Question

Find the maximum input voltage of a two cavity klystron amplifier with the following parameters:

V₀ = 1 KV, I₀ = 25 mA, R₀ = 40 kΩ, f = 3 GHz

Gap spacing in either cavity: 1 mm

Spacing between two cavities: 4 cm

Effective shunt impedance excluding beam loading = 30 kΩ

J1(X) = 0.582, X = 1.481

1. 1.85 V
2. 96.5 V
3. 105 V
4. 75 V

Answer 1

Correct Answer - Option 2 : 96.5 V

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

\({v_0} = 0.593 \times {10^6}\sqrt {{V_0}} \)

\({v_0} = 0.583 \times {10^6}\sqrt {1000} \)

v₀ = 18.75 × 10⁶ m/sec.

The gap transit Angle (θ_g)  is = \(\frac{{\omega d}}{{{v_0}}} = {\theta _g}\)

\( = 2\pi \left( {3 \times {{10}^9}} \right) \times \frac{{{{10}^{ - 3}}}}{{1.88 \times {{10}^7}}} = 1\;rad\)

The Beam coupling coefficient is

\({\beta _i} = {\beta _0} = \frac{{\sin \left( {\frac{{{\theta _g}}}{2}} \right)}}{{\frac{{{\theta _g}}}{2}}} = \frac{{\sin \frac{1}{2}}}{{\frac{1}{2}}} = 0.952\)

The Dc transit angle between cavities is

\({\theta _0} = \omega {T_0} = \omega \frac{L}{{{v_0}}}\)

\( = 2\pi \left( {3 \times {{10}^9}} \right)\left( {\frac{{4 \times {{10}^{ - 2}}}}{{1.88 \times {{10}^7}}}} \right)\)

= 40 rad

The maximum input voltage V₁ is given by-

\({V_1}_{max} = \frac{{2\;{V_0}\;X}}{{{\beta _i}{\theta _0}}}\)

\({V_1}_{max} = \frac{{2{{\left( {10} \right)}^3}\left( {1.841} \right)}}{{\left( {0.952} \right)\left( {40} \right)}} = 96.5\;V\)