Hey everyone! If you want to read through the transcript at your own pace, with the animations embedded, check it out on our site: www.circuitbread.com/tutorials/how-does-a-diode-work-part-2-forward-bias Have a great day!
95% of the explanation in this video is pretty much the same explanation that everyone learns in uni, but that extra 5% in your explanation really makes everything way way more clear. THANKS!!!!
😂 This is the special principle, what are you trying to do differently? It’s just that the microscopic process at the PhD stage is more complicated than this.
I love how decently you explain! My teacher confused me a bit (he was working with conventional current so everything just clashed in my head). Thank you for clearing it all up so expertly!
I admit, the conventional current versus electron movement still confuses me at times, especially when switching back and forth. I'm glad we were able to help!
2:40 what happens if you take off the negative terminal at this point? The valence electrons in the P region keep going to ground, but no more electrons break thru the depletion region? Also, what happens to the electrons in the depletion region at that point.
Hi and thanks for the video! Are you sure the representation at 2:00 is correct? From my PV books, the quasi-fermi level for electrons drops linearly when entering the p-region, and vice versa for holes. This would mean the electrons recombine, in contrast to your animation in which they pass all the way through the p-junction. While I'm saying this I realise that you mention that they recombine and are now in the valence band. This reconciles with my explanation of the fermi level, so thanks and nevermind what I said ;)
Thanks for the comment! I appreciate you sharing your concerns and then your own resolution as that may help someone else with a similar question. And we have found mistakes in the past, so if you see something else that doesn't seem right, don't hesitate to reach out again!
In the animations the only missing things are the little current due to the minority charge carriers in the p and n zones. The voltage barriers over the metal semiconductor junctions are also important (it is the principle about the Shottky diode)
sir...i have one doubt..in the end ..the animation suggested that electrons can more easily go from conduction band of n side to p side in forward bias ...and then suddenly they go to valence band of p side by losing the energy..so how are they able to cause current ..as current can only exist if electrons are in conduction band..???
I think the challenge here is to remember that there is electron current and hole current. Even as electrons fall into the valence band, there is still hole movement on that side of the junction. I hope that helps!
Yes, you can! That's actually the basis of schottky diodes - they use different materials that let them have a lower forward voltage and a faster response time. As far as I'm aware, the materials of the diode is the only way you can change the forward voltage.
This is the infamous "current flow versus electron flow" confusion. When Benjamin Franklin was establishing the direction that electrons seemed to be flowing, he made a guess, had a 50/50 chance, and it was wrong. Even though we assume that current flows from positive to negative, in reality, electrons flow from negative to positive.
This is really well explained and the most helpful source I've got so far. But I still have trouble understanding why the electrons do not recombine with holes when they pass through the depletion region and drop from conduction band to valence band ? Why do they not recombine and instead pass directly to the positive side of the bias ?
A question sir the depletion region has no free charge carriers neither the conduction electrons nor the valence holes so how current is passed through it?? (By which band conduction or valence) please explain.
Just because no free charge carriers *reside* in the depletion region, they are still able to pass through when reverse biased (which is why you get the small reverse current). When forward biased, and there is still a depletion region, electrons still can pass through. Though if the diode is biased enough, the depletion region basically disappears completely.
@@CircuitBread when the diode reverse biased then the almost all the atoms in p region have gain an another electron and completed its valence shell and conduction band is empty(but the p type will become slighlty negative becausr now it has 1 electron more than proton) similarly n type will lose the 1 conduction electron(and becomes negative) so the covalent bonds are complete on both sides now the leakage current is due to the some of the bands which have been broken due to the temperature as in intrinsic and now they are free conduction electrons as same leakage current passes through the depletion region in forward bias if applied voltage is less than 0.7 volts its ok its all that i have learnt the pn junction in reverse bias acts like intrinsic semiconductor. But how the current is passed when the forward biase voltage will be increased beyond 0.7 volts will this current passes through conduction or valence band kindly help i will be really thankful to you 😃
Hy, thank you for your explanation in this video. let me ask you about Voc in solar measurement. If the Vocs are 0.84, 0.81, 0.87, what does it means in terms of forward bias..? let me hear from you soon. donny
1. when electron from conduction band of n side enter conduction band of p side why it recombines with holes in valence band, y is it not going in the outer circuit with in conduction band only 2. when electrons that have entered p side recombines with valence band holes isn't it emitting photons
Hi Vrinda! For 1) when the electron crosses the PN junction, it loses energy in that transition, so that decrease in energy means they drop energy levels - energy levels that most likely put them in the valence band. 2) Actually, sometimes they do. While non-LED diodes aren't optimized for photon emission, that's actually how the concept of LEDs was originated. They noticed that some exposed PN junctions glowed when forward biased - this is briefly mentioned in our history of LEDs video, if I remember correctly. A lot of the energy is released as heat, though.
Thank you! My understanding is that they're related but cut-in voltage is a bit more "handwavy". Cut-in voltage is pretty straightforward - it's the point where we assume the diode is conducting. It's straightforward conceptually but difficult to know at what point we consider it to be conducting. For general diodes (the assumptions we use when we talk about diodes at CircuitBread) we assume the cut-in voltage to be .7V on standard diodes, which is kinda the industry standard as well. Built-in voltage is the calculated voltage potential created by the depletion zone, the electric field that will need to be overcome before current can flow. From my perspective (and differing or clarifying comments would be appreciated) they are talking about the same thing but cut-in voltage is by looking at the point in which the amount of current flowing passes a semi-arbitrarily selected value. Built-in voltage is the calculated result (from knowing the dopant levels of the diode) that yields the voltage needed to overcome that barrier in the diode.
@@CircuitBread Aha... Thanks for the info. I had used Sketch App many years ago but I can cant see how it is doing these unless its where you create the geometries. This brings the question of how do you simulate the moving electrons? Illustrator?? Best
Movement is done in After Effects, with more complicated animations (rarely, usually with client videos) we'll use Blender. But most of what you see on our channel is After Effects.
@@CircuitBread Thanks for the clarification. I have strted to learn Blender mydelf for other purposes but what you described is really a whole production process... The next thing for such skills is full cartoon animations but for such good scientific animation productions, your doing a wonderful job. My respects for all the excellent work that your doing... Well done
Then a group of 6 LEDs should consume 4.5 watts. It would be easier to say that each LED is consuming .75W so however you have it setup, you should be able to multiply the number of LEDs by .75W and reach the total amount of power consumed.
Your animation is good, but if it were me, I would add the hole flow description and the eletrostatic potential as well. Because you are only explaining electrons and energy bands.
In reverse bias, there's always a small leakage current and then, if you put too large of a voltage in reverse bias, you'll have a breakdown and your diode will start to conduct a lot of current. If you haven't checked out Part 3 of these videos on diodes, I recommend it!
Why does the forward current increase exponentially, instead of going linearly or parabolically...I mean, the number of available electrons are same...so where's this current increasing so exponentially
Hey Jonathan - it's not related to the amount of electrons available, it's because of the depletion region. A diode is non-linear, so its resistance isn't fixed, and as the voltage goes up, the "apparent" resistance drops as it gets easier and easier for the electrons to cross the depletion region as more voltage is going toward moving the electrons versus overcoming the barrier potential. There's almost certainly a better, chemistry-level explanation, but I'm hoping that clarifies things at the EE level.
why dont you mention why you caled cathode and anode to opposing regions. cathode generally be the electron acceptor and anode is the electron donor and they can change in differen situations. why dont mention and pass it like cathode is always the negative side ffs. I really dont understand why a person making a video about that dont mention it.
This is what happens when you go to college. No one knows what the fuck you’re talking about because while you’re explaining how it works, you’re using terms that nobody understands. Nobody knows what or where the forward bias is.
Legitimate concern! We're in the process of putting our current (and future) tutorials in more explicit and linear courses so that, if you work through them, by the time you reach a video like this (that we roughly label as an intermediate topic) you're already comfortable with the terms being used. Without any background (or at least, with limited background) in semiconductor devices, this would be extremely confusing.
95% of the explanation in this video is pretty much the same explanation that everyone learns in uni, but that extra 5% in your explanation really makes everything way way more clear. THANKS!!!!
Hey everyone! If you want to read through the transcript at your own pace, with the animations embedded, check it out on our site: www.circuitbread.com/tutorials/how-does-a-diode-work-part-2-forward-bias Have a great day!
Sorry, “ what is external bias voltage must greater than barrie potential”
95% of the explanation in this video is pretty much the same explanation that everyone learns in uni, but that extra 5% in your explanation really makes everything way way more clear. THANKS!!!!
Awesome, thanks for making my day!
😂 This is the special principle, what are you trying to do differently? It’s just that the microscopic process at the PhD stage is more complicated than this.
uni? i am here learning this for my high school test😭
I love how you illustrate the movement of current through animation. I really appreciate your videos! It's helped a bunch thank you.
I love how decently you explain! My teacher confused me a bit (he was working with conventional current so everything just clashed in my head). Thank you for clearing it all up so expertly!
I admit, the conventional current versus electron movement still confuses me at times, especially when switching back and forth. I'm glad we were able to help!
Hey man really appreciate your work , even though you have just started out on your journey , the quality is great !!
Thanks, I appreciate it! We're always learning and (hopefully) improving!
Thank you!!! This cleared up some holes in my understanding that my professor did not manage to fill after 4 hours of lecture.
One of the best video for understanding PN junction I came across 👏👏👏
These videos are so relevant for CBSE and JEE at the same time
This is by far the best explanation video on this subject
High quality video and really nice voice!
this is perfect explanation! Danke schön.
Man u explained it in the best way
Thnx
This is just awesome... I mean really satisfying description.
Those stupid smart classes animation is nothing before this! Thankyou Sir, for the amazing animation and explanation!
Thank Avani, glad it helped you!
Thank Avani, glad it helped you!
Thanks. Ive been searching for a video with a simple explanation for this. And you definitely delivers. Thanks a lot!!
You're welcome!!
Thank you sir
And ur explanation is understandable
I officially understand this concept!!! Thank you❤
Glad we helped!
This is super amazing.... Keep making more videos.... It will be great for you someday..
Thanks!
this is the best explaination I ve seen
I'm so glad that I found your channel💘💕💖
U r the beesttt
I've never seen clearer !
2:40 what happens if you take off the negative terminal at this point? The valence electrons in the P region keep going to ground, but no more electrons break thru the depletion region? Also, what happens to the electrons in the depletion region at that point.
Awesome and great quality by the guy ❤
Thanks pal for this useful explanation. Hope you will success in this way in future.
Thank you!
Completely Clear sir🖤🖤🖤🇧🇩
Very clear explanation 👌👏👍
💯💯💯Greatly explained.
An important video ❤
Hi and thanks for the video! Are you sure the representation at 2:00 is correct? From my PV books, the quasi-fermi level for electrons drops linearly when entering the p-region, and vice versa for holes. This would mean the electrons recombine, in contrast to your animation in which they pass all the way through the p-junction. While I'm saying this I realise that you mention that they recombine and are now in the valence band. This reconciles with my explanation of the fermi level, so thanks and nevermind what I said ;)
Thanks for the comment! I appreciate you sharing your concerns and then your own resolution as that may help someone else with a similar question. And we have found mistakes in the past, so if you see something else that doesn't seem right, don't hesitate to reach out again!
Amazing explanation!
3:26 Where does this energy go?
In forward bias why dont the width of the depletion region becomes zero??
In the animations the only missing things are the little current due to the minority charge carriers in the p and n zones.
The voltage barriers over the metal semiconductor junctions are also important (it is the principle about the Shottky diode)
Thanks for the video man,your great
Can you tell me what is recombination current
aahhhhh... what a heavenly voice you've got sir...
Thanks Jassi! 😂
thank you it was beautiful
Very helpful indeed👏🏼👍🏻Thanks alot
More more upcoming videos
Glad to hear, thanks!
sir...i have one doubt..in the end ..the animation suggested that electrons can more easily go from conduction band of n side to p side in forward bias ...and then suddenly they go to valence band of p side by losing the energy..so how are they able to cause current ..as current can only exist if electrons are in conduction band..???
I think the challenge here is to remember that there is electron current and hole current. Even as electrons fall into the valence band, there is still hole movement on that side of the junction. I hope that helps!
I LOVED THIS !!!!
Thanks for these vids
You're welcome!
Can you control this 0.7 voltage drop for example with the materials of the diode, or anything else?
Yes, you can! That's actually the basis of schottky diodes - they use different materials that let them have a lower forward voltage and a faster response time. As far as I'm aware, the materials of the diode is the only way you can change the forward voltage.
@@CircuitBread waaaaaat this is amazing definitely gonna add this to my report thanks 😊
i dont understand how electrons are moving from the negative side of the applied voltage instead from the positive side??????????????????????????????
This is the infamous "current flow versus electron flow" confusion. When Benjamin Franklin was establishing the direction that electrons seemed to be flowing, he made a guess, had a 50/50 chance, and it was wrong. Even though we assume that current flows from positive to negative, in reality, electrons flow from negative to positive.
Why is the energy levels or bandgap for the n and p region different,i don't understand the energy diagram part .
Thank you, I benefited a lot
This is really well explained and the most helpful source I've got so far. But I still have trouble understanding why the electrons do not recombine with holes when they pass through the depletion region and drop from conduction band to valence band ? Why do they not recombine and instead pass directly to the positive side of the bias ?
Same doubt🙄
I want to know this too. It's confusing
BESTTT!! Bring more please
A question sir the depletion region has no free charge carriers neither the conduction electrons nor the valence holes so how current is passed through it?? (By which band conduction or valence) please explain.
Just because no free charge carriers *reside* in the depletion region, they are still able to pass through when reverse biased (which is why you get the small reverse current). When forward biased, and there is still a depletion region, electrons still can pass through. Though if the diode is biased enough, the depletion region basically disappears completely.
@@CircuitBread when the diode reverse biased then the almost all the atoms in p region have gain an another electron and completed its valence shell and conduction band is empty(but the p type will become slighlty negative becausr now it has 1 electron more than proton) similarly n type will lose the 1 conduction electron(and becomes negative) so the covalent bonds are complete on both sides now the leakage current is due to the some of the bands which have been broken due to the temperature as in intrinsic and now they are free conduction electrons as same leakage current passes through the depletion region in forward bias if applied voltage is less than 0.7 volts its ok its all that i have learnt the pn junction in reverse bias acts like intrinsic semiconductor. But how the current is passed when the forward biase voltage will be increased beyond 0.7 volts will this current passes through conduction or valence band kindly help i will be really thankful to you 😃
what application do you use to illustrate the movement of forward bias and reverse bias sir?
What video software? I think he was using Element3D.
Hy, thank you for your explanation in this video. let me ask you about Voc in solar measurement. If the Vocs are 0.84, 0.81, 0.87, what does it means in terms of forward bias..?
let me hear from you soon.
donny
1. when electron from conduction band of n side enter conduction band of p side why it recombines with holes in valence band, y is it not going in the outer circuit with in conduction band only
2. when electrons that have entered p side recombines with valence band holes isn't it emitting photons
Hi Vrinda! For 1) when the electron crosses the PN junction, it loses energy in that transition, so that decrease in energy means they drop energy levels - energy levels that most likely put them in the valence band.
2) Actually, sometimes they do. While non-LED diodes aren't optimized for photon emission, that's actually how the concept of LEDs was originated. They noticed that some exposed PN junctions glowed when forward biased - this is briefly mentioned in our history of LEDs video, if I remember correctly. A lot of the energy is released as heat, though.
@@CircuitBread Thanks A LOT!
Thanks brother I appreciate your help! Please help me with this knowledge
This's good. THX
thank u sir........superb explaination
Thank you for the feedback, I'm glad we were able to help!
@CircuitBread.. Excellent explanations ... Can u please provide me a clarity on the difference btw built-in voltage and cut in voltage
Thank you! My understanding is that they're related but cut-in voltage is a bit more "handwavy". Cut-in voltage is pretty straightforward - it's the point where we assume the diode is conducting. It's straightforward conceptually but difficult to know at what point we consider it to be conducting. For general diodes (the assumptions we use when we talk about diodes at CircuitBread) we assume the cut-in voltage to be .7V on standard diodes, which is kinda the industry standard as well.
Built-in voltage is the calculated voltage potential created by the depletion zone, the electric field that will need to be overcome before current can flow.
From my perspective (and differing or clarifying comments would be appreciated) they are talking about the same thing but cut-in voltage is by looking at the point in which the amount of current flowing passes a semi-arbitrarily selected value. Built-in voltage is the calculated result (from knowing the dopant levels of the diode) that yields the voltage needed to overcome that barrier in the diode.
Dear CircuitBread:
Just a quick question. What software are you using to make these lovely graphics?
It depends, it’s usually either In Sketch App, Illustrator, or After Effects.
@@CircuitBread Aha... Thanks for the info. I had used Sketch App many years ago but I can cant see how it is doing these unless its where you create the geometries.
This brings the question of how do you simulate the moving electrons? Illustrator??
Best
Movement is done in After Effects, with more complicated animations (rarely, usually with client videos) we'll use Blender. But most of what you see on our channel is After Effects.
@@CircuitBread Thanks for the clarification. I have strted to learn Blender mydelf for other purposes but what you described is really a whole production process... The next thing for such skills is full cartoon animations but for such good scientific animation productions, your doing a wonderful job. My respects for all the excellent work that your doing... Well done
Ihave a9watt led lamp inside it there is12diod in series ,ldevide them to 6and6groub.what about its electric consumption?
If there are 9 watts being consumed by 12 diodes, then 6 diodes (with the appropriate voltage) would consume half. Unless I'm missing something?
@@CircuitBread l mean 2groubs of led.
Then a group of 6 LEDs should consume 4.5 watts. It would be easier to say that each LED is consuming .75W so however you have it setup, you should be able to multiply the number of LEDs by .75W and reach the total amount of power consumed.
Your animation is good, but if it were me, I would add the hole flow description and the eletrostatic potential as well. Because you are only explaining electrons and energy bands.
Thanks 🙏
You're welcome!
When does diode conduct current in reverse bias,Sir?
In reverse bias, there's always a small leakage current and then, if you put too large of a voltage in reverse bias, you'll have a breakdown and your diode will start to conduct a lot of current. If you haven't checked out Part 3 of these videos on diodes, I recommend it!
@@CircuitBread Thanks for the quick reply,Sir.It means a lot.
can you teach my solid state device course lol
Why does the forward current increase exponentially, instead of going linearly or parabolically...I mean, the number of available electrons are same...so where's this current increasing so exponentially
Hey Jonathan - it's not related to the amount of electrons available, it's because of the depletion region. A diode is non-linear, so its resistance isn't fixed, and as the voltage goes up, the "apparent" resistance drops as it gets easier and easier for the electrons to cross the depletion region as more voltage is going toward moving the electrons versus overcoming the barrier potential. There's almost certainly a better, chemistry-level explanation, but I'm hoping that clarifies things at the EE level.
Wow!
मोज कर दि l
Teach me my 4th semester courses :p
😂
why dont you mention why you caled cathode and anode to opposing regions. cathode generally be the electron acceptor and anode is the electron donor and they can change in differen situations. why dont mention and pass it like cathode is always the negative side ffs. I really dont understand why a person making a video about that dont mention it.
kind off confused
This is what happens when you go to college. No one knows what the fuck you’re talking about because while you’re explaining how it works, you’re using terms that nobody understands. Nobody knows what or where the forward bias is.
Legitimate concern! We're in the process of putting our current (and future) tutorials in more explicit and linear courses so that, if you work through them, by the time you reach a video like this (that we roughly label as an intermediate topic) you're already comfortable with the terms being used. Without any background (or at least, with limited background) in semiconductor devices, this would be extremely confusing.
Not enough illustration, hate to break it to you but you're explaining it like Uni prof would, and thats not a compliment :/
95% of the explanation in this video is pretty much the same explanation that everyone learns in uni, but that extra 5% in your explanation really makes everything way way more clear. THANKS!!!!