Question

Show that theoretical efficiency of a two cavity klystron amplifier is 58%​

Answer 1

Answer:

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Answer 2

The equivalent circuit of Klystron is shown in the diagram.

Where,

Rsho = Wall resistance of catcher cavity

Rb = Beam loading resistance

Rl = External load resistance

Rsh = Effective shunt resistance

Rsh = Rsho || Rb || Rl

$i_r_m_s =\frac{ i_f__m_a_x}{\sqrt{2} } =\frac{I_2}{\sqrt{2} } =\frac{2I_0{\beta_0J_1(X)}}{\sqrt{2} }$

Poutput = i²rms Rsh = 2I₀²β₀²J₁²(X)Rsh = β₀I₀J₁(X)V₂

Maximum output voltage = V₂ = $i_f__m_a_xR_s_h$ = 2β₀I₀J₁(X)Rsh

Efficiency = η = P₀ / Pdc

$\eta = \frac{2I_0^{2}\beta_0^{2}J_1^{2}(X)R_s_h }{V_0I_0}$

$\eta = \frac{\beta_0I_0J_1(X)V_2}{V_0I_0} = \beta_0J_1(X)\frac{V_2}{V_0}$

Maximum theoretical efficiency of Klystron is :

β₀ = 1 , J₁(X) = 0.582 , V₂ = V₀

η_max = 1 × 0.582 × V₂/V₀ × 100       (∵V₂ = V₀, we can cancel them)

η_max = 58.2%          (theoretical)

Therefore, the theoretical efficiency of a two cavity klystron amplifier is 58%.