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As I understand during forward bias more electrons get diffused to p side,
and more holes get diffused to the n side. This will be maintained by the external bias.

Now if we suddenly change the polarity of the bias,
what happens to these excess charge? How do they go back to their respective sides? (electrons back to n side and holes back to p side)?

My specific question is
During reverse recovery phase, the p side already has more electrons. Isn't the external battery also putting more electrons into the p side? The p side is already trying to get rid of the diffused electrons. Wouldn't the battery be putting more electrons into the p side? So how can the external reverse bias help in getting rid of the stored charge quicker?

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  • \$\begingroup\$ I don't think the P-side (or N-side for that matter) is inherently and independently (i.e. without influence from the externally applied electric field) trying to do anything but recombine electrons into the holes. \$\endgroup\$ – DKNguyen Jun 14 at 13:45
  • \$\begingroup\$ @DKNguyen ok the moment the bias is changed from forward to reverse, the p side has more electrons. Wouldn't the p side try to get rid of these ? Also recombination removes the free carriers, but it doesn't get rid of the excess electrons, it just puts those electrons in the covalent bonds. The excess charge is still there on the p side itself. Is my understanding wrong? \$\endgroup\$ – across Jun 14 at 13:54
  • \$\begingroup\$ As answered below, it can't since the reverse bias is stuffing more electrons onto the P-side. Like I said above it doesn't "want" to do anything but recombine. I believe the "charge to be removed" they are referring to is specifically the charge stored in the cap that the depletion region forms when forward biased which keeps the diode forward biased rather than any and all charge in general (if that makes sense). \$\endgroup\$ – DKNguyen Jun 14 at 13:55
  • \$\begingroup\$ @DKNguyen but the recombination doesn't help in getting rid of the excess electrons right? \$\endgroup\$ – across Jun 14 at 13:56
  • \$\begingroup\$ the initial state is more electrons on the p side, the final state is no more electrons on the p side. im just trying to see how the transition happens... \$\endgroup\$ – across Jun 14 at 13:57
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How is the stored charge removed during reverse recovery time?

During reverse recovery, the negative voltage at the p terminal attracts holes away from the depletion zone, and the positive voltage at the n terminal attracts electrons away from the depletion zone. The remaining minority carriers drift across the depletion zone and either recombine or, now as majority carriers in a reverse biased diode, get swept to the terminals. The carriers that get swept to the terminals form reverse recovery current.

Wouldn't the battery be putting more electrons into the p side?

Very few new minority carriers are introduced into either the n or p zones, because the ohmic metal-semiconductor junction doesn't allow that to any appreciable degree. (The minority carriers in a diode are almost all present because during forward bias, majority carriers crossed the depletion zone and became minority carriers.)

huge current flows during the reverse recovery phase right? This means the battery is putting more electrons into the p side. Is my understanding ok?

The current may be large, but only for a very short time, while the existing charge is "removed". So, no, "large" current does not imply that electrons are entering the p-semiconductor from the terminal (to any appreciable extent). Thermal energy does allow some electrons to enter the p side from the terminal. If the diode is at room temperature, there will not be many. If the diode is hot, a great deal more will flow. But strictly speaking, that is not reverse recovery current, but reverse leakage current.


Addendum: The original question was about how stored charge was removed during reverse recovery. However, discussion in the comments has revealed also an interest in how the charge that exists on both sides of a reverse biased junction (or even an unbiased junction) comes about. This discussion reveals some misconceptions that I would like to address.

The misconception is around the notion that electrons may leave a negatively charged wire connected to the p region, enter the p region, cross over the junction to the n region, and then enter the positively charged wire.

Mostly that does not happen. When electrons enter the p-region from a wire, it is almost always to recombine with holes in the p-region. It is possible for electrons to enter the p region from the wire and become minority carriers, but that is rare. (It increases with temperature, but it is rare). Almost all of the minority carriers in a diode begin their career as majority carriers in the oppositely doped region, and cross the junction. More about that in a minute.

Another answer states:

As holes flow into the n-side from the positive terminal of the battery, some of them are lost to recombination. These can be replaced by the stored minority carrier holes on the n-side.

This is the same mistake. Holes very rarely flow from a wire into the n-region.

However, when an pn junction is unbiased (or reverse biased) there are holes in the n region near the junction, and electrons in the p region near the junction. How does this state of affairs come to be?

The answer is diffusion current. We are all familiar with drift current, which is the flow of carriers due to an electric field. But all carriers have thermal energy associated with them, and there is a random motion caused by this thermal energy. If there is an region which in which the concentration of one species of carrier is higher than in a neighboring region, then those carriers will tend to diffuse from the region of higher concentration to the region of lower concentration. This is called diffusion current. Diffusion current can occur even against a voltage gradient. This in fact happens at a pn junction. Electrons, which are plentiful in the n region diffuse across the junction to the p region. Holes, which are plentiful in the p region diffuse across the junction to the n region. This sets up a negative charge on the p side and a positive charge on the n side. These charges in turn create an electric field. The electric field causes charges to flow in the opposite direction to the diffusion current. This flow, under the influence of an electric field is a drift current. When the drift current and the diffusion current are equal and opposite, a dynamic equilibrium is established. It is important to note that the equilibrium point does not occur when the electric field is zero, but when there exists a significant and important non-zero electric field.

It is also important to understand that the field is not created by currents from outside the diode, but from currents internal to the diode. Currents that basically only exist near the junction. No external current is required to create this separation of charges. In a reverse biased diode, almost all the free electrons in the p region came from the n region, and not from the negatively charged wire connected to the p region. Similarly, the holes in the n region came from the p region, and not from the positively charged wire connected to the n region.

This applies both to minority carriers that are left over from forward conduction, and also to the minority carriers that flank both sides of the junction when it is blocking current.

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  • \$\begingroup\$ Ty but huge current flows during the reverse recovery phase right? This means the battery is putting more electrons into the p side. Is my understanding ok? \$\endgroup\$ – across Jun 14 at 14:09
  • \$\begingroup\$ I'm referring to this graph eeweb.com/wp-content/uploads/… \$\endgroup\$ – across Jun 14 at 14:10
  • \$\begingroup\$ The current may be large, but only for a very short time, while the existing charge is "removed". Thermal energy does allow some electrons to enter the p side from the terminal. If the diode is at room temperature, there will not be many. If the diode is hot, a great deal more will flow. \$\endgroup\$ – Math Keeps Me Busy Jun 14 at 14:13
  • \$\begingroup\$ Ty so much this cleared up so much after reading multiple times. You said 'The remaining minority carriers drift across the depletion zone and either recombine or, now as majority carriers in a reverse biased diode, get swept to the terminals. The carriers that get swept to the terminals form reverse recovery current.'..(1/2) \$\endgroup\$ – across Jun 21 at 15:21
  • \$\begingroup\$ I don't get why they are swept to the battery terminals. The 'n' side already has less electrons at the moment we reversed the bias. So why don't the electrons just stay on 'n' side after crossing the junction? (2/2) \$\endgroup\$ – across Jun 21 at 15:24
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This is how I think about it.

  1. Start with current continuity. Whatever reverse current flows into the n-side, must also flow out of the p-side.

  2. As holes flow into the n-side from the positive terminal of the battery, some of them are lost to recombination. These can be replaced by the stored minority carrier holes on the n-side.

  3. Next, the reverse current flows into the p-side. Here some of its holes will recombine with the stored minority electrons on the p-side. The holes lost to recombination will be replaced by majority carrier holes on the p-side, such that the current entering the device equals the current exiting the device.

In this way the total stored charge is removed via the reverse current. At the same time, and independently of the reverse current, once the supply switches to put the device in reverse bias, the stored charge is depleted via normal recombination.

  1. Note that the first step (holes recombining with electrons on the n-side), as well as the last step (majority carrier holes exiting the p-side) both serve to uncover fixed dopant charge around the junction. This begins the removal of forward bias.

  2. As reverse current continues to flow, the bias moves from forward to zero. Once zero bias is reached the same flow continues except now without stored charge. At this point we can see it as the flow of majority carrier electrons pulled from the n-side to the positive side of the supply, and the loss of majority carrier holes on the p-side through recombination with those same electrons as they enter the p-side to complete the current loop. This continues until full reverse bias voltage is placed across the depletion region.

  3. There are thus two phases to the switch from forward bias to reverse bias: the removal of stored charge and the establishment of reverse bias. The sum of these times is the called "the reverse recovery time."

  4. During the first phase, when stored charge is being removed, diode power dissipation is negative (the diode is delivering energy to the circuit). HOWEVER, in a buck SMPS, the reverse current can be a major loss of converter efficiency because of the flow of reverse current through the high-side switch. The longer the stored charge removal time, the larger the power loss in the high-side switch. Also note that during this time, the full input voltage is across the high-side switch. So beware of reverse current.

  5. Finally, diode's can be manufactured with nonuniform doping in order to eliminate the reverse bias charge up time. Such diodes are called "step recovery" or "snap-back" diodes. The others are called "soft recovery". One draw back to the snap-back diode is increased switching noise.

I had the same question that was originally posed here. The thought in that question is that you can't push out stored charge simply by pushing in the same (reverse) current that comes out. Exchange mechanisms must exit to get rid of the stored charge.

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