p-n junction confusion. (1 Viewer)

[benji_10]

If the p-type semiconductor was connected to the positive terminal, and n-type to the negative terminal, then how can current flow from the p-type to the n-type?

As I understand, the electric field across the depletion zone is from right to left (n-type to p-type). This prevents any more electrons from crossing into the p-type semiconductor (negative charge on the p-type semiconductor after transfer repels electrons). But if the semiconductors were connected as outlined above, then why does conventional current flow from p-type to n-type?
Conventional current is described as the opposite direction of the flow of electrons. Since there is no flow of electrons due to the force exerted on the electrons by the electric field, how is there any current flowing?

Thanks in advance.

 

[genius12345676]

obvious there are times like this when it would be quite advantagous to have someone doing electrical engineering at uni to help you , but hey you guys screwed with me , thought I wasnt good enough to tutor so you obviously wouldnt like my help

 

[hscishard]

obvious there are times like this when it would be quite advantagous to have someone doing electrical engineering at uni to help you , but hey you guys screwed with me , thought I wasnt good enough to tutor so you obviously wouldnt like my help

You're a genius. Still here to annoy people, yep. Not realising how much life you're losing,yep. Not realising that noone gives a crap about you here no more, yep. Not realising you have excess fat molecules, yep true genius

 

[hscishard]

are you going to answer the question ?? .

Nop, I haven't gotten up to that yet and nop this doesn't mean I can't answer it by the end of the hsc

 

[genius12345676]

are you going to answer the question ?? .

Interesting isnt it ? You act just like all the other hsc students on here , I was good while I was helping you and then when I stop helping you become an asshole lol. Use me for a free ride huh?

Was interesting to see the reaction when pictures of myself were posted on this site , up until then I was a faceless genius , but now you have a go at me because you have seen pictures etc, i wonder what the response would have been like if everyone knew what I looked like from day 1 when I was helping people here .

 

[hscishard]

That's not interesting really. What is interesting is you still can't see how we didn't use you. You just developed this thinking via constantly bumping your tutor thread because noone ever replied onto your thread. Well that changed didn't it? Yea. BTW I don't think 90% of those tutoring threads ever get replied.

Basically, people wouldn't have found you if you realised...this^
cya

 

[Fizzy_Cyst]

If the p-type semiconductor was connected to the positive terminal, and n-type to the negative terminal, then how can current flow from the p-type to the n-type?

As I understand, the electric field across the depletion zone is from right to left (n-type to p-type). This prevents any more electrons from crossing into the p-type semiconductor (negative charge on the p-type semiconductor after transfer repels electrons). But if the semiconductors were connected as outlined above, then why does conventional current flow from p-type to n-type?
Conventional current is described as the opposite direction of the flow of electrons. Since there is no flow of electrons due to the force exerted on the electrons by the electric field, how is there any current flowing?

Thanks in advance.

So, we know that a depletion region exists. The electric field runs from the N to P. This depletion region opposes the diffusion of the electrons to the P and holes to the N.
So without and applied potential difference, the electric field in the depletion region acts as an insulator as the electrons cannot acquire enough energy to get across this depletion region.

Now if the P side is connected to the positive terminal of the power supply, essentially this is making the P side 'less negative', thus reducing the electric field within the depletion region and hence the electrons need less energy to move across from the N to the P side.

Basically, the external power supply has decreased the size or width of the depletion region, which makes conduction easier. This is called 'forward biasing'

Reverse biasing is the opposite, where you make the P side more negative making conduction harder.

Basically, A P-N junction will only allow current to flow in one direction. If you forward bias the P-N junction, current will flow. If you reverse bias the P-N junction (trying to make current flow the other way) current will not flow.

Such a device which ensures current only flows in one direction termed 'diodes' which is how LED's (Light Emitting Diodes) work. If you are familiar with LED's you will notice that they generally have two prongs and one is a little longer than the other. If you connect it one way, it will glow, connect the other way, it will not glow

 

[benji_10]

@Fizzy

So without an applied potential difference, the p-n junction insulates after the exchange of electrons and holes, right?

So if there is an applied potential difference, and when the p-type is connected to the positive terminal vice versa, why does it close the gap?
Is it because the holes are being pushed away from the positive terminal towards the electrons, while the electrons are also being pushed towards the holes away from the negative terminals, effectively closing the gap, thus making conduction easier? I didn't quite understand your explanation of how the depletion zone's size is reduced

Thanks for your explanation anyway! I didn't know that connecting it to a power supply actually made a difference.

 

[Fizzy_Cyst]

@Fizzy

So without an applied potential difference, the p-n junction insulates after the exchange of electrons and holes, right?

So if there is an applied potential difference, and when the p-type is connected to the positive terminal vice versa, why does it close the gap?
Is it because the holes are being pushed away from the positive terminal towards the electrons, while the electrons are also being pushed towards the holes away from the negative terminals, effectively closing the gap, thus making conduction easier? I didn't quite understand your explanation of how the depletion zone's size is reduced

Thanks for your explanation anyway! I didn't know that connecting it to a power supply actually made a difference.

Without the potential difference, the p-n junction does act like an insulator after it has reached a 'steady state' or 'equilibrium'

To answer the second question properly, let me explain a little further =)

First consider the process of diffusion. Diffusion is when particles move from areas of high concentration to areas of lower concentration. Consider you 'sandwich' an n-type crystal (more electrons that p-type) and a p-type crystal (more holes than n-type). Now at the junction where these crystals are in contact, the electrons will naturally want to diffuse to the p-type as the electron density of its conduction band is lower. The holes will naturally want to diffuse to the n-type. This creates an electrical current, which is quickly stopped due to the build up of an electric field between the p and n crystals. The electric field runs from n --> p. As n is positively charged and p is negatively charged (electric field lines run from positive to negative). A large part of this region of diffusion results in electrons combining with holes, leaving it depleted of charge, this is called the depletion region.
Now, if the positive terminal is connected to the p-type and negative terminal is connected to the n-type, this effectively creates an electric field which opposes the electric field in the depletion region, this lowers the strength of the electric field in the depletion region and therefore allows more charge to diffuse across the region.

I suppose the main thing you need to understand is that by connecting p to +ve terminal and n to -ve terminal, this lowers the strength of the electric field in the depletion region, which means that more diffusion can occur, hence current will flow.

 

[benji_10]

Aaahhhh that did the trick (Y) Cheers Fizzy.