/*-------------------------------------------------------------------------- Author: Thomas Nowotny Institute: Institute for Nonlinear Dynamics University of California San Diego La Jolla, CA 92093-0402 email to: tnowotny@ucsd.edu initial version: 2005-08-17 --------------------------------------------------------------------------*/ #ifndef CN_ECNEURON_CC #define CN_ECNEURON_CC #include "CN_neuron.cc" ECneuron::ECneuron(int inlabel, double *the_p= ECN_p): neuron(inlabel, ECN_IVARNO, ECNEURON, the_p, ECN_PNO) { } ECneuron::ECneuron(int inlabel, vector inpos, double *the_p= ECN_p): neuron(inlabel, ECN_IVARNO, ECNEURON, inpos, the_p, ECN_PNO) { } inline double ECneuron::E(double *x) { assert(enabled); return x[idx]; } void ECneuron::derivative(double *x, double *dx) { Isyn= 0.0; forall(den, den_it) { Isyn+= (*den_it)->Isyn(x); } // differential eqn for E, the membrane potential dx[idx]= -((pw3(x[idx+1])*x[idx+2]*p[0]+p[2]*x[idx+4])*(x[idx]-p[1]) + pw4(x[idx+3])*p[3]*(x[idx]-p[4])+ p[7]*(x[idx+5]*0.65+x[idx+6]*0.35)*(x[idx]-p[8])+ p[5]*(x[idx]-p[6])-Isyn+2.85)/p[9]; // diferential eqn for m, the probability for one Na channel activation // particle _a= -0.1*(x[idx]+23)/(exp(-0.1*(x[idx]+23))-1); _b= 4*exp(-(x[idx]+48)/18); dx[idx+1]= _a*(1.0-x[idx+1])-_b*x[idx+1]; // differential eqn for h, the probability for the Na channel blocking // particle to be absent _a= 0.07*exp(-(x[idx]+37)/20); _b= 1/(exp(-0.1*(x[idx]+7))+1); dx[idx+2]= _a*(1.0-x[idx+2])-_b*x[idx+2]; // differential eqn for n, the probability for one K channel activation // particle _a= -0.01*(x[idx]+27)/(exp(-0.1*(x[idx]+27))-1); _b= 0.125*exp(-(x[idx]+37)/80); dx[idx+3]= _a*(1.0-x[idx+3])-_b*x[idx+3]; _a= 1/(0.15*(1+exp(-(x[idx]+38)/6.5))); _b= exp(-(x[idx]+38)/6.5)/(0.15*(1+exp(-(x[idx]+38)/6.5))); dx[idx+4]= _a*(1.0-x[idx+4])-_b*x[idx+4]; // differential equation for the Ihf activation variable _a= 1/(1+exp((x[idx]+79.2)/9.78)); _b= 0.51/(exp((x[idx]-1.7)/10)+exp(-(x[idx]+340)/52))+1; dx[idx+5]= (_a-x[idx+5])/_b; // differential equation for the Ihs activation variable _a= 1/(1+exp((x[idx]+71.3)/7.9)); _b= 5.6/(exp((x[idx]-1.7)/14)+exp(-(x[idx]+260)/43))+1; dx[idx+6]= (_a-x[idx+6])/_b; } #endif