如何在IronPython執行CPython運算

首先在環境變數引入AEDT內建的CPython路徑，再用system()執行python "xxx.py"即可。

 os.environ['PATH'] = oDesktop.GetExeDir() + "commonfiles/CPython/3_7/winx64/Release/python"

os.system('python "alignCS_Calculation.py"')

交換兩條bondwire的落點

前一篇的分散bondwires的落點地script無法很好處理兩排die pad bondwire問題。可以透過以下script來交換bondwire的落點以解決bondwire空中交錯的問題。將其設為熱鍵，可以簡化操作。

import ScriptEnv
ScriptEnv.Initialize("Ansoft.ElectronicsDesktop")
oDesktop.RestoreWindow()
oDesktop.ClearMessages("", "", 2)
oProject = oDesktop.GetActiveProject()
oDesign = oProject.GetActiveDesign()
oLayout = oDesign.GetActiveEditor()

seles = oLayout.GetSelections()

A, B = seles[0:2]

Ax, Ay = oLayout.GetPropertyValue('BaseElementTab', A, 'Pt1').split(',')
Bx, By = oLayout.GetPropertyValue('BaseElementTab', B, 'Pt1').split(',')
oLayout.ChangeProperty(
    [
        "NAME:AllTabs",
        [
            "NAME:BaseElementTab",
            [
                "NAME:PropServers", 
                A
            ],
            [
                "NAME:ChangedProps",
                [
                    "NAME:Pt1",
                    "X:="        , str(Bx) + 'mm',
                    "Y:="        , str(By) + 'mm'
                ]
            ]
        ]
    ])
oLayout.ChangeProperty(
    [
        "NAME:AllTabs",
        [
            "NAME:BaseElementTab",
            [
                "NAME:PropServers", 
                B
            ],
            [
                "NAME:ChangedProps",
                [
                    "NAME:Pt1",
                    "X:="        , str(Ax) + 'mm',
                    "Y:="        , str(Ay) + 'mm'
                ]
            ]
        ]
    ])

oLayout.Select(seles[0:2]) 

(圖一)切換前後排交叉 bondwires的落點來避免交叉

分離bondwire腳本

以下腳本對單排的diepad可以適用。首先選擇Pin，執行腳本，落在其上的bondwires的Pt1會分離到不同的位置。可持續執行腳本直到到滿意的位置。將腳本連接到熱鍵可更快速的執行工作。目前只能確保2D平面直線不會交叉，因為立體結構，仍可能有交叉，仍需使用者肉眼判斷。

# -*- coding: utf-8 -*-
from itertools import combinations
from math import sqrt
import random


def ccw(A, B, C):
    Ax, Ay = A
    Bx, By = B
    Cx, Cy = C
    return (Cy - Ay) * (Bx - Ax) > (By - Ay) * (Cx - Ax)


def intersect(A, B, C, D):
    return ccw(A, C, D) != ccw(B, C, D) and ccw(A, B, C) != ccw(A, B, D)


def checkintersection(segments):
    for (A, B), (C, D) in combinations(segments, 2):
        if intersect(A, B, C, D):
            return True
    return False


import ScriptEnv

ScriptEnv.Initialize("Ansoft.ElectronicsDesktop")
oDesktop.RestoreWindow()
oDesktop.ClearMessages("", "", 2)
oProject = oDesktop.GetActiveProject()
oDesign = oProject.GetActiveDesign()
oLayout = oDesign.GetActiveEditor()

sele = oLayout.GetSelections()
AddWarningMessage(str(sele))


def getPkgGrid(pin_name):
    layer = oLayout.GetPropertyValue('BaseElementTab', pin_name, 'Start Layer')
    x0, y0 = oLayout.GetPropertyValue('BaseElementTab', pin_name, 'Location').split(',')
    x0, y0 = float(x0), float(y0)

    grid = []
    for i in range(-10, 11):
        for j in range(-10, 11):
            x = (x0 + 0.05 * i) * 1e-3
            y = (y0 + 0.05 * j) * 1e-3
            pt = oLayout.Point()
            pt.Set(x, y)
            if pin_name in oLayout.FindObjectsByPoint(pt, layer):
                x = (x0 + 0.04 * i) * 1e-3
                y = (y0 + 0.04 * j) * 1e-3
                grid.append((x, y))
    return grid


pkg = getPkgGrid(sele[0])
AddWarningMessage('Pkg Locations: {}'.format(len(pkg)))


def getDieGrid(pin_name):
    layer = oLayout.GetPropertyValue('BaseElementTab', pin_name, 'Start Layer')
    grid = {}
    for i in oLayout.FindObjects('Type', 'bondwire'):
        p1 = oLayout.Point()
        x, y = oLayout.GetPropertyValue('BaseElementTab', i, 'Pt1').split(',')
        pt = p1.Set(float(x) * 1e-3, float(y) * 1e-3)
        obj = oLayout.FindObjectsByPoint(p1, layer)

        if pin_name in oLayout.FindObjectsByPoint(pt, layer):
            x, y = oLayout.GetPropertyValue('BaseElementTab', i, 'Pt0').split(',')
            x, y = float(x) * 1e-3, float(y) * 1e-3
            grid[(x, y)] = i
    return grid


die = getDieGrid(sele[0])
AddWarningMessage('die Locations: {}'.format(len(die)))

pair = {}
N = 0
while (True):
    N += 1
    if N > 1000000:
        AddWarningMessage('Failed')
        segments = []
        break
    segments = []
    random.shuffle(pkg)
    for (pt0, pt1) in zip(die.keys(), pkg[0:len(die)]):
        segments.append((pt0, pt1))
    if checkintersection(segments) == False:
        AddWarningMessage('Successful')
        break

for pt0, pt1 in segments:
    pair[die[pt0]] = pt1

AddWarningMessage(str(pair))
try:
    for bw_name in pair:
        x, y = pair[bw_name]
        oLayout.ChangeProperty(
            [
                "NAME:AllTabs",
                [
                    "NAME:BaseElementTab",
                    [
                        "NAME:PropServers",
                        bw_name
                    ],
                    [
                        "NAME:ChangedProps",
                        [
                            "NAME:Pt1",
                            "X:="     , str(x),
                            "Y:="        , str(y)
                        ]
                    ]
                ]
            ])
except:
    pass
oLayout.Select(sele[0])

如何選擇落在某個pad上面的bondwires

選擇位於'L1_pkg'層的Pin，執行下列腳本便可以選擇落在選中的Pin的所有Bondwires。這個功能可以用來協助開發分離bondwires腳位的工具(見上一篇)。

import ScriptEnv
ScriptEnv.Initialize("Ansoft.ElectronicsDesktop")
oDesktop.RestoreWindow()
oProject = oDesktop.GetActiveProject()
oDesign = oProject.GetActiveDesign()
oLayout = oDesign.GetActiveEditor()

sele = oLayout.GetSelections()
AddWarningMessage(str(sele))
#polygon1 = oLayout.GetPolygon(sele[0])
toselect = []
for i in oLayout.FindObjects('Type', 'bondwire'):
    p1 = oLayout.Point()
    x, y = oLayout.GetPropertyValue('BaseElementTab',i, 'Pt1').split(',')
    p1 = p1.Set(float(x)*1e-3 ,float(y)*1e-3)
    obj = oLayout.FindObjectsByPoint(p1, 'L1_pkg')

    if set(obj).intersection(set(sele)):        
        AddWarningMessage(str(i))
        toselect.append(i)

oLayout.Select(toselect)

(圖一)選擇Pin，接著執行腳本

(圖二)落在被選中的Pin上的bondwires被選擇出來

Bondwire直線連接問題探索

有人問是否能自動調整bondwire pt2的位置，避免bondwire重疊導致網格失敗。所以先簡化成2D直線問題。在已知die端的座標點及pkg端的座標點且Npkg>=Ndie的前提下，採用隨機算法+交叉判斷，七條以內的直線不交叉配對可以在1sec內完成。雖然不代表bond-wires不會碰撞(因立體且有厚度)，但是或許可以從這裡出發來探索一個可行的方法。 

from itertools import combinations
from math import sqrt
import random

w = 0.1


def getparallel(A, B, d):
    x1, y1 = A
    x2, y2 = B
    r = sqrt((x2 - x1) ** 2 + (y2 - y1) ** 2)

    deltax = (d / r) * (y1 - y2)
    deltay = (d / r) * (x2 - x1)
    x3 = x1 + deltax
    y3 = y1 + deltay
    x4 = x2 + deltax
    y4 = y2 + deltay
    return (x3, y3), (x4, y4)


def ccw(A, B, C):
    Ax, Ay = A
    Bx, By = B
    Cx, Cy = C
    return (Cy - Ay) * (Bx - Ax) > (By - Ay) * (Cx - Ax)


def intersect(A, B, C, D):
    return ccw(A, C, D) != ccw(B, C, D) and ccw(A, B, C) != ccw(A, B, D)


def checkintersection(segments):
    for (A, B), (C, D) in combinations(segments, 2):
        _A, _B = getparallel(A, B, w / 2)
        A_, B_ = getparallel(A, B, -w / 2)
        _C, _D = getparallel(C, D, w / 2)
        C_, D_ = getparallel(C, D, -w / 2)

        if intersect(_A, _B, C_, D_) or intersect(A_, B_, _C, _D) or intersect(_A, B_, C_, _D) or intersect(A_, _B, _C,
                                                                                                            D_):
            return True
    return False


import matplotlib.pyplot as plt

y0 = -2
die = [(0, i) for i in range(y0, y0 + 8)]
for i, j in die:
    plt.scatter(i, j, marker='s')
x0, y0 = 3, 0
ds1 = [-1, -0.5, 0, 0.5]
ds2 = [-0.4, 0, 0.4, ]
pkg = [(x0 + i, y0 + j) for i in ds1 for j in ds2]

for i, j in pkg:
    plt.scatter(i, j)

while (True):
    segments = []
    random.shuffle(pkg)
    for (pt1, pt2) in zip(die, pkg[0:len(die)]):
        segments.append((pt1, pt2))
    if checkintersection(segments) == False:
        break

for pt1, pt2 in segments:
    x1, y1 = pt1
    x2, y2 = pt2
    plt.plot([x1, x2], [y1, y2])
plt.show()

如何簡化Python啟動步驟

不管是透過Run Script或ToolKit啟動python script都需要多個步驟。我們可以在Definitions/Script加入一段VB腳本，當中再呼叫Python Script。AEDT環境當中透過Run Script執行該VB就可以執行Python Script了。VB碼如下：

Dim oAnsoftApp

Dim oDesktop

Set oAnsoftApp = CreateObject("Ansoft.ElectronicsDesktop")

Set oDesktop = oAnsoftApp.GetAppDesktop()

oDesktop.RunScript("d:/demo/test.py")

(圖一)建立VB Script

(圖二)透過執行VB Script間接執行Python Script

HFSS Djordjevic Sarkar材料模型頻率變化公式字串輸出

多謝 Gregory, Liao的糾錯，程式計算的解果與HFSS完全一致。

from math import tan, pi, sqrt, atan, log, exp

'''
3.85309+0.0265244*ln((2.53303e+22+Freq*Freq)/(6.10915e+09+Freq*Freq))
1e-12+2.95123e-12*Freq*(atan(Freq/78161.1)-atan(Freq/1.59155e+11))
'''


def Djordjevic_Sarkar(dk=4, df=0.02, f0=10e9):
    epson0 = 8.8541878128e-12
    epson1 = dk
    delta1 = df
    f1 = f0
    fb = 159.15494e9
    sigma_dc = 1e-12

    K = (epson1 * delta1 - sigma_dc / (2 * pi * f1 * epson0)) / atan(fb / f1)
    epson_inf = epson1 - K * log(sqrt(fb ** 2 + f1 ** 2) / (f1))

    delta_epson = 10 * delta1 * epson_inf
    fa = fb / exp(delta_epson / K)

    formula1 = '{:.5f}+{:.7f}*ln(({:.5e}+Freq*Freq)/({:.5e}+Freq*Freq))'.format(epson_inf, K / 2, fb ** 2, fa ** 2)
    formula2 = '{}+{:.5e}*Freq*(atan(Freq/{:.5e})-atan(Freq/{:.5e}))'.format(sigma_dc, 2 * pi * epson0 * K, fa, fb)

    return formula1, formula2


dk, df = Djordjevic_Sarkar()
print(dk)
print(df)

(圖一)輸出Dk, Df字串

找出特定目錄底下最早及最近被修改過的檔案

 import os

import datetime
import pathlib

data = []
for path, subdirs, files in os.walk(r'C:\Program Files\AnsysEM\AnsysEM20.2\Win64'):
    for name in files:
        fname = pathlib.Path(os.path.join(path, name))
        mtime = datetime.datetime.fromtimestamp(fname.stat().st_mtime)
        data.append((mtime, fname))
data.sort()
print(data[0])
print(data[-1])

輸出結果

D:\Customer2020\pythonProject1\venv\Scripts\python.exe D:/Customer2020/pythonProject1/main.py
(datetime.datetime(1999, 12, 31, 20, 0), WindowsPath('C:/Program Files/AnsysEM/AnsysEM20.2/Win64/common/ROMViewer/resources/app/services/System.Diagnostics.Process.dll'))
(datetime.datetime(2021, 1, 4, 9, 18, 48, 61165), WindowsPath('C:/Program Files/AnsysEM/AnsysEM20.2/Win64/syslib/Toolkits/Icepak/Geometry/Packages/DUAL.py'))

Process finished with exit code 0