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Theoretical Deposition Rate based on RPM - Help ARRGGG

  1. Aug 3, 2005 #1

    minger

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    Theoretical Deposition Rate based on RPM -- Help ARRGGG!!

    Hey guys, what's going on. I have recently started doing research for a professor here in the field of nanotechnology. It turns out that it's not all small robots and stuff. Right now I'm working with thin film coatings. ANYWAYS, we are using Ion Sputtering as our method of deposition right now. The prof just asked me to do this big experimented, and part of it involves heating the substrate. Now the way the machine works is that is first takes pressure in the chamber down to around 5x10^-7 torr, then we start. If we want to heat the piece, we have to rotate it. It rotates under the ion heater, then under the targer which coats it, then continues to rotate around.

    ANYWAYS, he wants to know what the optimal rpm is to maximize coating. So, lets say we have an even just flux of whatever coming down into a circular area. You have another circular area rotating around underneath it. See the attached diagram for a better explanation. I just have no idea where to start. The deposition is a certain nanometers per meter squared per second.

    I guess another way to think of it is if you have a large flow of water (say a waterfall). Then you have a large bucket that rotates underneath of it. What is the optimal rpm which to fill the bucket up the fastest.

    Thanks a lot for the help if any. I'm thinking the final solution will be rather simple. Something in terms of radius and areas and rpm of course. I think getting there will involve some complicated integrals though.
     

    Attached Files:

  2. jcsd
  3. Aug 3, 2005 #2
    He must be pulling your leg. Unless I'm missing something it shouldn't matter.
    The material flux rate on average is independent of the RPM.
     
  4. Aug 3, 2005 #3

    PerennialII

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    ... initially makes want to concur with Antiphon, or what does "maximize the coating" mean in this case? If we use it in a broader sense, how thick of a coating you can grow depends on the manufacturing parameters (development of coating microstructure, residual stresses, the temperature the substrate and subsequent coating are at during deposition or if you've criteria for acceptable coating properties etc.) and as such indirectly can link to the rpm .... sort of.
     
  5. Aug 4, 2005 #4

    Danger

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    If I'm reading this correctly, and there's a pretty damned good chance that I'm not, then you need to minimize the time between the heater part of the orbit and the deposition part. Normally, I would think that the maximum rpm that you can achieve would be proper. On the other hand, it has to spend enough time under the heater to actually get heated. If you can stage the thing, I'd think that it would be best to slow down on the way into the heat stage, then accelerate like mad to the deposition stage and slow down long enough for a good soaking. Since that probably isn't possible, I'd consider determining how long it takes under the heat section to reach the proper temperature, and setting the rpm's to attain that exposure.
     
  6. Aug 4, 2005 #5

    minger

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    Yes, the problem is that the rotation must be constant. I guess it's kind of a hard thing to explain. I've tried reading some papers on ohiolink, and it seems that people have done this type of thing, but they haven't really explained how they got their answer.

    Antiphon - Of course the actual material flux rate will be constant, but how much of it lands on the substrate?

    This problem reminds me of the age old question of whether or not you get wetter by running or walking in the rain. At this point, we are ignoring heating. We haven't even really gotten all of the heater kinks worked out anyways. Just focus on this damn rotation.

    I appreciate the responses by the way.
     
  7. Aug 4, 2005 #6
    The same fraction will strike the target independent of the platter's RPM,
    except for the special case of no rotation.

    As a matter of fact, you get wetter by running if you are upright.
    But if you are lying flat it doesn't matter. The deposition rate on the substrate will be
    independent of the RPM.
     
  8. Aug 4, 2005 #7

    Danger

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    Since I don't know anything about the specific equipment or the size of the material that you're coating, this could be a bit off the wall. If the heater is giving you trouble anyhow, do you have to stick with ion heating? Also, how long does the deposition device take to spool up? What crossed my mind was the question as to whether or not you could use a stationary stage under the depositer and use laser pulses in between deposition stages to heat the thing. That way your heat and deposition exposures could be controlled independently.
     
  9. Aug 4, 2005 #8

    PerennialII

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    ... and thinking about the rpm dependency material wise, what kind of a coating you've ... something with columnar microstructure, characteristic microstructural length scales etc. which can link to the rpm ?
     
  10. Aug 5, 2005 #9

    minger

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    I just can't imagine this. I imagine a bucket of water passed underneath a faucet. Surely if I move it through the faucet quickly, only a little bit of water will get into the bucket, but if it moves slowly, then the water will have more time to flow into it.

    I have actually derived an equation for a linear system. I will post it here in a second to see if anything sees any problems with it.
     
  11. Aug 5, 2005 #10

    minger

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    OK, here's what I have. H is the height of the specimen. I ignore the height of the target area and just say that it's larger than the height of the speciment, so that the area exposed to deposition will simply be the substrate height times the length which it is exposed.

    Velocity times time is displacement. Displacement times height is that exposed area. (I assume the substrate starts at x=0, right at the edge of being deposited on) Then I take the differential area with respect to time and integrate it. I integrate from 0 (which I just explained) to (L1+L2)/V. L1 and L2 are the lengths of the substrate and target area respectively. Dividing by velocity gives me the time that it takes for the substrate to pass through the target area.

    **Now that I think of it, I am not addressing the issue that the substrate will also "fade" out of the area. To compensate for this, I will divide the limit by two (so that I am integrating from x=0 to the point where the substate is in the middle of the target area). Then I will just multiply the "exposed area" by two to get the whole thing. (OK image is now changed)**

    Now after integrating, I assign constants and graph it. Then I told myself, well this "exposed area" doesn't mean a whole lot without knowing how many times your going to do that per unit time. Then I assigned another constant which is the distance between cycles. The time it takes for another cycle to come around is going to be the velocity divided by that length. Obviously that is going to be a linear graph with increasing frequency with increasing velocity.

    I then graphed them on the same graph and they hit about 22, and I was using inches and seconds for my constants, so 22 in/s. Converting to angular gives about 1/2 rev/s based on a ~7" radius.

    Can anyone tell me if my analysis is correct of where I need to reevaluate things. Thanks for the help so far.

    edit: Realizing that where they meet doesn't mean jack, I changed things up. Realizing that the units on my "exposed area" are in²s (which makes sense), I decided to multiply by frequency (Hz, 1/s) to give me in², which will be the total area deposited on per unit time. Multiplying these two together gives me a straight line which corresponds to my initial gut reaction that RPM shouldn't matter. If anyone can confirm or find a mistake in my calculations, I'd really be greatful, thanks again.
     

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    Last edited: Aug 5, 2005
  12. Aug 5, 2005 #11
    Hmm....


    Anyway, yes if the RPM goes up the dwell time under the source
    is shorter, but the substrate is exposed that much more often.

    It all cancels out and unless the substrate is sitting still under the
    source, or unless you are making a very short deposition in time,
    the RPM shouldn't matter.
     
  13. Aug 8, 2005 #12

    minger

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    For some reason the way you worded your post I was under the impressions that you were telling me that per revolution, RPM didn't affect deposition. Anyways, I think it's all settled as long as nobody can find an error in my calculations and assumptions.
     
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