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


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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

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