Converting a SPICE subcircuit to Simscape component
Hi everyone,
I have a serious problem, which I cannot solve myself.
Task: I want to convert a SPICE subcircuit to a Simscape component using the matlab syntax “subcircuit2ssc(filename,target)” The SPICE subcircuit are only related to Inductors and mutual inductors.
SPICE-model:
.subckt Ckt_model_series 1 2 3 4 5 6
L1 1 4 3.93e-09
L2 2 5 5.35e-09
L3 3 6 5.39e-09
K12 L1 L2 0.38
K13 L1 L3 0.45
K23 L2 L3 0.19
.ends Ckt_model_series
Problem: After converting I get the following message in the .ssc file:
% Users should manually implement the following SPICE commands in order to
% achieve a complete implementation:
% ckt_model_series: K12 l1 l2 0.38
% ckt_model_series: K13 l1 l3 0.45
% ckt_model_series: K23 l2 l3 0.19
This basically means all three inductor coupling are not taken into consideration. According to this link my syntax should be right.
Question: Where is my mistake and how can I manually implement the SPICE commands afterwards? I need this solution for even bigger circuits
ANSWER
Matlabsolutions.com provide latest MatLab Homework Help,MatLab Assignment Help for students, engineers and researchers in Multiple Branches like ECE, EEE, CSE, Mechanical, Civil with 100% output.Matlab Code for B.E, B.Tech,M.E,M.Tech, Ph.D. Scholars with 100% privacy guaranteed. Get MATLAB projects with source code for your learning and research.
Could you please confirm which release you are using?
Support for the coupling factor component was added in R2019a, I believe. If you have access to that release or later, it should just work for you.
If you don’t have access to that release, then the best solution is to write the relevant equations for yourself. For this simple case, I can give you an example. Basically, it just involves writing out the differential equations. Note that L1, L2 and L3 are variables representing the currents through those inductors, so I’m just writing something like v1 = L1*i1.der + k12*sqrt(L1*L2)*i2.der + etc.
component ckt_model_series
% ckt_model_series nodes
node_1 = foundation.electrical.electrical; % node_1
node_2 = foundation.electrical.electrical; % node_2
node_3 = foundation.electrical.electrical; % node_3
node_4 = foundation.electrical.electrical; % node_4
node_5 = foundation.electrical.electrical; % node_5
node_6 = foundation.electrical.electrical; % node_6
end variables
L1 = {value={0,'A'},priority=priority.none};
L2 = {value={0,'A'},priority=priority.none};
L3 = {value={0,'A'}
SEE COMPLETE ANSWER CLICK THE LINK
