Full Solution to the Power Grid Matrix Discussed in my Previous Articles

Full Solution to the Power Grid Matrix Discussed in my Previous Articles

Here is the full solution to the power grid I discussed in my previous article: https://www.dhirubhai.net/pulse/introduction-solving-power-grid-matrices-deen-kotturi-msee-mba-bpapc/?trackingId=E3OoWecuHw6hnKZFUCG3hw%3D%3D

Fig 1., shows the power grid I used previously. We will use this same power grid through out the rest of the discussion.

Fig 1. A small 16-node power grid

This is going to be a bit long but I promise you it is well worth your time if you really want to know what EDA tools do behind the scenes to solve power grid matrices. I will use the same power grid I discussed in my previous articles. Here is the complete solution to solving the matrix for the power grid in Fig 1.

Fig 2. Original matrix and operations on Row 2


Fig 3. Operations on Row 3 and Row 5


Fig 4. Operations on Row 4 and Row 5


Fig 5. Operations on Row 6 and Row 5


Fig 6. Operations on Row 6, Row 7, and Row 6


Fig 7. Operations on Row 7, Row 8, and Row 7


Fig 8. Operations on Row 8, Row 9, and Row 8


Fig 9. Operations on Row 9, Row 10, and Row 9


Fig 10. Operations on Row 10, Row 12, and Row 10


Fig 11. Operations on Row 12, Row 13, and Row 12


Fig 12. Operations on Row 13, Row 14, and Row 12


Fig 13. Operations on Row 13, Row 14, and Row 14


Now that the matrix is an upper triangular matrix, we can solve for v16 and subsequent voltages using this upper triangular matrix. Here is the value of v16:

v16 = -0.35/0.75 = -0.467v        

Using backward substitution, we can solve for v15:

2.04*v15 = -0.2+1.23*v16
v15 = -0.380        

We can use the same approach to subsequently solve by backward substitution, for V14 through v1 voltages. Note that we already know the voltages at node 5 and node 9 to be 1V each. By using better precision while solving the matrix, we will get the following voltage values for all of our node voltages:

v1  =  0.5597V
v2  =  0.1194V
v3  = -0.2952V
v4  = -0.4071V
v6  =  0.0936V
v7  = -0.5977V
v8  = -0.5191V
v10 = -0.1472V
v11 = -0.6702V
v12 = -0.5524V
v13 =  0.4939V
v14 = -0.0123V
v15 = -0.3835V
v16 = -0.4679V        

I hope this matrix-solving full solution using an example power grid is useful in your quest to learn steady state IR drop analysis.

If you want to know the finer details on how to accurately solve power grids for static and dynamic analyses, I recommend looking into Voltus power grid analysis solution. Voltus gives excellent accuracy for full flat and hierarchical analyses on large highly dense multi-billion node power grids.

Feel free to repost it or use it for your own needs. Happy learning!


Pankaj Baghmar

Sr Staff Engineer @ Infineon Technologies| Synthesis to GDS, PNR, STA ,Sign off,EM/IR

6 个月

Well said!..

Mrutunjay Singh

Senior Lead Engineer @ Qualcomm

6 个月

Quiet easy and detailed explanation Deen!! Love reading your articles.

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