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Cathode Ray Tubes

  1. Dec 1, 2004 #1
    How does a CRT focus an electron beam? I get the basics of it. But I'm a little confused on the details.
    For electric focusing, how does accelerating the electrons cause it to focus?
    I'm looking for a more mathematical treatment of the subject. Everything I've found is just descriptive. For example, I supposed I'd like to know what the electric field looks like between the plates, grids, anodes, etc?
    What are the shapes of the plates, grides and anodes? How is the potential difference set up and measured?
    I know there are lots of variations of CRTs, but I am interested in the specifics of how they work (so pick your favorite type to discuss).
    Any information you could provide would be greatly appreciated!
     
  2. jcsd
  3. Dec 1, 2004 #2

    Astronuc

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    See this on Scanning electron microscope basics
    http://www.uleth.ca/emf/2001/lecture_notes/lecture_1.pdf


    Also check the library for the following reference or others like it:
    P. W. Hawkes and E. Kasper, Principles of Electron Optics, vol. 3: Wave Optics, Academic Press, London, 1994.
     
  4. Dec 1, 2004 #3

    krab

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    Electrons pass thru apertures which are held at various electric potentials. This can be thought of as causing a potential V(z) on the axis joining the aperture centres. If you solve Laplaces equation in cylindrical coordinates for this axial symmetry, you find the potential can be expanded as follows:
    [tex]\Phi(r,z)=V(z)-(V''(z)/4)r^2+...[/tex]
    Differentiate this and you will find that the electric field has a radial component, and so there is a radial force. This varies along z, being sometimes positive and sometimes negative, but the dynamics is such that the net effect is to focus.
     
  5. Dec 2, 2004 #4
    What considerations lead you to the Laplace equation?
     
  6. Dec 3, 2004 #5

    pervect

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    In the absence of any charges, maxwell's equations implies that the electric potential must satisfy Laplace's equation.

    [tex] E = -\nabla V [/tex]
    [tex] \nabla \cdot E = \frac{\rho}{\epsilon} = 0[/tex]

    therefore

    [tex] \nabla^2 V = 0 [/tex]

    See for instance this link

    I don't know anything specifically about electron optics, though. It's not as obvious as one would think that the charge density is zero - space charge is a real possibility in vacuum tubes. I'd assume from the reference to Laplace's equation that it's avoided when designing electron optics - I'm not surprised, I would expect that it would be hard to control precisely.
     
  7. Dec 4, 2004 #6
    I guess I have two questions.
    (1) Wouldn't the fact that there are electrons in a CRT imply the the change density is nonzero.
    (2) How do you go about solving this equation in the case of a CRT?


    In plane cylindrical coordinates,
    [tex] \nabla^2 V= \frac{\partial ^2 V}{\partial r^2}+\frac{1}{r}\frac{\partial V}{\partial r}=0[/tex]

    The solution is V=a+blnr.
    How do I determine the constants a and b?
     
    Last edited: Dec 4, 2004
  8. Dec 4, 2004 #7
    I'd also like to solve it in the 3d case, but the plane view will show how the beam focuses. I want to know E along the z axis however to estimate how much energy the electrons gain in passing through the anodes.
     
  9. Dec 4, 2004 #8

    krab

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    (1) Yes, but in general, the fields due to the electrons themselves are much smaller than those due to the electrodes. You can estimate it from Gauss' law. It will probably only be a few volts compared with kV for the electrodes. In any case, it is something that can be added later.
    (2) You forgot
    [tex]\partial^2V/\partial z^2[/tex]
    V is a function of z, not just r. That's the whole point of using electrodes to focus the electrons. Your formula is applicable to the case where the electrons are between coaxial cylinders, with a potential difference between the cylinders. It is not applicable to the case at hand.
     
  10. Dec 4, 2004 #9
    My thought is that I want to ignore the z part of V. In any event, the z part of V has no effect on the focusing of the beam. The focusing effect is mainly from the E field in the radial direction. I was hoping to show that E(r)=-b/r in the radial direction (ignoring the z). This way I can give an explanation for why the beam focuses. The problem is showing that b>0 (it seems to me that b will depend on z, however, for what I'm doing I think it's sufficient to find b at some z).
     
  11. Dec 4, 2004 #10

    krab

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    The radial fields and the z-variation are related through Laplace's equation. Read post #3 again. In a very real sense, it is the z-variation that causes the radial fields. You cannot show E=-b/r because it is not true. You were expecting an infinite field for r=0?
     
  12. Dec 5, 2004 #11

    pervect

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    Basically, the solution is determined by the boundary condtions. The boundary conditions are that the potential on the focusing electrodes must be constant. Also, the potential at infinity must be zero, but that will be mainly a concern for the function of the potential outside the focusing electrodes.

    To have a unique solution to Laplaces equation within a region, you have to specify the potential or the electric field (the normal derivative of the potential) everywhere on a closed boundary.

    I don't quite see how to setup the boundary conditoins for this problem though, it's easy enough to specify the potential on the electrodes is constannt, but if the electrodes aren't solid I'm not sure what to do exactly. (What do the focusing electodes look like, anyway?)

    Your solution is not general, as was pointed out. Your solution will work if the focusing electode is an infinitely long cylinder, or some approximation thereof (a cylinder much longer than it's length). If the focusing electrode doesn't have this shape, you need a more general solution.

    As far as methods go, I'd suggest a computer solution, but you can also do some problems with conformal transforms (you should be able to find some info on this with a web search).
     
  13. Dec 5, 2004 #12
    Ok, what is the difference between phi and V? Are they both the potential? What are the terms after the '...'? Can they be neglected? If I have this right, does this mean that the electric field in the radial direction is
    E=(-1/2)V''(z)r ?
    How do I determine V''(z)? Can you point me to a source that might discuss this in more detail?
     
  14. Dec 5, 2004 #13

    krab

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    1. phi and V are related by the following:
    [tex]V(z)=\Phi(0,z)[/tex]
    2. Yes, for a first pass, ignore the terms implied by "...". They are just a further expansion; terms are even powers of r. Their presence makes the optics nonlinear and so are undesirable, because they make it impossible to get a point focus. Generally, you can ignore those powers if the apertures in the electrodes are large compared with the size of the electron beam.
    3. Yes, that is the radial electric field.
    4. V'' is not easy to calculate. It is found numerically by solving Laplace's equation using a technique called over-relaxation. Some (complicated) formulas exist for specialized geometries.
    5. Look up "electron optics", "einzel lens", "unipotential lens", etc.
     
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