Why do NPN BJT's work the way they do?

I have been trying to learn about how NPN BJT's work and I cant seam to understand a few things.

I have attached a diagram of my setup



Vcc = 12v
Vb = 5v
R1 (Collector resistor) 3.935K
R2 (Emitter resistor) 2.94K

I understand that the Ve (Emitter voltage) will be .6v less than Vb (Base voltage). What I dont understand is why. When the transistor is switched on why is Vcc not felt at Vb. It is baffling to me because the current that flows through the collector (Ic) is approximately Ie (emitter current). Ie is completely determined by R2 and Vb.

Ie = (Vb - .6v)/R2 = (5v - .6v)/2.94K = 1.5mA

How is this possible? Once the transistor is on should it not just act like a tiny resistor (the transresistance) and Ve feel the voltage from both Vcc and Vb. I know this is not true but I am struggling to figur out why.

You are making several assumptions that are leading astray. The transistor is not a tiny resistor. You need to understand that the transistor is a semiconductor. The Base-emitter junction looks like a diode. Because of the physics of the semiconductor, you need at least 0.6volts to forward bias that diode (for a silicon transistor; for a germanium transistor it's about 0.2volts). The base-collector junction also looks like a diode, but that diode is REVERSE biased; thus, there is (essentially) no current flowing between the collector and base

barry said:

You are making several assumptions that are leading astray. The transistor is not a tiny resistor. You need to understand that the transistor is a semiconductor. The Base-emitter junction looks like a diode. Because of the physics of the semiconductor, you need at least 0.6volts to forward bias that diode (for a silicon transistor; for a germanium transistor it's about 0.2volts). The base-collector junction also looks like a diode, but that diode is REVERSE biased; thus, there is (essentially) no current flowing between the collector and base

Click to expand...

Thank you for your reply Barry, I understand that the transistor is not a tiny resistor. I am asking why it does not act like one once a base voltage of at least .6v is applied. Specifically from Vcc to ground. Current is flowing (when the transistor is on) from Vcc to ground through the transistor. What I don't understand is why is the current through R1 ultimately dictated by Vb. Also why in my diagram do I only see 4.4v at Ve when clearly there is a complete circuit form both Vb and Vcc through the transistor to ground. Therefore, the voltage from Vcc should be seen at Ve. Again, I have tested this on a bread board and seen that indeed Ve is 4.4v, yet I don't understand why Vcc is not influencing Ve.

Again, I know that
Ve = Vb -.6v = 5v - .6v = 4.4v
Ie = Ve/R2 = 4.4v/2.95K = 1.5mA
Ic is approxatmatly Ie = 1.5mA
Vc (measured after R1) = Vcc - (Ic)(R1) = (12v - 1.5ma)(3.935K)=5.9025v

Indeed, through experimentation these numbers are correct. Yet, as you can see from the equations, Vc is determined by Ic which is determined by Ie which is determined by Ve which is determined by Vb. Why is Vcc not affecting Ve when the transistor is on?

Because that's kinda the whole point, not letting every
terminal control the current. Just one. Ideally, anyway.

Vce's influence on Ic is considered a nonideality and is
given a name (Early voltage, VA) and optimized as far
as possible (toward high numbers meaning little influence).

Yes, it does seem counter-intuitive to think that VCC is a minor factor in determining the amount of current flow.

There is the analogy of a transistor being like a coal chute, controlled by a hinged flap. We pull down on the flap (apply bias current). The harder we pull, the wider the chute (collector) opens, and the more coal drops through.

Ordinarily we would expect that it makes a difference how high the column of coal is (voltage at collector leg). And that the greater the pressure, the more coal will pour through. However we find this is not the case.

And although it is not a resistor, it acts like one where Ohm's law is concerned, since the transistor's NP junction needs to assume a higher ohm value if we raise VCC, in order to maintain current at the same amperage.

BradtheRad said:

And although it is not a resistor, it acts like one where Ohm's law is concerned, since the transistor's NP junction needs to assume a higher ohm value if we raise VCC, in order to maintain current at the same amperage.

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BradtheRad, for my opinion, this is a rather "dangerous" explanation. Following this explanatin I am afraid that newcomers could get a complete false understanding of the real BJT operation. A rising Vcc will not cause any internal pn resistance to rise as well. The currents Ic and Ie will remain nearly uneffected because of known physical reasons (the number of carriers collected by the collector depend only a little on the collector voltage, if it is above a certain value).
Thus we see, that the BJT acts as a curent source. This is a non-linear effect which does not follow Ohm`s law.

LvW said:

BradtheRad, for my opinion, this is a rather "dangerous" explanation. Following this explanatin I am afraid that newcomers could get a complete false understanding of the real BJT operation. A rising Vcc will not cause any internal pn resistance to rise as well. The currents Ic and Ie will remain nearly uneffected because of known physical reasons (the number of carriers collected by the collector depend only a little on the collector voltage, if it is above a certain value).
Thus we see, that the BJT acts as a curent source. This is a non-linear effect which does not follow Ohm`s law.

Click to expand...

Of course the last thing I want to be is 'dangerous'!

Nevertheless, are we supposed to say Ohm's law does not apply sometimes? We know that it is not broken. And it is not broken in this instance.

To clarify it in my mind, I had to try seeing it in simple >electrical< terminology (even if not advanced >electronic< terminology). Thus the NP junction assumes, adopts, etc., an ohmic value that will carry unchanging current regardless of which voltage we apply at the upper N terminal. The NP junction can be seen as a 'self-adjusting resistance'.

The reason I revert to simple >electrical< terms, is to focus on a detail of the advanced >electronic< concept. The aim is clarification. In fact I would not be surprised if some electronics instructor has said 'There is an elf inside the transistor, who turns a little potentiometer so the transistor passes a current equal to bias times gain'.

Until I looked at it in this way, I was confused (same as the OP, and a lot of other people, I imagine) about what was going on.

I had to look at it in simple >electrical< terms, because trying to see through the advanced >electronic< terms did not clarify for me.

Similarly I had to look at a diode as a self-adjusting resistor (instead of a 0.6V drop), in order to reach clarification about how it behaves in accordance with Ohm's law.

BradtheRad, I know what you mean:
With reference to the Ic=f(Vce) set of curves you speak about the DC resistance which could be identified (better: defined) as the inverse slope of a line connecting the origin with the operating point (Ic,Vce) on one of these curves, right? I repeat: This is a DC resistor.

However, if this resistor could be treated as an ohmic one it should have the same resistance also for the ratio d(Vce)/dIc. But - as you know - this is not the case. This dynamic resistor is much larger than the corresponding static resistor. This is no surprise because for each non-linear I=f(V) relation we must discriminate between static and dynamic (differential) resistances.
Therefore my objections.
In summary, I think, it does not help at all to follow your "model" - in contrast, I am afraid, it will prevent a correct understanding of the BJT principle.

Surely the simple explanation of a BJT at this level is that it acts to control Ic such that Vbe is maintained at a constant volatge of ~0.7V, and that both base and collector current appear at the emitter.

Once this very simple model (which actually suffices for most things) is understood then things like the early voltage and the fact that there are much more sophisticated models can be introduced (Ebbers-Moll, Gummel-Poon all that good stuff), but for the OPs question, the above plus the fact that there is effectively a diode in series with the base should suffice.

73 Dan.

LvW said:

BradtheRad, I know what you mean:
With reference to the Ic=f(Vce) set of curves you speak about the DC resistance which could be identified (better: defined) as the inverse slope of a line connecting the origin with the operating point (Ic,Vce) on one of these curves, right? I repeat: This is a DC resistor.

However, if this resistor could be treated as an ohmic one it should have the same resistance also for the ratio d(Vce)/dIc. But - as you know - this is not the case. This dynamic resistor is much larger than the corresponding static resistor. This is no surprise because for each non-linear I=f(V) relation we must discriminate between static and dynamic (differential) resistances.
Therefore my objections.
In summary, I think, it does not help at all to follow your "model" - in contrast, I am afraid, it will prevent a correct understanding of the BJT principle.

Click to expand...

I guess it becomes a question which hat we feel called on to wear when handling such questions... the physicist's hat or the educator's hat?

Is the question in regard to the physics of how a transistor works, or is it how we make practical use of transistors?

Is the question basic, or is it advanced?

Does it arise from going beyond the textbook model of a current source?

To my mind, if we typically used transistors as a current source most of the time, then I would emphasize the current source model (for that is the correct concept). However we mostly use transistors to pinch current on and off to another device, in varying amounts. This usage resembles a dynamic resistance. (If I suggested it was static, I did not intend that.)

So I think the models found in textbooks can only go a certain length toward edifying the student. Those models do not allow us easily to predict transistor behaviors, namely the kind that show up only in practical hands-on experience.

For this reason the educator must depart from the textbook, and draw on any illustration, analogy, etc., in his experience, in order to help the student get a grasp as to practical usage.

No one model or analogy can explain it all. Over time, the student realizes where this or that illustration is imperfect. Nevertheless our learning process needs that repertoire of incomplete illustrations, to reach some kind of comprehensive knowledge.

BradtheRad said:

Is the question in regard to the physics of how a transistor works, or is it how we make practical use of transistors?

Click to expand...

Is it really necessary always to decide between both approaches?

BradtheRad said:

So I think the models found in textbooks can only go a certain length toward edifying the student. Those models do not allow us easily to predict transistor behaviors, namely the kind that show up only in practical hands-on experience.
For this reason the educator must depart from the textbook, and draw on any illustration, analogy, etc., in his experience, in order to help the student get a grasp as to practical usage.

Click to expand...

....must depart?
OK - I see, everybody has made his own experience in educating students. And that`s OK.

karden,

I have been trying to learn about how NPN BJT's work and I cant seam to understand a few things

Click to expand...

PNP's work the same way, but their charge carriers are of opposite polarity.

What I dont understand is why. When the transistor is switched on why is Vcc not felt at Vb.

Click to expand...

Because Vc is connected to Vb through a reverse biased diode, which inhibits charge flow.

It is baffling to me because the current that flows through the collector (Ic) is approximately Ie (emitter current). Ie is completely determined by R2 and Vb.

Click to expand...

That is because you know something about what a transistor does, but not how it does it. Unlike the current in a resistor, which is controlled by an electric field produced by the voltage across the resistor, the base-emitter of a BJT is a diffusion device. The more heavily doped emitter diffuses its charge carriers into the base until it reaches equilibrium with the back voltage caused by the ion accumulation. Applying a Vbe lowers the back-voltage, and allows the charge carriers to flow until they reach an equilibrium current. So Vbe controls the emitter current by enabling more or less diffusion, not by moving the charge carriers by an electric field. The collector is biased so as to attract the charge carriers that come from the emitter by applying an electric field. However, its supply of charge carriers is limited by the diffusion controlled by Vbe. So within reason, applying a higher voltage to the collector will not increase the collector current, because there is a limited amount of charge carriers available, and that amount is controlled by Vbe. And that is why Vc does not primarily control Ic, and why Ic is almost equal to Ie. At the voltages in your example, the Early effect does not apply.

How is this possible? Once the transistor is on should it not just act like a tiny resistor (the transresistance) and Ve feel the voltage from both Vcc and Vb. I know this is not true but I am struggling to figur out why.

Click to expand...

Already explained above. A transistor can be modelled as a current amplifier, voltage amplifier, transresistance amplifier, or a transconductance amplifier. Remember, a model only shows what a device does, not how it works. A transistor connected to no circuit acts like a transconductance device, not a current amplifier like some books would have you believe.

Ratch

Ratch said:

Unlike the current in a resistor, which is controlled by an electric field produced by the voltage across the resistor, the base-emitter of a BJT is a diffusion device. The more heavily doped emitter diffuses its charge carriers into the base until it reaches equilibrium with the back voltage caused by the ion accumulation. Applying a Vbe lowers the back-voltage, and allows the charge carriers to flow until they reach an equilibrium current. So Vbe controls the emitter current by enabling more or less diffusion, not by moving the charge carriers by an electric field. The collector is biased so as to attract the charge carriers that come from the emitter by applying an electric field. However, its supply of charge carriers is limited by the diffusion controlled by Vbe. So within reason, applying a higher voltage to the collector will not increase the collector current, because there is a limited amount of charge carriers available, and that amount is controlled by Vbe. And that is why Vc does not primarily control Ic, and why Ic is almost equal to Ie. At the voltages in your example, the Early effect does not apply.
Ratch

Click to expand...

Hi Ratch, probably you have noticed that in this thread it was recommended that your physically exact description of the BJT working principle would overstrain the students.
Thus, it would be better to "depart from the textbook" and treat the BJT as a "self-adjusting resistor ...in order to reach clarification about how it behaves in accordance with Ohm's law".

May I ask what is your opinion?

LvW said:

Hi Ratch, probably you have noticed that in this thread it was recommended that your physically exact description of the BJT working principle would overstrain the students.
Thus, it would be better to "depart from the textbook" and treat the BJT as a "self-adjusting resistor ...in order to reach clarification about how it behaves in accordance with Ohm's law".

May I ask what is your opinion?

Click to expand...

My opinion is that since the OP desires to know why Vc does not control the Ie and Ic very much, it is necessary to explain how a BJT works. It is necessary to aver that the transistor is a voltage controlled current source and explain why diffusion makes that happen. I believe I have departed from the textbook in explaining things better, but not with respect to facts. I still used basic principles and not advanced concepts. The OP can ask further questons if he does not understand something.

Ratch

Guess I'll weigh in again.

If we were using transistors in the role of current sources more of the time, then I believe we would think of them more as current sources.

Instead our practical thinking has us look more often at the On-resistance of transistors and mosfets. We wonder whether it is dissipating too much power due to a high On-resistance, especially when using it to switch heavy current. (I have a suspicion that you, LvW, have looked at transistors in this manner, on another day.)

Notice also the OP schematic has resistors in the C and E legs. The transistor appears to be in the role of something other than current source.

So I see the initial question not as a case of a student being overstrained by the textbook explanation (to use LvM's words)...

But rather it is from a bright student sensing (perhaps intuitively) there is an important point about transistor behavior which is not covered in the textbook.

(Textbooks are reknowned for their uncanny ability to hold a curtain of obscurity in front of the very thing they are expounding. The reasons behind this are for another thread.)

So, Bright Student goes to Professor with his question.

Professor (who wears the physicist hat) taps the textbook and says "Read the definition again. And if it overstrains you, read it again."

Student walks away, feeling stupid.

The next day he goes to another professor, one who has on the educator's hat.

Professor listens to the question, and says "You can use your trusty VOM to get a better sense of what is going on. Take volt measurements across C-E. Measure current into the collector terminal. Apply different supply voltages. See how the measurements change. Make a table of your results and extract information about On-resistance, and power dissipated."

Student goes home, does hands-on experiments. Calculates operating parameters.

Calculates resistance through C-E and finds it goes up as the supply voltage goes up.

He also feels the transistor heat up. Discovers this could be the way the greater VCC is 'felt', thus justifying his sense of intuition telling him VCC ought to be felt in some way.

Student has a more complete grasp of transistor behavior.

Student no longer feels stupid.

The professor wearing his educator's hat did not contradict the textbook, but he did depart from it, in order to further a student's learning process.

Should one who puts on the educator's hat be called 'dangerous' by one wearing the physicist's hat?

BradtheRad said:

So, Bright Student goes to Professor with his question.
Professor (who wears the physicist hat) taps the textbook and says "Read the definition again. And if it overstrains you, read it again."
Student walks away, feeling stupid.

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Hi Bradthe Rad, I was able to follow your story - however, I hope you can agree that the second part is relevant only under the condition that
* the first prof did reply as assumed by you (a very bad teacher, by the way)
* the student really was overstrained by the description in the textbook (I think, your term "definition" is not quite appropriate).

For my opinion, this is a "worst case situation" - and the student should make up his mind if it is not better to study social sciences or something non-technical.
That`s my position as far as this scenario is concerned.

LvW

EDIT: I do not mean the prof would be a bad teacher because he wears "the physicist hat", but because he is not able or willing to explain things.
In contrary, to explain electronic phenomena to students (not to 15 year old boys) based on physical effects is - for my opinion and according to my experience - the best way (if not a necessity).