Why there are so many positive and negative ions in N and P types?

AI Thread Summary
The discussion centers on the presence of positive and negative ions in N-type and P-type semiconductors, particularly questioning their existence outside the depletion region. It is clarified that while the depletion region is devoid of free carriers, the bulk of the semiconductor remains neutral with free-moving electrons and holes. Doping introduces extra charge carriers, with N-type semiconductors having additional electrons and P-type having holes, which are created by impurities. The concentration of doping atoms is relatively low compared to silicon atoms, yet they significantly influence conductivity. Overall, the ions' distribution is consistent throughout the semiconductor, not limited to the depletion region.
anhnha
Messages
179
Reaction score
1
The image bellow is the PN junction under equilibrium.
I wonder why there are so many positive and negative ions in N and P types respectively.
For me, I think that these ions should only exist in the depeletion region not outside the region.
attachment.php?attachmentid=59368&stc=1&d=1370674622.jpg

Thanks for help.
 

Attachments

  • PN junction in equilibrium.JPG
    PN junction in equilibrium.JPG
    17.4 KB · Views: 592
Physics news on Phys.org
Do you know what we call n type and p type semiconductor?

ehild
 
ehild said:
Do you know what we call n type and p type semiconductor?

ehild

Yes, I do.
I have just watched it carefully now I see. For example in N type right next to each positive ion is an electron.
I know that the electron and positive ion are separated but I usually consider that they is an unit beacuse its net charge still zero.
 
The p and n type regions are neutral, but the electrons and holes are free to move in the whole crystal, like the conduction electrons in a metal. The pure (intrinsic) semiconductor has only few free charge carriers, as the forbidden band, the gap between valence and conduction band is broader than the thermal energy. But adding impurity atoms which have one electron more than the host atoms (for example phosphorus in silicon) these extra electrons are very loosely connected to their atoms and the thermal energy (kT) is enough to remove them from the atom. We say that the impurities make a donor level just under the conduction band. Similarly, atoms less than one electron than silicon capture an electron from an other atom, so the "hole", the place of the missing electron moves as free positive charge carrier in the crystal. The number of doping atoms can be about 1015-1020 atoms/cm3, so they are not so many with respect to the silicon atoms.

In the depletion region, the free carriers move to the other side of the junction by diffusion, electrons to the p part and holes to the n part, leaving the ions alone. The depletion region is free from free carriers, but the concentration of the ions is the same as in the bulk of the semiconductor.

ehild
 
  • Like
Likes 1 person
Thread 'Collision of a bullet on a rod-string system: query'
In this question, I have a question. I am NOT trying to solve it, but it is just a conceptual question. Consider the point on the rod, which connects the string and the rod. My question: just before and after the collision, is ANGULAR momentum CONSERVED about this point? Lets call the point which connects the string and rod as P. Why am I asking this? : it is clear from the scenario that the point of concern, which connects the string and the rod, moves in a circular path due to the string...
Thread 'A cylinder connected to a hanged mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top