nice lecture explanation is very good, thanks sir , kal mera exam hai aur aaj mai aapka semiconductor ka pura course dekha liya , bahut hi accha laga sab samajh aa gya
Amazing explanation and fantastic series! Can you though explain how to calculate the length of the space charge/depletion region? I found a question on that and you didn't mention it
A difference of Fermi level means a difference in overall potential energy for the electrons. If one side of the junction had a higher Fermi level than the other side, the electrons would move from the higher level to the lower level. As electrons leave the region of higher Fermi level, the Fermi level there decreases as states become available. Eventually the Fermi levels in both regions equalize, and the overall flow of electrons stops. This is the equilibrium state of the p-n junction, where the Fermi level everywhere across the junction is the same, i.e. a constant.
Due to diffusion of carries, charges are developed near the junction which result in electric field. This electric field gives rise to the drift current.
When p-n junction is reversed biased, the positive terminal of the external battery is connected to n-side of p-n junction and its negative terminal to p-side of p-n junction. The positive terminal of the external battery attracts the majority carrier electrons from the n-region and its negative terminal attract the majority carrier holes from p-region. Due to it, the majority charge carriers move away from the junction. This increases the width of the depletion layer.
Yes. but if you understand this energy band diagram concept, then it will be very easy for you to understand the properties of any semiconductor device. Just understand that It is difficult for electrons to climb up the energy hill. And the energy bands on the side connected to negative terminal of battery will rise up.
Thanks for saving my semester 🙏
Most welcome and best luck..
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Never once had to by heart derivations and formulas ..after your lects.. ty sir 💯
Glad to know it helped you.. all the best👍
nice lecture explanation is very good, thanks sir , kal mera exam hai aur aaj mai aapka semiconductor ka pura course dekha liya , bahut hi accha laga sab samajh aa gya
All the best for your exam..👍
Crystal clear lecture...much needed such lectures ..thank you sir
So nice of you
thank you all this lectures are very helpful you're a saviour
Amazing explanation and fantastic series! Can you though explain how to calculate the length of the space charge/depletion region? I found a question on that and you didn't mention it
keep going on sir, you're doing a great job :))
Thank you.. keep watching and share with your friends..
You might save me in this course. Thank you
Happy to help!
5:45 Why Fermi level have to be of same level on. Both side
If possible can you please reply with centre of gravity concept of Fermi level
A difference of Fermi level means a difference in overall potential energy for the electrons. If one side of the junction had a higher Fermi level than the other side, the electrons would move from the higher level to the lower level. As electrons leave the region of higher Fermi level, the Fermi level there decreases as states become available. Eventually the Fermi levels in both regions equalize, and the overall flow of electrons stops. This is the equilibrium state of the p-n junction, where the Fermi level everywhere across the junction is the same, i.e. a constant.
Great video sir it helped a lot!
Glad it helped.. best luck..👍
sir what will happen to the bend of the energy band if the concentration of doping is increased
Sir @ 16:20
Q1.where is the Fermi level of each n and p type semiconductor?
Q2. If they are not aligned what does it represents?
When we forward bias, Fermi levels split again and transfer occurs when Vb is such that Ef(n)-Ef(p)=qVb
why is the energy in n side lower than in p side?
Thank you Sir?
Thank you sir 🙏
Most welcome and best luck..
8:27 how drift current flows if pn junction is unbiased(not connected to in a circuit)
Due to diffusion of carries, charges are developed near the junction which result in electric field. This electric field gives rise to the drift current.
Nice explanation sir...thank you
You're most welcome
Thank you Sir!!!
Most welcome and best luck 👍
Nice❤❤
Amazing sir
Thank you.. keep watching
Thank you very much sir
Welcome
thanks a lot, sir
Most welcome...
Sir can you please tell why width of depletion layer increase when we increase reverse bias voltage
When p-n junction is reversed biased, the positive terminal of the external battery is connected to n-side of p-n junction and its negative terminal to p-side of p-n junction. The positive terminal of the external battery attracts the majority carrier electrons from the n-region and its negative terminal attract the majority carrier holes from p-region. Due to it, the majority charge carriers move away from the junction. This increases the width of the depletion layer.
@@EngineeringPhysicsbySanjiv OK thank you very much sir
Welcome..
thank you sir
Welcome
Sir, can you share this ppt?
17:29
Diagram is so difficult to understand
Yes. but if you understand this energy band diagram concept, then it will be very easy for you to understand the properties of any semiconductor device. Just understand that It is difficult for electrons to climb up the energy hill. And the energy bands on the side connected to negative terminal of battery will rise up.