Forward and reverse current mechanism | Class 12 (India) | Physics | Khan Academy
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- เผยแพร่เมื่อ 18 เม.ย. 2018
- In this video, let's go deeper and explore the mechanism of the current in both forward and reverse bias. We will see that's way different than what happens inside a conductor.
Learn the next topic here --www.khanacademy.org/science/i...
Class 12 Semiconductors: We cannot imagine our life without computers today. But what makes a computer tick? What's making this technology grow at such an exponential rate? It's all due to semiconductors. As unbelievable as that might sound, altering the properties of semiconductors allows us to build these computers. In this topic, we will explore the world of semiconductors. It's a journey from what semiconductors are all the way to creating building blocks of these computers.
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Created by Mahesh Shenoy
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This video answered like 5 questions I've had for years
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I am an electronics engineer since years and we've never concentrated on this interesting topic.. thanks a lot.. you're amazing
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That's because he doesn't need no crown for people to acknowledge him🙂
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Khan Academy is an Emotion ❤️
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At first I want to really thank you for these lessons, I have work many years with electronic without understanding the physics behind this. Thanks!
Now to one question: Shoulden't the drifting current increase when the depletion region increases causing a higher chance of hole/electron to be thermal generated there?
As is explained in the last part, widened depletion region just increases kinetic energy of the minority carriers. Their numbers are still fixed and depends on thermal excitation. As numbers are fixed, so the current is also fixed
That's a good point actually!
@@devanshatray6545 but looking at the example, won't the 5 holes turn to like more than 5? Cause now the probability of them being thermally generated increases?
@@shubhamgupta-qi9nk As its name, the number of thermally generated holes would be only related to temperature. So under reverse bias, widening the depletion region won't increase thermally generated holes.
Increase in electric field across the depletion region ≠ increase in temperature.
An analogy: the speed limit on a highway is independent from how many cars get on it.
Great job... I'm only going or search your videos..
The only useful video i found here
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THE Quality education is here
Very constructive explanation!!! Awesome
I can feel enlightenment.
Great efforts never dies ✌️
I must admit, i had never watched any video that explains about pn diode this well.
Thank you so much man, you are amazing!✨
Excellent explanation sir
Nice explanation bro ...
thank you sooo much
very good!
I love you sir. You are one of the best teacher.
I also want to know does really length of the depleted region increases or it is just the effective length which increases due to the external applied voltage .Does really the no. of positively and negatively charged ions in the depleted region increases?? Plzz answer...
When we increase the voltage the holes will swept faster and due to which it takes less time to cross the depletion region hence the value of current should increase as current i=q/t, for the constant charge the time taken is less and hence current should increase. Kindly explain this also as it is confusing my mind. BTW your way of teaching is excellent. Keep up the good work.
I had the same doubt, this makes sense if the charge carriers are continuously swept across, but that's not the case. Say 10 e-h are being generated every second, but they take only a microsecond to get swept across. What's the current? It's not 10e/1 microsecond, because for almost another second, there are no charges being swept. Hence, the average current is 10e/1 second.
Now even if they got swept faster, say in half a microsecond, the current would be the same right?
So here t in q/t is the generation time, and not the time it took to get swept (which could be orders of magnitude smaller and can be neglected).
Thank you very much
I'm in a masters program right now and this video cleared up so many fundamental questions I had about pn junctions.
So, A) thank you so very much for uploading this, and B) God help me in my program lmao.
Same dude , which university?
@@EEShyama Purdue
@vinaykumarreddy8988 No. The number of charge carriers remains a strong function of *mobile* charge carriers. Just because there are more atoms in the depletion region, does NOT necessarily mean that there are more charge carriers. So the number of charge carriers in the depletion region remains a strong function of temperature (as stated at 10:47 in the video).
IT ALL MAKES SENSE NOW
I dont really have words to say... I spent sooo much time on this topic..referring to various articles, lectures..but was always confused..
Now finally I understood what's this pn junction biasing currents all about !
Thanks a llloottt Khan academy.
This video deserves much more likes.
I see you are shitting your pants about the physics exam.
Metoo
You are great sir , a student from iraq ❤
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Thank you
Ye manusya devta hai physics ka 😊
¡Thank you!
Also in the depleted region all charge carrier have recombined then there is no free electron in the conduction band in the depleted region to conducted the current from N side .Then how the conduction occurs?
This channel is best for night studies !
I have one question why not the electrons from the neutral region in N side comes in the positively depleted region in N side and hence the positive charge would get distributed over all the N side. Because the positively depleted region of N side should attracts the electrons from the N side???plzz answer..
Doesn't the depletion region widen upon reverse biasing the junction...
Which implies that there will be a higher probability and hence a higher no. of holes generated in the depletion region and hence a higher drift current?
I have got a doubt here.. 10:04, although the no. of holes generated remains the same, but the time gap between 2 different generation reduces right? So in one second, many such time gaps will get included and eventually the current should increase right? This might really be a noob question, but thats getting me confused! Please help me, thankyou!
It is injection or diffussion of charge carriers under influence of field
Nice
Dear sir how does the thermal generation was created in semiconductors.
What about those electrons which are being supplied by the battery to the n-region??
Won't it make the concentration of minority charges high in the n-region??
Is this 'generation current' and 'leakage current' same?
When current is flowing, shouldn't a new electron and hole pair be generated for every one that is recombine , since if the TOTAL number of charge carriers is decreasing, the current would eventually fall to zero as there would be no charge carriers left? In that case, does the applied voltage provide excitation energy to electrons?
Same doubt yaar
v good explanation
that moment when stuff finally makes sense, and so many myths are busted from a single video, the only channel thats actually dedicated to teaching theory properly in a way that makes it make senseee
If we increase voltage in reverse bias, the depletion region increases , which causes more probability of holes getting generated in depletion region. Explain?
The stronger the current, the wider the depletion region gets; hence holes having a higher probability of spawning in this region.
The increase in depletion region does not generate more holes. It is independent of voltage change. Now the generation of charge carriers in the depletion region is due to the increase in kinetic energy of the minorty charge carriers under high voltage which can knock out electrons situated on the inside. This is how more carriers are generated.its know as the avalanche effect.
Sir please complete 12th physics
Sir at 07:32
Won't the electrons from the metallic end (to the left of the "P" part get sucked into the holes (which are huge in number) at the left most end of the "P" part? As there are a loooot of holes to get sucked into at the left most end of the "P" part?
Did you ever find an answer to this question?
Bolo?
Current in forward biased is also independent of voltage provided by battery because it is recombination current ???
10:14 When you increase the reverse bias voltage, doesn't the depletion region also widen? So if the drift current is caused by the thermal generation of minority charge carriers in the depletion region, shouldn't the probability of this increase if you increase the reverse bias voltage because there is now a greater area in which thermal generation would contribute to the drift current (so the drift current should increase)?
I think you're right, but the current would get saturated when the reverse bias keeps increasing because the depletion region reaches it maximum of width. Then it would go through Zener breakdown.
That's my point, if i'm wrong please let me know.
Yes, it does. In most datasheets of diodes there is usually a logarithmic plot showing the increase of the reverse current (y-axis) while the reverse bias increases on the x-axis.
Are you saying that there are much more free electrons in the n-side that the holes that are introduced from the p-side and therefore most of the holes travelling in the n-side are sucked in?
Not really.
We are assuming that the doping concentration is same for both P and N and hence the majority carrier number will be same in both. (Same no of holes in P as in electrons in N).
The point is when they cross the junction, there is a very high chance of them recombining.
And hence very low chance of them making their way to the opposite ends.
So chances of holes making their way to the edge of N is very low. (Similarly chances of electrons reaching to the edge of P is also very low).
Hope this clears it?
Why TF doesnt this have more views?
Selfless actions
in the forward bias, won't recombination destroy the election-hole pair? this might maintain the concentration on both sides, but won't the concentration eventually become zero with no electrons or holes left?
Same doubt!
Nah bro, because the external potential applied prevents the depletion region to increase.
For each electron whole combination that takes place near the junction of covalent bond breaks in the p section near the positive pole of the. Of the electron and the whole produced the electron is captured by the positive terminal while the whole moves towards the junction. on the other hand as soon as the whole is created in the peace action due to breaking of covalent bond and electron is released from negative terminal of battery into n section to replace the electron lost by the combination with a hole at the junction.
Sir but you never talked about thermal generation from last 12-13 videos and now suddenly talking about it without any explaination why?
Anyway, I really thank you for making these teaching videos for we students.
If anyone thinks that will the holes be destroyed in the process then the reason is when the hole will recombine in N type then p+ will be exposed leading to charge imbalance and an electron will flow from the battery for compensation and a hole will be created in the P type hope I am correct please know me if I am wrong at the point
Sir at 04:52
Holes travel when an electron occupies it. but recombination is defined as an electron occupying a hole, so how do holes exactly "travel"? Because if an electron occupies a hole it is "destroyed" and hence there will be no hole which can travel
Recombination (destruction) happens when the free electrons (from conduction) band recombines with the hole. When this happens, as you mentioned, the hole-electron pair is destroyed.
However, covalently bonded electrons (in the valence) band can also occupy their neighboring holes. When that happens, they leave behind a hole, and this is how one can visualise holes 'traveling'.
@@KhanAcademyIndiaEnglish But sir will a hole get formed when a free electron moves? As a free electron too occupies space, and that space will be vacant when a free electron moves
@@niranjan7457 Holes are gaps in covalent bonds. I recommend watching previous videos on holes & electrons
10:50 doesnt look accurate, my understanding is that by increasing the battery voltage we will end up increasing the deplition zone and thus we will indeed end up with higher generation of holes/electrons. So even with reverse bias, sounds like it should increase when battery voltage is increased
Current is no. Of holes generated per second and if we increase the electric field in the depletion region then also the current remains the same even if the holes acceleration increases..
If acceleration increases then time taken by these holes to reach the negative plate will decrease hence the value of current must increase as I = q/T..??
I had the same doubt. But here is the thing. Charges are not continuously flowing. Once the minority charges get swept across we still have to wait for more electron-hole pairs to get generated right? For example.
Let's say it 1 hole is generated every second (just an example). Then, even if the holes get swept across within milliseconds, the no. of holes getting collected will still be 1 hole per second right?
In other words, charges get swept almost instantly compared to how long it takes to generate them. Hence, T in our I =q/T should be waiting time for a generation, and hence it only depends on the generation rate and not the speed (or the field)
Let me know if you need more clarity
@@KhanAcademyIndiaEnglish please explain deeply
7:52... click 💯
@Khan Academy India - English You’ve said in your videos that holes are actually electrons from valence band that contribute to net current in circuit. But isn’t that the point of band theory, that only the electrons in conduction band are free to move, hence only they can generate current? Please answer...😟
so the charges in the pn junction move due to diffusion but when the depletion region is created a potential barrier is created so the charges cant diffuse anymore so when we connect the battery the charges get enough energy for diffusion ?
Yes! The probability of diffusion increases, as the built in potential barrier becomes lower!
@@ricomajestic right thank u very much
@Khan Academy India
I agree that due to forward current, every hole that is recombined is replaced by another hole. But the holes are still finite. I mean that the new hole will also get recombined. So finally all holes will get destroyed.
I know i am missing something very obvious, but still please help me out.
Well the holes are continuously supplied from the external circuit.
For every hole that recombines, a hole is supplied back to the edge of the P side.
Now ocourse what's really happening is that the electrons from the VB in the P side move into the wire leaving behind a hole.
@@KhanAcademyIndiaEnglish Can you please explain in greater detail why for every hole that recombines, a hole is supplied back to the edge of the P side? Maybe there is something I didn't get from your explaination. Anyway, these semiconductor lessons are amazing!
@@KhanAcademyIndiaEnglish how holes are continuously supplied from ext circuit
@@KhanAcademyIndiaEnglish please explain why
@@KhanAcademyIndiaEnglish since u have previously said that for 10^ 16 atoms only one electron is produced.. so how it will compensate the huge loss of holes
What about the number of holes lost due to recombination nearby junction, how the number of holes will be maintained? Plzz, anyone here answer my question.
I too have that same question !? Finally all holes will be disappeared due to recombination and whole things gonna stop ? Please anyone clear this doubt !
Sir I have the same doubt plz explain this
@Khan Academy India - English I have a question, say 3 holes reach somewhere in the n-side(under fwd bias) and 2 recombine then 1 moves forward, wouldn't this reduce the net current in the transistor, but the current should be the same throughout the circuit, how is this possible?????
I think there is one mistake, at the end of the junction, majority charge carriers should cause the current flow in the wire as otherwise the animation suggests that there is high current at the center and low at the end of the junction .
Amazing question. You are absolutely right, the hole current and the electron current will NOT bet the same everywhere. But their sum (which is the total current) will be the same everywhere.
For example, the hole current will be maximum near the P end (because of minimum recombination chance) and minimum near the N end (maximum recombination chance). But the electron current will be the opposite. It will be max near the N end and min. near the P end. So when you add both the electron and hole current, you will see the value is the same at all positions.
That makes sense right? Let me know if you have more doubts :)
Sir I have a doubt that's always confusing me how the holes are being continuously at p type...plz reply sir..I want detailed explanation 😊
Sir I really need the answer 🙏
The holes are not actually continuous there are some electrons in p side (ie minority charge carriers)
You said that holes recombine with the electrons while traveling from p to n
But that means the holes and electrons are vanishing. So eventually there will be no charge carriers left. So how the current continues??
Good question. The point is all holes and electrons won't destroy each other.
The chances of recombination near the edge is very low so there will always be high majority charge concentration there. Chances of recombination is max near the junction and hence we have almost zero majority charge carriers there (depletion region).
This difference in concentration gets maintained. It won't change with time.
And this difference in concentration drives the diffusion of the majority charge carriers giving rise to a steady forward current.
And in turn this forward current is the reason why charge carriers at any particular place will remain constant. For every recombination that destroys one majority carrier, say a hole for example, a hole will move into that place due to this forward current.
Feel free to further comment if you have more doubts
@@khanacademyindia when forward biasing there are more holes diffuse to N , wouldn't it increase the depletion region width as holes and electrons are cancelling each other at the N side edge of depletion region?
@@jingmiao4140 For every electron and hole that recombine in the N-region a new electron is supplied by the battery (to the N-region), keeping charge neutrality in the N-region and preventing the depletion region from widening.
So, reverse current and drift current are one and the same!? No? Can someone please clear my doubt?
Each video feels like removing a gray part in an RPG game
Why does the current doesn't increases as the holes move faster
Doubt:
In the starting we learnt that diffusion creates an depletion region
Now in forward bias, when voltage is applied, diffusion increases; then how can depletion layer decrease
Should'nt depletion region increase even more?
Nah bro, because the external potential applied prevents the depletion region to increase.
For each electron hole combination that takes place near the junction of covalent bond breaks in the p section near the positive pole of the. Of the electron and the whole produced the electron is captured by the positive terminal while the whole moves towards the junction. on the other hand as soon as the whole is created in the peace action due to breaking of covalent bond and electron is released from negative terminal of battery into n section to replace the electron lost by the combination with a hole at the junction.
Better than
Byju
Vedantu
What does he mean by thermal generation?
Heat is supplied to the atom which rips apart the electron from the atom. This generates a hole. Hence, it is called thermal generation
The generation of an electron and a hole (jumping of electron from valence band to conduction band) due to thermal energy of the atom.
How do you forward or reverse bias a PN diode in AC supply , as there are no positive and negative terminals in AC supply?
As current is flow of electrons, how will P side of a PN diode get positive charge. 😆.. so confusing
Through DC supply...batteries are the source of dc current which is unidirectional!!! We have used battery over here
so...what is the use of the battery now?...if ... what you've said...that recombination now does the walk.you forgot to tell the rule of the battery.what you have said will not happened with out the battery..
What's with holes? They just make it hard to understand. Feels like you are studying an abstract concept
I agree with you... it just feels like an unnecessary addition
When electrons leave a site it generates a hole. That's how I understood.
here you are saying , external voltage doesn't change much .it actually the diffusion which causes the generation and recombination process but in previous videos u said by increasing voltage we can get higher current now here external voltage is helping to increase current .
that's two of your sayings confusing my mind. please anyone clearify this
In forward bias when u increase the voltage, current will increase till the potential barrier is zero and depletion layer becomes zero
And in reverse bias, change in voltage does not change current because minority charge carriers are less
How wrong i was about diode currents 😭
What's with the accent?
What is your problem? It's his original accent, watch his other and very old videos on his own channel "float head physics", it's the same accent there too! Instead of focusing on people's accents, you could rather appreciate the work he is doing for free..it would really motivate him.
Grow up......what kind of comment is this??...@ Adrija
Yaaa absolutely bro u are a brilliant teacher 🏓🏓