- #1
PainterGuy
- 940
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hi everyone,
i have discussed this problem with some other people but couldn't understand it well because of my own limited knowledge and understanding of such things. then i sent an email to my very dear teacher and a friend. email copy and my teacher's reply is below. now the problem is i couldn't exactly understand my teacher's reply too and he won't be available for some time. i would highlight the parts which i find more problematic and perhaps you can shed some light from different angle. and i could reach a understanding. i will be very much grateful for any help you can come up with. many thanks.
copy of e-mail sent:--
please checkout the fig. 42.25 “a” and “b” on this linked document:
https://docs.google.com/viewer?a=v&p...thkey=CKP-7OsD
i can understand the fig. 42.25a and why the current flows. but I have some questions about fig. 42.25b.
in the fig. 42.25b, the positive terminal of the battery will suck electrons out of the n-side (because this side is rich in -ve charge) which will leave behind 'positive' holes. the negative terminal of the battery will feed the electrons to holes in the p-side. As the n-side is deficient in electrons because the positive terminal of the battery has sucked them toward itself so this n-side will attract electrons out of the p-side which is now rich in electrons because the negative terminal of the battery has supplied electrons to the p-side (or, in other words the p-side has sucked electrons out of the negative terminal). in my opinion this will result in an electric current and this current won’t be small, if not larger than the current when the diode is forward-biased. that would mean there would also be current when battery terminals are reversed as in the fig 14.25b, or, when the diode is reverse-biased
i'm trying to understand from physics point of view, how a P-N junction works. help me please.
the reply received:-In the P material, the holes must outnumber the electrons by roughly 10^8 to one, just as in the N material the electrons must outnumber the holes by a similar ratio.
This is very analogous to the chemistry of aqueous solutions. In such a solution at 25C, the product of concentration of H+ ions with the concentration of OH- ions must be 10^-14. In an acid solution, by forcing a high concentration of H+ ions, you necessarily suppress the concentration of OH- ions.
In an undoped material, none of the sites in the crystal lattice has an excess electron or is deficient in electrons (similar to water at pH 7). Yet just like neutral water, which has both H+ and OH- ions in it, the neutral crystal has both electrons and holes in roughly equal numbers. The product of their concentration is fixed by the crystal and its temperature (just as it is with water).
In the N material, you are forcing a high concentration of mobile e-carriers by placing atoms with an excess electron into the crystal lattice. The excess electron becomes mobile, and the atom of dopant becomes a fixed positively charged site. So the net charge spread out over the material remains neutral. But it is conductive because you have mobile negative charges and fixed positive charges.
In the P material you have dopant atoms that are deficient in electrons -- that is there are not enough of them to complete all the covalent bonds with neighboring atoms. But the tendency of each site in the crystal to form covalent bonds with each of its four neighbors is very strong -- strong enough so that when there is a deficiency at one site, the deficient site might steal an electron from a neighbor to complete its four bonds -- even though this leaves the thief negatively charged. The neighbor that gave up the electron is now positively charged. And indeed this "hole" is almost as likely to be at a silicon node as it is to be at, say, an aluminum node. So now the hole is mobile and the negative charge is fixed.
In each of the materials, the product of the concentration of electrons and holes is the same as it would be in an undoped material.
On the P side, the electrons from the negative terminal of the battery cannot enter the P material (at least not in significant numbers) simply because there is no place for them. The P type crystal does not admit very many mobile negative carriers.
When the P side is connected to the positive battery terminal, the holes are repelled from the positive voltage. But the P material still needs to have its required concentration of holes. So it produces new ones by pulling electrons from the lattice at the positive electrode, and those electrons flow out that electrode toward the battery.
Recall that a metal is a lattice of fixed positive charges in a sea of very mobile negative charges. Electrons from that sea can pass into an N material because there is room for them there. Electrons can be introduced to that sea from a P material because in the P material, electrons are the byproduct of producing new holes, and there is no room for them in the P material.
cheers
i have discussed this problem with some other people but couldn't understand it well because of my own limited knowledge and understanding of such things. then i sent an email to my very dear teacher and a friend. email copy and my teacher's reply is below. now the problem is i couldn't exactly understand my teacher's reply too and he won't be available for some time. i would highlight the parts which i find more problematic and perhaps you can shed some light from different angle. and i could reach a understanding. i will be very much grateful for any help you can come up with. many thanks.
copy of e-mail sent:--
please checkout the fig. 42.25 “a” and “b” on this linked document:
https://docs.google.com/viewer?a=v&p...thkey=CKP-7OsD
i can understand the fig. 42.25a and why the current flows. but I have some questions about fig. 42.25b.
in the fig. 42.25b, the positive terminal of the battery will suck electrons out of the n-side (because this side is rich in -ve charge) which will leave behind 'positive' holes. the negative terminal of the battery will feed the electrons to holes in the p-side. As the n-side is deficient in electrons because the positive terminal of the battery has sucked them toward itself so this n-side will attract electrons out of the p-side which is now rich in electrons because the negative terminal of the battery has supplied electrons to the p-side (or, in other words the p-side has sucked electrons out of the negative terminal). in my opinion this will result in an electric current and this current won’t be small, if not larger than the current when the diode is forward-biased. that would mean there would also be current when battery terminals are reversed as in the fig 14.25b, or, when the diode is reverse-biased
i'm trying to understand from physics point of view, how a P-N junction works. help me please.
the reply received:-In the P material, the holes must outnumber the electrons by roughly 10^8 to one, just as in the N material the electrons must outnumber the holes by a similar ratio.
This is very analogous to the chemistry of aqueous solutions. In such a solution at 25C, the product of concentration of H+ ions with the concentration of OH- ions must be 10^-14. In an acid solution, by forcing a high concentration of H+ ions, you necessarily suppress the concentration of OH- ions.
In an undoped material, none of the sites in the crystal lattice has an excess electron or is deficient in electrons (similar to water at pH 7). Yet just like neutral water, which has both H+ and OH- ions in it, the neutral crystal has both electrons and holes in roughly equal numbers. The product of their concentration is fixed by the crystal and its temperature (just as it is with water).
In the N material, you are forcing a high concentration of mobile e-carriers by placing atoms with an excess electron into the crystal lattice. The excess electron becomes mobile, and the atom of dopant becomes a fixed positively charged site. So the net charge spread out over the material remains neutral. But it is conductive because you have mobile negative charges and fixed positive charges.
In the P material you have dopant atoms that are deficient in electrons -- that is there are not enough of them to complete all the covalent bonds with neighboring atoms. But the tendency of each site in the crystal to form covalent bonds with each of its four neighbors is very strong -- strong enough so that when there is a deficiency at one site, the deficient site might steal an electron from a neighbor to complete its four bonds -- even though this leaves the thief negatively charged. The neighbor that gave up the electron is now positively charged. And indeed this "hole" is almost as likely to be at a silicon node as it is to be at, say, an aluminum node. So now the hole is mobile and the negative charge is fixed.
In each of the materials, the product of the concentration of electrons and holes is the same as it would be in an undoped material.
On the P side, the electrons from the negative terminal of the battery cannot enter the P material (at least not in significant numbers) simply because there is no place for them. The P type crystal does not admit very many mobile negative carriers.
When the P side is connected to the positive battery terminal, the holes are repelled from the positive voltage. But the P material still needs to have its required concentration of holes. So it produces new ones by pulling electrons from the lattice at the positive electrode, and those electrons flow out that electrode toward the battery.
Recall that a metal is a lattice of fixed positive charges in a sea of very mobile negative charges. Electrons from that sea can pass into an N material because there is room for them there. Electrons can be introduced to that sea from a P material because in the P material, electrons are the byproduct of producing new holes, and there is no room for them in the P material.
cheers
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