% ROGOWSKI COIL DESIGN AND SIMULATION clc; % Code for simulation of rogowski coil u0=4*pi*10^-7; e0=8.85*10^-12; d=0.01; r1=1.591*10^3; cf=0.01*10^-6;
%Permeability of free space %Permittivity of free space % Diameter of coil in meters %R1 of integrator(assumed) %Cf of integrator(assumed)
area=pi*(d^2)/4;
%Area of coil
for n=5:20 for n=5:20 %n is the no of turns for current=0:10000:500*10^3 for current=0:10000:500*10^3 m=u0*n*area; %Mutual inductance v_out=(m/(r1*cf))*current;%v_out v_out=(m/(r1*cf))*current; %v_out is the final output of integrator subplot(2,2,1); stem(current,v_out); xlabel('Current xlabel('Current in Amperes'); Amperes'); ylabel('Output ylabel('Output Voltage in Volts'); Volts'); hold on on; ; subplot(2,2,2); stem(n,v_out); xlabel('Nunber xlabel('Nunber of turns'); turns'); ylabel('Output ylabel('Output Voltage in Volts'); Volts'); hold on on; ; end; end ; end; end ; % Code for design current=10000; for r=0:0.1:0.5 for r=0:0.1:0.5 B=((u0*current)/(2*pi*r)) % Magnetic flux density fprintf('radius=%d\n' fprintf('radius=%d\n',r); ,r); subplot(2,2,3); stem(r,B); xlabel('distance xlabel('distance from conductor in Metres'); Metres'); ylabel('Magnetic ylabel('Magnetic flux density(in Tesla) for 10kA'); 10kA' ); fprintf('current=%d' fprintf('current=%d',current); ,current); hold on on; ; end; end ; current=500000; for r=0:0.1:0.5 for r=0:0.1:0.5 B=((u0*current)/(2*pi*r)) % Magnetic flux density fprintf('radius=%d\n' fprintf('radius=%d\n',r); ,r); subplot(2,2,4); stem(r,B); xlabel('Distance xlabel('Distance from conductor in Metres'); Metres'); ylabel('Magnetic ylabel('Magnetic flux density(in Tesla) for 1.5MA'); 1.5MA');
% % % %
fprintf('current=%d',current); hold on; end;
From the above design: Clearly, The magnetic flux density for a current of 10kA is significant upto a distance of 20 cm from the current carrying conductor. So the toroid must be wound within a range of 20cm from the conductor.
% The Mutual inductance is also given by the formula % M21= (u0*n/2)*(a+b-sqrt(a*b)) % Where a and b are inner and outer diameters respectively. % Also, diameter of coil is 0.01m.Hence, % b-a= 0.01m. % Therefore, by calculating for a, we get, % a=15.42 cm and ; % b=16.42 cm %------------------------------------------------------------------------%
Calculation for lumped parameters
a=0.1542; b=0.1642; d=0.01; u0=4*pi*10^-7; pc=17.2*10^-9; radius=0.0005;
% Inner diameter of coil % Inner diameter of coil
% Resistivity of copper wire %Radius of wire
n=input('Specify the number of turns (not more than 16)\n'); %Optimum number of turns fprintf('Length of coil'); l=n*pi*d fprintf('resistance of coil'); r1=(pc*l)/(pi*radius*radius) fprintf('Inductance of coil'); l1=(((u0*n*n*d)/(2*pi))* log10(b/a)) fprintf('Capacitance of coil'); c1=((4*pi*pi*e0*2*(b+a))/ log10((b+a)/(b-a))) fprintf('Self inductance of coil'); L2=(u0*n*n/2)*(a+b-2*sqrt(a*b))
% Length of coil
% Resisitance of coil %Inductance of coil %Capacitance of coil
%Self Inductance of coil
