Thank you so much sir! I was very confused as to how a current would flow in reverse biased condition. But your videos and energy level diagrams along with animation really made me understand what actually happens and what causes the current at breakdown voltage to be really high. Thank you so much.
Thank you for the feedback! This video goes into the underlying mechanics, the VI characteristics are discussed in the diode introduction tutorial: www.circuitbread.com/tutorials/introduction-to-diodes
Great video and it explained almost everything, but i am still confused about 1 thing. At 0:52 why electrons move to holes(near depletion region) and get fixed there? Why those electrons just do not go through depletion region and then further?(intuitively battery's positive terminal attracts electrons more than depletion region repels them, therefore they should move forward right?)
In the animation it would be useful to represent the electrons, from the left conductor, passing through the p zone in the valence band, until they reach che depletion region increasing it thickness
Thank you so much !!! your videos have cleared many confusion. I would love if you could explain about photodiodes, like how they work and how they are different from regular diodes :D Again, thank you!!!
thank you very much for your effort. But to me i still finde some questions which are not allow me to understand it fully: 1:09 In reversed biased mode the conduction band has in general no electrons left. According to your picture there would be a possibility to flow from the p-conduction band to the n-conduction band. Only electrons which are thermaly generated from the valence band exists there, and they are the reason for re reversed diode current. Am i correct? Same picture: Actually all Explanations say that there are no wholes left in de p - valence band, so there are no current carriers left. But actually there are electrons filling the wholes. And they are still electrons which still can produce a current. I can try to press more and more ectrons into the p-valence band. So I can increase the potential of the Electrons until they have enough energy to overcome the depletion region. Am i also correct?
Here's what I don't understand about reverse bias. It says "holes move towards the left side because they are attracted by the negative voltage." But there is no such thing as holes moving. The only way a hole moves is if an electron is moving in the opposite direction. It's all electrons moving. So now the question is, exactly which electrons are moving? Are electrons from the voltage source, maybe a battery, moving into the P side, combining with the holes, making it look like holes are moving to the left into the battery? Or is it electrons in the P material itself moving towards the junction, making it look like holes are moving away from the junction? Also, if it is electrons in the P side moving to the junction, then why don't they cross the junction and go into the N side? Why do they stop once they hit the junction, or maybe they stop a few atoms away from the junction, since the atoms right next to the junction don't have any holes for the electrons to join with. All I can think of is that it requires a small amount of energy to get the electrons in the P side to move from hole to hole towards the junction, but then the electron, once it's in a hole, does not want to leave that hole and jump into empty space, where the N side is. There are no holes in the N side for the electron to jump into, so they stay put in the P side. Unless you apply the breakdown voltage, and that is so much energy that it does force those electrons to jump into empty space in the N side, and current flows in the reverse direction.
my way of understanding this is that the negative terminal connected to the p region will release electrons and fill up the holes existing in the p region, thus expanding the depletion layer and increases the junction voltage. Hope that helps even though its after a year
@@zclee7325 I actually saw some videos on UA-cam from a channel called CircuitBread that cleared the whole thing up for me. On the p side, electrons are are in the valence band, so they can only jump from hole to hole. They can't jump out of the holes into empty space because nothing is supplying enough energy for them to do that. On the n side, electrons are in the conduction band, so they can move around freely anywhere. The electrons moving away from the negative terminal on the p side make it to the depletion region, but then they can't move any further away because there are no more holes for them to jump to, so they end up widening the depletion region.
I love the completeness that the CircuitBread videos offer. Very intuitive. Very comprehensible. What I don't understand about reverse bias is this: The high concentration of majority charge carriers is the driving force responsible for diffusion (hence the name being analogous to diffusion from high concentration to low in liquids). Yet reverse bias reduces those high concentrations, which would seem to reduce the force propelling diffusion and thus reduce the depletion voltage. But it increases the depletion voltage. This seems counter-intuitive. Can someone help me visualize that?
Hey Ted - thanks for the feedback! I am a bit confused what you're referring to with the term "depletion voltage" - could you clarify? Then we'll see what we can do to help your understanding.
I mean the width of the depletion region and thus the barrier that must be overcome to cross it. Some guy at EE Stack Exchange electronics.stackexchange.com/questions/512322/are-there-two-band-contributions-to-diffusion-current-at-the-pn-diode-junction/512479?noredirect=1#comment1314727_512479 is saying "Reverse biasing a diode does not change the electron or hole majority carrier concentrations in the neutral region. It does however increase the barrier height they must overcome. This reduces the current. That video is either very confusing with its animation or outright wrong." I referred him to this point in your video (he's responding after having watched it), which would seem to suggest the opposite of what he claims: ua-cam.com/video/C6Ctnl5RYD0/v-deo.html&list=PLfYdTiQCV_p7sDswtLZKK43BWOd2mTmHC&t=38 That point in your video makes intuitive sense in that attaching a positive electrode to the N-type would make sense in drawing away majority carrier electrons and their absence stripping that side of the depletion region of more electrons (exposing more positive ions and widening the depletion region). It just seems counterintuitive to the action of diffusion across the junction in reducing their concentration. See what I mean?
Hey Ted! The depletion region is not formed due to the high concentration of the majority charge carriers but because of the positive and negative charge layers created when the concentration of the majority charge carriers is reduced (reduced during the formation of the depletion region as free electrons and holes combine, no external voltage applied, just the heat energy at room temperature). So when in reverse bias, the concentration of the majority charge carriers is reduced but the number of positive and negative ions (which creates the depletion region) is increased and this is completely independent of diffusion current. Is this helpful?
How do you made animations? Explanation with animation is really excellent. I want to learn to make these kind of animation for different concepts. Can you please tell me how did you do that? Thank you.
These were made in Element3D which hasn't been updated in a long, long time. Our more recent animations have either been in AfterEffects (for the most simple animations) and then Blender for the most complex animations.
So will a diode show some potential drop which is equal to the breakdownvoltage like 0.7 volt in forward biasing and the current here passes by breaking simple covalent bonds by high energy and attract them into the conduction band but how the current passed in that depletion region in forward biasing?
Hi Raza, this question isn't very clear but I think I answered it in the comments on the Part 2 video. Please clarify this question if it hasn't already been answered. Thanks!
@@CircuitBread my question is simple sir that the leakage current in both reverse and forward bias below breakdown and the knee voltage is due to the same mechanism and the charge carriers are conduction electrons but as current passes after breakdown by giving much energy to the covalent bonded electrons they become conduction electrons its ok but how the current passes through the forward bise depletion region which are the charge carriers when electrons passes through depletion region ?
When the electrons move from the N-region to the P-region do the electrons first go to the conduction band of the P-region and then lose energy (heat or light) and go to the valence band of the P-region? I would appreciate if u could answer this question fast as im working on a project
1:15 you probably meant the opposite; free electrons can't go up, and holes can't go down, because the free electrons are in N region, not P. This is how you tought us in previous videos.
Why does the positive terminal of battery attract s elctrons. Isn't the positve terminal and the negative terminal of battery just a convention. And what about the current which battery supplies to diode. Where is that? Plz answer
First, thank you for the compliment, I appreciate it! Breakdown current is when the electrons flowing backward gain enough energy that they start to knock things loose. Imagine a bowling ball slowly bumping into a pin, it doesn't actually knock it over. But if a bowling ball hits a pin hard enough, not only is that pin knocked over, but it hits into other pins, knocking them out. Those energetic electrons knock other electrons away from their atoms, raising their energy so that they're in the conduction band and can move around freely. The odd thing is that, from a devices point of view, the mechanics are very different from what's happening when forward-biased, though from the perspective of a user, it doesn't really matter. Just that once you get to a certain reverse voltage, it starts to conduct in the "wrong" direction. I hope this helps!
I couldn't understand one thing.Holes cannot move like electrons but in reverse bias they move towards negative terminal.Do they break their covalent bonds?
No it's just that electrons from the - terminal that fills up the holes. And if u see it from the other way it is like the hole are attracted to the - terminal.
I understand that the depletion layer gets wider and narrower for forward and reverse biased diodes respectively. But why does the widening of the depletion layer makes the energy hill steeper and vice-versa in case of the depletion layer getting narrower?
Excellent question! The energy hill represents how hard it will be for the electron to move from one point to another. So as the increased voltage increases the size of the depletion layer, that means there are more fixed ions for the charges to pass through. This larger layer of fixed ions means it's harder for both the electrons and the holes to pass through the barrier. Focusing on the electrons, this means that the electrons need to gain even more energy to overcome that potential barrier, which is represented by a steeper energy hill. Let me know if this helps!
@@CircuitBread Thank you so much sir! Yes, I did help a lot. I would again and again mention, how great videos you make, and how equally your helpfulness in the comment sections have made our life easier :') Apart from that, yes, I had a kind of deeper follow-up question to above confusion, and I guess I have put it under the next video on forward biased diode (sorry, coz the question came into mind while i was watching that lol). So if you please would like to check that there, it would be super awesome! Thanks again!
if I google, I find reverse bias diagram and it's p type fermi level is higher than n type fermi level. Does this mean the voltage is higher than break voltage? I think only forward bias will have current. Then why fermi level difference doesn't make current in reverse bias? (fermi level is energy level at which 50% of electrons are in at 0K as my knowledge. It's hard to understand for me)
Forward biased diodes can also break if you put too much current through them, you'll melt them or catch something on fire. But you won't have a reverse voltage breakdown because they're designed to have their forward voltage "breakdown" at as low of a voltage as reasonably possible.
There are both positive and negative ions in the depletion region, which, strangely, repels both positive and negative charges. But "electrons flow opposite to electric field" is a bit vague as electric fields, depending on their polarity, and either attract or repel electrons.
I mean, electrons usually go from negative terminal to positive terminal but in this case of potential barrier in p-n semiconductor, it stops the electrons or potential is steeper up and not deeper down? Maybe there is diffusive current (concentrations of electrons independent with electric field potential) that counter drift current (dependent on a electric field potential). Sorry for this question. This is only my confusion about this topic.
Hi Vishal! We have already shot a video about BJT transistors and it is in the queue for editing and animation. I'll try to remember to come back and let you know when it has been posted.
Hey Nicola, probably someday but it's not in the queue at the moment. We'll stick it in but it will probably be awhile as we work through our current backlog.
I both agree and disagree. I agree because a hole is... well, it's a hole. The lack of something doesn't really move because it doesn't really exist. However, I disagree because it's much, much easier to understand and to model the movement as a hole moving than it is to imagine and model the movement of a series of discrete electron movements that are filling gaps in such a way that makes it look like a hole is moving. That's why, in the semiconductor topics, I've always seen and heard it discussed as "holes moving." But, it is weird, you won't get any argument from me about that.
Thank you so much sir! I was very confused as to how a current would flow in reverse biased condition. But your videos and energy level diagrams along with animation really made me understand what actually happens and what causes the current at breakdown voltage to be really high. Thank you so much.
Insanely underrated.
Literally UA-cam is place of wisdom for the younger generation, as usual awesome video and explanation ❤
Explanation is mind blowing 🤯
Woh!
just cleared the confusion in such a great way!!!
And your presentation skills are top notch professor!!!!
Thanks
Thank you
Your videos are amazing.
Keep it up.
this was the most amazing video i have even seen,, I just wish he explains the VI characteristics too
Thank you for the feedback! This video goes into the underlying mechanics, the VI characteristics are discussed in the diode introduction tutorial: www.circuitbread.com/tutorials/introduction-to-diodes
@@CircuitBread thnx for the help sir
Great video and it explained almost everything, but i am still confused about 1 thing. At 0:52 why electrons move to holes(near depletion region) and get fixed there? Why those electrons just do not go through depletion region and then further?(intuitively battery's positive terminal attracts electrons more than depletion region repels them, therefore they should move forward right?)
Amazing short and concise video. These videos will steadily increase in view count for eternity!
In the animation it would be useful to represent the electrons, from the left conductor, passing through the p zone in the valence band, until they reach che depletion region increasing it thickness
Thank you so much !!! your videos have cleared many confusion.
I would love if you could explain about photodiodes, like how they work and how they are different from regular diodes :D
Again, thank you!!!
thank you very much for your effort. But to me i still finde some questions which are not allow me to understand it fully:
1:09
In reversed biased mode the conduction band has in general no electrons left. According to your picture there would be a possibility to flow from the p-conduction band to the n-conduction band. Only electrons which are thermaly generated from the valence band exists there, and they are the reason for re reversed diode current. Am i correct?
Same picture:
Actually all Explanations say that there are no wholes left in de p - valence band, so there are no current carriers left. But actually there are electrons filling the wholes. And they are still electrons which still can produce a current. I can try to press more and more ectrons into the p-valence band. So I can increase the potential of the Electrons until they have enough energy to overcome the depletion region.
Am i also correct?
this is amazing, thank you!
I will share your channel with my friends
Thank you! Glad you're enjoying it!
Here's what I don't understand about reverse bias. It says "holes move towards the left side because they are attracted by the negative voltage." But there is no such thing as holes moving. The only way a hole moves is if an electron is moving in the opposite direction. It's all electrons moving. So now the question is, exactly which electrons are moving? Are electrons from the voltage source, maybe a battery, moving into the P side, combining with the holes, making it look like holes are moving to the left into the battery? Or is it electrons in the P material itself moving towards the junction, making it look like holes are moving away from the junction? Also, if it is electrons in the P side moving to the junction, then why don't they cross the junction and go into the N side? Why do they stop once they hit the junction, or maybe they stop a few atoms away from the junction, since the atoms right next to the junction don't have any holes for the electrons to join with.
All I can think of is that it requires a small amount of energy to get the electrons in the P side to move from hole to hole towards the junction, but then the electron, once it's in a hole, does not want to leave that hole and jump into empty space, where the N side is. There are no holes in the N side for the electron to jump into, so they stay put in the P side. Unless you apply the breakdown voltage, and that is so much energy that it does force those electrons to jump into empty space in the N side, and current flows in the reverse direction.
my way of understanding this is that the negative terminal connected to the p region will release electrons and fill up the holes existing in the p region, thus expanding the depletion layer and increases the junction voltage. Hope that helps even though its after a year
@@zclee7325 I actually saw some videos on UA-cam from a channel called CircuitBread that cleared the whole thing up for me. On the p side, electrons are are in the valence band, so they can only jump from hole to hole. They can't jump out of the holes into empty space because nothing is supplying enough energy for them to do that. On the n side, electrons are in the conduction band, so they can move around freely anywhere.
The electrons moving away from the negative terminal on the p side make it to the depletion region, but then they can't move any further away because there are no more holes for them to jump to, so they end up widening the depletion region.
@@firstlast1947 i see thats pretty insightful, thanks
I love the completeness that the CircuitBread videos offer. Very intuitive. Very comprehensible. What I don't understand about reverse bias is this: The high concentration of majority charge carriers is the driving force responsible for diffusion (hence the name being analogous to diffusion from high concentration to low in liquids). Yet reverse bias reduces those high concentrations, which would seem to reduce the force propelling diffusion and thus reduce the depletion voltage. But it increases the depletion voltage. This seems counter-intuitive. Can someone help me visualize that?
Hey Ted - thanks for the feedback! I am a bit confused what you're referring to with the term "depletion voltage" - could you clarify? Then we'll see what we can do to help your understanding.
I mean the width of the depletion region and thus the barrier that must be overcome to cross it. Some guy at EE Stack Exchange
electronics.stackexchange.com/questions/512322/are-there-two-band-contributions-to-diffusion-current-at-the-pn-diode-junction/512479?noredirect=1#comment1314727_512479
is saying "Reverse biasing a diode does not change the electron or hole majority carrier concentrations in the neutral region. It does however increase the barrier height they must overcome. This reduces the current. That video is either very confusing with its animation or outright wrong."
I referred him to this point in your video (he's responding after having watched it), which would seem to suggest the opposite of what he claims:
ua-cam.com/video/C6Ctnl5RYD0/v-deo.html&list=PLfYdTiQCV_p7sDswtLZKK43BWOd2mTmHC&t=38
That point in your video makes intuitive sense in that attaching a positive electrode to the N-type would make sense in drawing away majority carrier electrons and their absence stripping that side of the depletion region of more electrons (exposing more positive ions and widening the depletion region). It just seems counterintuitive to the action of diffusion across the junction in reducing their concentration. See what I mean?
Hey Ted! The depletion region is not formed due to the high concentration of the majority charge carriers but because of the positive and negative charge layers created when the concentration of the majority charge carriers is reduced (reduced during the formation of the depletion region as free electrons and holes combine, no external voltage applied, just the heat energy at room temperature). So when in reverse bias, the concentration of the majority charge carriers is reduced but the number of positive and negative ions (which creates the depletion region) is increased and this is completely independent of diffusion current. Is this helpful?
An amazing teacher
How do you made animations? Explanation with animation is really excellent. I want to learn to make these kind of animation for different concepts. Can you please tell me how did you do that? Thank you.
These were made in Element3D which hasn't been updated in a long, long time. Our more recent animations have either been in AfterEffects (for the most simple animations) and then Blender for the most complex animations.
So will a diode show some potential drop which is equal to the breakdownvoltage like 0.7 volt in forward biasing and the current here passes by breaking simple covalent bonds by high energy and attract them into the conduction band but how the current passed in that depletion region in forward biasing?
Hi Raza, this question isn't very clear but I think I answered it in the comments on the Part 2 video. Please clarify this question if it hasn't already been answered. Thanks!
@@CircuitBread my question is simple sir that the leakage current in both reverse and forward bias below breakdown and the knee voltage is due to the same mechanism and the charge carriers are conduction electrons but as current passes after breakdown by giving much energy to the covalent bonded electrons they become conduction electrons its ok but how the current passes through the forward bise depletion region which are the charge carriers when electrons passes through depletion region ?
When the electrons move from the N-region to the P-region do the electrons first go to the conduction band of the P-region and then lose energy (heat or light) and go to the valence band of the P-region?
I would appreciate if u could answer this question fast as im working on a project
1:15 you probably meant the opposite; free electrons can't go up, and holes can't go down, because the free electrons are in N region, not P. This is how you tought us in previous videos.
Thank you so much sir!
Why does the positive terminal of battery attract s elctrons. Isn't the positve terminal and the negative terminal of battery just a convention. And what about the current which battery supplies to diode. Where is that?
Plz answer
great work sir thanks for this enjoyable
video
excellent videos. Great
thank you for your work
in riverse biased why dont electrons of n side attracts to battery and move in circiuit
thank you . it was really helpful
Please what is a breakdown current? I want your definition, you're a great teacher.
First, thank you for the compliment, I appreciate it! Breakdown current is when the electrons flowing backward gain enough energy that they start to knock things loose. Imagine a bowling ball slowly bumping into a pin, it doesn't actually knock it over. But if a bowling ball hits a pin hard enough, not only is that pin knocked over, but it hits into other pins, knocking them out. Those energetic electrons knock other electrons away from their atoms, raising their energy so that they're in the conduction band and can move around freely. The odd thing is that, from a devices point of view, the mechanics are very different from what's happening when forward-biased, though from the perspective of a user, it doesn't really matter. Just that once you get to a certain reverse voltage, it starts to conduct in the "wrong" direction. I hope this helps!
I couldn't understand one thing.Holes cannot move like electrons but in reverse bias they move towards negative terminal.Do they break their covalent bonds?
Hi Muhammad! We did a video talking about hole movement that may help things be a bit more clear - ua-cam.com/video/N8MuD_xu6L4/v-deo.html
No it's just that electrons from the - terminal that fills up the holes. And if u see it from the other way it is like the hole are attracted to the - terminal.
@@godkiller4020 are you talking about reverse bias?
Holy shit this is a godsend!
Needed animation in 2:10 part
I understand that the depletion layer gets wider and narrower for forward and reverse biased diodes respectively. But why does the widening of the depletion layer makes the energy hill steeper and vice-versa in case of the depletion layer getting narrower?
Excellent question! The energy hill represents how hard it will be for the electron to move from one point to another. So as the increased voltage increases the size of the depletion layer, that means there are more fixed ions for the charges to pass through. This larger layer of fixed ions means it's harder for both the electrons and the holes to pass through the barrier. Focusing on the electrons, this means that the electrons need to gain even more energy to overcome that potential barrier, which is represented by a steeper energy hill. Let me know if this helps!
@@CircuitBread Thank you so much sir! Yes, I did help a lot. I would again and again mention, how great videos you make, and how equally your helpfulness in the comment sections have made our life easier :')
Apart from that, yes, I had a kind of deeper follow-up question to above confusion, and I guess I have put it under the next video on forward biased diode (sorry, coz the question came into mind while i was watching that lol). So if you please would like to check that there, it would be super awesome! Thanks again!
if I google, I find reverse bias diagram and it's p type fermi level is higher than n type fermi level. Does this mean the voltage is higher than break voltage?
I think only forward bias will have current. Then why fermi level difference doesn't make current in reverse bias?
(fermi level is energy level at which 50% of electrons are in at 0K as my knowledge. It's hard to understand for me)
Can u plzz plzz plzz do a video on pnp transistor and how it works based on electron movement.
So only reverse biased diodes are intended to break? And forward biased diodes possibly wont break right?
Forward biased diodes can also break if you put too much current through them, you'll melt them or catch something on fire. But you won't have a reverse voltage breakdown because they're designed to have their forward voltage "breakdown" at as low of a voltage as reasonably possible.
@@CircuitBread thank you so much sir 🙏👍
Electrons flows opposite to electric field right? Then why this electric field stops electrons in potential barrier?
There are both positive and negative ions in the depletion region, which, strangely, repels both positive and negative charges. But "electrons flow opposite to electric field" is a bit vague as electric fields, depending on their polarity, and either attract or repel electrons.
I mean, electrons usually go from negative terminal to positive terminal but in this case of potential barrier in p-n semiconductor, it stops the electrons or potential is steeper up and not deeper down? Maybe there is diffusive current (concentrations of electrons independent with electric field potential) that counter drift current (dependent on a electric field potential). Sorry for this question. This is only my confusion about this topic.
can I used your animation of the diode?
Definitely, just let people know where you got it, that's totally fine!
Sir can you explain about transistor as a switch
Hi Vishal! We have already shot a video about BJT transistors and it is in the queue for editing and animation. I'll try to remember to come back and let you know when it has been posted.
Hey Vishal, this is very belated, but we did post that BJT video - sorry about taking so long to come back and let you know!
How to use a Bipolar Junction Transistor (BJT) as a Switch?: www.circuitbread.com/tutorials/how-to-use-a-bipolar-junction-transistor-bjt-as-a-switch
Thanks!
Will you ever upload a video on solar cells? thanks anyway
Hey Nicola, probably someday but it's not in the queue at the moment. We'll stick it in but it will probably be awhile as we work through our current backlog.
I still can grasp
妙！
Holes don't move#wrong concept
I both agree and disagree. I agree because a hole is... well, it's a hole. The lack of something doesn't really move because it doesn't really exist. However, I disagree because it's much, much easier to understand and to model the movement as a hole moving than it is to imagine and model the movement of a series of discrete electron movements that are filling gaps in such a way that makes it look like a hole is moving. That's why, in the semiconductor topics, I've always seen and heard it discussed as "holes moving." But, it is weird, you won't get any argument from me about that.
Thank you so much 🤍