transitor: the base pin.

[PlagueDoctor]

I've always been to embarrassed to ask this question, since I've designed & built several circuits, beambots, etc using transistors, even helped other people out... Any ways.

(Assuming NPN type) when applying a current to the base pin to allow the current to flow from the collector to the emmiter(or vice versa) where does the current from the base pin go?

I've always assumed either one of two things:

The current of the base pin shares the ground with the emmiter pin. (Which doesn't make sense to me)

Or the current doesn't actually flow to the base pin, but rather puts it under an electrical tension. (Which makes a lot more sense to me)

Or am I way off base? Surprisingly, despite having read quite a few articles about them. Even going as far to learn how they are manufactured. I haven't crossed any text that explains this clearly.

Thank you for your time"

[Simon]

The base current flows from the base and out of the emitter. Why does this not make sense ?

The "electrical tension" you refer to is how MOSFET's work where no current flows.

Between the base and emitter is basically a diode, the current through it x gain = collector current.

[LvW]

It is the the voltage Vbe that causes a current Ie out of resp. into the emiitter  - and this current is split into one large current Ic and a much smaller current Ib (which is a more or less fixed percentage of Ic). Therefore, the current Ib of course goes into (or out of) the emitter node according to KCL: Ie=Ib+Ic.

[Tandy]

No need to be embarrassed, the best way to learn is ask questions about things you are unsure about. It is better to admit you don't know and learn the answer than pretend you do. There are plenty of people pretend to know and then give out misinformation to others.

As others have said the current applied to the base travels to the emitter.

When I first started to learn electronics I connected a circuit where I had tied the collector to +v and the emitter to a buzzer. I couldn't understand why the transistor would not switch reliably but eventually realised that the small current couldn't flow between the base and the emitter due to the resistance introduced by the buzzer. This is why you typically use a common emitter circuit connecting the emitter to ground with an NPN transistor and control the amount of current applied to the base using a resistor.

A somewhat inaccurate but easy to visualise illustration of a transistor is a big pipe filled with large balls connected to the collector, and a tiny pipe with little balls connected to the base. Each tiny ball that passes through from the base to emitter carries a big ball with it from the collector. The more small balls you pass through the base the more big balls travel from the collector. Both the small and big ball drop out of the emitter.

[Zero999]

Yes, the current flows into the base and out of the collector.

The base-emitter junction forms a diode and this is why, in a common emitter amplifier the base current needs to be limited, normally by a resistor, otherwise an unlimited current will flow, causing it ti overheat.

A base resistor is not required in an emitter follower because when the transistor turns on, the load between the emitter and 0V drops a voltage across it which lifts the emitter voltage up, causing the base current to fall, so the circuit balances itself; this is an example of negative feedback and is why the input impedance is high.

Normally the base is the input but it's also possible to fix the base at a steady voltage and change the emitter voltage, as is done in a common base amplifier.

[LvW]

A somewhat inaccurate but easy to visualise illustration of a transistor is a big pipe filled with large balls connected to the collector, and a tiny pipe with little balls connected to the base. Each tiny ball that passes through from the base to emitter carries a big ball with it from the collector.

OK, this visualization looks rather good but one should mention that it models the BJT as a current-controlled device. No great problem - such a model works for many applications. However, you always should know what you are doing and, thus, realize that this is a model only. The physical reality is that the BJT is, of course, a voltage-controlled device (Ic=f(Vbe)).  This is a proven fact.

[rsjsouza]

As others have said, the base current flows to the emitter.

Physically speaking, the base region is very narrow when compared to the emitter and the collector regions, and the flow of current occurs when a positive (NPN) or negative (PNP) voltage (in relationship to the emitter) is applied to it.

This causes electrons (NPN) or holes (PNP) to be attracted to it at a great speed and, being narrow, the majority of these charges (electrons or holes) will completely bypass the base region and jump to the collector, which on itself has a positive (NPN) or negative (PNP) voltage applied to it in relationship with the emitter. This voltage on the collector ends up accelerating the charges even more, which causes a large current flow.

The higher the base voltage, the larger the current flow between emitter and collector, but only up to a certain limit. This limit is the transition voltage between the base and the emitter, which for silicon is typically 0.6~0.7V.

The base current itself is the small percentage of charges that end up staying in the base region (recombination) and flow outside of the device. The relationship between the base current and the large collector current is the transistor gain.

(I am almost sure I got all the facts correct, but it is 6AM and I may have missed something)

[Simon]

OK, this visualization looks rather good but one should mention that it models the BJT as a current-controlled device. No great problem - such a model works for many applications. However, you always should know what you are doing and, thus, realize that this is a model only. The physical reality is that the BJT is, of course, a voltage-controlled device (Ic=f(Vbe)).  This is a proven fact.

really ?

[deephaven]

OK, this visualization looks rather good but one should mention that it models the BJT as a current-controlled device. No great problem - such a model works for many applications. However, you always should know what you are doing and, thus, realize that this is a model only. The physical reality is that the BJT is, of course, a voltage-controlled device (Ic=f(Vbe)).  This is a proven fact.

really ?

http://www.vanderbilt.edu/AnS/physics/brau/H182/Theory%20of%20Transistors.pdf section 3.1.3

[macboy]

A somewhat inaccurate but easy to visualise illustration of a transistor is a big pipe filled with large balls connected to the collector, and a tiny pipe with little balls connected to the base. Each tiny ball that passes through from the base to emitter carries a big ball with it from the collector.

OK, this visualization looks rather good but one should mention that it models the BJT as a current-controlled device. No great problem - such a model works for many applications. However, you always should know what you are doing and, thus, realize that this is a model only. The physical reality is that the BJT is, of course, a voltage-controlled device (Ic=f(Vbe)).  This is a proven fact.

For most people here, how it works is far more important and useful than why it works. The voltage developed at the Vbe junction may be why current flows in the collector, but the function "f" in Ic=f(Vbe) is non-trivial. It is defiantly complex.

This is where the how it works is so important. The Vbe voltage corresponds to a Ib current, and reciprocally, any Ib current gives a Vbe voltage... just like with any forward biased PN junction. But what is great is that in Ic=f(Ib), the function f is much, much, much simpler than in the above case. At its simplest, a first order approximation works very well over a relatively wide operating range. So Ic=G*Ib where G is the current gain or hFE of the transistor. It is a simple linear relationship (understanding of course that this is an approximation). So, how a BJT works is simple, current Ic is current Ib times gain. Can you imagine trying to design a BJT amplifer with only the knowledge that Ic is controlled by Vbe, and with no knowledge of the relationship with Ib? It would be a nightmare. Engineers design BJT circuits with Ib, and account for some Vbe that is implicitly created by that Ib. Physicists know that Vbe controls the flow of electrons through both the base and collector, but it stops there. While Vbe controls the current flow, explicitly controlling Vbe in order to control current flow is not practical. The concept is ridiculous.

So, in short, stop preaching about Vbe. This is an engineering forum, not a physics forum, and furthermore, it is the "beginners" section. You are confusing people.

[dannyf]

The physical reality is that the BJT is, of course, a voltage-controlled device (Ic=f(Vbe)).  This is a proven fact.

Only to those who agree with such "fact" of yours.

where does the current from the base pin go?

Emitter. You will often see that emitter current is (1+beta)*Ib, vs. beta*Ib for collector current.

Or the current doesn't actually flow to the base pin, but rather puts it under an electrical tension.

For BJT, there will always be a base current, however low it is.

For FET, there is no "gate current" (in theory and in DC analysis). The tension is the opening / closing of the gate area (channel as it is called).

However, in real life, there is gate current, due to leakage and more importantly in AC analysis, as the gate of a FET is essentially a capacitor.

[LvW]

So, in short, stop preaching about Vbe. This is an engineering forum, not a physics forum, and furthermore, it is the "beginners" section. You are confusing people.

Macboy - why so sensitive? Is it really necessary to answer this way?

1.) I am not "preaching" - I gave a pure technical comment .

2.) In particular, for an engineer it is important to know WHAT he is doing and WHY this is allowed (or not). If you read my reply again you will notice that I did NOT mention any charge carrier behaviour or somethink like that. Hence, why do you mention a "physic forum"?
I am sure, everybody who knows the exponential law for the pn junction of a diode will have no problems to accept a similar relation for the transistor.
Hence, where do you seee a problem?

3.) I am confusing people? Just because I have mentioned the physical reality ? Do you really need some justification derived solely from CIRCUIT behaviour? 

4.) What people really confuses (and this is my experience in teaching electronics at a university for 25 years): 
Teaching current-control of BJT`s for beginners and being NOT able to explain all effects to be observed in electronic circuits (without using the voltage-control properties).

5.) Did you ever realize WHY the input resistance of a circuit with Re feedback increases? This is theory of voltage feedback! Is this physics or electronics?
How do you explain the difference between class-A amplifiers and class-AB or class-B amplifiers? 
Where do YOU select the desired operational point? Not on the Ic=f(Vbe) characteristics?
I hope you now recognize that I am writing as an engineer and not as a physicist.

Regards
LvW

[LvW]

The physical reality is that the BJT is, of course, a voltage-controlled device (Ic=f(Vbe)).  This is a proven fact.

Only to those who agree with such "fact" of yours.

Is this a technical comment? Do you need references?
I think, it is really a pity that in such a forum like this a pure technical discussion seems to be problematic. Why so polemic?

[dannyf]

not a physics forum

his/her comments are more wrong from a physics point of view, stemming I guess a complete lack of understanding of what a "current control" vs. "voltage control" mechanism means.

[c4757p]

Instead of having another tedious current-vs-voltage pissing contest, why not just go read the last twelve we had?

[IanB]

not a physics forum

his/her comments are more wrong from a physics point of view, stemming I guess a complete lack of understanding of what a "current control" vs. "voltage control" mechanism means.

I am not following your comment. The collector current in a BJT is controlled by the base-emitter voltage. This is how the device works and can be represented with considerable accuracy by device models.

The current gain of a transistor is a simplifying abstraction that can be used to construct many aspects of a circuit design, but as a simplifying abstraction it has limitations and is not sufficient to handle all design considerations.

[zapta]

OK, this visualization looks rather good but one should mention that it models the BJT as a current-controlled device. No great problem - such a model works for many applications. However, you always should know what you are doing and, thus, realize that this is a model only. The physical reality is that the BJT is, of course, a voltage-controlled device (Ic=f(Vbe)).  This is a proven fact.

This looks correct to me. Ic = f(Ib),  Ib = g(Vbe) an therefore Ic = f(g(Vbe)).  In a similar way, the light emitted  by a LED is a function of its voltage. These are not useful models because of their sensitivity and instability but correct nevertheless.

[Tandy]

IMHO this kind of thing is very off putting to a beginner. If I had been the original poster who asked the question I would probably now be thinking bah I don't understand what these guys are talking about I'll just forget it.

I appreciate that people are just trying to be helpful in giving a more detailed explanation but too much detailed analysis can be detrimental.

For example, when a diode was first explained to me it was using a 1 way water valve example. Such an example does not show things like voltage drop or recovery times or any other aspect of a diode. Using water as an example also quickly breaks down when we need to explain things like EMF. However as a complete beginner this was perfectly adequate to set me on a path of understanding the basic purpose of a diode and an idealised explanation of how it functioned. Later when I had this understanding I was able to progress to a more detailed understanding of how a diode works.

I had a recent similar experience not related to electronics. I was considering getting a CNC machine and a vacuum forming machine to make some plastic parts. I joined a CNC forum and asked a few beginners questions trying to establish how to get started only it set off big debates among the knowledgable people over the merits of certain machine types and construction techniques etc that in the end I felt overwhelmed and decided I was unable to understand what they were saying enough to make an assessment of the merit of their opinions. In the end I just gave up.

[LvW]

Instead of having another tedious current-vs-voltage pissing contest, why not just go read the last twelve we had?

Had we? Why do you qualify this as a "pissing" contest (instead of giving a technical reply)?
Is this not a forum for discussing technical matters?
Don`t you agree that students/beginners can learn a lot from such discussions (and also how engineers should NOT interact with each other!)

[c4757p]

Instead of having another tedious current-vs-voltage pissing contest, why not just go read the last twelve we had?

Had we? Why do you qualify this as a "pissing" contest (instead of giving a technical reply)?
Is this not a forum for discussing technical matters?
Don`t you agree that students/beginners can learn a lot from such discussions (and also how engineers should NOT interact with each other!)

Exhibit: How I thought about this the last twelve times we had this discussion. This time, I return with a bit more experience actually trying to teach people. It just confuses them. These arguments have been trotted out loads of times before and most are tenuous at best. We're all bashing each other over the heads with our perspectives and it doesn't get anyone anywhere.

[LvW]

IMHO this kind of thing is very off putting to a beginner. If I had been the original poster who asked the question I would probably now be thinking bah I don't understand what these guys are talking about I'll just forget it.
I appreciate that people are just trying to be helpful in giving a more detailed explanation but too much detailed analysis can be detrimental.

Tandy - in the following I repeat one sentence from post#1:
Or am I way off base? Surprisingly, despite having read quite a few articles about them. Even going as far to learn how they are manufactured. I haven't crossed any text that explains this clearly.

Don`t you agree that this question deserves a bit more than only a simple analogy?
I agree with you that - in explaining things - we always should try to find a good trade-off regarding the details of techical/physical effects. However, are you afraid that a simple exponential law for the pn junction will overstrain the questioner (who did ask for a "clear" explanation) ?   

[Tandy]

Don`t you agree that this question deserves a bit more than only a simple analogy?

Perhaps this is down to a different understanding of the meaning of the question. I felt that the OP had read information explaining transistors, some quite technical but none of the explanations answered the (relatively simple) question.

Most transistor descriptions I read explain things like the doping of semiconductors to create P & N regions, the way the electrons overcome the barrier, the hFE etc but don't explain in a simple way for a beginner that current is flowing from base to emitter when controlling the transistor. So a simple illustration would provide a simple way of answering the question without complicating it. Only the OP will be able to clarify the meaning of their question and their level of current understanding. I however prefer to start simple in a beginners forum and then they can always turn round and say, the explanation is to basic I want to know more about the detailed workings etc.

[IanB]

(Assuming NPN type) when applying a current to the base pin to allow the current to flow from the collector to the emmiter(or vice versa) where does the current from the base pin go?

I've always assumed either one of two things:

The current of the base pin shares the ground with the emmiter pin. (Which doesn't make sense to me)

Or the current doesn't actually flow to the base pin, but rather puts it under an electrical tension. (Which makes a lot more sense to me)

If we reduce this question to its core, it asks "Where does the base current go?" Answer: it goes out at the emitter.

Reading the rest of the question betrays some misconceptions, however. For instance, you do not "apply a current" to anything. You apply a voltage and current flows.

Current tends to flow between two different points if there is a voltage difference between them and if there is a path for the current to take.

In a normally biased NPN transistor the base voltage is higher than the emitter voltage. Therefore a current will tend to flow from the base to the emitter. An NPN transistor provides a path for this current and so a current does in fact flow into the base and out from the emitter.

When the transistor is operating normally, current also flows into the collector and out of the emitter.

Since all current must be accounted for, there is an equation that can be written for the transistor:

      I_c + I_b = I_e

This equation says that the sum of the base current and the collector current is the emitter current.

[Tandy]

IanB I like the explanation, very clear.

[free_electron]

It is the the voltage Vbe that causes a current Ie out of resp. into the emiitter  - and this current is split into one large current Ic and a much smaller current Ib (which is a more or less fixed percentage of Ic). Therefore, the current Ib of course goes into (or out of) the emitter node according to KCL: Ie=Ib+Ic.

nonsense

It is an electrical field that causes electrons to flow into the emitter and out of the base ( for an NPN transistor , using electron model , not conventional model ).
in order for electrons to start moving you need to breakdown the recombination zone (often wrongly called the depletion zone. A depletion zone is an area that doesnt have any free electrons. (note : do not confuse this with depletion in depletion mode mosfets. that is classic model , not electron model !)) between the N and P material. The field required for that is in the order of 0.6 to 0.7 volts. once the recombination zone is gone current flows. Vbe is the field across this band. the bipolar transistor is NOT voltage controller. the relation is Ic = Ib x Beta. Pump more current in the base and you let more current flow in the collector.
Simple as that. Bipolar transistors are current controlled.

i gave a detailed electron model explanation including diagrams already on the forum . search for it.

trust me, i've spent 23 years of my life taming electrons to run where i want them to run. i got a pretty good grasp on that stuff.