Does this mean that Ic will affect base current and consequently Vbe will change?

Not directly. The cause and effect relationship for a given circuit is that changes in Ib will change Ic. Changes in Ic may or may not cause other changes depending on the external circuit. For example will the collector load resistor allow the changes in Ic to occur given the value of the resistor and Vcc.

 

Does this mean that Ic will affect base current and consequently Vbe will change?

In fact, the other way around is true. The changess in Vbe will change the Ic.
https://forum.allaboutcircuits.com/blog/fun-with-the-diode-equation.589/

 

Does this mean that Ic will affect base current and consequently Vbe will change?

No,Vbe changes only slightly with Ib.
And Ib affects Ic, not the other way around.

Below are the characteristic curves of a BJT for a 0.2mA step change in the base current to 1mA maximum, with a 0 to 1V sweep of the collector voltage.
Note the relatively small change in the base voltage, V(b), with change in base current steps.
You can also see how the collector current, Ic(Q1), changes with collector voltage (X-axis scale), for the step changes in base current.

The collector voltage less than about 0.25V is the transistor saturation region.
To the right of this voltage the collector current essentially equals the base current times the transistor current gain (beta).

 

In fact, the other way around is true. The changess in Vbe will change the Ic.
https://forum.allaboutcircuits.com/blog/fun-with-the-diode-equation.589/

Yes, and that is how I understand it.

In reality the Vbe will go up by a 60mV or down by 60mV for every upward /down change in Ic by a factor of ten (decade change).

The way it was worded implied that Ic change would cause Ib to change. So, let me know if the better way to put it would be to say that the Ic will change by a factor of 10 for every 60mV change in Vbe. Correct?

 

https://en.wikipedia.org/wiki/Early_effect

 

Yes, and that is how I understand it.
The way it was worded implied that Ic change would cause Ib to change. So, let me know if the better way to put it would be to say that the Ic will change by a factor of 10 for every 60mV change in Vbe. Correct?

Aleksey....there are several proofs and indications that the collector current Ic is determined solely by the base-emitter voltage Vbe (as mentined already by Jony130).
This is according to the well-known Shockley equation which describes the voltage-current relationship for any pn junction (pn diode and BJT).
There is not a single proof that Ib would control Ic.
Ib is nothing than an (unwanted) by-product of the emitter current Ie.
The misunderstanding simply results from a misinterpretation of the relation Ib=Ic/beta.
This equation tells you that Ib is a small fraction of Ic - and the other form Ic=beta*Ib may be used for calculation purposes but must NOT be interpreted as a cause-and-effect sequence.

Of course, you can use a large series resistor Rb (and a corresponding DC voltage Vb) for biasing the base (instead of a voltage divider at the base node) - and it is common practice, in this case, to speak about "current injection" into the base because the current is primarily determined by the Vb-Rb combination.
However, this is just "laboratory slang" - in reality we have nothing else than a voltage divider consisting of the resistor Rb and the B-E path resistance (which is non-linear and allows a voltage of app. 0.6...0.7 volts only).
Therefore, also in this case we have , of course, voltage control Ic=f(Vbe).

Here is another example for "laboratory slang": We say that in a chain of several resistors the voltage "drop" across one particular resistor would be caused by the current Ic. But this is wrong! A current cannot "produce" a voltage. Each current is the RESULT of a voltge. As far as the voltage-current relation is concerned, some people speak of a "chicken-egg" problem. But this is wrong: No current without a driving voltage !!

 

But as I said, to make your life easier, including being able to understand datasheets use the model in which the transistor provides current gain where both input and output are current.

 

There are a few pedantic members of these forums who make a large point of stating that a BJT is voltage controlled, not current controlled.
While this is technically true at the solid-state physics level, it's not generally a useful model for determining the bias point of amplifiers, or in the design of large signal digital/switching circuits.
For those, a current-controlled, black-box functional model of the BJT is more useful.
The only place the voltage controlled (transconductance) BJT model is really helpful for manual calculations in circuit design, is in determining the small-signal AC gain of a transistor.

So the current-controlled BJT black-box model is more than just "laboratory slang", it is a very useful and commonly used model in the design of circuits.
That is why the current gain (Beta or hFE) is given in all BJT data sheets.

 

Let me see if I am starting to get it. Here is what happens as I see it.

The base-emitter is really just like any common diode (p-n junction for npn transistor). Let's ignore current leaks, temperatures and all other minor things for now and I'll try to concentrate on the concept.

So, initially as I start applying voltage the base-emitter is practically not conducting any electricity and there is no base current, i.e. Ib=0. This means that Vbe will start to increase and this is where transistor is considered to be in cut-off mode , i.e Ic=0.

However things start to change once I hit Vf (about 0.6V). At this point base-emitter starts to conduct electricity and current starts to flow from collector as well. As I increase voltage pn junction resistance continues to drop which increases Ib and causes Vbe to rise only slightly. This current-voltage relationship is not linear. This is also where transistor is in active mode and where amplification can happen and where hFE is used. And I guess here is where its easier to look at current gain and Ib <-> Ic relationship.

At this point one would realize that we need a current limiting resistor since base-emitter really now practically a short circuit.

Now, assuming circuit is setup with current limiting resistor and if you continue applying voltage while still keeping Ib current within acceptable range transistor is considered to be in saturation mode and collector-emitter can be practically considered like a closed circuit.

Am I getting closer and on the right track?

 

Thanks for posting this thread.

I, too, am learning about transistors and it's been very helpful

 

Now, assuming circuit is setup with current limiting resistor and if you continue applying voltage while still keeping Ib current within acceptable range transistor is considered to be in saturation mode and collector-emitter can be practically considered like a closed circuit.

That's true if there's a collector resistance to limit the maximum collector current also.
If the collector is directly connected to a voltage source, then the collector current will just keep increasing with the base current (until it blows from overcurrent).

 

That's true if there's a collector resistance to limit the maximum collector current also.
If the collector is directly connected to a voltage source, then the collector current will just keep increasing with the base current (until it blows from overcurrent).

Yes, this part I understand. But did I get whole concept right otherwise?

 

Yes, this part I understand. But did I get whole concept right otherwise?

Looks good to me.

 

Looks good to me.

Perfect! Thanks!

Now let's see if I can apply this using actual BC550 transistor with a datasheet. Here are parameters for convenience.


And here is the basic schematic.

So, to get this transistor into saturated mode I got to provide at least 5mA into base - second line for Vbe(sat). To calculate R1 resistor value I got to subtract 900mV voltage drop from 5V and divide that by 5mA. (5-0.9)/0.005. This gets me 820Ω. If there are no resistors of that value there are two ways. One way is to find next closest available resistor. If you are ok with transistor not being fully saturated but close to that you can take 1KΩ resistor. Otherwise lets assume next lower value resistor I have is 470Ω. If I use that resistor I will get approximately Ib≈8mA.

Is my thinking correct?

 

820 is actually a standard value, as id 560 amd 680. Not sure why ypu would use 470.

Bob

 

820 is actually a standard value, as id 560 amd 680. Not sure why ypu would use 470.

Bob

Even better. I just don't have it handy in real life, but I do have 470. That is the only reason.

 

The base current needed for saturation is determined by the value of the collector resistor, RLoad, which I don't see.

Why do you say 5mA?

 

I used condition column for Vbe(sat) which indicates 100mA (which is a max current) Ic with 5mA Ib. I was assuming this is where it will be open at maximum. And I thought that all I had to do is to limit current at collector to avoid frying it with higher Vbe.

I guess its not that simple and that I'm still missing something.

 

I guess its not that simple and that I'm still missing something.

5mA is for the maximum rated transistor current, which is likely overkill for a typical application.

You determine the maximum collector current that the RLoad allows at the operating supply voltage.
You then provide a base current about 5 to 10% of that current to insure saturation.
Any base current above that value, is just a waste of current and power.

 

There are a few pedantic members of these forums who make a large point of stating that a BJT is voltage controlled, not current controlled.
While this is technically true at the solid-state physics level, it's not generally a useful model for determining the bias point of amplifiers, or in the design of large signal digital/switching circuits.
For those, a current-controlled, black-box functional model of the BJT is more useful.
The only place the voltage controlled (transconductance) BJT model is really helpful for manual calculations in circuit design, is in determining the small-signal AC gain of a transistor.

So the current-controlled BJT black-box model is more than just "laboratory slang", it is a very useful and commonly used model in the design of circuits.
That is why the current gain (Beta or hFE) is given in all BJT data sheets.

Thank you very much for classifying myself as "pedantic". (Why such a purely personal remark/offense ?)
I like to remind you on the original question: " I am relative beginner in electronics and want to try to really understand transistors. I do understand how it works basically but wanted to get some more in-depth knowledge."

Hence, the questioner is not asking for some "models" or handy formulas for calculation purposes - he wants to "really understand" transistors and he wants to get some "in-depth knowledge".
So - is it really "pedantic" to give him an answer that even you consider as "technically true"?
However, this is not true only on "solid.state physics level" - it is simply physical/technical reality and can be verified by measurements !!

It is a complete other question, which model seems to be convenient for some particular design steps.
And I consider it as very important that a beginner is able to strictly distinguish between (1) physical mechanisms/functional principles and (2) equivalent small-signal models which are nothing else than visual representations of known formulas for a given DC quiescent current.

My point of view results from several years experience in teaching electronics. And - believe it or not - there were always students which were able to see the contradiction between (a) the current-control (physically wrong) model and (b) some observable/measureable effects.

Example 1: According to the 4-pole theory, the input resistance of a system with feedback increases for VOLTAGE feedback only!
As you know, an emitter resistor RE causes negative feedback - and drastically increases the input resistace at the base.
What does this mean? Yes - it is a clear indication of VOLTAGE feedback (the voltage Vbe=Vb-Ve decreases for increasing emitter current).

Example 2: Who is able to explain the function of a simple two-transistor current mirror - based on current-control model?

Finally - nobody will doubt the existance of the base current Ib=Ic/beta. But it is a complete other question if the form Ic=beta*Ib could give you any information on the physical control mechanism. This would be just a misinterpretation.
We must, continuously, realize that a small-signal model is a nice tool to support calculations - but it does not necessarily tell us something about physical working principles.

(Who can really believe how 2 additional charged carriers in the base region should be able to release 500 additional carriers from the emitter, asuming beta=250 ?)