What determines the thickness of depletion region of a given p-n junction diodegiven

[ovais]

Hello all,
I am somewhat confused in understanding the depletion region of a p-n junction diode.Everywhere a depletion region is defined as a region near the p-n junction which is devoid of charge carriers.And the formation of this region is said is explain that it is due to the diffusion of majority charge carriers of p-type crystal into n-type(movement of hole from p and electrons from n), and the diffusion of the carriers(electrons and holes) results into a region which is devoid of charge carriers(there however are ions in the depletion region i.e an anions(-ve) in p and cations(+ve) in n, as such in the formation of p-type intrinsic semiconductor it is doped with trivalent atom which creates hole as majority carrier by accepting an electron thereby become -vely charged ion or anion and the n type has electrons and +vely charged ion or cation)Thus the width of depletion region should be a space within the junction which has no charge carriers(electron or hole) but charged ions(cations on one side and anions on the other side of the junction)As the loss of holes in p side and loss of electrons in n side(during the formation of junction, i.e. on sandwiching them) are equal and the number of anions in p side of the depletion region is also equal to the loss of holes on that side similarly number of cations on the n side of depletion region is equal to the loss of electrons on that side.It means the region on right of depletion(p-side) has equal number of holes(after diffusion) and anions(so the region is neutral) and also that to left of depletion region(n side) will have equal number of electrons(after diffusion) and cations.So the only area of negative and positive charge is the depletion region in which on one side we have charged ions of one type(-ve ions) and on the other side ions of another type(+ve ions).This what is studied, further i studied the concept of potential barrier, forward biasing and reverse biasing.There is however coming a gap between formation of depletion layer and development of potential barrier.The formation of depletion region is clear(I explained in detail what i understand how it forms(concept of diffusion) and what it is made of(+ve and -ve ions)and that region to right and left of depletion region is neutral, this makes me think the following- 1-As the right and left of depletion is neutral, the only source of formation of potential; barrier is the ions in the sides of the deplettion region. 2-As these ions build up the potential, the higher their number the higher the potential barrier. 3-In forward biasing the hole and electron migrate toward the depletion region.Therefore a larger portion on the sides near the junction will be devoid of charge(due to further combination of hole and electron due to applied electric field or voltage)So the thickness of depletion region should increase, while it is well known that during forward biasing depletion layer thickness decreases.This is what i am not understanding. I know is must missing something.please make me understand why deletion layer decreases.Also let me make clear that on forward biasing, do the holes and electron(which moved due to applied voltage) cross the junction or just remain in the depletion region(which was formed during the formation of p-n junction after they were sandwiched)? [/B

 

[sokrates]

Depletion region decreases when you apply forward bias because you decrease the energy barrier that carriers experience in the junction. Conversely, if you reverse bias the diode, the energy barrier (for both types of carriers) will increase - hence there must be a HIGHER voltage drop across the junction. As you said, the only way you can have voltage drop across the junction is by changing the width of depletion region (because the other parts are neutral). Hence, depletion region expands in this case.Try to look at it from an energy-band perspective, drift-diffusion and internal mechanics can be understood from that vantage point.

 

[ovais]

Thanks sokrates for your reply.This fine to say that on forward biasing the externally applied field makes the potential barrier weak.There is however another fact(if this really a fact) associated with the potential barrier that the cause of potential barrier is the ions of the depletion region(anions on side of depletion region and cations on the other side of depletion region in between is the junction of the diode) therefore magnitude of potential barrier is directly proportional to the thickness of the depletion region.Also nothing disturbs the mind when it is said that when depletion layer decreases the potential barrier decreases.I mean the value of potential barrier(of a junction diode) depends directly on the depletion layer thickness.The problem where my mind is baffling is that what happens(during forward biasing) in side the pre-exist depletion region that makes its thinner.By definition depletion region is the region which is devoid of free charge carriers(that is electrons and holes), so decrease of depletion layer thickness will mean that some part of the pre-exist depletion region has now(on forward biasing) some free charge carriers due to this that portion will be out of depletion region i.e width decreases.I wonder what do happen when holes from p side and electrons from n side(on forward biasing) migrate towards the depletion region(due to forward biasing).There are three possibilities comes to my mind,(1)-the electrons and holes move toward the depletion region, enter into it , combines with each other and vanish.(2)-the holes from p side and electrons from n side get accelerated toward the depletion region but can not penetrate into the depletion region and remains accumulated on either side of the depletion region(that is holes get accumulated near p side and electrons near n side of the depletion region). (3)-the holes from p side not only penetrate the depletion region but also cross the junction and reached into the n side, and these hole become the minority carriers of n side, similarly electrons of n side comes to p side by crossing the junction and becomes minority carriers of p side. As u said consulting energy band will be useful, yes it will surely be helpful to get into greater in sight but i am having problem in understanding the basic idea of what happens to those electrons and holes when they move towards the junction.Any help in this part will be highly appreciated. THANKS A BUNCH

 

[geekpassion]

The same question ran through my mind ... and your question is very good and worth asking...here's what my school book and wikipedia has to say on this-

In forward bias, the p-type is connected with the positive terminal and the n-type is connected with the negative terminal.
In forward bias the holes in the P-type region and the electrons in the N-type region are pushed toward the junction. This reduces the width of the depletion zone.
As electrons and holes are pushed toward the junction (remember, they are not crossing the junction and increasind its length), the distance between them decreases. This lowers the barrier in potential and the width of the depletion zone decreases.
The depletion zone eventually becomes thin enough that the zone's electric field cannot counteract charge carrier motion across the p–n junction, as a consequence reducing electrical resistance.
Now p-n junction is nothing but a mere resistor. The electrons due to the connected battery can now flow through this resistor. Now even the holes and electrons of the diode take part in this current flow. Electrons from the battery, flow from one end of n- side to the junction and eventually to the other end of p side, back to the battery. The charge flow is through the circuit and does not stop at the junction and effect its length.

reply if you understood

 

[bmaggy]

This is what I think and please correct me if I'm wrong.
There is no change in the number of atoms in the depletion region in either forward or reverse bias.When not connected to a voltage source i.e in equilibrium the +ve and -ve ly charged ions are stable cause their octates are complete. So even when voltage source is connected there is no change in the barrier potential as these ions don't combine with either e's or holes.

 

[marcusl]

In forward biasing the hole and electron migrate toward the depletion region.Therefore a larger portion on the sides near the junction will be devoid of charge(due to further combination of hole and electron due to applied electric field or voltage)So the thickness of depletion region should increase, while it is well known that during forward biasing depletion layer thickness decreases.This is what i am not understanding.

During forward bias, the energy bands bend under the applied potential and the height of the potential barrier drops. Charge carriers begin to flood into the depletion region, eventually (with sufficient forward bias) turning it into an electrically neutral plasma. Some carriers are lost to recombination but most diffuse through the junction region to the opposite sides (holes to the n side and electrons to p), supporting a current.