Lecture 24 (CHE 323) CVD, part 1

แชร์
ฝัง
  • เผยแพร่เมื่อ 16 ก.ย. 2024

ความคิดเห็น • 16

  • @gdk870
    @gdk870 8 ปีที่แล้ว +3

    Amazing stuff Sir.
    Now please forgive my troll level of physics ignorance, i've only had physics until my senior year in high school. Nevertheless, I was so intrigued by CVD diamond growing when first i heard about it, that it got me thinking. Let's say you had a diamond disc of radius 5cm placed in you CVD chamber. And as you were depositing the methane gas which builds the lattice for the growing diamond, you sequentially (by first switching off the microwave source..or not) introduce all the chemical elements necessary to make a neodymium magnet. Making sure to deposit only about a 2-atom-thick layer of these neodymium magnet ingredients. However, the neo-magnet ingredients should be deposited within a smaller radius to the diamond disc, something like 4cm instead of the disc's 5cm. So that the next deposition of methane (with accompanying microwaves) at a radius equal to the disc, would essentially (and simultaneously) encase and cook the neo-magnet ingredients into a rudimentary neodymium block ready for magnetization, albeit it being 2 atoms thick. Once the second round of methane deposition is done, we continue to deposit the neo-magnet ingredients again, and so on. The result would be a diamond cylinder block that looks like a stack of CD's stacked on top of each other, since if you looked at the stack along its length, you'd be able to peer to the other side of the room. But look at the stack from the top, and you'd see nothing but the Musician's album cover/art.
    My thinking for depositing such ultrathin layers of neo ingredients was that the diamond lattice which has to form on top could grant the magnet lattice some of their immovable properties (being diamond after all) to prevent the the finished diamond from hopefully tolerating very high temperature, physical, and other forms of stress.
    Now that you have an idea of what i am getting at, i want you to imagine the CVD chamber being fitted with a powerful electromagnet that would induce and maintain polar orientation of the neodymium layer at select times during the entire CVD process. So regarding the placement of the electromagnet poles, perhaps the "North" pole could be just atop the Chamber, while the "South" pole is fit just below the chamber. It look like you pinching a AAA battery at its ends between your thumb and index finger.
    Now of course, this is probably flawed to humorous level, but what the heck.
    This is just an entertaining thought for times in the shower or when trying to fall asleep, but whatever you do after reading this, please think about what such a magnet infused diamond could be capable of. Cheers, and thanks for reading :)

  • @PwrElec
    @PwrElec 10 ปีที่แล้ว +2

    As always.... Excellent!!!

  • @reasonandevidence
    @reasonandevidence 4 ปีที่แล้ว

    Excellent lecture. Concise and logical. Love it.

  • @pavelkolesnichenko3210
    @pavelkolesnichenko3210 8 ปีที่แล้ว

    Why is the slope of pseudo-Arrhenius plot in the reaction-controlled regime proportional to -E_a/k_B? Arrhenius equation for reaction rate constant is k_T = A*exp(-E_a/(k_B*T)). Substituting it into the formula for deposition rate we can find the natural logarithm of the deposition rate being represented as
    ln(v) = ln(k_T/N*(1/(k_B*T))*P_g) = ln(A*P_g/(k_B*T)) + ln(exp(-Ea/(k_B*T))) - ln(T), so
    ln(v) = const - E_a/(k_B*T) - ln(T)
    So in addition to linear dependence on 1/T, there is additional dependence of deposition rate on temperature in the form of ln(T) or -ln(1/T).

    • @ChrisMack
      @ChrisMack  8 ปีที่แล้ว +1

      +Zumbi Azul Your derivation assumes that the partial pressure of the reactant in the bulk gas, P_g, is kept constant as the temperature is changed. Another possibility is that the concentration of the reactant in the bulk gas, C-g, is kept constant as the temperature is changed. In this second scenario, the pseudo-Arrhenius plot is linear with 1/T.

    • @pavelkolesnichenko3210
      @pavelkolesnichenko3210 8 ปีที่แล้ว

      +Chris Mack You are right, I had to express the partial pressure as P_g = k_B*T*C_g. Thanks!

  • @pavelkolesnichenko3210
    @pavelkolesnichenko3210 8 ปีที่แล้ว +1

    I guess there is a misprint at the slide numbered 7 in the bottomest formula. The denominator should be (h_g+k_s), not (h_g+C_g).

    • @ChrisMack
      @ChrisMack  8 ปีที่แล้ว +3

      +Zumbi Azul You are correct! Thank you for finding the error. I have updated the pdf of the slides on the course website to show the correct formula. Please note that all of the formulas on the rest of the slides are correct.

    • @pavelkolesnichenko3210
      @pavelkolesnichenko3210 8 ปีที่แล้ว

      +Chris Mack Thanks for your precious lectures! They helped me a lot to get into the understanding of CVD process. I'm doing my PhD now and I have to write my first year report. And I wanted to ask if I could use the material in the report with a reference to your name?

    • @ChrisMack
      @ChrisMack  8 ปีที่แล้ว +2

      +Zumbi Azul Yes, you may do so!

    • @pavelkolesnichenko3210
      @pavelkolesnichenko3210 8 ปีที่แล้ว

      Thanks!

    • @saad3677
      @saad3677 5 ปีที่แล้ว +1

      So this is how you can also get permissions now :)

  • @debashisdasjnvp
    @debashisdasjnvp 4 ปีที่แล้ว

    CVD is an aqua-based or solvent-based coating process?

    • @ChrisMack
      @ChrisMack  4 ปีที่แล้ว +2

      Neither. Gas phase.

  • @ChrisMack
    @ChrisMack  9 ปีที่แล้ว +4

    PDF copies of all the slides in this course are available at:
    www.lithoguru.com/scientist/CHE323/course.html

  • @nirmalbabu9570
    @nirmalbabu9570 8 ปีที่แล้ว +3

    too complicated for me