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A reflex klystron operates under following conditions: $V_0=625 \mathrm{~V}, \mathrm{~L}=1 \mathrm{~mm}, \mathrm{R}_{\text {sh }}=14 \mathrm{~K} \Omega$ and $\mathrm{f}=8 \mathrm{GHz}$. ..

A reflex klystron operates under following conditions: $V_0=625 \mathrm{~V}, \mathrm{~L}=1 \mathrm{~mm}, \mathrm{R}_{\text {sh }}=14 \mathrm{~K} \Omega$ and $\mathrm{f}=8 \mathrm{GHz}$. If the device is operating at the peak of $1 \frac{3}{4}$ mode, Calculate a) Repeller Voltage and b) Efficiency.

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Solution:

Reflex klystron,

Given:

$$ \begin{aligned} & V_0=625 \mathrm{~V}, L=1 \mathrm{~mm} \quad R_{s h}=14 \mathrm{k} \Omega \\ & f=8 \mathrm{CH}_2 \end{aligned} $$

the device operating mode is $13 / 4$ Hence $\quad n=2$

calculation,

$$ \begin{aligned} & \text { a) Reseller voltage }\left(v_\gamma\right) \\ & \frac{v_0}{\left(v_r+v_0\right)^2}=\frac{e}{m} \frac{(2 n \pi-\pi / 2)^2}{8 \omega^2 L^2} \\ & \end{aligned} $$

$$ \frac{625}{\left(V_0+625\right)^2}=\frac{1.759 \times 10^{11} \times(2 \times 2 \times \pi-\pi / 2)^2}{8 \times(2 \pi \times 8 \times 109)^2\left(1 \times 10^{-3}\right)^2} $$

on simplification, we set

$$ \left[v_\gamma=145.701 \text { volt }\right] $$

b) Efficiency:

The efficiency is expressed as -

$$ \eta=\frac{2 x^{\prime} J_1\left(x^{\prime}\right)}{2 n \pi-\pi / 2}=2 x i $$

where $x^{\prime}=2.408$

$$ J_1\left(x^{\prime}\right)=0.52 $$

$$ =\frac{2 \times 2.408 \times 0.52}{2 \times 2 \times \pi-\pi / 2} \times 100=22.7 \% $$

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