Cathode Ray Tube

Introduction

The cathode ray tube or "Braun tube" is a vacuum tube that contains electron guns pointed at a fluorescent screen. They can be found in old television or computer monitors which are replaced nowadays by lcd monitors.

Structure and theory of operation

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The thermionic cathode electrode is heated by the heating voltage $$U_H$$ and emits electrons (thermionic emission). These are accelerated by the acceleration voltage $$U_B$$ towards the ring anode. When the electrons passed the ring anode the move with constant velocity further to the capacitor. There they are directed towards to screen by the electric field of the capacitor and finally arrive at the screen. The screen consists of material that produces light when hit by electrons.

The wehnelt cylinder takes care that the electron beam of the thermionic cathode electrode does not divert in all directions and as such misses the ring anode.

Acceleration phase

The electrons are accelerated on the distance $$m$$, from the thermionic cathode electrode to the ring anode by the acceleration voltage $$U_B$$. Thereby the electrons are supplied electric energy $$E_{el} = U_B \cdot e$$ which is transformed into kinetic energy. As such the following holds:

\begin{aligned} E_{kin} & = E_{el} \\ \dfrac{m_e}{2} \cdot v^2 & = U_B \cdot e \\ \end{aligned}

As the mass and the electric charge of an electron are known the velocity $$v$$ can be determined:

\begin{aligned} \dfrac{m_e}{2} \cdot v^2 & = U_B \cdot e \\ v^2 & = \dfrac{2 \cdot U_B \cdot e}{m_e} \\ v & = \sqrt{ \dfrac{2 \cdot U_B \cdot e}{m_e} } \\ \end{aligned}

Direction phase

We assume that the electrons arrive at the capacitor with the velocity $$v_0$$. They will continue to move with this constant speed along the x-axis beacuse the capacitor only changes their speed along the y-axis.

$$x = v_0 \cdot t$$

The acceleration along the y-axis of the capacitor leads to a uniformly accelerated motion of the electrons along the y-axis:

$$y = \dfrac{1}{2} \cdot a_y \cdot t^2$$

From the capacitor voltage $$U_A$$ and the distance of the capacitor plates $$d$$ the electric force $$F$$ of the field can be determined. Then the acceleration of the electrons can be calculated:

$$F = \dfrac{U_A}{d} \cdot e \qquad \Rightarrow \qquad a_y = \dfrac{F}{m_e} = \dfrac{U_A \cdot e}{d \cdot m_e}$$

Put in:

$$y = \dfrac{1}{2} \cdot \dfrac{U_A \cdot e}{d \cdot m_e} \cdot t^2$$

Direction at the end of the capacitor

We can now determine the displacement $$y_1$$ of the electrons along the y-axis after passing the the capacitor. This is done by putting the time $$t_1$$ that an electron needs to travel through the capacitor into the formula above:

$$t_1 = \dfrac{s}{v_0} \qquad \Rightarrow \qquad y_1 = \dfrac{1}{2} \cdot \dfrac{U_A \cdot e}{d \cdot m_e} \cdot \left( \dfrac{s}{v_0} \right)^2$$
Velocities at the end of the capacitor

The velocity along the x-axis did not change:

$$v_x = v_0$$

The velocity along the y-axis can be calculated with the formular for accelerated motion. We also use $$t_1$$ here:

$$v_y = a \cdot t_1 = \dfrac{U_A \cdot e}{d \cdot m_e} \cdot \dfrac{s}{v_0}$$
Position on the screen

In the area after the capacitor, the electron moves uniformly with the velocity $$v_y$$ during the time $$t_2$$ along the y-axis:

$$t_2 = \dfrac{l}{v_0} \qquad \Rightarrow \qquad y_2 = v_y \cdot t_2 = \dfrac{U_A \cdot e}{d \cdot m_e} \cdot \dfrac{s}{v_0} \cdot \dfrac{l}{v_0}$$

The position on the screen can be determined by the addition of $$y_1$$ and $$y_2$$:

\begin{aligned} y_{Screen} & = y_1 + y_2 \\ & \\ y_{Screen} & = \dfrac{1}{2} \cdot \dfrac{U_A \cdot e}{d \cdot m_e} \cdot \left( \dfrac{s}{v_0} \right)^2 + \dfrac{U_A \cdot e}{d \cdot m_e} \cdot \dfrac{s}{v_0} \cdot \dfrac{l}{v_0} \\ & \\ y_{Screen} & = \underset{\text{Ausklammern}}{\underbrace{  \dfrac{U_A \cdot e}{d \cdot m_e} \cdot \dfrac{s}{(v_0)^2}  }} \cdot \dfrac{s}{2} + \underset{\text{Ausklammern}}{\underbrace{  \dfrac{U_A \cdot e}{d \cdot m_e} \cdot \dfrac{s}{(v_0)^2}  }} \cdot l \\ & \\ y_{Screen} & = \dfrac{U_A \cdot e}{d \cdot m_e} \cdot \dfrac{s}{(v_0)^2} \cdot \left( \dfrac{s}{2} + l \right) \\ \end{aligned} Position dependant of $$U_B$$

We can determine the position dependant of the acceleration voltage $$U_B$$:

$$v_0 = \sqrt{ \dfrac{2 \cdot U_B \cdot e}{m_e} }$$

As such:

\begin{aligned} y_{Screen} & = \dfrac{U_A \cdot e}{d \cdot m_e} \cdot \dfrac{s}{(v_0)^2} \cdot \left( \dfrac{s}{2} + l \right) \\ & \\ y_{Screen} & = \dfrac{ U_A \cdot \cancel e}{d \cdot \cancel m_e} \cdot \dfrac{s}{\dfrac{2 \cdot U_B \cdot \cancel e}{\cancel m_e}} \cdot \left( \dfrac{s}{2} + l \right) \\ & \\ y_{Screen} & = \dfrac{1}{2} \cdot \dfrac{U_A}{d} \cdot \dfrac{s}{U_B} \cdot \left( \dfrac{s}{2} + l \right) \\ & \\ \end{aligned}

We observe that the position on the screen does not depend on the electron charge to mass quotient $$\dfrac{e}{m_e}$$.

Sources

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