The timestamps for the different topics covered in the video: 0:15 Avalanche breakdown effect 4:18 Zener diode and Zener breakdown effect 7:41 Effect of temperature of Avalanche and Zener breakdown voltages
It should've been mentioned that, in reality, the Zener effect is tunneling of charges across the (narrow) potential barrier. Valuable the mention about the reduction of the mean free path of carriers, with the temperature increase in the avalanche effect. Also worth mentioning the fact that in the region where zener and avalanche effect have similar effect, the temperature coefficient becomes nearly zero and devices are very good to make stable voltage regulators. Very good job however.
The strong built-in electric field results in a narrow depletion width. Zener breakdown occurs due to tunneling and is largely independent of the applied electric field. Even a small applied voltage can induce tunneling, as the majority carriers do not need to overcome the built-in potential.
Watch Neso Academy, they are good for digital electronics and coding related tutorials. All About Electronics does only analog electronics and circuits guess
Your videos are awesome bro, such care is taken to create the images and graphs! Very amazing 😁 I have one suggestion though: please consider hiring a narrator to speak the text because I am very sorry to say but the spoken text is borderline incomprehensible, and requires so much thought to understand that it becomes distracting. I tried to turn on the subtitles (thanks for adding those!) but having to read the subtitles while following along with the images and graphs AND committing the information to memory at the same time is impossible for me. Maybe I am dumb? Who knows 🤪🤯 please consider my request. I am very sorry if this was in someway offensive because your content is epic and deserves nothing but the highest praise!
Personally, I think if he hired a narrator that would make things easier for people who aren't used to his accent, but if he just wants to put his own voice out there, I respect that. I found it hard to understand his voice at first too, but I think he should have his voice on them as a symbol of taking credit for his work.
Actually I find it only takes a moment to tune into his accent, his words don't run into each other nor do they drop off as I find with many American voices, I'm Scottish and people often seem to make little effort to overcome an unfamiliar accent whilst ignoring their own accent!@@Kevin-jz9bg
I don't understand why the depletion zone gets narrower in Zener Breakdown? I am just a beginner. But I think when the PN junction is formed in a heavily doped region, the carriers through the PN junction should become more and penetrate deeper, so the depletion region should become wider, but the electrons and holes around the depletion region should be so dense that a large electric field is generated.
Because the Electric field E= V/d. So, as the depletion region width reduces, the electric field will become stronger. In addition to that now as both p and n regions are heavily doped, there will be more immobile ions in the same depletion region. So, the electric field will be much stronger. I hope it will clear your doubt.
In zener breakdown already the diode is heavily doped.and in avalanche if it is done then it will function same as zener , so rather read texts thoroughly than mugging up videos
Sir but in the zener breakdown when u said that the depletion layer is going to decrease then how the no. of immovable ions are increasing? They should be decreasing as u said and hence the E.F will also be reducing....?
Due to the heavy doping, the electrons and holes will recombine very near to the junction itself. But the number of electrons or holes which are recombined will remain the same. So, the immobile ions will remain the same. But due to the heavy doping, the width of the depletion region will reduce. Means the same immobile ion now existing in the small depletion region. And due to that the built-in electric field is strong in the depletion region. (E= V/d) I hope it will clear your doubt.
SIR ONE DOUBT , WHEN TEMPERATURE INCREASE THEN SILICON ATOMS EXXITE AND GAIN KINETIC ENERGY CAUSING ATOMS TO IONIZE,THEN SHD AVALANCHE ALSO HAVE NEGATIVE TEMPERATURE THING BECAUSE HEAT CAUSES ATOM TO IONIZE AS ELECTRONS ALSO GAIN KINETIC ENERGY( YOU VIDEO HELPED ME A LOT SIR, THANK YOU SOO MUCH SIR) CLEAR EXPLANATION SIR BUT ONLY ONE DOUBT, PLEASE HELP ME SIR
Yess, but i think it is not, because the field effect is much more dominant on the effect of thermal ionization.... again its just a guess😁 btw, good question✌
@@samarthtandale9121 ooh so the temepratures they mention is it like 100 degrees or 50 degree celsius sir? if the temperatures are not too high i guess the field effect wioll dominate. Understood!!!!!!!!!!!!!!!! thabkyoiu so much
breakdown voltage is a voltage at which the diode has elctrons in depletion region excite due to increase in reverse voltage and therefore the electric field is also increased and electron in depletion region strike on silicon and there fore the charge increases and current starts to flow and due to high current flow diode may damage. is this right
You shouldn't use the term "knock out the electrons". The separation of valence electron from a atom solely depends on the energy of atom(determined by its temperature) which is provided by the minority charge carriers that have gained high energy and speed due to increase in reverse voltage. Better to say there is transfer of energy from minority charge carriers to the atoms on their path of movement.
How do immobile ions increase if depletion width increases? I thought immobile ions are formed only by the diffusion current. And the sole purpose of depletion width increasing is to prevent the flow of diffusion current. 🤔
In avalanche breakdown, if the depletion width increases under reverse bias condition, how can the electric field increase in depletion field ? Because E=V/d. If d increases because of reverse bias, then E should decrease right ? But you said E would increase in depletion region during avalanche breakdown . Can you clarify this ?
The depletion region width d ∝ √ V. That means as the V increases, electric filed increases. I hope it will clear your doubt. If you still have any doubt, let me know here.
@@ALLABOUTELECTRONICS But sir there in zener breakdown you told that since E=V/d, as d decreases due to heavy doping , E increases. So there was an inverse relationship between d and E in zener breakdown. But here you are saying that they are directly proportional . Is there one generic relationship between E and d that can be followed throughout ?
the concentration of majority charge carriers in the depletion region is also increasing here, in reverse bias condition. So, hopefully may be the dependency is more predominant on the conc. of charges rather than width of space charge region.
thank you :) For pn noise generators: how do I know that I picked the right diode or transistor? Is any diode or transistor okay which has a breakdown voltage about 6V or are there models out there which are super low-noise? Thx!
The timestamps for the different topics covered in the video:
0:15 Avalanche breakdown effect
4:18 Zener diode and Zener breakdown effect
7:41 Effect of temperature of Avalanche and Zener breakdown voltages
👍👍👍
sir please make a vedio on transistors please
transistor class 12 as you explain awesome
Bacha ye transister he hai
0:34 to 0:40 . Depletion region of reverse biased p-n junction should be greater! But it is showing it lesser than forward biased!
It should've been mentioned that, in reality, the Zener effect is tunneling of charges across the (narrow) potential barrier.
Valuable the mention about the reduction of the mean free path of carriers, with the temperature increase in the avalanche effect.
Also worth mentioning the fact that in the region where zener and avalanche effect have similar effect, the temperature coefficient becomes nearly zero and devices are very good to make stable voltage regulators.
Very good job however.
You know, sometimes I zener to the corner
So, is this electric field explanation of the zener effect wrong?
he mentioned already!
The strong built-in electric field results in a narrow depletion width. Zener breakdown occurs due to tunneling and is largely independent of the applied electric field. Even a small applied voltage can induce tunneling, as the majority carriers do not need to overcome the built-in potential.
This saved me a lot of time. Very precise.
Respect🙌👏
very clean explanation. Helped me for my exam preparation
Thanks you🙏
Great explanation!! But too many sentence connectors.
You mean conjunction.. 😅
thank you so much sir, U're such an amazing teacher! I was struggling through this concept throughout the sem your 1 video made it clear!
Great best lecture to have knowledge on electronics at school level. Sir please put video on transistor working as an amplifier
Yes, very soon that will also be covered. Just wait for a week or so.
please explain digital electronics too! you're doing a fantastic job with the Analog part
Watch Neso Academy, they are good for digital electronics and coding related tutorials. All About Electronics does only analog electronics and circuits guess
Sir you explain with full detail so it becomes easy for us to understand the concept in a better.. thank you sir for these types of videos 🙏🏻🙏🏻
Very nice vedio
Your videos are awesome bro, such care is taken to create the images and graphs! Very amazing 😁 I have one suggestion though: please consider hiring a narrator to speak the text because I am very sorry to say but the spoken text is borderline incomprehensible, and requires so much thought to understand that it becomes distracting. I tried to turn on the subtitles (thanks for adding those!) but having to read the subtitles while following along with the images and graphs AND committing the information to memory at the same time is impossible for me. Maybe I am dumb? Who knows 🤪🤯 please consider my request. I am very sorry if this was in someway offensive because your content is epic and deserves nothing but the highest praise!
Personally, I think if he hired a narrator that would make things easier for people who aren't used to his accent, but if he just wants to put his own voice out there, I respect that. I found it hard to understand his voice at first too, but I think he should have his voice on them as a symbol of taking credit for his work.
@@Kevin-jz9bg I hadn't looked at it that way before, but now that you've suggested that it very much seems like a no brainer to me!
Actually I find it only takes a moment to tune into his accent, his words don't run into each other nor do they drop off as I find with many American voices, I'm Scottish and people often seem to make little effort to overcome an unfamiliar accent whilst ignoring their own accent!@@Kevin-jz9bg
Thank you so much sir!!❤️...you are such a great teacher
I can understood it easily without any doubts thank you sir
Sir bhot gajjab,,,,,kya animation🙌🙌🙌🙌🙌
Tq u so much sir because of u I understand this question which is one's difficult for me now it is an easy question sir tq u so much
Bahut gajab samjaya👍👍👍👍🤠🤠😊😊
I don't understand why the depletion zone gets narrower in Zener Breakdown? I am just a beginner. But I think when the PN junction is formed in a heavily doped region, the carriers through the PN junction should become more and penetrate deeper, so the depletion region should become wider, but the electrons and holes around the depletion region should be so dense that a large electric field is generated.
This is because zener diodes are heavily doped
thank you very much
Thank you very much 👍🏼🙏🙏
Crystal clear 🙏
Zener breakdown starts at 4:45 time
A great video again sir. good reason behind every facts.
very clear explanation.thank you very much sir
I understand perfectly now tq very much sir😇
Thank you so much for this explanation ☺️
Can u suggest some books for electronics sir ?
thanku for this viedo nice explanation with animation
Very well explained. Hats off
Awesome video! Very clear concepts and awesome animations!!
grt. sir bahut aasani se samajhe aa gaya
Everything is very well explained.Sir,I want notes for these lectures..
Sir, At 5:40 why the electric field is much stronger even though the depletion region width decreases
Because the Electric field E= V/d. So, as the depletion region width reduces, the electric field will become stronger.
In addition to that now as both p and n regions are heavily doped, there will be more immobile ions in the same depletion region.
So, the electric field will be much stronger.
I hope it will clear your doubt.
@@ALLABOUTELECTRONICS Does this mean that the barrier voltage for a diode will change with the doping concentration?
@@divyakaur9516 Yes.
Thank uuu so much...well explained👍👍👍😊😊😊
Thank u for the video it helps me a lot
I really enjoy this starting music every time
Me too
Thanks for the simple explanation!
You Sir, are truly brilliant!
Amen, That is why he is my God.
Osm explanation, Thanks dear
Can you try to solve all Difficult GATE analog questions from ECE, EE and IN ? make a separate series.
There is a seperate playlist. You can check that on playlist page.
Awesome video👍👍👍
What's the effect of increased doping in case of avalanche breakdown voltage and zener breakdown voltage sir?
In zener breakdown already the diode is heavily doped.and in avalanche if it is done then it will function same as zener , so rather read texts thoroughly than mugging up videos
You Explain it extraordinarily sir thank you
Nicely explained ...thnx
Thanks for the informative video
super
Great explanation. Thank you
04:19 Zener Breakdown 👍❤
amazing explanation
Very clear concept
thnx bruhhhh
Great videos👍
please make video on transistor........
Helped me a lot.....thankyou sir❤️
Good
Thank u😇💕
Thanks for the concept
Sir but in the zener breakdown when u said that the depletion layer is going to decrease then how the no. of immovable ions are increasing? They should be decreasing as u said and hence the E.F will also be reducing....?
Due to the heavy doping, the electrons and holes will recombine very near to the junction itself. But the number of electrons or holes which are recombined will remain the same.
So, the immobile ions will remain the same. But due to the heavy doping, the width of the depletion region will reduce. Means the same immobile ion now existing in the small depletion region.
And due to that the built-in electric field is strong in the depletion region. (E= V/d)
I hope it will clear your doubt.
Sir Very thanks
Awesome sir
Sir what is electric field ? How it can effect electrons and holes ?
Nice video, , would be nice to get temperature curves in function of temperature to illustrate the explanation
Very nice information
nicely covered all
Please speed slow and effectiveness will increases, write at screen to more effictive
Well explained
Thank you sir
Gr8 explanation ..
Helped a lot thank u
good explaination bro
very nice explanation
Good explanation...
this was masterpiece
Useful 🤗
SIR ONE DOUBT , WHEN TEMPERATURE INCREASE THEN SILICON ATOMS EXXITE AND GAIN KINETIC ENERGY CAUSING ATOMS TO IONIZE,THEN SHD AVALANCHE ALSO HAVE NEGATIVE TEMPERATURE THING BECAUSE HEAT CAUSES ATOM TO IONIZE AS ELECTRONS ALSO GAIN KINETIC ENERGY( YOU VIDEO HELPED ME A LOT SIR, THANK YOU SOO MUCH SIR) CLEAR EXPLANATION SIR BUT ONLY ONE DOUBT, PLEASE HELP ME SIR
Yess, but i think it is not, because the field effect is much more dominant on the effect of thermal ionization.... again its just a guess😁 btw, good question✌
@@samarthtandale9121 ooh so the temepratures they mention is it like 100 degrees or 50 degree celsius sir?
if the temperatures are not too high i guess the field effect wioll dominate.
Understood!!!!!!!!!!!!!!!! thabkyoiu so much
@@akshinbarathi8914 Great! 👍
😅Just don't call me Sir, I'm a First Year Engg. student 🙌
@@samarthtandale9121 oooh iam studying for neet finished 12th now only
thnaks a lot friend
@@akshinbarathi8914 😄🙌👍🤞
Thanks sir
Oh my god I fell asleep listening to this video but good info
😂😅
breakdown voltage is a voltage at which the diode has elctrons in depletion region excite due to increase in reverse voltage and therefore the electric field is also increased and electron in depletion region strike on silicon and there fore the charge increases and current starts to flow and due to high current flow diode may damage.
is this right
I want to explain to me further about the electric field in the zener breakdown
nice video sir
nice vedio
but try to use a bigger pointer
You shouldn't use the term "knock out the electrons". The separation of valence electron from a atom solely depends on the energy of atom(determined by its temperature) which is provided by the minority charge carriers that have gained high energy and speed due to increase in reverse voltage. Better to say there is transfer of energy from minority charge carriers to the atoms on their path of movement.
Sir in zener diode when depletion region is very less in unbiased condition then them how immobile ion will will be more
Nice video
GREAT!!!!
Nice
Good explanation sir thank you🙏🙏
nice...!💚
What is the difference between diffusion and tunneling
nice one.
Bahut acche
How do immobile ions increase if depletion width increases? I thought immobile ions are formed only by the diffusion current. And the sole purpose of depletion width increasing is to prevent the flow of diffusion current. 🤔
Or maybe it does because the majority charge carriers move away from the depletion region on both sides?
In avalanche breakdown, if the depletion width increases under reverse bias condition, how can the electric field increase in depletion field ?
Because E=V/d. If d increases because of reverse bias, then E should decrease right ? But you said E would increase in depletion region during avalanche breakdown . Can you clarify this ?
The depletion region width d ∝ √ V. That means as the V increases, electric filed increases. I hope it will clear your doubt.
If you still have any doubt, let me know here.
@@ALLABOUTELECTRONICS But sir there in zener breakdown you told that since E=V/d, as d decreases due to heavy doping , E increases. So there was an inverse relationship between d and E in zener breakdown. But here you are saying that they are directly proportional . Is there one generic relationship between E and d that can be followed throughout ?
the concentration of majority charge carriers in the depletion region is also increasing here, in reverse bias condition.
So, hopefully may be the dependency is more predominant on the conc. of charges rather than width of space charge region.
I get my answer 😌
👏
Why does the zener diode does not break above zener voltage?
thank you :)
For pn noise generators: how do I know that I picked the right diode or transistor? Is any diode or transistor okay which has a breakdown voltage about 6V or are there models out there which are super low-noise? Thx!
I have a doubt??
Depletion region lo atoms yela untayi
No
Immobile ions are present not the atoms!
Good explanation but kindly repeat some sentences to take make understanding better
What is the forward bias voltage drop of an avalanche diode?
Forward voltage drop is similar to the normal diode.( 0.6V to 0.7V)
@@ALLABOUTELECTRONICSshould it not be larger due to the doping
At reverse bias condition there is a strong current
But Zener diode also has large sudden current why it not damage the diode