Thank You, Sir, I never understood it so clearly as I did today. You started from fundamentals and explained E-beam lithography in the best possible way. I am a big fan of yours from now onwards...Thanks again!!
Thank you very much for a great video, introducing the theoretic concepts before explaining the actual lithography process is essential for the full process understanding, and most explanatory articles or videos do not provide this vital information. Great job!
Dear Chris, thank you for your contribution. Would love to know/hear more from you regarding EBL and its architecture/physics in future. It would be nice to discuss system specific discussions like Raith, JEOL etc. systems.
I'm not sure I understand your question. Are you asking about the wavelength of an electron (which depends on energy) compared to the dimensions currently being fabricated using EUV lithography? EUV lithography is capable of printing about 16 nm lines and spaces today. At 100keV, the wavelength of the electron is 0.0037 nm, which is much, much smaller than any feature being fabricated.
Hi Chris, thanks for the nice presentation. What does "Throughput" mean in E-Beam lithography? Does it mean the speed of the system in completing the pattern?
@@ChrisMack Thanks! Are you sure the wavelengths are correct at 4:05 ? I tried to calc. it myself for 100eV using the formula: E = (h*c)/(lambda). This was my result: bfy.tw/OBwx I am two orders of magnitude bigger for some reason. 10^(-8)
@@MrsFrau Hey! I hope that you found the answer to your question. You are using here an expression of the energy representing a photon, rather than an electron of mass m and momentum P = mv, where v is its velocity. The kinetic energy of that electron is E = P^2/2m. To calculate lambda, you must apply De Broglie's postulate of the wave-particle duality: lambda = h/P, where h is Planck's constant. P = (2mE)^0.5 Make sure that you express h in eV.sec and E in eV in the formula P = (2mE)^0.5. You will end up with the relation: lambda(E) = Ko/(E)^0.5 Where: Ko = 12.265 E is the energy expressed in eV lambda(E) is expressed in Angstrom
Hey Chris, could you help me understand 1 thing? Papers/people seem to throw the word 'mask' around alot as such I am not quite sure what it means. In our e-beam example, as you said, we don't need a mask. We have a program which tells the electron beam where/how to print. Ok, good - maskless writing. Now I am reading papers where they argue that they use this technique to CREATE masks. Ok, again good. We so we create a mask, through maskless e-beam litography. But what can someone do with such a mask? Can this mask, be then passed on to a computer-chip company which works with photo-lithography and they will use regular light to illuminate that mask such that they are able to print its features on a wafer and create micro chips? Or what industrial use can such a mask have? Please help, I was nowhere able to finde the answer. Thanks
I will learn it after my invention will become public for all the world. By the way my profession is a dentist and I am not related to AI, Quantum, Encryption.
Extremely useful video on UA-cam. Surprised it exist. Thanks.
Thank You, Sir, I never understood it so clearly as I did today. You started from fundamentals and explained E-beam lithography in the best possible way. I am a big fan of yours from now onwards...Thanks again!!
PDF copies of all the slides in this course are available at:
www.lithoguru.com/scientist/CHE323/course.html
Thank you very much for a great video, introducing the theoretic concepts before explaining the actual lithography process is essential for the full process understanding, and most explanatory articles or videos do not provide this vital information. Great job!
Dear Chris, thank you for your contribution. Would love to know/hear more from you regarding EBL and its architecture/physics in future. It would be nice to discuss system specific discussions like Raith, JEOL etc. systems.
Thank you Dr. Mack.
4:40 how does or does the electrons wavelength relates to current (13.5 EUV)nm structures? i.e. 0.037≈3.7nm?
I'm not sure I understand your question. Are you asking about the wavelength of an electron (which depends on energy) compared to the dimensions currently being fabricated using EUV lithography? EUV lithography is capable of printing about 16 nm lines and spaces today. At 100keV, the wavelength of the electron is 0.0037 nm, which is much, much smaller than any feature being fabricated.
@@ChrisMack thats is what i was asking. Si according to the specs, at 100Kav an EB can etch 3.7nm?
Hi Chris, thanks for the nice presentation. What does "Throughput" mean in E-Beam lithography? Does it mean the speed of the system in completing the pattern?
Throughput would something like area written per unit time.
@@ChrisMackthank you so much 🙏🏻
Thank you so much for this lecture.
Thank you very much for the great videos!
Anyone knows what is meant by lens NA (9:40) in this video?
NA = Numerical Aperture, the sine of the maximum half-angle that can make it through the lens.
@@ChrisMack Thanks! Are you sure the wavelengths are correct at 4:05 ? I tried to calc. it myself for 100eV using the formula: E = (h*c)/(lambda). This was my result:
bfy.tw/OBwx
I am two orders of magnitude bigger for some reason. 10^(-8)
@@MrsFrau
Hey! I hope that you found the answer to your question. You are using here an expression of the energy representing a photon, rather than an electron of mass m and momentum P = mv, where v is its velocity. The kinetic energy of that electron is E = P^2/2m. To calculate lambda, you must apply De Broglie's postulate of the wave-particle duality: lambda = h/P, where h is Planck's constant. P = (2mE)^0.5
Make sure that you express h in eV.sec and E in eV in the formula P = (2mE)^0.5.
You will end up with the relation:
lambda(E) = Ko/(E)^0.5
Where:
Ko = 12.265
E is the energy expressed in eV
lambda(E) is expressed in Angstrom
Excelente, Thanks
Hey Chris, could you help me understand 1 thing? Papers/people seem to throw the word 'mask' around alot as such I am not quite sure what it means. In our e-beam example, as you said, we don't need a mask. We have a program which tells the electron beam where/how to print. Ok, good - maskless writing. Now I am reading papers where they argue that they use this technique to CREATE masks. Ok, again good. We so we create a mask, through maskless e-beam litography.
But what can someone do with such a mask? Can this mask, be then passed on to a computer-chip company which works with photo-lithography and they will use regular light to illuminate that mask such that they are able to print its features on a wafer and create micro chips? Or what industrial use can such a mask have? Please help, I was nowhere able to finde the answer. Thanks
Yes, masks are used in projection imaging in photolithography. See lecture 39 (and later lectures) in this series.
9:00 by "more than examine", you might also mean "fry the observed sample" :-P
I will learn it after my invention will become public for all the world.
By the way my profession is a dentist and I am not related to AI, Quantum, Encryption.