Verilog-A does not support node numbers, so you cannot use 0 to represent ground (I try to make that clear in my book).
You can declare a node to be ground using:
Code:electrical gnd;
ground gnd;
If you do that, you cannot also add the following:
Code:analog V(gnd) <+ 0;
That would create a loop of shorts.
Now, lets think about a simplified version of what you proposed.
Code:analog begin
@(initial_step)
V(vdd) <+ 0 ;
V(vdd) <+ transition(1.25,10n,20m);
end
The
initial_step triggers on the first timepoint, meaning that on the first time point, this code simplifies to:
Code:analog begin
V(vdd) <+ 0 ;
V(vdd) <+ transition(1.25,10n,20m);
end
The contribution operators sum all contributions made at the same time point, so this is equivalent to:
Code:analog begin
V(vdd) <+ 0 + transition(1.25,10n,20m);
end
Thus, the
initial_step block does not do any thing.
Finally, the transition filter only operates upon changes in its arguments. In this example, the argument never changes, so the transition filter can be removed without changing the behavior. In the final simplification, the above is equivalent to:
Code:analog begin
V(vdd) <+ 1.25;
end
So, I would not expect your supplies to ramp up like you want.
Perhaps you can try something like the following:
Code:real Vdd;
analog begin
@(timer(10n))
Vdd = 3.3;
V(vdd) <+ transition(Vdd,0,20m);
end
One more thing. You should
never put a contribution statement in an event block. Fortunately, when you did it you contributed zero, which really doesn't do anything. If you have used a non-zero value you would have driven the circuit with impulses, which is always problematic.
-Ken
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TB ground issue? (Read 1687 times)