{"id":875,"date":"2019-11-28T22:11:46","date_gmt":"2019-11-28T20:11:46","guid":{"rendered":"https:\/\/www.hirehamva.hu\/wpblog\/?p=875"},"modified":"2019-11-28T22:11:48","modified_gmt":"2019-11-28T20:11:48","slug":"ampermero-arduino","status":"publish","type":"post","link":"https:\/\/www.hirehamva.hu\/wpblog\/archivum\/875","title":{"rendered":"Amperm\u00e9r\u0151 arduino"},"content":{"rendered":"\n<p class=\"wp-block-paragraph\">Mint \u00edg\u00e9rtem az el\u0151z\u0151 bejegyz\u00e9sben, k\u00f6zz\u00e9teszem, hogy lehet amperm\u00e9r\u0151t k\u00e9sz\u00edteni Arduinoval. Ahogyan a voltm\u00e9r\u0151 eset\u00e9ben is , a hogyanr\u00f3l lesz sz\u00f3, nem k\u00e9sz megold\u00e1ssal szolg\u00e1lok.<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">Az \u00e1ramk\u00f6r v\u00e9gtelen\u00fcl egyszer\u0171, egyetlen s\u00f6nt ellen\u00e1ll\u00e1sra (R1=15 ohm) lesz sz\u00fcks\u00e9g\u00fcnk, egy Arduino Unora \u00e9s az el\u0151z\u0151 cikkben is haszn\u00e1lt TM1637 kijelz\u0151re. Jelen estben n\u00e1lam ezek fi\u00f3kb\u00f3l el\u0151kotort alkatr\u00e9szek lesznek, csak a mik\u00e9ntj\u00e9t fogom demonstr\u00e1lni.<\/p>\n\n\n\n<figure class=\"wp-block-image size-large\"><img loading=\"lazy\" decoding=\"async\" width=\"1024\" height=\"606\" src=\"https:\/\/www.hirehamva.hu\/wpblog\/wp-content\/uploads\/2019\/11\/Amperm\u00e9r\u0151-1024x606.png\" alt=\"\" class=\"wp-image-884\" srcset=\"https:\/\/www.hirehamva.hu\/wpblog\/wp-content\/uploads\/2019\/11\/Amperm\u00e9r\u0151-1024x606.png 1024w, https:\/\/www.hirehamva.hu\/wpblog\/wp-content\/uploads\/2019\/11\/Amperm\u00e9r\u0151-300x178.png 300w, https:\/\/www.hirehamva.hu\/wpblog\/wp-content\/uploads\/2019\/11\/Amperm\u00e9r\u0151-768x455.png 768w, https:\/\/www.hirehamva.hu\/wpblog\/wp-content\/uploads\/2019\/11\/Amperm\u00e9r\u0151.png 1203w\" sizes=\"auto, (max-width: 767px) 89vw, (max-width: 1000px) 54vw, (max-width: 1071px) 543px, 580px\" \/><\/figure>\n\n\n\n<!--more Tov\u00e1bbi Inform\u00e1ci\u00f3k-->\n\n\n\n<p class=\"wp-block-paragraph\">Mivel az Arduino anal\u00f3g-digit\u00e1l \u00e1talak\u00edt\u00f3ja (0-8 l\u00e1b) fesz\u00fclts\u00e9get tud m\u00e9rni, ez\u00e9rt Ohm t\u00f6rv\u00e9nye lesz a bar\u00e1tunk. Teh\u00e1t a fogyaszt\u00f3 \u00e1ltal felvett \u00e1ramot ezzel tudjuk kisz\u00e1molni. I=U\/R ahol I- a felvett \u00e1ram, U &#8211; a m\u00e9rt fesz\u00fclts\u00e9g az A\/D \u00e1talak\u00edt\u00f3 \u00e1ltal (rajzon U2), R &#8211; a s\u00f6nt ellen\u00e1ll\u00e1sunk (rajzon R1=15 ohm)<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">Az ellen\u00e1ll\u00e1s eset\u00e9ben min\u00e9l nagyobb teljes\u00edtm\u00e9ny\u0171t \u00e9s kis \u00e9rt\u00e9k\u0171t \u00e9rdemes v\u00e1lasztani, f\u0151leg ha nagy \u00e1ramokat szeretn\u00e9nk m\u00e9rni. (meleged\u00e9s)<strong> Ha magasabb fesz\u00fclts\u00e9g\u0171 \u00e1ramk\u00f6r\u00f6n \u00e1tfoly\u00f3 \u00e1ramot m\u00e9r\u00fcnk akkor ahogyan a voltm\u00e9r\u0151 eset\u00e9ben egy fesz\u00fclts\u00e9goszt\u00f3t kell az A\/D \u00e1talak\u00edt\u00f3 el\u00e9 helyezz\u00fcnk \u00fagy sz\u00e1molva hogy a legmagasabb osztott fesz\u00fclts\u00e9g 5V legyen. (ennyit b\u00edr ki az Arduino bemenete) Teh\u00e1t U2 max 5V lehet, k\u00fcl\u00f6nben f\u00fcst lesz. \ud83d\ude00<\/strong><\/p>\n\n\n\n<p class=\"wp-block-paragraph\">Ebben az esetben milliamperekr\u0151l lesz sz\u00f3, a kijelz\u0151n is milliamperben fog megjelenni az eredm\u00e9ny.<\/p>\n\n\n\n<p class=\"wp-block-paragraph\"><strong>Arduino programja:<\/strong><\/p>\n\n\n\n<pre class=\"wp-block-code\"><code>#include &lt;TM1637Display.h>\n\nconst int CLK = 2; \/\/Kjelz\u0151 CLK l\u00e1ba\nconst int DIO = 3; \/\/Kijelz\u0151 DIO l\u00e1ba\nTM1637Display display(CLK, DIO); \/\/Kijelz\u0151 inicializ\u00e1l\u00e1s.\n\nint be = A0; \/\/0 analog bemenet \nint dc = 0;\nfloat voltbex; \nfloat voltki = 0.00;\nfloat aram = 0.00;\nfloat R1 = 15; \/\/  R1 ellen\u00e1ll\u00e1s \u00e9rt\u00e9ke ohm-ban\n\nvoid setup() {\n  \/\/ put your setup code here, to run once\nSerial.begin(9600);\npinMode(be, INPUT);\ndisplay.setBrightness(0x0a); \/\/kijelz\u0151 f\u00e9nyer\u0151 max\n}\nvoid loop() {\n  \/\/ put your main code here, to run repeatedly:\n  for(int m=0;m&lt;=40;m++){        \/\/40 mintav\u00e9tel\n dc = analogRead(be); \/\/ad \u00e1talak\u00edt\u00f3 leolvas\u00e1sa\n    voltki = (dc * 5.0) \/ 1024.0; \/\/ 5.0 referencia fesz\u00fclts\u00e9g \n    voltbex=voltbex+voltki; \n    delay(5);\n  }\n  voltbex=voltbex\/40;\/\/ \u00e1tlag sz\u00e1m\u00edt\u00e1s\n  aram=voltbex\/R1;\n    aram = aram * 1000;\/\/mA-\u00e9 alak\u00edt\u00e1s\n    Serial.println(aram);\/\/m\u00e9rt \u00e9rt\u00e9k ki\u00edrat\u00e1sa soros portra\n    display.showNumberDec(aram);  \/\/ki\u00edrat\u00e1s kijelz\u0151re  \n    voltbex=0;\n    delay(200);\n}<\/code><\/pre>\n\n\n\n<p class=\"wp-block-paragraph\"><strong>Vide\u00f3 a m\u00e9r\u00e9sr\u0151l:<\/strong><\/p>\n\n\n\n<figure class=\"wp-block-embed-youtube wp-block-embed is-type-video is-provider-youtube wp-embed-aspect-16-9 wp-has-aspect-ratio\"><div class=\"wp-block-embed__wrapper\">\n<iframe loading=\"lazy\" title=\"Amperm\u00e9r\u0151 arduino\" width=\"525\" height=\"295\" src=\"https:\/\/www.youtube.com\/embed\/BeeCMOWyBH8?feature=oembed\" frameborder=\"0\" allow=\"accelerometer; autoplay; clipboard-write; encrypted-media; gyroscope; picture-in-picture; web-share\" referrerpolicy=\"strict-origin-when-cross-origin\" allowfullscreen><\/iframe>\n<\/div><\/figure>\n","protected":false},"excerpt":{"rendered":"<p>Mint \u00edg\u00e9rtem az el\u0151z\u0151 bejegyz\u00e9sben, k\u00f6zz\u00e9teszem, hogy lehet amperm\u00e9r\u0151t k\u00e9sz\u00edteni Arduinoval. Ahogyan a voltm\u00e9r\u0151 eset\u00e9ben is , a hogyanr\u00f3l lesz sz\u00f3, nem k\u00e9sz megold\u00e1ssal szolg\u00e1lok. Az \u00e1ramk\u00f6r v\u00e9gtelen\u00fcl egyszer\u0171, egyetlen s\u00f6nt ellen\u00e1ll\u00e1sra (R1=15 ohm) lesz sz\u00fcks\u00e9g\u00fcnk, egy Arduino Unora \u00e9s az el\u0151z\u0151 cikkben is haszn\u00e1lt TM1637 kijelz\u0151re. Jelen estben n\u00e1lam ezek fi\u00f3kb\u00f3l el\u0151kotort alkatr\u00e9szek lesznek, &hellip; <\/p>\n<p class=\"link-more\"><a href=\"https:\/\/www.hirehamva.hu\/wpblog\/archivum\/875\" class=\"more-link\"><span class=\"screen-reader-text\">&#8222;Amperm\u00e9r\u0151 arduino&#8221;<\/span> b\u0151vebben<\/a><\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"closed","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[1],"tags":[84,5,86,85],"class_list":["post-875","post","type-post","status-publish","format-standard","hentry","category-alapok","tag-ampermero","tag-arduino","tag-ohm","tag-tm-1637"],"_links":{"self":[{"href":"https:\/\/www.hirehamva.hu\/wpblog\/wp-json\/wp\/v2\/posts\/875","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/www.hirehamva.hu\/wpblog\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/www.hirehamva.hu\/wpblog\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/www.hirehamva.hu\/wpblog\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/www.hirehamva.hu\/wpblog\/wp-json\/wp\/v2\/comments?post=875"}],"version-history":[{"count":17,"href":"https:\/\/www.hirehamva.hu\/wpblog\/wp-json\/wp\/v2\/posts\/875\/revisions"}],"predecessor-version":[{"id":893,"href":"https:\/\/www.hirehamva.hu\/wpblog\/wp-json\/wp\/v2\/posts\/875\/revisions\/893"}],"wp:attachment":[{"href":"https:\/\/www.hirehamva.hu\/wpblog\/wp-json\/wp\/v2\/media?parent=875"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.hirehamva.hu\/wpblog\/wp-json\/wp\/v2\/categories?post=875"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.hirehamva.hu\/wpblog\/wp-json\/wp\/v2\/tags?post=875"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}