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How to avoid Hidden State in VerilogA model for SpectreRF

RFStuff

 Dear All,

I am a beginner in verilogA .

I wrote a code but when I ran PSS it showed hidden state in the code and didn't run.

My behavioural model is as below:-

 

 

// VerilogA for VERILOG_A_MODEL, HARD_LIMIT_GM, veriloga

`include "constants.vams"
`include "disciplines.vams"

module HARD_LIMIT_GM(in,out);
  inout in,out;
  parameter real vtrans = 0;
  parameter real tdelay = 0 from [0:inf);
  parameter real trise = 1p from (0:inf);
  parameter real tfall = 1p from (0:inf);
  parameter real Gm=-5m;
  electrical in,out;
  real vout_val;
  analog begin
 
         @ (cross(V(in) - vtrans, 1))  vout_val = 1;
         @ (cross(V(in) - vtrans, -1)) vout_val = 0;
 
         I(out) <+ Gm * transition( vout_val, tdelay, trise, tfall); 
  end   
endmodule

 

Could anybody please tell how I can avoid the hidden state  (vout_val ) ?

 

Kind Regards,

Parents

  • The reason you have a hidden state is because vout_val is not updated on every iteration. It's only set within the @cross blocks - and is held (and hence retains state) between cross events.

    If you replace the two @cross statements with:

         @ (cross(V(in) - vtrans)) ;
         vout_val = V(in)>vtrans;

    Then it will do what you want. The first @cross will force there to be a timestep close to either the positive or negative transition - but doesn't have any action within the @cross itself. The second line is simply computing the vout_val based on whether V(in) is greater than vtrans - but does this at every timestep and hence is not a hidden state.

    Kind Regards,

    Andrew.

Reply

  • The reason you have a hidden state is because vout_val is not updated on every iteration. It's only set within the @cross blocks - and is held (and hence retains state) between cross events.

    If you replace the two @cross statements with:

         @ (cross(V(in) - vtrans)) ;
         vout_val = V(in)>vtrans;

    Then it will do what you want. The first @cross will force there to be a timestep close to either the positive or negative transition - but doesn't have any action within the @cross itself. The second line is simply computing the vout_val based on whether V(in) is greater than vtrans - but does this at every timestep and hence is not a hidden state.

    Kind Regards,

    Andrew.

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