Mike wrote:
? ?"IC=-100.8u
? ?"I understand this is to establish an initial current. But how do we know which circuit it applies to, and how do we tell it is intended to set the current in L1?"
We know it's not a comment because it isn't in blue. ?It is an actual SPICE parameter or directive. ?Check for that first.
Use Ctrl-right-click on L1. ?You will see that "IC=-100.8u" is one of the parameters of that element. ?Alternatively, generate the SPICE Netlist for the schematic (View > SPICE Netlist) and then search it for "IC=-100.8u" and see which element it applies to. ?(Third method: Use the "Move" operation (F7) and momentarily "pick up" L1 and see which text moves with it. ?Then right-click to cancel the Move operation.)
L2 also has an IC parameter ("IC=0"), but that one was not set to be visible.
It would be 'nice' if LTspice had a way to display the attachments between text and the components they belong to. ?Someone could create a very messed up schematic by moving component values around.
? ?"When you set an initial condition in this manner, how do you make a mental image of what is happening? For example, if I imagine a current is going through L1, I would think it has to charge C1. This would create an initial voltage across C1, but I have no way to tell how much that voltage will be."
Here is my take on the question:
The initial condition was used in conjunction with "UIC" on the ".TRAN" statement. ?This instructs LTspice to use only the specified initial conditions as the complete initial operating point. ?In this case I(L1)=-100.8uA, and V(C1)=0 because it was not specified. ?In the instant AFTER t=0, the current through L1 would begin to charge C1. ?Right at t=0, C1 has zero voltage ... but note that this transient simulation does not start saving waveform data until 10us have passed, so you need to edit the .TRAN statement and remove that, to see what happened at the actual start of the simulation.
? ?"Do we consider the current is applied for zero time and terminates at the start of the transient analysis?"
Depending on what you're asking: ?Not exactly, because that would violate how inductors behave.??Its current can't drop to zero instantly.??We consider that somehow the inductor had 100.8uA flowing up until t=0; and then at t=0+, L1 starts behaving like a normal inductor.
But you might think of it as if a current source had been placed across only L1, and it is removed at t=0+. ?(Maybe that was what you meant.)
? ?"BTW, I think you have provided an outstanding service to tens of thousands of engineers through your steadfast dedication to the LTspice group. I don't know how you manage to maintain your sanity over such a long time, but I offer my heartfelt gratitude for what you have accomplished."
Helmut is pretty amazing! ?He is one of the more valuable resources the LTspice community has.
Regards,
Andy
