Time step too small

Discussion in 'Homework Help' started by hisotaso, May 21, 2013.

  1. hisotaso

    Thread Starter New Member

    May 21, 2013
    1
    0
    Before I begin I would like to say I have spent about 5 hours today reading about others with this problem, none of the standard steps have worked.

    I am working on a term-long project to design a VCO. The project is broken up in bi-weekly labs, each lab focusing on another aspect. One of the labs was to design a circuit capable of reading the frequency of an input signal up to ~10 MHz with accuracy >= 0.01%. My frequency measurement block meets the specifications up to 12 MHz. The frequency measurement block was tested using a provided VCO block.

    I should add that the components I used to create my frequency measurement block were created by a recently retired professor, using as far as I can tell, undocumented LTSpice functions. This professor was known as the "hard-ass that never gives an A. Ever." The components consist of an edge counter, a cross detect, a first cross detect, and a behavioral source.

    Now, I hook up my VCO to the frequency measurement block and I receive the error "Analysis: Timestep too small; initial timepoint: trouble with node "U5:shi". U5 is the edge counter referred to above, and shi corresponds to one of the nodes defined in the library file as "SAMPLEHOLD" as far as I can tell. This is the text corresponding to the component from the library file:

    * edge_count
    .subckt edge_count in out ref
    .params td=10n vtrig=0 ic=0 tripdt=10p tripdv=10u tholdoff=1u
    A1 shi 0 clk 0 0 0 out 0 SAMPLEHOLD vhigh=10g tripdt={tripdt} ic={ic}
    B1 shi 0 V= int(1.5 + delay(v(out) - v(ref), {td})) tripdt={tripdt}
    B2 clk 0 V=if(time < {tholdoff}, 1k, v(in) - v(ref) - {vtrig} + 0.5) tripdt={tripdt} tripdv={tripdv}
    .ends edge_count

    I have included the netlist, please let me know what else I need to provide. Thanks in advance.

    JC

    EDIT: I wanted to add that there is no wild oscillation at the output of the VCO, by measuring the period with the cursors I can see the frequency is about 915 kHz, so the measurement block should have no trouble with this. I'm thinking this has something to do with the hysteresis.
     
    Last edited: May 21, 2013
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