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) 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?

 

To understand the junctions of a transistor, you need to start out with understanding a PN junction, DIODE.

A diode is built with 2 substrates, one is a material that is doped with extra electrons, while the other is doped to have a deficiency of electrons, producing pos. charged holes, in the lattice structure.

These by themselves make neither a good conductor, nor a good insulator.

But when they are combined the form a SEMI conductor.

Now during manufacturing process, the two materials P and N type, are fused together, the moment that happens, then the majority carriers (electrons) in the N type material, combine with the holes in the P type material.

After this recombination, then a net charge is produced where the PN junction now has a potential barrier, means all the holes in the P type material NEAR THE JUNCTION ONLY, have electrons in there gaps, that now gives them a neg. charge, this neg. charge repells any more electrons from crossing the junction, so now the diode is inactive.

This barrier potential , can only be overcome by an external voltage source great enough to pull electrons out of the holes, this potential is around 0.6 to 0.8 volts, (silicon) known as the junction barrier voltage, this junction voltage breaks down the barrier, to allow current to flow readily through both materials.

Once the external voltage is removed, then recombination takes place and the barrier is established once again.

 

As a start, you may view the following:

http://ocw.mit.edu/NR/rdonlyres/Ele...D46-A37A-A6C9A4EFCF56/0/MIT6_012s09_lec14.pdf

For an overall understanding of such subject mater, you may browse around the following link:


http://ocw.mit.edu/OcwWeb/Electrica...ience/6-012Spring-2009/LectureNotes/index.htm

 

To understand the junctions of a transistor, you need to start out with understanding a PN junction, DIODE.

A diode is built with 2 substrates, one is a material that is doped with extra electrons, while the other is doped to have a deficiency of electrons, producing pos. charged holes, in the lattice structure.

These by themselves make neither a good conductor, nor a good insulator.

But when they are combined the form a SEMI conductor.

Now during manufacturing process, the two materials P and N type, are fused together, the moment that happens, then the majority carriers (electrons) in the N type material, combine with the holes in the P type material.

After this recombination, then a net charge is produced where the PN junction now has a potential barrier, means all the holes in the P type material NEAR THE JUNCTION ONLY, have electrons in there gaps, that now gives them a neg. charge, this neg. charge repells any more electrons from crossing the junction, so now the diode is inactive.

This barrier potential , can only be overcome by an external voltage source great enough to pull electrons out of the holes, this potential is around 0.6 to 0.8 volts, (silicon) known as the junction barrier voltage, this junction voltage breaks down the barrier, to allow current to flow readily through both materials.

Once the external voltage is removed, then recombination takes place and the barrier is established once again.

I understood your explanation on working of pn junction diode.
But you didnt answer my 3 questions. You have given an explanation on how pn diode works which i understood after reading your explanation.

 

holes in the base may diffuse into the emitter and combine with electrons

The P material has holes due to a manufacturing process that injects atoms into the base silicon. Those atoms create the P characteristic. They are embedded, and do not move. They can't migrate across the junction into the N material.

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

As above, that can't happen. That also covers both your #3 questions.

 

The P material has holes due to a manufacturing process that injects atoms into the base silicon. Those atoms create the P characteristic. They are embedded, and do not move. They can't migrate across the junction into the N material.

But i didnt give my own assumption this site about BJTS in Semiconductors only tells that the holes in base can diffuse into the emitter.See for yourself
http://www.allaboutcircuits.com/vol_3/chpt_2/8.html

Also i have one more important question.
Does the negative terminal of Vcc in common emitter configuration also bias the emitter.
Also why is Vbb ie the supply connected to base emitter is always quite small. What happens if it is large ?

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) 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?