Thank you so much for making these videos. I would not have done nearly as well in my semiconductor courses without this. Please make more videos! These are the best videos I've seen. You cover essentially an entire semiconductors course with such detail and clarity and I have never seen that before.
at some point in history there must have been a person who had the idea that these diagrams were a good way to visualize semiconductor physics. If only that person had been murdered as a child, people in 2016 might have a different visualization to draw from and I might actually understand this.
Sir, in the reverse bias , the saturation current is the drift component of the depletion region. Will there be a component because of the movement of the electrons and holes away from the junction region (the region which was neutral in equilibrium condition) which was responsible for the increase in the depletion width.
You mentioned that the distance of uncovered ions on P side is less than on the N side (Width of P is less than Width of N) because the P side is highly doped. Can you elaborate? Or maybe can you give more explanations with this relationship?
Hi Darwin, Nd Xn = Na Xp (charge neutrality). To keep this true and satisfy Na > Nd, Xp should be less than Xn. Dopant density is high on p-side w.r.t n-side. n-side should be depleted for more distance to expose same amount of charge equal to p-side. As charges exposed on p-side should be equal to charges exposed on n-side (charge neutrality). For same amount of charge to be exposed on both sides, more dense doping side needs to deplete less distance w.r.t low doping density side. Hope this helps! Thanks, Techgurukula.
respected sir holes do not diffuse from p to n type under unbiased condition of pn junction,, but only electrons holes recombine at junction so please correct it sir
I would still support the diffusion - because of the following reason: diffusion current will be in one direction and drift will be in the opposite - if we don't consider diffusion, it cannot be explained. Both currents will be present in equilibrium - only that they are equal and opposite. Hence, net current is zero. Under forward bias, Electric field decreases. Hence drift decreases relative to diffusion, and resultant current is due to Diffusion. check the other videos. Thanks, Techgurukula.
That equation is discussed in this video - ua-cam.com/video/1g02VFOPq9I/v-deo.html Please watch the playlist in sequence, they are put in logical sequence - shouldn't get questions. Playlist: ua-cam.com/play/PL311CA3047EB6EC58.html
Sir according to theory discussed....the drift current flows after the diffusion stops ...so there will be some time difference between drift and diffusion current so why we are taking them equal??
E max pata hai aur xp ya xn pata hai to right angled triangle bana ke xp se xn tak kahi bhi electric field nikal sakte hai. Proportionality rule lagakar
We can use (Nd*Xn= Na*Xp) this equation to find the width of n and p side directly right? Why do we have to Xn=WNp/Na+Nd for n side and the same for p side??? Pls help me...
if Total depletion width is given (W) and doping concentrations are given - finding Xn and Xp directly: Xn = W * (Na/(Na+Nd)), Xp = W * (Nd/(Na+Nd)). if 3 out of 4 are given in the equation (Nd * Xn = Na * Xp), we can use this.
If fermi level is not same through out the device, there would be net current flow - which any way would drive the fermi level to be same through out the device. This is when we say the diode is in equilibrium. I believe it's better explained in the video - hope that helps.
+arkabsu The bent in the Band, represents Electric field present, which is directly proportional to the slope of the Ec (or) EV (or) Ei => Electric field is related to the variation of potential. Hence the slope should be equal to the q times the potential as the Ec/Ev/Ei are in electron volts. Hope this helps! Suggestion: If you see the videos in sequence as put up in the playlist, you wouldn't get these doubts. If you still get doubt's, feel free to ask. Thanks, Techgurukula.
+David Ei is Intrinsic Fermi Energy level (Fermi energy level for Intrinsic semiconductor). For silicon, it is approximately in the mid of the Energy band gap ((Ec+Ev)/2). Refer to this Video: ua-cam.com/video/lc8R-wqEnk0/v-deo.html for more detailed explanation and derivation. Hope this helps. Thanks, Techgurukula.
Thank you so much for making these videos. I would not have done nearly as well in my semiconductor courses without this. Please make more videos! These are the best videos I've seen. You cover essentially an entire semiconductors course with such detail and clarity and I have never seen that before.
Before this video I didn't understand anything about electric field and electric voltage relationship in pn junction. Thanks for this video...
You are much much better than ace and made easy prof specially Ramesh sir
Yes ramesh sir only dances in class
I owe my success in my semiconductors module to you, can't thank you enough!...Please keep going and make videos for MOSFET
holy.......you just super awesome! so clear! thank you very much!
at some point in history there must have been a person who had the idea that these diagrams were a good way to visualize semiconductor physics. If only that person had been murdered as a child, people in 2016 might have a different visualization to draw from and I might actually understand this.
William shockely, and these diagram are pretty intuitive man
i can feel u bro
to be fair, this was first taught in the 1950s and you couldn't do much better then
lol
Excellent video, cleared so many doubts so easily, thank you very much
It's very clear and helpful! Such a nice and short video. Thank you so much for making it!
I usually have difficulty understanding indian accents, but this has been really helpful.
thx for the video sir, gods bless you
Sir, in the reverse bias , the saturation current is the drift component of the depletion region. Will there be a component because of the movement of the electrons and holes away from the junction region (the region which was neutral in equilibrium condition) which was responsible for the increase in the depletion width.
thanx a lot sir , God bless u.
Dear Kula, at the time 13:30. You didn't explain why Xn is more intense.
10:40 Why dont we take into the account not just negative acceptor ions, but also diffused electrons that came from n side?
have you got it bro?
@@lies4212 I think not :D
You mentioned that the distance of uncovered ions on P side is less than on the N side (Width of P is less than Width of N) because the P side is highly doped. Can you elaborate? Or maybe can you give more explanations with this relationship?
Hi Darwin,
Nd Xn = Na Xp (charge neutrality). To keep this true and satisfy Na > Nd, Xp should be less than Xn.
Dopant density is high on p-side w.r.t n-side. n-side should be depleted for more distance to expose same amount of charge equal to p-side. As charges exposed on p-side should be equal to charges exposed on n-side (charge neutrality).
For same amount of charge to be exposed on both sides, more dense doping side needs to deplete less distance w.r.t low doping density side.
Hope this helps!
Thanks,
Techgurukula.
Why does the electric field point from the +ve side of the depletion region to the -ve side ?
property of field lines.it moves away from positive charge.
goodness how do you do to explain so well !!
Wonderful explanation sir ,thanks a lot
I could not understand how u drawn the electric field-vs-x curve from density-vs-x curve,
Pls explain.
Why charge is neutral outside the junction
It is all great! BUT, I would like to have notes, of your blackboard :) thanks a lot !
respected sir holes do not diffuse from p to n type under unbiased condition of pn junction,, but only electrons holes recombine at junction so please correct it sir
I would still support the diffusion - because of the following reason:
diffusion current will be in one direction and drift will be in the opposite - if we don't consider diffusion, it cannot be explained. Both currents will be present in equilibrium - only that they are equal and opposite. Hence, net current is zero.
Under forward bias, Electric field decreases. Hence drift decreases relative to diffusion, and resultant current is due to Diffusion. check the other videos.
Thanks,
Techgurukula.
Is "diffusion potential" and "built-in potential" same thing?
Sir how u written equation for concentration... because in some privious lectures we have derived some other formula
That equation is discussed in this video - ua-cam.com/video/1g02VFOPq9I/v-deo.html
Please watch the playlist in sequence, they are put in logical sequence - shouldn't get questions. Playlist: ua-cam.com/play/PL311CA3047EB6EC58.html
Sir according to theory discussed....the drift current flows after the diffusion stops ...so there will be some time difference between drift and diffusion current so why we are taking them equal??
Plzz reply
Which sources did you use?
Why conduction band and valance band of p-type and n-type are not at same level.
Great lectures
Why should Ef to be constant throughout the length.....
Hello sir,
How to find electric field at particular point in depletion region.
E max pata hai aur xp ya xn pata hai to right angled triangle bana ke xp se xn tak kahi bhi electric field nikal sakte hai. Proportionality rule lagakar
@@prashantpandey2299 thik h pandu
is the type of equilibrium here static or dynamic ? thanks
We can use (Nd*Xn= Na*Xp) this equation to find the width of n and p side directly right? Why do we have to Xn=WNp/Na+Nd for n side and the same for p side??? Pls help me...
if Total depletion width is given (W) and doping concentrations are given - finding Xn and Xp directly: Xn = W * (Na/(Na+Nd)), Xp = W * (Nd/(Na+Nd)).
if 3 out of 4 are given in the equation (Nd * Xn = Na * Xp), we can use this.
what about energy band diagram of biased p type semi conductor?
Sir all is fine except you have not use volume in the formula
You are good
For separate p and ntype Efn and Efp have different energy,but in pn junction diode how they have same energy level.please tell me
If fermi level is not same through out the device, there would be net current flow - which any way would drive the fermi level to be same through out the device. This is when we say the diode is in equilibrium. I believe it's better explained in the video - hope that helps.
@@techgurukula thank u sir for it reply .i will see the video now.
Sir formula of Vbi has a mistake
please share the name of software usedd?
can u explain Poisson's E and charge density relation in simple ..
Sir please tell me book name?
how the difference in Ec calculated as qVbi? at 21:10
+arkabsu
The bent in the Band, represents Electric field present, which is directly proportional to the slope of the Ec (or) EV (or) Ei => Electric field is related to the variation of potential. Hence the slope should be equal to the q times the potential as the Ec/Ev/Ei are in electron volts.
Hope this helps!
Suggestion: If you see the videos in sequence as put up in the playlist, you wouldn't get these doubts.
If you still get doubt's, feel free to ask.
Thanks,
Techgurukula.
Sir, please upload bjt,mosfet,jfet also
what is ni ?
10:45 What is Ei exactly?
+David My understanding is it is the mid point between energy needed for thermal generation and energy at 0 Kelvin.
+David
Ei is Intrinsic Fermi Energy level (Fermi energy level for Intrinsic semiconductor). For silicon, it is approximately in the mid of the Energy band gap ((Ec+Ev)/2).
Refer to this Video: ua-cam.com/video/lc8R-wqEnk0/v-deo.html for more detailed explanation and derivation.
Hope this helps.
Thanks,
Techgurukula.
techgurukula Thanks
I quit, it's too hard
thanku sir
Did not get it...not even a single point
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