Exercise 4.12 A pentode vacuum tube

a)

An answer:
A partial answer: you should be able to map this on a corresponding SSEC. In the end, that’s what you did for an MOS transistor and/or for a BJT and/or for the triode tube previously. $$\begin{array}{rcll}& & \frac{\partial i_G}{\partial v_{\mathit{GC}}}=0\to \mathrm{\text{open}}& \\ & & \frac{\partial i_G}{\partial v_{\mathit{AC}}}=0\to \mathrm{\text{open}}& \\ & & \frac{\partial i_A}{\partial v_{\mathit{GC}}}=K\cdot \frac{3}{2}\cdot {\left(v_{\mathit{GC}}+\frac{v_{\mathit{AC}}}{{\mu }_c}+\frac{V_{\mathit{HIGH}}}{{\mu }_s}\right)}^{1∕2}\to \mathrm{\text{voltage controlled current source }}& \\ & & \frac{\partial i_A}{\partial v_{\mathit{AC}}}=K\cdot \frac{3}{2{\mu }_c}\cdot {\left(v_{\mathit{GC}}+\frac{v_{\mathit{AC}}}{{\mu }_c}+\frac{V_{\mathit{HIGH}}}{{\mu }_s}\right)}^{1∕2}\to \mathrm{\text{resistor}}& \end{array}$$