1
\$\begingroup\$

My textbook says something like this :
When a diode is switched from forward bias to reverse, the current changes direction but the magnitude doesn't decrease immediately. The charges continue to flow and the diode draws the power from the battery through out the reverse recovery time.enter image description here

During forward bias, there exist some "excess" minority carriers near the junction. These contribute to the current during reverse recovery time.

In forward bias, negative plate of the battery pushes electrons from n side to the junction, then the electrons recombine with holes and travel in the valence band through the holes of p side and reach the positive plate of the battery. It seems electrons only travel in valence band in p side. I'm happy having this mental picture for the charge movement during forward bias. But I'm failing to visualize a similar picture for the charge movement during reverse recovery time. I keep getting questions like :

1) Does the negative terminal of the battery put electrons at the edge of the p side, then these electrons fall into holes, the electrons then travel through the holes to the junction and they see depletion region! Then what happens ?
2) What happens to the excess minority carries on the p side near the junction ? Do they feel the battery and move to the n side ? Even if they move to the n side, the excess electrons on p side are never going to disappear because an equal number of electrons seem to enter from the negative terminal of the battery... Highly appreciate any help. Thanks!

\$\endgroup\$
  • \$\begingroup\$ current flow in a diode is a mixed flow of holes and electrons, you seem to be concentrating on electrons so only getting half the picture. Check out this link as a general background before getting into the specifics of a semiconductor diode \$\endgroup\$ – Neil_UK Jun 7 '17 at 4:38
  • \$\begingroup\$ Ahh @Neil_UK amasci was my favorite site when I was in high school. I read almost everything in that site! That link doesn't answer the question I've posted. I'm not looking for the accurate quantum mechanical model at this time; if possible I simply want to treat hole as a missing electron and keep going with the cheap model given by my textbook... I know hole is much more than a missing electron, but I don't know enough quantum mechanics to study diodes using fermi dirac stats and other clever ideas :( \$\endgroup\$ – Hiiii Jun 7 '17 at 4:58
  • 1
    \$\begingroup\$ I know the link doesn't answer your specific question. However 'a hole being a missing electron' is just enough information to get you stuck. Treat it as a moving charge carrying entity with mass just like you treat an electron, you don't need the full QM treatment. Think about it, why do you treat an electron as a charge carrying particle? You've never seen one, don't know how they exist in the orbitals of an atom in fixed energy levels, how they get emitted from neutrons in beta radiation, all weird. Why get weirded out by a moving charge carrying hole? Just go with it, as for electrons. \$\endgroup\$ – Neil_UK Jun 7 '17 at 5:09
  • 1
    \$\begingroup\$ Excellent. Now if you do want to get slightly more into holes, bear in mind that the contacts to a diode junction are heavily doped N and P type, so the majority (not the only) carrier is electrons in one, and holes in the other. The ohmic junction with metals, in which the only carrier is electrons, occurs way away from the junction, and is a different issue from the diode junction itself. That's where you can, with profit, start thinking about holes being 'missing electrons', at the metal interface. \$\endgroup\$ – Neil_UK Jun 7 '17 at 5:26
  • 1
    \$\begingroup\$ Where an electric current flows in a material, the current is the sum of all mobile carriers. In a hot plasma (like the arc of an electric welder), it's a mix of electrons and positive ions, in semiconductors it's a mix of electrons and holes, in ice it's protons, in metals it's electrons. If you connect your battery with metal wires, it's all electron flow in them. If you connect your battery with ice blocks, it's all protons. At the interfaces between different types of material, that's where it's worthwhile to consider the mechanism whereby one type of flow is continuous with the other. \$\endgroup\$ – Neil_UK Jun 7 '17 at 5:30
1
\$\begingroup\$

So, you reverse the bias and the battery begins pulling electrons out of the N type silicon and holes out of the P-type silicon *, eventually the area around the junction becomes charged up like a capacitor and the current stops flowing.

(*) you could picture this as filling the holes with electrons, but either way because of the electric field of the battery the holes move to the metal, and then accepting an electron from the metal, they disappear.

\$\endgroup\$
  • \$\begingroup\$ I think I get that now, I'm trying to visualize what happens to the excess minority carries at the junction... Isn't the current possible only because of those excess carriers on either side near the junction ? Battery is pulling electrons out of N type and holes out of P type. I get so far! Thanks to you and Neil :) With this knowledge, now I'm trying to visualize how electrons and holes travel inside the diode during reverse recovery period.. \$\endgroup\$ – Hiiii Jun 7 '17 at 5:52
  • 1
    \$\begingroup\$ Right at the time of changing the bias, it seems the P-side is negatively charged and N-side is positively charged due to the excess minority carriers and ions in the depletion region. Since battery pulls out equal amount of electrons and holes, it seems the battery cannot remove that excess charge.. \$\endgroup\$ – Hiiii Jun 7 '17 at 5:59

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.