I'm hooked on these lectures. When I first started my studies for MSc I hit a brick wall with the level of physics analog circuit design requires (simply because I didn't know how to learn something that isn't immediately obvious - up until that point everything had been obvious). I decided I'll just stick to digital instead (and was quite a successful designer/architect with a multinational tech giant that was at one point one of the most valuable companies and brands in the world, partly due to the ICs we designed). Then I got bored with working in the office and went off to study something completely different. Now I'm an emergency physician who's bored to death by the everyday bulk of patients who have absolutely no emergency and I dream of the times I was working with something really challenging - in tech, in an office or a lab. And when I now watch these lectures, I realize that I would no longer need to stick to digital but can actually understand the fundamental physics. How I wish I was a young man again...
I've been working in the eletronics sector for years and currently I'm watching your course (and also that of Prof.Razavi) as refresh of fundamentals of circuit design. I have to say that your teaching style is absolutely brilliant. I like the way in which you make students reason on how and why things work, using at the same time logical thinking and all the maths embedded in electronics and physics.
@@tharun541 Hi, I don't know exactly the book he followed. To be honest, I blindly followed this lecture before giving my lecture in class. You can follow Neamen or Umesh Mishra's book.
I have a question. While discussing the forward biased diode we arrived at the conclusion that the current grows exponentially with lowering the potential barrier ψ0 because of the probability of the carriers having large enough energy (according to Boltzmann’s distribution) facilitates that relation. However, if we increase bias voltage to VD = ψ0 the barrier is reduced to zero when *every* electron in the conduction band has enough energy to leap over the “barrier” (which is then basically nonexistent). If that is indeed the case, how may further increase of VD lead to increase of diode current (according to the diode equation) ? Thank you in advance!
This is a good question. What often in reality is that you practically do not get there, since as you lower the barrier the current grows exponentially and there is larger and larger I*R drop within the bulk of the diode, so the junction with experience a smaller fraction of the excess external voltage. The current associated with the potential barrier fully lowered often cannot be supported by the series ohmic resistance of the diode without a significant external voltage.
Sir, the fermi energy level of n-type and p-type semi-conductor at 0 Kelvin is at Ec and Ev respectively. Does this fact change for p-n junction at 0 K as you have shown at 7:45 in the video?
"electrons are pretty dumb".... i dont know why but that triggered me haha. Many thanks for this lecture series (from a physics major at Penn in Philly)!
Why do the Conduction and Valence energies get bent in the PN Junction? I thought they were fundamental properties of the silicon atom's shells. Is it because Ec and Ev actually mean "the electrostatic potential energy of an electron in the conduction and valence bands at any given position while considering the contribution of electrostatic potential due to the depletion layer's E-field? The difference between a PN junction and the previous examples of semiconductors in these videos (which all had the same Ec and Ev for the same element) is that there was no net E-field we had to account for that would otherwise influence the electrostatic potential energy of electrons in the conduction or valence bands. Is this non-0 E-field the cause of the bending in the Ec and Ev? If so, then the idea that the Conduction and Valence band edge energies are fixed values, as initially introduced in most videos & books (ex: "The minimum conduction band energy of element x is y Joules"), is not general because these electrostatic potential energies can change depending on if the atom is isolated or if it is in the presence of an E-field, right? Thanks.
The conduction and valance bands are energy levels for the local electronics. If there is an electric field between region 1 and region 2, there is an electric potential difference there and therefore and energy difference for the electrons in the conduction (or valance) bands in those two regions. This exhibits itself as the bending of the bands. The intrinsic Fermi level is essentially half-way point between the valence and conduction bands, which will bend with them (but not theorem energy level, which is parameter of the distribution) Hope this helps.
@@AliHajimiriChannel thank you very much. Im watching the BJT videos in this playlist right now. I cannot thank you enough for your practical and theoretical insights which offer a comprehensive understanding of how first principles can be leveraged to design practical semiconductor devices.
sir at 11.38, you told this is called diffusion.but before that u told temperature is increased then u explained the whole process.but in diffusion ,no external energy is applied,its only due to difference in concentrations,but as per ur explaination when the temperature is increased then diffusion happens ,i did not get it sir ,can u tell how please.
Why the Vd is smaller than Si0 after applying that external electric feild the internal potential was reduced by Vd then it must be negative because if i want to apply an external voltage to the diode it must be larger than the internal potential in the diode in order to have current but according to what is written the Vd is less than the internal potential and it has reduced a little without making it negative I wish you understood my question sir
At 34.26, you mentioned "net increase" in electrons and to calculate that you subtract electrons ""before" forward bias from electrons "after" forward bias. But then while calculating the "net flow", you say the "net current" (total diode current ) is only proportional to net increase. Shouldn't the net flow be only proportional to electrons after forward bias?
Yes, the diode current equation is Is*(e^(VD/Vt)-1). The '-1' is to count for that difference. However, in most practical case of using the forward bias, the exponential is much larger than the '1', so it is often ignored to get a simpler more tractable equation.
No. The potential drop is due to work function difference. When you make two contacts to external lead metals of the diode, you form two more juctions with their own work function differences, so if you measure the potential drop between the two metal terminals of the diode, you will see the algebraic sum of these three work function differences, which is zero (unless you use two different metals for less(wires) of the two terminals.
Is it okay to see this as, pn built-in potential is created by losing energy (electrons holes combined to get to lower energy state, near the junction?), so just connecting a wire across it can't give current. Because flow of current will imply energy transfer from the diode to the wire?
(Referring to the instant 9:18 in the video), one electron per donor atom is loosely in n-type material, but not completely free from its attraction. How come it is so likely that these loosely bound electrons form bonds with acceptor atoms in p-type region, leaving their bonds with donor atoms?
The required energy for breaking a bond and forming a new one with the acceptor is much less than the one associated with breaking a bond and not forming a new one. So the thermal energy will allow for this to happen frequently at room temperature. There are other ways think about it. The process is really a quantum mechanical one similar to tunneling through a large potential barrier.
Based on the energy band diagram, the energy level EA is much lower than the energy level ED, EA is right above the valence band while ED is right below the conduction band; electrons as the temperature is decreasing would like to be in the lowest possible energy state, and that is why they would leave the phosphorus atoms and go into the vacancies available in the covalent bonds of Boron atoms.
I'm hooked on these lectures. When I first started my studies for MSc I hit a brick wall with the level of physics analog circuit design requires (simply because I didn't know how to learn something that isn't immediately obvious - up until that point everything had been obvious). I decided I'll just stick to digital instead (and was quite a successful designer/architect with a multinational tech giant that was at one point one of the most valuable companies and brands in the world, partly due to the ICs we designed). Then I got bored with working in the office and went off to study something completely different. Now I'm an emergency physician who's bored to death by the everyday bulk of patients who have absolutely no emergency and I dream of the times I was working with something really challenging - in tech, in an office or a lab. And when I now watch these lectures, I realize that I would no longer need to stick to digital but can actually understand the fundamental physics. How I wish I was a young man again...
I've been working in the eletronics sector for years and currently I'm watching your course (and also that of Prof.Razavi) as refresh of fundamentals of circuit design. I have to say that your teaching style is absolutely brilliant. I like the way in which you make students reason on how and why things work, using at the same time logical thinking and all the maths embedded in electronics and physics.
you are making rules like free education and education to all;became a huge fan for ur teaching style and for ur kind nature.
Impeccable and beautiful teaching. Thanks Professor
Hi, can you please tell me the book to be followed for this lecture series?
@@tharun541 Hi, I don't know exactly the book he followed. To be honest, I blindly followed this lecture before giving my lecture in class. You can follow Neamen or Umesh Mishra's book.
Thanks a Lot professor Lots of Love and Respect from India!!
Appreciate it.
Thank you. This is an incredible series.
@11:30 diffusion current is due to concertation gradient.
I have a question.
While discussing the forward biased diode we arrived at the conclusion that the current grows exponentially with lowering the potential barrier ψ0 because of the probability of the carriers having large enough energy (according to Boltzmann’s distribution) facilitates that relation. However, if we increase bias voltage to VD = ψ0 the barrier is reduced to zero when *every* electron in the conduction band has enough energy to leap over the “barrier” (which is then basically nonexistent).
If that is indeed the case, how may further increase of VD lead to increase of diode current (according to the diode equation) ?
Thank you in advance!
This is a good question. What often in reality is that you practically do not get there, since as you lower the barrier the current grows exponentially and there is larger and larger I*R drop within the bulk of the diode, so the junction with experience a smaller fraction of the excess external voltage. The current associated with the potential barrier fully lowered often cannot be supported by the series ohmic resistance of the diode without a significant external voltage.
I cant understand why the fermi energy has to be constant along the pn junction?
Sir, the fermi energy level of n-type and p-type semi-conductor at 0 Kelvin is at Ec and Ev respectively. Does this fact change for p-n junction at 0 K as you have shown at 7:45 in the video?
"electrons are pretty dumb".... i dont know why but that triggered me haha. Many thanks for this lecture series (from a physics major at Penn in Philly)!
Why do the Conduction and Valence energies get bent in the PN Junction? I thought they were fundamental properties of the silicon atom's shells. Is it because Ec and Ev actually mean "the electrostatic potential energy of an electron in the conduction and valence bands at any given position while considering the contribution of electrostatic potential due to the depletion layer's E-field? The difference between a PN junction and the previous examples of semiconductors in these videos (which all had the same Ec and Ev for the same element) is that there was no net E-field we had to account for that would otherwise influence the electrostatic potential energy of electrons in the conduction or valence bands. Is this non-0 E-field the cause of the bending in the Ec and Ev? If so, then the idea that the Conduction and Valence band edge energies are fixed values, as initially introduced in most videos & books (ex: "The minimum conduction band energy of element x is y Joules"), is not general because these electrostatic potential energies can change depending on if the atom is isolated or if it is in the presence of an E-field, right? Thanks.
The conduction and valance bands are energy levels for the local electronics. If there is an electric field between region 1 and region 2, there is an electric potential difference there and therefore and energy difference for the electrons in the conduction (or valance) bands in those two regions. This exhibits itself as the bending of the bands. The intrinsic Fermi level is essentially half-way point between the valence and conduction bands, which will bend with them (but not theorem energy level, which is parameter of the distribution)
Hope this helps.
@@AliHajimiriChannel thank you very much. Im watching the BJT videos in this playlist right now. I cannot thank you enough for your practical and theoretical insights which offer a comprehensive understanding of how first principles can be leveraged to design practical semiconductor devices.
the depletion region has E_f > E_i ?
Why is it so obvious that the Fermi level is constant in thermal equilibrium in a PN junction? I feel like I missed something.
Why cannot electrons flow in EA band in n type material? Why should we excite them to flow into conduction band ?
sir at 11.38, you told this is called diffusion.but before that u told temperature is increased then u explained the whole process.but in diffusion ,no external energy is applied,its only due to difference in concentrations,but as per ur explaination when the temperature is increased then diffusion happens ,i did not get it sir ,can u tell how please.
Why the Vd is smaller than Si0 after applying that external electric feild the internal potential was reduced by Vd then it must be negative because if i want to apply an external voltage to the diode it must be larger than the internal potential in the diode in order to have current but according to what is written the Vd is less than the internal potential and it has reduced a little without making it negative
I wish you understood my question sir
Do you mean that the diode will take a small amount of the external voltage of the battery and leaves the rest to cross the diode ??
Where is he delivering these lectures? Anyone ?
At 34.26, you mentioned "net increase" in electrons and to calculate that you subtract electrons ""before" forward bias from electrons "after" forward bias. But then while calculating the "net flow", you say the "net current" (total diode current ) is only proportional to net increase. Shouldn't the net flow be only proportional to electrons after forward bias?
Yes, the diode current equation is Is*(e^(VD/Vt)-1). The '-1' is to count for that difference. However, in most practical case of using the forward bias, the exponential is much larger than the '1', so it is often ignored to get a simpler more tractable equation.
Diode has built-in potential, so if we connect terminals of a diode with a metal wire, can we expect some current to flow?
No. The potential drop is due to work function difference. When you make two contacts to external lead metals of the diode, you form two more juctions with their own work function differences, so if you measure the potential drop between the two metal terminals of the diode, you will see the algebraic sum of these three work function differences, which is zero (unless you use two different metals for less(wires) of the two terminals.
Is it okay to see this as, pn built-in potential is created by losing energy (electrons holes combined to get to lower energy state, near the junction?), so just connecting a wire across it can't give current. Because flow of current will imply energy transfer from the diode to the wire?
Dr Ali Hajimiri, May I ask why fermi level wont change with voltage applied?
(Referring to the instant 9:18 in the video), one electron per donor atom is loosely in n-type material, but not completely free from its attraction. How come it is so likely that these loosely bound electrons form bonds with acceptor atoms in p-type region, leaving their bonds with donor atoms?
The required energy for breaking a bond and forming a new one with the acceptor is much less than the one associated with breaking a bond and not forming a new one. So the thermal energy will allow for this to happen frequently at room temperature. There are other ways think about it. The process is really a quantum mechanical one similar to tunneling through a large potential barrier.
Thank you Prof.! Got it.
Based on the energy band diagram, the energy level EA is much lower than the energy level ED, EA is right above the valence band while ED is right below the conduction band; electrons as the temperature is decreasing would like to be in the lowest possible energy state, and that is why they would leave the phosphorus atoms and go into the vacancies available in the covalent bonds of Boron atoms.
sir while hearing with earphones only left side working right not working
Yes. we had a problem in recording of a couple of the lectures. It was correct in the later lectures. I am sorry about that.
No Audio ??
Please double check. There is audio.
sir why Fermi-level is constant through out the junction ?
Because any Slope in the Fermi level
implies there is a current flow
I am not able to understand EA and ED energy levels. Can some one please explain what they are?
EA= acceptor energy level, ED= Donor energy level
awsome! thnx professor!