I have discovered will doing a PID sub circuit that the default parallel resistance in an inductor isn’t infinite as it is said in the Help.
Indeed, we can see below that there is a limitation at the frequency of R/L in the differentiator section of the PID bode plot, and if the RPAR value was infinite we wouldn’t see it.
For now I had to set the RPAR value to 1e15 Ohm in the Inductor’s parameter :
Maybe an update could fix that
That’s not the point at all. Gmin works strangely.
Here is an example where I define the actual part - a parallel resistor. I find the conductivity, then the real (actual) conductivity and by finding the inverse I determine which resistor is in parallel with the inductance. My opinion is that something is wrong with Gmin.
I suggest you use .opt Gmin=0 for now.
@LoicL.TPS could you please share the spice model for your PID symbol ?
.subckt pid_comp in out
XU1 in prop Gain A={km}
XU2 in mid1 Gain A={1/tau_i}
XU3 mid1 integ Integ initcond=0
XU4 diff prop integ out Adder3
XU5 in mid2 Gain A={tau_d}
XU6 mid2 diff Diff
.ends pid_comp
.subckt Diff in out
G1 0 int in 0 1
L1 int 0 1 Rpar=1e15
E1 out 0 int 0 1
.ends Diff
.subckt Integ in out
.param initcond=0
G1 0 int in 0 1
C1 int 0 1 ic=initcond
R1 int 0 1e18
E1 out 0 int 0 1
.ends Integ
.subckt Adder3 in1 in2 in3 out
E3 out int2 in3 0 1
E2 int2 int1 in2 0 1
E1 int1 0 in1 0 1
.ends Adder3
.subckt Gain in out
E1 out 0 in 0 A
.ends Gain
