2
$\begingroup$

When no external voltage is applied,

  • After the formation of depletion region, why the free electrons are not captured by the positive donor ions near the junction in the N side?
  • How minority carriers (electrons in P side) drift through the junction against the negative acceptor ions in the P side?
  • Are the electrons not repelled by the negative ions?
$\endgroup$
2
$\begingroup$

The process of ionization of the donor is temperature dependent. Since here one deals with a large number of atoms statistical physics is applied. At high temperature the fraction of donors ionizing is large compared to unionized donors. The process of diffusion makes the ionized electron move to the p-region thus creating the depletion region (similar conditions hold for the n region with holes migrating). Coming to your question as to why electrons are not captured by the positive ions formed in the depletion region they are in fact but thermal energy leads to dissociation. To answer your second question note that the E field formed in the depletion region points from the n-region to the p-region and thus support the movement of negative charge from the p-region to the n-region.

$\endgroup$
  • $\begingroup$ Thank you sir for your explanation. My one more doubt is that first the election has to enter into the electric field crossing the negative ions. How is this possible? $\endgroup$ – Ayyanar Mar 18 '17 at 15:46
  • $\begingroup$ You are welcome. Actually when there are so many particles involved there is a finite probability that a very small number would make it to the E field. $\endgroup$ – SAKhan Mar 18 '17 at 16:42
  • $\begingroup$ Sir, I would like to ask some more doubts. $\endgroup$ – Ayyanar Mar 20 '17 at 2:25
  • $\begingroup$ Can ask. No problem. $\endgroup$ – SAKhan Mar 20 '17 at 17:39
  • $\begingroup$ After a PN junction is formed, first majority carriers move that create diffusion current. After sometime depletion region is formed which creates an electric field. This field makes the minority carriers to move which creates drift current. Then these two currents become equal and opposite thus the net current in diode is zero. My doubt is that after this equilibrium these two currents exist but cancel each other (or) no movement of carriers (no drift or diffusion current) occur till it is connected to a circuit. To ask simply whether this equilibrium is static (or) dynamic equilibrium? $\endgroup$ – Ayyanar Mar 21 '17 at 2:53

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.