Question

A two cavity klystron amplifier has the following specifications: Beam voltage Vo = 900 V Beam Current Io = 30mA Frequency f = 8 GHz Gap spacing in either cavity d = 1 mm Spacing between centre of cavity L = 4 cm Shunt Impedance Rsh= 40 K Output Impedance Ro = 30 K Determine: (i) The Electron Velocity (ii) The dc transit time of maximum of electron (iii) The input voltage for maximum output (iv) The voltage gain in decibels.​

Answer 1

Answer:

96.5V

Explanation:

Given:

V0 = 1 kV

I0 = 25 mA

R0 = 40 kΩ

f = 3 Ghz

d = 1 mm

L = 4 cm

Rsh = 30 kΩ

For maximum voltage V2, J1 (x) must be maximum. This means J1 (x) = 0.582 at x = 1.841

Electron velocity just leaving the cathode is

v0 = 18.75 × 106 m/sec.

The gap transit Angle (θg) is =

The Beam coupling coefficient is

The Dc transit angle between cavities is

= 40 rad

The maximum input voltage V1 is given by-

Answer 2

The specifications of a two cavity klystron amplifier is given , to find

•  The Electron Velocity
• The dc transit time of maximum of electron
• The input voltage for maximum output
• The voltage gain in decibels.​

Given,

$V_{0}$= 1 kV

$I_{0}$ = 25 mA

$R_{0}$= 40 kΩ

f = 3 Ghz

d = 1 mm

L = 4 cm

$R_{sh}$= 30 kΩ

For maximum voltage , $V_{2}$J1 (x) must be maximum.

This means J1 (x) = 0.582 at x = 1.841

Electron velocity just leaving the cathode is

 $V_{0}$= 18.75 × 106 m/sec.

The gap transit Angle (θg) is =

The Beam coupling coefficient is

The Dc transit angle between cavities is

= 40 rad