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Analysis was skipped due to inability to compute operating point.

AQoutb
AQoutb over 4 years ago

I am trying to simulate a very simple design using a dc current source and a switch.  I keep getting Analysis was skipped due to inability to compute operating point.

Could you please help me with this?

try to select any parameter to the design, it doesn't mater and it will keep getting errors as long as there is a dc current source.

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  • Andrew Beckett
    Andrew Beckett over 4 years ago

    To be sure, it would be best to see the input.scs file from your simulation as then I could see the parameters and how it is set up.

    However, fundamentally you need to think about what the poor simulator is going to have to deal with here. You have an ideal current source feeding into a switch. When the switch is open, there is nowhere for that current to flow, yet the simulator is going to need to make it flow (otherwise the equations can't be solved). In some cases it might put a "gmin" conductance so there's somewhere for the current to flow (not sure it would here - I'd need to try it and I'd sooner wait until you have provided the netlist) - assuming this is the case, the voltage would end being enormous across that gmin conductance (assuming it's 1e-12 Siemens -  i.e. 1TOhm).

    So that's likely to be the issue - an ideal current source driving a current into nothing.

    Andrew

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  • AQoutb
    AQoutb over 4 years ago in reply to Andrew Beckett

    I thought it should behave like it is driving an open circuit so nothing should pass through. Do you recommend changing the open resistance of the switch to some other value?

    Here is the input.scs


    simulator lang=spectre
    global 0
    parameters R=1K
    I52 (net5 net16) isource dc=5m type=dc
    R14 (net6 0) resistor r=R
    W5 (net16 net6 net10 0) relay vt1=0 vt2=3.3 ropen=1T rclosed=1.0
    V16 (net5 0) vsource dc=1 type=dc
    V0 (net10 0) vsource type=pulse val0=0 val1=3.3 period=100n
    simulatorOptions options reltol=1e-3 vabstol=1e-6 iabstol=1e-12 temp=27 \
        tnom=27 scalem=1.0 scale=1.0 gmin=1e-12 rforce=1 maxnotes=5 maxwarns=5 \
        digits=5 cols=80 pivrel=1e-3 sensfile="../psf/sens.output" \
        checklimitdest=psf
    tran tran stop=800n write="spectre.ic" writefinal="spectre.fc" \
        annotate=status maxiters=5
    finalTimeOP info what=oppoint where=rawfile
    modelParameter info what=models where=rawfile
    element info what=inst where=rawfile
    outputParameter info what=output where=rawfile
    designParamVals info what=parameters where=rawfile
    primitives info what=primitives where=rawfile
    subckts info what=subckts where=rawfile
    save W5:1
    saveOptions options save=allpub

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  • Andrew Beckett
    Andrew Beckett over 4 years ago in reply to AQoutb

    I see in that example you've changed the open resistance to 1Tohm, which means you end up with 5GV at the left side of the relay when it is open. If you were expecting an ideal current source to magically become non-ideal and turn itself off when the circuit is open, it won't do that. In this case a 5mA current source is always a 5mA current source and it tries to ensure that this constraint is always met (in the same way that an ideal voltage source always drives that voltage regardless of the load).

    Perhaps rather than having a switch on the output of an ideal current source, you could use vccs (Voltage Controlled Current Source) and have the current switched on or off?

    It's rather unclear to me what you're actually trying to do here anyway, so it's hard to give advice - I can't quite see what the point of this circuit is.

    Andrew.

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  • Andrew Beckett
    Andrew Beckett over 4 years ago in reply to AQoutb

    I see in that example you've changed the open resistance to 1Tohm, which means you end up with 5GV at the left side of the relay when it is open. If you were expecting an ideal current source to magically become non-ideal and turn itself off when the circuit is open, it won't do that. In this case a 5mA current source is always a 5mA current source and it tries to ensure that this constraint is always met (in the same way that an ideal voltage source always drives that voltage regardless of the load).

    Perhaps rather than having a switch on the output of an ideal current source, you could use vccs (Voltage Controlled Current Source) and have the current switched on or off?

    It's rather unclear to me what you're actually trying to do here anyway, so it's hard to give advice - I can't quite see what the point of this circuit is.

    Andrew.

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