Hi everybody

I could really use some help with a circuit that i am trying to simulate to generate a sigmoid function. the description of the circuit is given below, but my problem has more to do with the coding part(i think). When i run a simulation with .rise(1m) for the pulse input, i find that the simulation stalls with an error message stating that the time step used is too small.

Increasing the rise parameter helps make the simulation run completely but the input voltage is not swept completely(ie: for a -2.5 to +2.5 range, i see -2.5 to -1.8v upon selecting the inputsignal in the waveform viewer).

I have posted a description of the circuit below and the core portion of my code. I'd be happy to give the complete code if somebody can help me.

Circuit Description:

This is a voltage comparator coupled to 2 inverters. the comparator is an opamp with input sweep voltage(-2.5 to +2.5v) given through a 40kohm resistor to the +ve terminal. there is a feedback resistance of 1000kohm from output to +ve , and the neg terminal is grounded. the output of the opamp is fed to an inverter through a 600kohm resistor and the output of the inverter is again fed to another inverter through a 600kohm. there is a 100kohm resistor in the feedback loop of each inverter(from output to input).

I'd like to know if I can simulate this circuit at all as such, because I read somewhere that opamps with positive feedback and grounded neg terminals are bi-stable devices with output driven to saturation state. Could somebody help me?

Verilog-AMS code:

opampcmos opamp1(vout1,vinp1,vinn1,vdd,vss,vref);
mosinverter inv2(vinv1,vout2,vss,vdd,vref);
mosinverter inv3(vinv2,vout3,vss,vdd,vref);

res #(.rn(40k)) resistor1(vinr,vinp1);
res #(.rn(1000k)) resistor2(vinp1,vout1);

res #(.rn(600k)) resistor3(vout1,vinv1);
res #(.rn(100k)) resistor4(vinv1,vout2);
res #(.rn(600k)) resistor5(vout2,vinv2);
res #(.rn(100k)) resistor6(vinv2,vout3);

//Voltage sources and inputs
vsource #(.type("pulse"),.val0(-2.5),.val1(2.5),.rise(1m))vin1(vinr,vref);

vsource #(.type("dc"),.dc(15))vin2(vdd,vref);
vsource #(.type("dc"),.dc(-15))vin3(vss,vref);

analog
V(vinn1,vref) <+ 0;

endmodule

My individual components(resistor,op-amp,inverter) work fine, so i am guessing that thats not the problem !

Yes, the op-amp reaches saturation state(14.something volts for a 15v supply) when tested individually. this op-amp model has been constructed using the Level-1 MOS transistor present in Cadence's verilog-A Library and using the op-amp layout described by the authors Baker,Li and Boyce in their book. I can supply the code if required.

I'm sorry Ken. I had not supplied the entire code in my posting earlier. This is the test bench I am using.

I did try adding a feedback capacitance(10pf first and then a 100 pf) but I got an error message to the effect that convergence could not be achieved.I am using the Cadence-AMS simulator for this.


`include "constants.vams"
`include "disciplines.vams"
`timescale 10ps/1ps

module cmossigvar(vinr,vinp1,vinn1,vout1,vinv1,vout2,vinv2,vout3,vdd,vss,vref);
inout vinr,vinp1,vinn1,vout1,vinv1,vinv2,vout2,vout3,vdd,vss,vref;
electrical vinr,vinp1,vinn1,vout1,vinv1,vinv2,vout2,vout3,vdd,vss,vref;

ground vref;
integer results;

//Writing data into file
initial
begin
results=$fopen("cmossigvar100.txt");
$fdisplay(results,"CMOS Sigmoid Variable Gain Model Test Bench Results:Gain=100");
forever
#1000000 $fdisplay(results,"%t %f %f",$time,V(vinr),V(vout3));
$fwrite(results,"\n");
$fclose(results);
end

//Component Instantiation
opampcmos opamp1(vout1,vinp1,vinn1,vdd,vss,vref);
mosinverter inv2(vinv1,vout2,vss,vdd,vref);
mosinverter inv3(vinv2,vout3,vss,vdd,vref);

res #(.rn(40k)) resistor1(vinr,vinp1);
res #(.rn(1000k)) resistor2(vinp1,vout1);
capa #(.c(100p)) capacitf(vinp1,vout1);

res #(.rn(1k)) resistor3(vout1,vinv1);
res #(.rn(100k)) resistor4(vinv1,vout2);
res #(.rn(1k)) resistor5(vout2,vinv2);
res #(.rn(100k)) resistor6(vinv2,vout3);

//Voltage sources and inputs
vsource #(.type("pulse"),.val0(-2.5),.val1(2.5),.rise(1m))vin1(vinr,vref);

vsource #(.type("dc"),.dc(15))vin2(vdd,vref);
vsource #(.type("dc"),.dc(-15))vin3(vss,vref);

analog
V(vinn1,vref) <+ 0;

endmodule

Hi Ken

This is the error message I receive when I simulate that
testbench

Trying `homotopy = gmin' for initial conditions.
Trying `homotopy = source' for initial conditions.
Trying `homotopy = dptran' for initial conditions.

Error found by spectre at time = 617.321 us during transient analysis `transient'.
   No convergence with minimum time step.  Last acceptable solution computed at 617.321 us.

The values for those nodes that did not converge on the last Newton iteration are given below.  Also given
       is the manner in which the convergence criteria were not satisfied in the following form:
           Failed test: | Value | > RelTol*Ref + AbsTol

   V(cmossigvar.capacitf.p) = 8.41434 mV, previously 8.41599 mV.
       update too large:  | 21.8846 uV | > 8.41434 uV + 1 uV
   I(cmossigvar.opamp1.resistor2:pos_neg_flow) = -9.16592 uA, previously -9.16767 uA.
       update too large:  | -23.2304 nA | > 9.16592 nA + 1 pA
   V(cmossigvar.opamp1.vd11) = -14.9998 V, previously -14.9998 V.
       residue too large: | 633.135 pA | > 276.256 pA + 1 pA

The resistor and capacitor models are linear, The opampcmos and mosinverter are gate-level descriptions of an op-amp and an inverter respectively. As such, these models work fine when tested independently.Hope this helps!

cmostsigvar is where you are having the convergence problems, but it is not found anywhere in the circuit you gave.

You are still having a time step too small problem. This is because you have a jump discontintuity, which occurs if you have a region of positive feedback. You must find this and put capacitors in to slow the jump so that the simulator can follow it. If you don't want to spend the time to track it down, and if you are using Spectre, you can always use the cmin option to place capacitors everywhere.

Don't play with the tolerances, the problem is with the jump.

-Ken

It is completely dependent on your model.

-Ken