Thank you for this very logical explanation and color coding! I really appreciate that you draw the graphs directly below to show how you go from one representation to the next.
Thank you so much for these videos about pn junction. When my professor talked about this I couldn't understand one bit of what that energy diagram had of information. Now I completely get it!!!!
the continuity of E(x) and V(x) could be explained using the integration constants that we get when integrating. If we define the Q(x) a piecewise function then E(x) and V(x) also could be expressed as piecewise functions. The integration constants could be set such that there is continuity at the boundary (which becomes the inflection point in V(x).
8:26 Is it correct that the energy is equal to -qV? Because now I am confused as according to electrostatics the energy is equal to +qV with q=+/-|e| depending on the carrier type.
Good lecture! Have you tried to calculate the fields and potential without doping constant assumption, taking the concentration of p as a function of x? I solved it (probably wrong) by assuming that potential of the base counteracts the greatest chemical potential of an emitter, by solving that equation I get xp, and then I solve an equation of electrical fields, finding xn.
Thank you so much for this explanation, but I've got a question: Shouldn't the electric field's strength increase as we approach the positive charges? Then why exactly does it start decreasing as it moves from the negative charges?
The electric field is proportional to the integral of the charge distribution, in other words the sum of all of the little sections of charge. So imagine you are walking along the x-axis in the +x direction and collecting the charges in a basket. Your basket starts electrically neutral, and as you walk along the region from x = -xp to x = 0 you notice your basket only contains negative charges. Once you reach the metallurgical junction (x=0) your basket contains the maximum number of negative charges - this corresponds to the strongest negative electric field. As you cross over the junction into the region of positive charges, when you place each successive positive charge into your basket, it offsets a negative charge, reducing the negative electric field strength. This continues until you reach x = xn where you have collected exactly enough positive charges to offset the negative charges you picked up on the p-side of the metallurgical junction and the electric field strength drops to zero.
charge neutrality, say in the n-region any positive ions has its charge balanced out by negative electrons (thats why the assumption of abrupt junction is made so that p region is simply a p-Si) you can also think of the fact that when doping we introduce atoms not ions so the overall charge must be 0.
Thanks for the fantastic videos. As a chemist, I'm confused about one thing; how is electric field decreasing? I understand the electric field is at a maximum at the pn junction interface, because that's where you 'feel' the full positive and negative charge on either side of that interface and within the depletion region. So why isn't electric field increasing as you approach the interface from either side? Another thing about using the word 'decreasing'; an electric field can't be negative, right?
Electric field is a vector, which means it has both magnitude and direction. Magnitude is always non-negative (this must be what you had in mind). Direction in this one-dimentional case can be (+) or (-). Here it is negative, because it's opposite of the direction of x axis (E is towards left from positive to negative ions, while x is towards right).
Could you tell me why is the graph of the potential flipped over even though in both the cases the p-side and the n-side were same relative to each other ?
Because you are integrating the electric field, which is always pointing in the same direction. It’s the same exact reason why when you have charge on two capacitor plates the voltage across them is nonzero.
Because if there was, the system would not be in equilibrium, and there would be a restoring force (an electric field) which drives the system to equilibrium (same charge on either side).
Nope, the charge density of any given slice of the depletion region is not zero, but if you add up all the charge in all the slices it will sum to zero.
Maria Thereza Great question! I believe I cover that in the prior video, you get that from integrating Gauss’ Law in 1D (which relates the derivative of the E-field to charge density).
Thank you for this very logical explanation and color coding! I really appreciate that you draw the graphs directly below to show how you go from one representation to the next.
Thanks :D
I want to thank you from the heart ! I was trying to understand this for a such a long time . Thank you so so much.
Thank you so much for these videos about pn junction. When my professor talked about this I couldn't understand one bit of what that energy diagram had of information. Now I completely get it!!!!
I just saw someone in the comments say you're a lifesaver, and you know what? he's absolutely right
🥰
facts!
This guy is crazy with his videos. THank you so much
I follow your videos from Italy! Thank you for very good contents and explanations!
You’re a lifesaver .. thank you 💙💙💙💙💙💙💙
I was having an existential crisis before I came across your channel:'))
Man I remember having an existential crisis in my device physics class too, glad you've been pulled back to the world of normal crises :)
Extraordinary explanation, thanks a lot
thanks for the video ,it helps a lot ! good good study ,day day up!
the continuity of E(x) and V(x) could be explained using the integration constants that we get when integrating. If we define the Q(x) a piecewise function then E(x) and V(x) also could be expressed as piecewise functions. The integration constants could be set such that there is continuity at the boundary (which becomes the inflection point in V(x).
8:26 Is it correct that the energy is equal to -qV? Because now I am confused as according to electrostatics the energy is equal to +qV with q=+/-|e| depending on the carrier type.
3:47 the subtle humor 🤣😂
adding to bucketlist - get gauss laws tattoed on my arm xD
get all of Maxwell's equations tattooed on you
@Shane Jericho Fucking scammers leave this holy channel!
Good lecture! Have you tried to calculate the fields and potential without doping constant assumption, taking the concentration of p as a function of x? I solved it (probably wrong) by assuming that potential of the base counteracts the greatest chemical potential of an emitter, by solving that equation I get xp, and then I solve an equation of electrical fields, finding xn.
Thank you so much for this explanation, but I've got a question:
Shouldn't the electric field's strength increase as we approach the positive charges? Then why exactly does it start decreasing as it moves from the negative charges?
The electric field is proportional to the integral of the charge distribution, in other words the sum of all of the little sections of charge. So imagine you are walking along the x-axis in the +x direction and collecting the charges in a basket. Your basket starts electrically neutral, and as you walk along the region from x = -xp to x = 0 you notice your basket only contains negative charges. Once you reach the metallurgical junction (x=0) your basket contains the maximum number of negative charges - this corresponds to the strongest negative electric field. As you cross over the junction into the region of positive charges, when you place each successive positive charge into your basket, it offsets a negative charge, reducing the negative electric field strength. This continues until you reach x = xn where you have collected exactly enough positive charges to offset the negative charges you picked up on the p-side of the metallurgical junction and the electric field strength drops to zero.
@erikerlandson3154 it makes sense now, thanks!
Why there is no net charge outside the junction
charge neutrality, say in the n-region any positive ions has its charge balanced out by negative electrons (thats why the assumption of abrupt junction is made so that p region is simply a p-Si) you can also think of the fact that when doping we introduce atoms not ions so the overall charge must be 0.
Thanks for the fantastic videos. As a chemist, I'm confused about one thing; how is electric field decreasing? I understand the electric field is at a maximum at the pn junction interface, because that's where you 'feel' the full positive and negative charge on either side of that interface and within the depletion region. So why isn't electric field increasing as you approach the interface from either side?
Another thing about using the word 'decreasing'; an electric field can't be negative, right?
Electric field is a vector, which means it has both magnitude and direction. Magnitude is always non-negative (this must be what you had in mind). Direction in this one-dimentional case can be (+) or (-). Here it is negative, because it's opposite of the direction of x axis (E is towards left from positive to negative ions, while x is towards right).
Could you tell me why is the graph of the potential flipped over even though in both the cases the p-side and the n-side were same relative to each other ?
at 8:00 how come the voltage doesn't go back down to 0? You said it stays high, but I'm not sure why
Because you are integrating the electric field, which is always pointing in the same direction. It’s the same exact reason why when you have charge on two capacitor plates the voltage across them is nonzero.
Thank you very much.
Thank you for this very logical
At 3:38 why there’s no net charge outside the 2 regions you drew?
Because if there was, the system would not be in equilibrium, and there would be a restoring force (an electric field) which drives the system to equilibrium (same charge on either side).
thank you!!!
One doubt, is the net (effective) charge density at each point of the depletion layer zero??
Nope, the charge density of any given slice of the depletion region is not zero, but if you add up all the charge in all the slices it will sum to zero.
Just amazing
3:18 I think you meant "the charge desnity is p as a function of x." You mixed up Q and p.
thanks doctor
You computed the magnitude of the E field for the opposite direction wrt to the E arrow you draw
why is E(x) the integral of the charge?
Maria Thereza Great question! I believe I cover that in the prior video, you get that from integrating Gauss’ Law in 1D (which relates the derivative of the E-field to charge density).
isn't energy W for work?
Yes that’s an unfortunate conflict of notation >
Ill get it tattooed, only if @Jordan Edmunds says so.
"I apologize to mathematicians watching this video" , Hahahahaha 4:20
I Tattooed it on my arm....😂😂😂
I was looking for this comment XD
2 mathematicians unliked the video
"if you haven't memorised this by now you should get it tattoed on your arm" 😂
I personally would suggest the memorization route... xD
should have a new tattoo on my arm haha
4:10 Why apologize! :) :) :)
if you dont memorise that equation just have a tottoo done it on your arm! ahahaha
XD
why did you apologized from mathematicians?😀
Noice
memegenerator.net/img/instances/68375583/noice.jpg
Rma😂