Mail Archives: geda-user/2012/12/12/00:38:15
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On Tue, 11 Dec 2012, Joshua Lansford wrote:
> He he, what if you considered each pin which belonged to net A be a source,
> each pin which belongs to net B be a sink, pass a virtual current from each
> source to each sink and then colored things by how 'hot' they become. :-P
> Might generally help identify the short location.
This may work with some modifications (or maybe I am just thinking over
the implied details).
First, what do we do in case more than 2 nets participate in a short? We
could run this for each pair once. I supsect this would cause more shorts
to be highlighted than the minimum number of cuts needed for separating
the networks, but this may be a minor issue.
Second, how do we do the current calculation exactly? Input is one of the
nodes in net1 (source) and one of the nodes in net2 (sink); we need to:
- identify _all_ paths between them
- connect them in parallel and clauclate a resistor network for
dissipation
- once we could do that reliably, I assume we'd sum dissipations got
from different paths for the same edge/node. The simple approach is
take each path and increase a 'current counter' for each edge the
path includes. A more complicated way is a real series/parallel
resistor network of 1 ohm resistors between every source-sink
combo.
So the method is:
1. select a source and a sink
2. replace all other nodes with simple junctions
3a. place a counter on each edge; or
3b. replace all edges with a 1 ohm resistor and calculate dissipation on
each resistor assuming a voltage source (or current source) between
source and sink
4. repeat on all source-sink combinations and sum the counter or
dissipation per resistor to find the hot spots
Using the usual net1=a,b net2=C,D examples, edges (aka resistors) numbered
now, it would work like this:
1 2
a--+--b
|3
C--+--D
4 5
Paths:
a-D: 1,3,5
a-C: 1,3,4
b-C: 2,3,4
b-D: 2,3,5
Dissipation per edge with the simple counter method (3a.):
edge 3: 4
edge 1, 2, 4 and 5: 2
So it indeed solves the simple case.
Let's find something trickier:
1
a---b
|2 |3
C---D
4
We expect to find 2 and 3 be hotter than 1 and 4.
a-D: 1,3 and 2,4
a-C: 2 and 1,3,4
b-C: 3,4 and 1,2
b-D: 3 and 1,2,4
With the simple counter all edges are referenced 4 times so no specific
hot points.
With resistor networks (sch attached with my calculations, please
veryify), R2 and R3 are highlighted, so the more complicated method works
on this case.
However, it's not very hard to construct an example where it would fail:
if there are 10 parallel instances of edge 2, power dissipation will drop
to 1/10 on them, cooling them down so much that even 1 and 4 seem to be
hotter. An obvious fix is merging parallel resistors "where it does not
matter", in other words find all connections which are not contributing in
the short, and deal with the rest (either find the best cut or just
highlight the whole set and let the user decide). In the first example
this would result in 3, on the last example in 2,3. But as far as I can
tell this was the original problem we tried to solve.
So my conclusion: if we can identify which set of edges are contributing
in the short, this method could help highlighting the best segment of each
set, but alone does not solve the problem.
Best regards,
Tibor Palinkas
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