What is the reverse recovery time in a diode?
1\$\begingroup\$ @Cell-o, what have you read so far, what do you already know about diodes? \$\endgroup\$– KortukMay 5, 2011 at 13:50
1\$\begingroup\$ @Kortuk - of course ,I have read.But,I do not exactly understand.So, What factors affect reverse recovery time in diode? \$\endgroup\$– Cell-oMay 5, 2011 at 13:55
1\$\begingroup\$ It's caused by the carrier recombination time, and the explanation requires a lot of difficult maths. \$\endgroup\$– Leon HellerMay 5, 2011 at 14:56
1\$\begingroup\$ If you don't have a good electronics book handy I recommend Sedra and Smith 4th edition, found here: amazon.com/… \$\endgroup\$– AngryEEMay 5, 2011 at 17:46
2\$\begingroup\$ I think no one has answered because the question is very vaguely worded and could mean anything and the poster has failed to change his wording or offer any hints as to what exactly he wants. That and general laziness. Plus I get the impression that even with a good answer the poster will ask something like 'what is this mean? how do i use it?!' \$\endgroup\$– AngryEEMay 6, 2011 at 14:21
If a diode is conducting in a forward condition and immediately switched to a reverse condition, the diode will conduct in a reverse condition for a short time as the forward voltage bleeds off. The current through the diode will be fairly large in a reverse direction during this small recovery time.
After the carriers have been flushed and the diode is acting as a normal blocking device in the reversed condition, the current flow should drop to leakage levels.
This is just a generic description reverse recovery time. It can affect quite a few things, depending on context, as mentioned in the comments.
1\$\begingroup\$ So with forward bias, the depletion region shrinks to nothing. If "instantly" reverse biased, the depletion region will require some finite amount of time to grow large enough to prevent conductance. Yeah? \$\endgroup\$– ajs410May 23, 2011 at 22:37
2\$\begingroup\$ That is the way I understand it. With it forward biased, the silicon is "on". So it has the ability to flow backwards until the flow causes it to turn off. \$\endgroup\$– JoeMay 24, 2011 at 1:48
A space charge within a P-N junction needs to be established before forward current can flow. (If the first sentence makes you ask why, that's really a separate question -- perhaps this can help. Let's just look at the dynamics of establishing and neutralizing that space charge.)
From zero, this space charge can be established quite quickly, because an externally applied forward bias voltage can route electrons externally around. Electrons diffuse from the n-type material into the edge of the p-type material, holes in the p-type material diffuse into the edge of the n-type material, and at the metal interfaces, new electrons are injected into the n-type end and holes are generated at the p-type end to produce free electrons that can flow in the external circuit. All of these flows are flows of majority carriers in their respective materials, so diffusion happens quickly driven by much larger concentration gradients. A space charge develops rapidly because majority carriers are flowing to turn the diode on -- electrons in the n-type material, and holes in the p-type material.
However, if the external voltage is then reversed to be a reverse bias, the space charge is attracted to itself to recombine. But this recombination only happens through the diffusion of minority carriers. This minority carrier diffusion has much smaller concentration gradients, and therefore diffuses orders of magnitude more slowly. An external circuit providing reverse bias can aid in speeding this recombination, as it can allow for faster neutralization of excess holes that migrated back to the p-type material, and removal of excess electrons that migrated back to the n-type material. This hole-electron recombination or charge neutralization is assumed to happen essentially instantaneously at the semiconductor-metal interfaces, so if the external current can supply and remove electrons under reverse bias, it will do so much faster than the "normal" hole-electron recombination rate in the bulk of the semiconductor. That's why there can be huge reverse currents during the reverse recovery time.
I put together a little simulation of reverse recovery time in a 1N4007 diode vs a 1N4148:
The demo shows the diodes being switched under a square wave, and shows that the 1N4007 takes a few microseconds to turn completely off!
(See also a PDF titled "Recombination Time in Semiconductor Diodes".)
If diode is forward biased and you want to turn it off, it takes a while to extinguish free carriers flowing across the junction (electrons have to get back to n-region and holes have to get back to p-region, then they can recombine at the anode and the cathode, respectively). This time is called "reverse recovery time" and the total current flowing across the diode is negative, because carriers flow in opposite directions with respect to forward bias. The charge flowing during reverse recovery time is called "reverse recovery charge" and the diode has to extinguish it ("recovery" from reverse-biased to neutral condition) before you can turn it on. In the end, reverse recovery phenomenon depends on silicon doping and geometry and is a parasitic effect in diodes, because energy involved in the process is lost. strong text
The time taken by a diode to switch its condition that is from forward biased (ON condition) to OFF condition is called “Reverse Recovery Time”. When a diode is forward biased and you turn it OFF, it takes a while to completely turn OFF; in this time first a diode will attain a reverse biased condition and then slowly reach to the OFF condition rather than it directly attain an OFF condition. During this time electrons go back to n-region and protons go back to p-region to attain OFF condition and the total current flowing across the diode is negative, because carriers flow in opposite directions with respect to forward bias. The charge flowing during reverse recovery time is called “reverse recovery charge”.
The charge flowing during reverse recovery time is called “reverse recovery charge”. When switching from the conducting to the blocking state, a diode or rectifier has stored charge that must first be discharged before the diode blocks reverse current.
When switching from the conducting to the blocking state, a diode or rectifier has stored charge that must first be discharged before the diode blocks reverse current. This discharge takes a finite amount of time known as the Reverse Recovery Time, or trr. During this time, diode current may flow in the reverse direction.
1\$\begingroup\$ I'm not able to understand what additional value this post brings, relative to the pre-existing answers to this question. \$\endgroup\$ Sep 18, 2013 at 5:38
When you turn off any diode then a reverse current will be flowing through diode for a particular time due to stored charges in depletion layer. so the time "when the reverse current start to flowing through diode and reached its peak value and again decaying and reached to 25% of its peak value" this time is known as reverse recovery time of diode.
when the diode is conducting in the forward bias condition suddenly if the diode is reverse biased,and the electrons which are about to get connected with the +ve terminal when it is forward biased now(reversed biased)cannot connect to -ve terminal and has to go back to the p region and settle as minority carrier. The time taken for this is called recovery time.
when forward diode current decays to zero ,the diode carries on to conduct in the reverse direction due to presence of stored charges in the two layers . "the reverse current flows for a time which is known as reverse recovery time".
\$\begingroup\$ is the stored charge in the p and n layers, or is it in the junction where both kinds of charge exist during conduction? \$\endgroup\$ Nov 17, 2018 at 18:13