Node combinations

Hi, I wanted to share an easy method to make combinations of nodes to measure voltage across. Very easy and I find it very useful.

Voltage is measured across two nodes, right?, ok, so when you are faced with a circuit that has many nodes like the one below, the question is:
----"How many combinations of two can I make?"



This is the method I use and it's pretty quick!:

Calculators have a button called "nCr":



You have to press the following keystrokes:
(since the above circuit has 6 nodes:)

"[6] --> [nCr] --> [2] ---> [nCr]"

(you probably have to press shift, "2nd function")

and the calculator should return 15.

That means you can make 15 combinations of 2 with 6 nodes.

But wait, there's more:

-----How do you do the combinations??

Simple:

Common nodes referenced to ground:

[1,0], [2,0], [3,0], [4,0], [5,0]

Other nodes not referenced to ground:

[1,2], [1,3], [1,4], [1,5]
[2,3], [2,4], [2,5]
[3,4], [3,5]
[4,5]

for the above: you grab all the non-zero nodes (1,2,3,4,5) and do the combinations just like you would with soccer teams in a competition:

--grab 1 and pair it with the rest of the others on the right:

[1,2], [1,3], [1,4], [1,5]

--grab 2, and do the same, always picking the ones on the right:

[2,3], [2,4], [2,5]

--we go to 3:

[3,4], [3,5]

--and finally:

[4,5]

And that's it, now you have ALL the possible combinations for a resistor circuit like the one above:

[1,0], [2,0], [3,0], [4,0], [5,0]
[1,2], [1,3], [1,4], [1,5]
[2,3], [2,4], [2,5]
[3,4], [3,5]
[4,5]

15 combinations!!

Total nodes: 6
Total combinations: 15
Common: 5
Extra: 10

You use Kirchhoff's Voltage Law to calculate the voltage drops across those combinations.

Spice by default returns all the voltage drops for nodes with respect to 0.
You have to enter the rest manually, like this:

.print dc v(2,5), v(3,5)...

Now you are sure you're not missing anything.

Hope it helps someone!

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