Why diode switching speed varies (p-n junction)?

[NuLED]

Hello - I am curious why diodes have a certain switching speed where the operation begins to break down. What is happening at the p-n junction? Is it because the electron/hole flow across the depletion layer not happening fast enough to catch up with the change of current direction? (If so, then will different ways of doping the silicon create different switching speeds?).

I understand generally the semiconductor at the atomic level (what I described above) so if anyone can explain along these lines I would appreciate the knowledge sharing.

Thanks!

 

[(*steve*)]

Imagine that the Diode has a small capacitor in parallel with it (the junction capacitance)

 

[NuLED]

right

 

[(*steve*)]

So did that answer your question? (I'm not sure from your answer if it did or if I have left you more confused than before)

 

[Laplace]

There are actually two effects involved in switching off a forward biased p-n junction: storage and transition. On sudden application of a reverse bias voltage a reverse current flows as excess minority carriers in the immediate vicinity of the junction are swept back across the barrier. The time for the stored-minority charge to become zero is called the storage time. The transition interval will be completed when the minority carriers which are at some distance from the junction have diffused to the junction and crossed it and when, in addition, the junction transition capacitance across the reverse biased junction has become fully charged. But I have no idea how different doping levels can affect the storage and transition times.

 

[NuLED]

Hi guys - thanks for checking in on me and also providing further info. My understanding now is a bit of both (more knowledge but also a little bit of confusion, or perhaps more curiosity, not confusion). But certainly I am progressing along with understanding this, thanks to your help.

Regarding the junction, with the reverse bias, my understanding is that the depletion layer will expand somewhat as ions are produced (to an equilibrium state of 0.7V for silicon, across the junction). And so it is across this potential that the minority carriers are moving from the P to the N type, as described by Laplace (unless I understood that wrong).

So the WIDTH of the ionized junction (the depletion layer) will have an effect on the frequency? i.e., the storage time described above? (more distance to cover?).

By the way one thing missing in my knowledge is how capacitance relates to frequency. I was going to try to google around about that part next to see how it relates to semiconductors.

UPDATE: It just occurred to me that I wrote the wrong information. The 0.7V does not occur because of reverse bias. It is already intrinsic to the p-n junction (of silicon). But with reverse bias, the depletion layer DOES grow wider. So, there are more ions. So won't the barrier potential still get larger and larger? Anyway it may be at this point that my understanding, and not the diode, breaks down.

 

[(*steve*)]

The width of the depletion layer depends on voltage, and this affects the capacitance.

Check out varicap diodes which use this technique to create a voltage controlled capacitance.

One form of breakdown is avalanche breakdown. Another is zener breakdown. One is caused by minority carriers, the other by electrons tunnelling through the barrier. The effect is similar, a rapid increase in current flow at some reverse potential.

Note that with tunnel diodes the electron tunneling effect occurs in the forward direction.

 

[NuLED]

Guys - I am revisiting the basics of the p-n junction because of this thread. Please see here the fundamental question I have now:

Let's look at the diode's p-n junction. Now it has 0.7v barrier potential.

We connect it in forward bias.

With anything less than 0.7v from the power cell, no current flows because the electrons from the battery do not have sufficient energy to cross the junction, because they are repelled from the NEGATIVE IONS on the p-type side of the junction (which have stabilized in number to reach 0.7v, relative to the positive ions on the n-type side).

OK now we increase voltage, so that it exceeds 0.7v just a bit. Current starts to flow.

HOWEVER my question now is:

If we increase the voltage, more electrons come from the wire connected on the n-type side. So, there is a net increase of new free electrons coming across from the n-type. These ostensibly knock off more electrons from the n-type, but the holes there are quickly filled back up, because we have so many new electrons coming in.

OK, so here is the question:

With the p-type side getting bombarded with more electrons that have cross the depletion layer, we are creating even MORE negative ions out of the doped p-type material, right?

So shouldn't these negative ions INCREASE the REPULSION of the new negative charge carriers (now the majority carrier electrons) and this should adversely affect current?

I assume that as voltage increases tremendously, any such repulsion should be overwhelmed even if the entire junction now is filled with negative charge.

But from the moment beyond 0.7v barrier potential, to full current flow, should there not be a DECREASE/resistance in current flow, because of the vast number of negative ions (and almost zero positive ions)?

 

[(*steve*)]

Remember that the electrons aren't piling up in there. They are exiting the other side of the diode at the same speed that they arrive.