MATLAB Answers

Converting a SPICE subcircuit to Simscape component

2 views (last 30 days)
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 ...

Accepted Answer

David John
David John on 3 Aug 2021
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'},priority=priority.none};
end
branches
L1: node_1.i -> node_4.i;
L2: node_2.i -> node_5.i;
L3: node_3.i -> node_6.i;
end
equations
value(node_1.v-node_4.v,'V') == 3.93e-09*value(L1.der,'A/s')+0.38*sqrt(3.93e-09*5.35e-09)*value(L2.der,'A/s')+0.45*sqrt( ...
3.93e-09*5.39e-09)*value(L3.der,'A/s');
value(node_2.v-node_5.v,'V') == 5.35e-09*value(L2.der,'A/s')+0.38*sqrt(5.35e-09*3.93e-09)*value(L1.der,'A/s')+0.19*sqrt( ...
5.35e-09*5.39e-09)*value(L3.der,'A/s');
value(node_3.v-node_6.v,'V') == 5.39e-09*value(L3.der,'A/s')+0.45*sqrt(5.39e-09*3.93e-09)*value(L1.der,'A/s')+0.19*sqrt( ...
5.39e-09*5.35e-09)*value(L2.der,'A/s');
end
end
  2 Comments
David John
David John on 4 Aug 2021
Glad to hear that you got unstuck. If you're happy with the answer provided, please feel free to "accept" it. This helps us to identify questions that still require further investigation.
Many thanks,
David

Sign in to comment.

More Answers (1)

Pranav Verma
Pranav Verma on 11 Dec 2020
Hi Florian,
The link states that since SPICE is a rich language, it is not always possible to perform a full conversion without some manual intervention.
You can refer to the below link to see if it helps:
Thanks
  1 Comment
Florian Sawallich
Florian Sawallich on 19 Jan 2021
Hi Pranav,
thank you for your reply. Do you have any ideas on how to manually implement the mutual inductors into the Simscape model?

Sign in to comment.

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!