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Título: Programa para calcular Choke de filtro Enviado por: xformer em 18 de Maio de 2024, as 14:38:07 Com base neste artigo publicado na revista Antena, escrevi um programa em Python para calcular um indutor de choque para filtro de fonte de tensão. Assim temos a coleção completa para calcular transformadores para valvulados e o choque de filtro.
Spoiler (clique para mostrar ou esconder) Listagem do programa: Citar import math def lookupWire(area): rc = -1 diametro = 2*(area/3.1415926535)**(1/2) for AWG in range(40,0,-1): Bitola = 0.005*92**((36-AWG)/39)*25.4 if diametro <= Bitola: rc = AWG break return rc # premissas LI2 = [0.02,0.03,0.04,0.05,0.06,0.07,0.08,0.09,0.1,0.125,0.15,0.175,0.2,0.25,0.3,0.4,0.5,0.6,0.7,0.8,0.9,1,1.5,2,2.5,3,3.5,4,5,5.5,6,7,8,9,10] SEC = [1.8,2.3,2.8,3.1,3.5,4,4.5,4.8,5,5.5,6,7,8.8,9,10,13,15,16,18.8,19,21,23,30,38,42,48,52,58,60,65,72,82,90,100,110] NI = [200,225,250,265,280,300,325,340,350,365,380,410,440,455,480,550,590,610,640,690,710,740,820,910,970,1050,1100,1150,1200,1250,1270,1350 ,1450,1500,1550] DC = 3.0 LX = float(input("Indutancia do Choke (H) ?")) IDC = float(input("Corrente DC maxima em A ?")) LXI2 = LX * IDC * IDC for indice in range(0,34,1): LI2N = LI2[indice] if LI2N >= LXI2: LXI2 = LI2N break NUCLEO = SEC[indice] ESPIRAS = int( NI[indice]/IDC) print("Secçao minima do nucleo: " + str(round(NUCLEO,2)) + "cm2 perna central de " + str(round(NUCLEO**(1/2),1))+"cm") print("Numero de espiras: " + str(ESPIRAS)) Sp = IDC/DC Bitolafio = lookupWire(Sp) print("Bitola do fio: AWG" + str(Bitolafio)) COMPRIMENTO = 7 * (NUCLEO**(0.5)) ENTREFERRO = 0.03 * COMPRIMENTO / 2 print("GAP do nucleo:" + str(round(ENTREFERRO,3)) + " mm") |