Question

9-8. A two-cavity klystron amplifier has the following parameters Vo 30 kV Beam voltage lo 3 A Beam current f 10 GHz Operating frequency Beam coupling coefficient: dc electron charge density Signal voltage Cavity shunt resistance Total shunt resistance including load: Vi = 15 V(rms) Rsht-10 kΩ Calculate a. The plasma frequency b. The reduced plasma frequency for R 0.4 c. The induced current in the output cavity I wish to help me in solving this question this the reference book is Microwave Devices and Circuits Third Edition SAMUEL d. The induced voltage in the output cavity e. The output power delivered to the load f. The power gain g. The electronic efficiency Y. LIAO Professor of Electrical Engineering California State University, Fresno sec 9

Answer 1

a. Given,

   dc electron charge density,

     
Po = 10"C/m2

The Plasma frequency,

   $w_{p}=\sqrt{\frac{\eta \rho _{o}}{\varepsilon _{o}}}$

   
| 1.759 × 1011 × 10-7 Wp8.854 x 10-12

4.46 x 10/rad's Wo

b. The reduced plasma frequency is-

  
-RWD 0.4 x 4.46x 10 1.784 x 10'rad/s Wa

c. The induced current in the output cavity is-

   $\left | I_{2} \right |= \frac{1}{2}\frac{I_{o}}{V_{o}}\frac{w}{w_{q}}\beta ^{2}\left | V_{1} \right |$

3 2π x 10 x 10912 1리 2 ЗО x 103 1,784 x 107

22.641A

d. The induced voltage in the output cavity is-

  $\left | V_{2} \right |=\left | I_{2} \right |R_{shi}$

Vs! = 2.641 x 10 x 103-26.41kv

e. The output power delivered to the load is-

$P_{out}=\left | I_{2} \right |^{2}R_{shi}$

$P_{out}=(2.641)^{2}\times 10\times 10^{3}=69.75kW$

f. The power gain is-

$power gain=\frac{1}{4}\left ( \frac{I_{o}w}{V_{o}w_{q}} \right )^{2}\beta _{o}^{4}R_{sh}.R_{shi}$

$power gain=\frac{1}{4}\left ( \frac{3\times 2\pi \times 10^{10}}{30\times 10^{3}\times 1.784\times 10^{7}} \right )^{2}1_{o}^{4}\times 10^{3}.10^{4}$$power gain=3.06\times 10^{5}=54.86dB$

g. The electronic efficiency is-

   $\eta =\frac{P_{out}}{P_{in}}=\frac{69.75\times 10^{3}}{3\times 30\times 10^{3}}$

   $\eta =$ 77.5 %