If there is no electron flow through R2 then there is no voltage across R2. Since R2 is in parallel with the base-emitter junction of the transistor, there is no V(be). This in turn causes the transistor to be off.

So, your statement that no electrons flow through R2 is incorrect.

You also say that the electrons do not automatically start from the emitter. The -terminal of the supply pushes it by being a source of electrons. It also pushes electrons through R2 as it has no preference for the transistor.

Considering only the transistor, the electrons from the emitter do then flow through the collector and the base. The amount of doping sets up the relationship and the controls needed to accomplish this. Because of the way the device is created, i.e. doped, etc..., the relationship of currents and voltages is established.

Ic = beta * Ib
Ie = (beta + 1) * Ib

I believe the Ebers-Moll model sets up the relationships of the currents to the external excitation voltages, V(be) and V(ce).

Given these detailed relationships of the currents and the voltages, the transistor can be used to control the amount of current that flows through it.

The model of the transistor needs to be inserted into the circuit and not take over the circuit. R1 and R2 are typically used to bias the transistor ON so the current will flow at all.

 

Transistor is comsidered "transisting resistor" which means is a transistor acts as a variable resistance between the emitter and collector, due to a base current activating it.

The doping of a transistor is to make it conduct under certain conditions, unlike a resistor, which conducts continuously.

A transistor is doped so that it acts instead like a SEMI conductor, while a resistor acts as a CONTINUOUS conductor.

So a transistor cannot supply electrons continuously, after the barrier junction has recombined electrons with holes.

You seem to have a good understanding of the junction barriers, so think about it, when the electrons recombine with holes at the junction, then there is a charge of opposite polarity at the junction, this is the barrier .

In order to overcome that junction barrier there has to be an external supply of electrons (charge) to push forward the electrons at that junction, that is the Base bias voltage Vbe, that everyone is telling you about.

Therefore the transistor will sit passively until this barrier is overcome, then it goes into conduction region, and becomes an active devise.

It is this doping that makes a semiconductor NON linear, as well.

Please don't PM me with any questions, because most of the time I don't log in but just read the posts as a guest, so I hardly ever read any personal mails, cause I don't know there in there, until late, after later time a log in.

The forum is an open forum so all questions please bring to the open thread.

Thankyou...

 

If there is no electron flow through R2 then there is no voltage across R2. Since R2 is in parallel with the base-emitter junction of the transistor, there is no V(be). This in turn causes the transistor to be off.

I cant understand it. I want a detail explanation on this. Perhaps my answer lies in this statement.Please give a detail explanation on this.

You also say that the electrons do not automatically start from the emitter. The -terminal of the supply pushes it by being a source of electrons. It also pushes electrons through R2 as it has no preference for the transistor.

I said that the -terminal of battery pushes electrons in emitter and not from -ve terminal.
Ok according to you let the -terminal start supplying electrons, then how does the emitter maintain the concentration of free electrons. IF we consider the flow of electrons starting from -terminal then we neglect the free electrons in emitter. THats why i asked why the transistor is doped and also kept insisting that electrons start from emitter due to push given by -terminal.

Are you are finally telling that in all the electronics circuits the -terminal of the batter only supplies the electrons and not the solid state device?
Perhaps does the batter supply more electrons that Ie = Ib + Ic. Then only the concentration of free electrons in emitter can be maintained. Otherwise if battery supplies electrons = Ib + Ic then some flows through R2 and rest to emitter and the emitter wont get complete Ie it has lost thereby affecting the property of the transistor. Is the electrons supplied by the battery greater than Ie?

Then could you please help on what does the transistor actually do and what for it used and how it accomplishes its duty.

 

This is a further question, sent by PM. It is posted by request of the OP -

http://www.allaboutcircuits.com/vol_3/chpt_2/8.htmlIn this website
The explanation on Bjt is very good . But i have some doubts in it.
It is mentioned in the site that

1) holes in the base may diffuse into the emitter and combine with electrons, contributing to base terminal current. I understand the diffusion process but what is that base terminal current.

2) Also if the holes have been diffused into the emitter wont the emitter acquire negative charge and start blocking the incoming electrons from emitter?

3) Also what will happen if we increase Vbe. Do the minority carriers i.e holes in emitter flow to the base thereby making base more positive and increasing the collector current ?

3) http://www.tpub.com/content/neets/14...s/14179_69.htm

If we increase Vce do the minority carriers i.e holes in collector flow to the base therby making the base more positve and increasing current flow?

 

If you remove the transistor from the circuit, do any electrons flow? Yes! They will flow through the remaining resistors that are connected. What is supplying the electrons in this case?

The current from the electron flow through the resistors causes a voltage drop across the resistors according to the relationship V = I * R. If the electron flow through R2 is zero, the voltage drop across R2 is zero. Now look at the original diagram. R2 is connected across the transistor V(be) and Re. Electron current flowing through Re and V(be) would cause a positive voltage being developed across both. However, it must be equal to the voltage drop across R2 since they are connected together at Vb. Therefore, you must either have an electron current through both paths or none through either path.

How does the battery push the electrons from the emitter, considering it must go through the Re that is connected to ground?

Perhaps does the batter supply more electrons that Ie = Ib + Ic. Then only the concentration of free electrons in emitter can be maintained. Otherwise if battery supplies electrons = Ib + Ic then some flows through R2 and rest to emitter and the emitter wont get complete Ie it has lost thereby affecting the property of the transistor. Is the electrons supplied by the battery greater than Ie?

Yes, the electrons supplied by the battery is greater than the number needed for just Ie, it is enough for Ie + I(R2).

The doping of the device tries to keep the concentration of free electrons a certain way so the device operates in a specified way. As free electrons move into the base and emitter, they must be replaced.

These devices are very useful devices whose properties of Vbe ~ 0.6V, Ic = beta * Ib, and Ie = Ic + Ib. These parameters are setup by the doping of the device. Consider small electron currents in Ib. They generate much larger currents in Ic. It amplifies the signal! Very useful indeed! And, it is accomplished by the way it is constructed and by the fact that its construction requires free electrons to move from the emitter to the base and collector.

There are, however, many more elements in the circuit that need to be considered as well.

 

If you remove the transistor from the circuit, do any electrons flow? Yes! They will flow through the remaining resistors that are connected. What is supplying the electrons in this case?

The current from the electron flow through the resistors causes a voltage drop across the resistors according to the relationship V = I * R. If the electron flow through R2 is zero, the voltage drop across R2 is zero. Now look at the original diagram. R2 is connected across the transistor V(be) and Re. Electron current flowing through Re and V(be) would cause a positive voltage being developed across both. However, it must be equal to the voltage drop across R2 since they are connected together at Vb. Therefore, you must either have an electron current through both paths or none through either path.

How does the battery push the electrons from the emitter, considering it must go through the Re that is connected to ground?




Yes, the electrons supplied by the battery is greater than the number needed for just Ie, it is enough for Ie + I(R2).

The doping of the device tries to keep the concentration of free electrons a certain way so the device operates in a specified way. As free electrons move into the base and emitter, they must be replaced.

These devices are very useful devices whose properties of Vbe ~ 0.6V,.

NOW I UNDERSTOOD WHY ELECTRONS SHOULD FLOW THROUGH R2 FROM YOUR SUPERB EXPLANATION (IN BLUE) . .THanks.

Since the you have stated ( in red) that the electrons supplied across the battery is greater tha Ie, then why are we neglecting the electrons in emitter. What is the purpose of the electrons in emitter and what do they accomplish if we consider only the electrons flowing from battery and not talking about movement of electrons in emitter.

From your explanation ( pink), thats what i am asking from the beginnning. You are telling that the concentration of electrons must be kept constant in emitter but we are never considering the action of electrons in emitter but we are considering only electrons from battery which simply move through the transistor as if it were like a resistor.
We are not giving respect to the free electrons in emitter .
My doubt regarding this topic is almost over since you beautifullly convinced me about why electrons should flow through R2.

PLease give a detailed explanation as to what will actually happen to the free electrons in emitter when the bias is applied and also when the battery electrons approach the emitter.

Also please explain this
http://www.allaboutcircuits.com/vol_3/chpt_2/8.htmlIn this website
The explanation on Bjt is very good . But i have some doubts in it.
It is mentioned in the site that

1) holes in the base may diffuse into the emitter and combine with electrons, contributing to base terminal current. I understand the diffusion process but what is that base terminal current.

2) Also if the holes have been diffused into the emitter wont the emitter acquire negative charge and start blocking the incoming electrons from emitter?

3) Also what will happen if we increase Vbe. Do the minority carriers i.e holes in emitter flow to the base thereby making base more positive and increasing the collector current ?

3) http://www.tpub.com/content/neets/14...s/14179_69.htm

If we increase Vce do the minority carriers i.e holes in collector flow to the base therby making the base more positve and increasing current flow?

 

The negative terminal of Vcc pushes the electrons in emitter to the base.

This is where you're wromg.

It should go like this:
The negative terminal of Vcc pushes the electrons towards emitter and the bottom pin of R2.

 

We do not neglect the free electrons in the emitter. The doping of the transistor determines how the electrons flow through the device. Once the free electrons begin to flow they are indistinguishable from the ones supplied by the battery.

The electrons in the emitter do not stay in the emitter. They move to help form the current. Some of the electrons from the battery stay in the emitter until they are needed to maintain the current flow. They can become the free electrons in the emitter.

So, maybe you can say the electrons from the battery push the emitter electrons to the base and collector. Just don't forget that the electrons from the battery will also flow through R2 as the electron is basically just looking for the path of least resistance.

I have not read the AllAboutCircuits section on BJT so I will refrain from answering those questions until I get more time...