where r_pi is the b-e junction small signal equivalent resistance. The input signal source voltage is divided among source resistance, rbb' internal bjt base spreading resistance, r_pi, re (internal emitter resistance), and Re2' (reflected un-bypassed emitter degeneration resistance). So we can obtain vbe by computing r_pi then using voltage divider circuit.

So how do we compute r_pi? Easy, r_pi = hfe/gm. Refer to my sheets. So the fraction of input voltage that is divided down to the b-e junction is dependent on hfe. That is my whole point. When computing current gain of the stage, we use hfe, R1, R2, Re1, Re2, re, etc. But no gm value appears in *current gain* of stage. Now to compute voltage gain we need stage transconductance value. To get this we need raw device value gm, and r_pi value. But r_pi varies directly with hfe (ac beta). So we cannot compute voltage gain w/o hfe. However, if degeneration is sufficient we can prove that gain asymptotically approaches -Rc'/Re2. There is loading of the input source by input bias resistors R1 & R2.

A bjt w/ higher hfe can allow us to increase R1 & R2 values resulting in less loading of input source. So good high hfe minimum worst case value provides high gain capability and good network design neutralizes hfe variation issues. My point - good practice involves knowing the influence of hfe, knowing how to immunize stage against hfe variations, and how to obtain maximum performance by selecting device with highest min worst case hfe w/o compromising other requirements.

You cannot disregard hfe, nor gm, and I can continue my computations with more detail on how this is done. Maybe the weekend will allow it.

9. ### MrAlWell-Known MemberMost Helpful Member

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Hello again,

Eric:
I think this thread finally gives the members a way to express their opinions. Also, my intent here was not to argue one over the other, but to show results from different analysis methods and maybe come up with some definite results that helps with these circuits.

Ratchit:
Ok you could not get the data sheet, but now i realize that Eric had posted the required curves from the data sheet anyway. If you cant find them let me know.

All:
Keep in mind that this thread was supposed to be about using the equations to calculate Ic. Arguments about control techniques could be done in the other thread. I expect to get results here rather than argue back and forth like little kids
See i dont think we have to figure out voltage control over current control or any other control, we just have to show equations and results.

10. ### WinterstoneBanned

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Hi Claude Abraham,
it seems that you did not recognize my reply#160 (wow - nearly 170 contribution in this thread!).
It is not necessary anymore to ask me to read your 4-page paper. I did - and I gave my comments in post#160.
Therefore, all the additional information given in the other parts of your last reply (not quoted above) are redundant (to me!).

But there is one point I must comment again: Ie can indeed be controlled externally and "...it is shown how to control Ie to a pretty good consistent value using emitter degeneration."

That is exactly the point SteveB was referring to: What means "to control"?
Of course, the emitter current Ie changes its value by varying Re. However, what happens really?

You know, the emitter degeneration is equivalent to negative feedback.
What means "feedback"? Derived from the output signal (in this case: currrent Ie) another signal is fed back to the input (in our case: node E) with a sign that reduces the input signal.
In our case, a voltage is developed across Re which reduces the B-E voltage. Thus, we have a current-controlled voltage feedback. That`s for sure and can be found in all books and papers dealing with this subject. (By the way: I know, that we never can establish a pure and clean classification "current controlled voltage feedback" - but that`s not the main point here)

With other words: The B-E junction is the input of this device that exhibits negative feedback. And the signal that controls Ie is the voltage Vbe (exploiting the voltage across Re).

Of course, you are right by saying "...to control Ie to a pretty good consistent value using emitter degeneration". But you are not allowed to forget that between emitter degeneration and consistent Ie value the main action takes place: Vbe change.
We have the same (or better: a similar) effect using operational amplifiers. The feedback signal reduces the input differential voltage and the gain is primarily (by opamps nearly up to 100%) determined by the external feedback elements. But only because the feedback works - in conjunction with the active element which establishes the feedback operation between input and output.

And the same situation exist for transistors. The gain is (for large transconductance values) primarily determined by the external elements. But only because the feedback in conjunction with the active element works!

You simply cannot deny the role of the base-emitter voltage, which in out case forms the input to the active element.
And look into your own calculation: You have used the relation ic=gm*v(pi) with v(pi)=v(b-e).
That means you have used the fact that the small-signal collector current is controlled by the voltage vbe. And that is correct!

If you cannot agree with me, please explain how the emitter degeneration resistor Re controls Ie (without using Vbe)

Please, if you intend to reply to this post, show me which claims or statements are wrong.

W.

Last edited: Jun 14, 2013
11. ### MrAlWell-Known MemberMost Helpful Member

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Hi,

Really though this thread is not about arguing over one type of control over another with words and more words. It's time for equations and more equations that show actual results, and the comparison of those equations.

When we look at the equations we will see what is happening and why the different views are possible. It's only that way that we can see the difference. Otherwise it will be a back and forth forever.

But we need to see all the equations at once on a single sheet. I think that will help enormously.

A light example is like this:

Ic=Is*exp(Vbe/VT-1)
Ic=Ib*Beta
Ic~=Ie=(Vb-Vbe)/Re

From this last one we can see that when Re swamps re we have a very predictable circuit, and yet we have not used Shockleys equation (#1).

12. ### ()blivionActive Member

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No Ratchit, I'm not going to let you lead me by the nose into your perpetual attempt to detract from common sense. Commenting with you is pointless, so I am going to ignore you. I'm suggesting everyone else ignore you as well, as it is painfully clear from your most recent comments that your only objective is to cause mischief and grief.

Sorry, it's partly my fault for the negativity. I understand what you are after, I would like to see the same things. I just have to scoff at the propaganda of the "be-all-end-all" of the voltage controlled BJT that this discussion is always going to bring about. It's like asking "what's the difference between democrats and republicans" in a room full of both. One can't expect a realistic, unbiased answer, no matter how sincere their intentions are. Edit: this is of course NOT your fault by any means MrAl :/

Winterstone, I'm not trying to argue with you here. Just giving you my opinion and understanding of how it works because you asked for an answer. You are of course free to disagree if you feel like it.

Re controls Ie by avoiding the exponential effects of diode law, Ic=Is*exp(Vbe/VT-1) In favor of Ohms law Ic=V/R. Since Re fills in R in ohms law, Re controls(keeps stable) Ib. Since Ic=Ib*Hfe, and Ic≈Ie*α, and α≈1, then Ie≈Ic, thus it can be concluded that, Ie≈Ib*Hfe. As such... the effects of Ie≈Ib*Hfe are not going to be drastic, because as was shown Ib is not dependant on Ic=Is*exp(Vbe/VT-1), but rather Re via I=V/R. Now, even though there is V in I=V/R, this V is NOT Vbe, it is the total voltage drop of Re plus the base-emitter impedance. The base-emitter impedance is equivalent to subtracting the approximate voltage drop of one diode, ≈0.65. So, the equation becomes Ie=(V-0.65)/Re. And as long as Re is large enough that the actual variation in the base-emitter impedance does not significantly change the base current, Ic will continue to follow I=V/R, rather than Ic=Is*exp(Vbe/VT-1).

I am not 100% confident with math, so I want to say it in words as well...

The emitter resistor prevents the base/emitter junction diode from dominating the current/voltage relationship. Such a relationship(diode law) is exponential, and not favorable to good stability. So we place an emitter resistor in the same path so that the large flat impedance swamps out the diode impedance characteristic, thus the current/voltage relationship is ultimately reduced to Ohms law instead of diode law. To fill in Ohm's law, we need to know some voltages. We use approximations for the forward voltage drop of the base-emitter diode, but this is not the real/actual voltage drop at any given time, because that is dependant of current.

The important distinction is that the base-emitter impedance, is not the same thing as the base-emitter voltage. They are two totally different things, but they do interact.

Alternate answer...

It can also be said, as Claude has pointed out, Re controls(stabilizes) Ic/Ie, simply because Re is stable and related via α.

Emitter degeneration is equivalent to negative feedback because when Ic/Ie dumps current through Re, Re develops a voltage across it that increases it's effective impedance by essentially raising the ground reference. The decrease in Ib, of course, lowers gain because Ic≈Ie≈Ib*Beta.

Comparison of equations

Ic=Is*exp(Vbe/VT-1)
This is clearly the Shockley diode equation in different skin(Ebers–Moll). Diode law certainly does describe a diode junction to a T, even if that diode is in a BJT, so it certainly applies. My major beef with the practicality of this equation is that 1) enough of the parameters are "unit specific", i.e. must be measured part for part. 2) It uses exponential math, which is simply annoying.

1) Is, the reverse saturation current, is small, but the exponential nature of the equation makes it have a large effect, and 10^[SUP]−15[/SUP] to 10^[SUP]−12[/SUP] ampere range for Is is not sufficient to apply the equation in real life. Vbe is going to be current dependant, generally speaking one does not choose Vbe, it is developed as a result of current. VT is temperature dependant, it can be rounded to ≈26 mV at room temp, but that alone will cause problems as ignoring that it changes with temp is not a great idea in a BJT.

2) Exponential math is objectively harder to do (for people and computers), causes more problems with smaller mistakes/approximations, and is generally less controlled when in real life. When it works for you, it works great, when it works against you... well... let's just say this is why most properly employed encryption is uncrackable.

Ic=Ib*Beta
This one is good, it works if you know or can measure Beta/β/Hfe and can control Ib. It's liner, which is better than the above equation. And all the datasheets will give you a good idea of what Beta/β/Hfe is, and you can measure Beta/β/Hfe fairly easy if need be. This doesn't mean it doesn't have problems.

Beta is a wild tiger, you can't choose it, and you can't really control it. If you need it to be a specific value for your circuit to work, you're almost certainly "doing it wrong". Worse still is it changes with temperature, and a bunch of other crap probably. In any case, it's a lot like nitroglycerin, very practical applications, but unstable in it's pure form.

Ic~=Ie=(Vb-Vbe)/Re
This is the beez kneez because it makes the variability of transistors parameters mostly irrelevant. Even the approximation Ic≈Ie is acceptable, as it's accurate to less than 1% by most transistors. Vb is trivial to measure or calculate. Vbe will certainly vary, but it is safe to just call it ≈0.65 in most situations. Finally, Re is under deliberate and direct control.

You really can't ask for more out of an equation/design tool. It gives you all you need, and really doesn't make you jump through hoops to use it.

This is the equation/philosophy people should be using to design circuits IMHO.

Last edited: Jun 14, 2013
13. ### RatchitWell-Known Member

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Claude,

Yes, I finally got around to looking at it. I agree with Winterstone. It looks good. I especially like the fact that the base resistance is much lower than the emitter resistance. That helps keep the Icbo out the the emitter base junction so it cannot be betatized. I calculated Vb by node analysis, and got the same result as you did using the superposition principle. Good example.

Agreed, large values of Re minimize variations of β, ΔVbe, and Icbo.

Ratch

14. ### WinterstoneBanned

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Isn`t it funny? I have explained my view as short as possible in post#169 - and with the aim to find the core of the different opinions I simply have asked "what is wrong"?
A simple question which requires a simple answer.
And what did I get? See above.

()blivion - why are you not able to answer such a simple question? Not able or not willing?
But I still have the hope that Claude A. does. Because the question was primarily directed to him.

W.

By the way: The following quotes from ()blivion demonstrate the level we have reached now in the course of our discussion:

* The important distinction is that the base-emitter impedance, is not the same thing as the base-emitter voltage. They are two totally different things, but they do interact.
(Interesting, is it not?)

* It can also be said, as Claude has pointed out, Re controls(stabilizes) Ic/Ie, simply because Re is stable and related via α.
(Aha - Re controls Ie because it is stable.)

Perhaps it would be wise to follow Eric`s advice and close the thread. No real discussion (question-answer) is possible.

15. ### WinterstoneBanned

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Oh I forgot to ask another question, which is even simpler to answer :

This is a typical newcomer question:

How do you explain the role of the emitter resistor Re in case of temperature increase?

I wonder if I get an answer.

Thank you
W.

16. ### ericgibbsWell-Known MemberMost Helpful Member

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hi All,

This dispute between the two opposing factions is getting nowhere and I strongly belief that no agreement will ever be reached on the Topic.

So I have Closed the Thread.

Eric

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