CRT Electron Beam Focusing: Math Treatment & Details

In summary, the electric field causes the electrons to focus and the electric potential difference between the anodes and the cathodes provides the focusing force. The shapes of the anodes, grids, and cathodes are important so that the electric field has a radial component.
  • #1
Euclid
214
0
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!
 
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  • #2
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.
 
  • #3
Euclid said:
For electric focusing, how does accelerating the electrons cause it to focus?
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.
 
  • #4
krab said:
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]
.
What considerations lead you to the Laplace equation?
 
  • #5
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.
 
  • #6
pervect said:
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}+\frac{{\partial^2 V}{\partial z^2}} = 0[/tex]

therefore

[tex] \nabla^2 V = 0 [/tex]
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?
 
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  • #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.
 
  • #8
Euclid said:
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?
(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.
 
  • #9
krab said:
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.
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).
 
  • #10
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?
 
  • #11
Euclid said:
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?

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).
 
  • #12
krab said:
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.
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?
 
  • #13
Euclid said:
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?
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.
 

What is CRT electron beam focusing?

CRT (Cathode Ray Tube) electron beam focusing is a process used in the display technology of CRT monitors. It refers to the precise control of the electron beam to produce a sharp and focused image on the screen.

Why is math treatment important in CRT electron beam focusing?

Math treatment is crucial in CRT electron beam focusing because it involves complex mathematical calculations to control the electron beam. These calculations take into account factors such as the size and shape of the beam, the distance from the electron gun to the screen, and the desired image resolution.

What are the key components involved in CRT electron beam focusing?

The key components involved in CRT electron beam focusing include the electron gun, deflection coils, and the focusing coil. The electron gun generates a beam of electrons, the deflection coils control the movement of the beam on the screen, and the focusing coil adjusts the strength of the beam to produce a sharp image.

How does CRT electron beam focusing work?

CRT electron beam focusing works by using magnetic fields produced by the deflection and focusing coils to control the path of the electron beam. The beam is directed to specific points on the screen, and the strength of the beam is adjusted to produce a focused image.

What are the advantages of CRT electron beam focusing?

CRT electron beam focusing has several advantages, including the ability to produce high-quality images with excellent color and brightness. It also allows for fast refresh rates, making it suitable for applications such as gaming and video editing. Additionally, CRT monitors are relatively inexpensive compared to other display technologies.

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