A pentode vacuum tube

Exercise 4.12 A pentode vacuum tube

The figure below shows the symbol of a pentode vacuum tube, identifying its anode (A), cathode (C), screen (S) and grid (G) nodes. The heater is not shown in this symbol. The I-V relations for this device are (simplified version of the Child-Langmuir law):

\begin{array}{rcl} i_{G} = 0 \\ i_{A} = K \cdot {(v_{GC} + \frac{v_{AC}}{μ_{c}} + \frac{v_{SC}}{μ_{s}})}^{3 ∕ 2} (for v_{GC} + \frac{v_{AC}}{μ_{c}} + \frac{v_{SC}}{μ_{s}} > 0) \end{array}

where $K$ , $μ_{c}$ and $μ_{s}$ are device properties.

a): In its most simple application, the screen voltage is fixed to some constant and high voltage, $v_{SC} = V_{HIGH}$ . This is also assumed in the rest of this question. The pentode then effectively reduces to a device having 3 terminals (A,G,C) that see current and/or voltage changes. Derive a proper small signal equivalent circuit for this (then two-port) device, including equations relating the small signal parameters to bias settings.

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