Question

A two-cavity amplifier klystron has the following parameters:

Beam voltage:

Beam current:

Frequency:

Gap spacing in either cavity:

Vo=900V

lo= 30 mA

f = 8 GHz

d = 1 mm

Spacing between centers of cavities: l = 4cm

Effective shunt impedance: R,h = 40 kfl

Determine:

a. The electron velocity

b. The de electron transit time

c. The input voltage for maximum output voltage

d. The voltage gain in decibels​

Answer 1

a. The electron velocity

Answer 2

Answer:

We have given the parameters for a two-cavity amplifier klystron:

Beam voltage, V₀ = 900 V = 0.9KV

Beam Current, I₀ = 30mA

Frequency, f = 8 GHz

Gap spacing in either cavity, d = 1 mm

Spacing between center of cavity, L = 4 cm

Shunt Impedance, $R_{sh}$= 40 KΩ

Output Impedance, R₀ = 30 KΩ

J₁(X) = 0.582 and X = 1.481

The Electron velocity: $v_{o}=0.583*10^{6}\sqrt{V_{o}}$

$v_{o}=0.583*10^{6}\sqrt{900}\\v_{o}=17.79*10^{6}m/s$

The gap transit angle is given by:

$\theta_{g}=\frac{\omega\; *\; d}{v_{o}}$

where ω = 2πf  where f = 8×10⁹ Hz and d = 1×10⁻³m

$\theta_{g}=2\pi (8*10^{9})\frac{10^{-3}}{17.79\;*10^{6}}$

$\theta _{g}=2.87\;rad$

Beam coupling coefficient:

$B_{i}=B_{o}=\frac{sin\;(\theta_{g}/2)}{(\theta_{g}/2)}$

$B_{o}=\frac{sin\;1.436}{1.436}$

$B_{o}=0.690$

The dc transit angle between two cavities:

$\theta_{o}=\omega T_{o}=\omega \frac{L}{v_{o}}$

$\theta_{o}=2\pi (8*10^{9})\frac{4*10^{-2}}{17.79*10^{6}}$

$\theta_{o}=113\;rad$

The input voltage for maximum output voltage:

$V_{max}=\frac{2V_{o}X}{B_{i}.\theta_{o}}$

$V_{max}=\frac{2* (900)(1.481)}{(0.690)(113)}$

$V_{max}=34.19V$

The dc electron transit time is , $t_{d}=\frac{L}{v_{o}}$

$t_{d}=\frac{4*10^{-2}}{17.79*10^{6}}$

$t_{d}=2.248*10^{-9}sec$