Re: Agenda for committee call - 30 May 2006

Resending my response to peter back to the list:

I would expect both of those expressions to fail a
syntax check, due to using the output value on both
the left hand and right hand side of the equation.. 
To do that I think you need to use the "IMPLICIT"
form..
V(n1) : 2*V(n1)+6 = 0;
which I would read:
(the Voltage on n1 is defined such that 2*V(n1)+6 = 0;

You could model a resistor divider from n1 to n2 in
two ways: 
conservative: 
V(n1,n2) <+ R1* I(n1,n2);
V(n2) <+ R2*I(n2); // gnd is implicit reference
// handles non-zero current flow out of n2 into other
circuitry

signal flow potential
V(n2) <+ V(n1)*R2/(R1+R2);
//Assumes I(n2) ~0
// but the voltage is the same even if its not.. 

But While I would expect a signal_flow (flow) nature
to follow KCL: without the accompanying potential we
can hardly call it conservative. 

its either CONSERVATIVE (Potential & FLOW)
or its SIGNAL FLOW (Potential or Flow).. never both. 

an example:
---
so if the SOURCE model has 

module top;
electrical extres;
voltage vbandgap;
current iref;
source REF (.n1(iref), .vbg(vbandgap) .res(extres));
resistor #(.r(12K)) Rext (extres,gnd!);
load DUT (.ibias(iref));
endmodule

module source (n1, vbg, res);
output current n1;
output voltage vbg;
output electrical res;
analog begin
  V(vbg) <+ 1.20;
  V(res) <+ 1.20; // 1.2v/12K = 0.1m = 100uA 
   // out => -100mA
  I(n1) <+ I(res)/2.0; // -50ua 
end

module load (ibias);
input current ibias;
real Ibias;
analog begin
 @timer(0,10m) begin
  $strobe("INFO: %M: Ibias = %g at %g\n", I(ibias),
$abstime);
 end
end
endmodule

I Should get  
INFO: top.DUT: Ibias = 5.00e-5 at 0.000
INFO: top.DUT: Ibias = 5.00e-5 at 1.00e-5
...
in the log file.

Showing that the current OUT of the Source = the
current INTO the load (and "into" is the positive
current direction, even for output pins..  )

In my view, a single module in Verilog is useless..
its the interconnection of multiple modules that is
useful.. 


Jonathan

--- Peter Liebmann <peterl@xpedion.com> wrote:

> I have a question about signal flow with a flow
> nature.  Is it conservative?
> The reason I ask if one has a signal flow potential
> nature , the solution
> to a simple equation with only one source at n1,
> 
>     V(n1) <+ 2*V(n1) +6;
> 
> is obviously -6.
> 
> However, if n1 is a signal flow flow nature and is
> conservative,
> 
>     I(n1) <+ 2*I(n1) +6;
> 
> has no solution since I(n1) must sum to 0 and there
> is only one source 
> at n1.
> 
> Is this what we want?
> 
> Peter Liebmann
> 
> 
> Sri Chandra wrote:
> > 
> > Time & Date: 30 May 2006, 3pm Pacific
> > Dialin: 1-877-346-8823 (US - toll free)
> >           1-203-320-0407 (Intl)
> > Pin: 602538
> > 
> > 
> > * Review of chapter 2 - lexical tokens with
> updated syntax (Graham)
> > 
> > cheers,
> > Sri
> > 
> 
>