$\mathrm{Na\_conductance}=\mathrm{g\_Na}hm^3\mathrm{i\_Na}=\mathrm{Na\_conductance}(V-\mathrm{E\_Na})$ $\mathrm{alpha\_m}=\frac{-0.1(V+50)}{e^{\frac{-(V+50)}{10}}-1}\mathrm{beta\_m}=4e^{\frac{-(V+75)}{18}}$ $\mathrm{alpha\_h}=0.07e^{\frac{-(V+75)}{20}}\mathrm{beta\_h}=\frac{1}{e^{\frac{-(V+45)}{10}}+1}$