{"id":1357,"date":"2017-11-03T11:45:26","date_gmt":"2017-11-03T11:45:26","guid":{"rendered":"http:\/\/bryceautomation.com\/?p=1357"},"modified":"2022-03-21T12:06:28","modified_gmt":"2022-03-21T12:06:28","slug":"voltage-dividers","status":"publish","type":"post","link":"https:\/\/bryceautomation.com\/index.php\/2017\/11\/03\/voltage-dividers\/","title":{"rendered":"Voltage Dividers"},"content":{"rendered":"<h2>Introduction to Voltage Dividers<\/h2>\n<p>Voltage dividers are very important to circuits monitored by microprocessors.\u00a0 \u00a0 Important to realize, microprocessors will usually handle 3.3 volts, or 5 volts.\u00a0 \u00a0 With this in mind, the voltage of the circuits they control can be much higher.<\/p><div id=\"bryce-3753987848\" class=\"bryce-afterfirst bryce-entity-placement\"><script async src=\"\/\/pagead2.googlesyndication.com\/pagead\/js\/adsbygoogle.js?client=ca-pub-8316758073402323\" crossorigin=\"anonymous\"><\/script><ins class=\"adsbygoogle\" style=\"display:block;\" data-ad-client=\"ca-pub-8316758073402323\" \ndata-ad-slot=\"7728240895\" \ndata-ad-format=\"auto\"><\/ins>\n<script> \n(adsbygoogle = window.adsbygoogle || []).push({}); \n<\/script>\n<\/div>\n<p>In this case, we want to measure the voltage of the 9v battery.\u00a0 \u00a0 Because the processor will only handle up to 5v, we need to reduce the voltage proportionally before we can measure it.\u00a0 \u00a0We have to reduce the voltage because of our hardware limitations.\u00a0 \u00a0With this in mind,\u00a0 once we take our reading in the processor, the answer has to be mathematically multiplied.\u00a0 \u00a0As a result, we will store the correct voltage into a variable in the processor.<\/p>\n<h2>Set up the Voltage Dividers<\/h2>\n<p>At this point, let&#8217;s look at the voltage divider circuit.<\/p>\n<p><img decoding=\"async\" class=\"aligncenter size-full wp-image-1359 lazyload\" data-src=\"https:\/\/bryceautomation.com\/wp-content\/uploads\/2017\/11\/initialckt.png\" alt=\"Voltage Divider Circuit\" width=\"1283\" height=\"500\" data-srcset=\"https:\/\/bryceautomation.com\/wp-content\/uploads\/2017\/11\/initialckt.png 1283w, https:\/\/bryceautomation.com\/wp-content\/uploads\/2017\/11\/initialckt-300x117.png 300w, https:\/\/bryceautomation.com\/wp-content\/uploads\/2017\/11\/initialckt-768x299.png 768w, https:\/\/bryceautomation.com\/wp-content\/uploads\/2017\/11\/initialckt-1024x399.png 1024w\" data-sizes=\"(max-width: 1283px) 100vw, 1283px\" src=\"data:image\/svg+xml;base64,PHN2ZyB3aWR0aD0iMSIgaGVpZ2h0PSIxIiB4bWxucz0iaHR0cDovL3d3dy53My5vcmcvMjAwMC9zdmciPjwvc3ZnPg==\" style=\"--smush-placeholder-width: 1283px; --smush-placeholder-aspect-ratio: 1283\/500;\" \/><\/p>\n<p>First, let&#8217;s concentrate on the two resistors.<\/p>\n<p><img decoding=\"async\" class=\"aligncenter size-full wp-image-1360 lazyload\" data-src=\"https:\/\/bryceautomation.com\/wp-content\/uploads\/2017\/11\/resistors.png\" alt=\"Resistors\" width=\"204\" height=\"410\" data-srcset=\"https:\/\/bryceautomation.com\/wp-content\/uploads\/2017\/11\/resistors.png 204w, https:\/\/bryceautomation.com\/wp-content\/uploads\/2017\/11\/resistors-149x300.png 149w\" data-sizes=\"(max-width: 204px) 100vw, 204px\" src=\"data:image\/svg+xml;base64,PHN2ZyB3aWR0aD0iMSIgaGVpZ2h0PSIxIiB4bWxucz0iaHR0cDovL3d3dy53My5vcmcvMjAwMC9zdmciPjwvc3ZnPg==\" style=\"--smush-placeholder-width: 204px; --smush-placeholder-aspect-ratio: 204\/410;\" \/><\/p>\n<p>Keep in mind that we need a MAXIMUM of 5 volts to measure at the processor.\u00a0 \u00a0At first, you might look at the circuit, and think this is easy to figure out.\u00a0 \u00a05 + 4 = 9.\u00a0 \u00a0We need a 5 volt drop to measure.\u00a0 Before jumping to any conclusion, let&#8217;s do the math to prove this.\u00a0 The reason you need to understand the math is because you won&#8217;t always be working with exact proportions of resistance.<\/p>\n<p>First, let&#8217;s calculate the current.\u00a0 \u00a0Provided that I=V\/R (current equals voltage over resistance), we can conclude that I=9\/9,000.\u00a0 \u00a0Therefore current = .001 Amps, or 1 mA.\u00a0 \u00a0 Secondly, let&#8217;s calculate the voltage drop of each resistor.\u00a0 For the Vd of the 4K resistor, V=IR.\u00a0 \u00a0Therefore, V=4,000*.001, or 4 volts.<\/p>\n<p>Secondly, there are two ways to calculate the voltage drop of the second resistance.\u00a0 \u00a0 We know that the sum of the voltage drops across the resistors much equal 9v.\u00a0 \u00a0Since we&#8217;ve already dropped 4 volts, we know the second resistor must drop the other 5 volts.\u00a0 \u00a0Let&#8217;s prove this with Ohms law as well.\u00a0 \u00a0 V=5,000 ohms * .001 mA.\u00a0 \u00a0 We see the voltage drop is 5 volts.<\/p>\n<p>Finally, let&#8217;s take a measurement to prove that we have a 5v drop across the second resistor.<\/p>\n<p><img decoding=\"async\" class=\"aligncenter size-full wp-image-1361 lazyload\" data-src=\"https:\/\/bryceautomation.com\/wp-content\/uploads\/2017\/11\/VoltageDrop.png\" alt=\"Voltage Drop\" width=\"697\" height=\"460\" data-srcset=\"https:\/\/bryceautomation.com\/wp-content\/uploads\/2017\/11\/VoltageDrop.png 697w, https:\/\/bryceautomation.com\/wp-content\/uploads\/2017\/11\/VoltageDrop-300x198.png 300w\" data-sizes=\"(max-width: 697px) 100vw, 697px\" src=\"data:image\/svg+xml;base64,PHN2ZyB3aWR0aD0iMSIgaGVpZ2h0PSIxIiB4bWxucz0iaHR0cDovL3d3dy53My5vcmcvMjAwMC9zdmciPjwvc3ZnPg==\" style=\"--smush-placeholder-width: 697px; --smush-placeholder-aspect-ratio: 697\/460;\" \/><\/p>\n<p>Lastly, it&#8217;s now safe to connect this point to the analog input of our processor.\u00a0 \u00a0We know that we are not exceeding 5 volts across this resistor (as long as the 9v battery does not exceed 9v).<\/p>\n<p><img decoding=\"async\" class=\"aligncenter size-full wp-image-1362 lazyload\" data-src=\"https:\/\/bryceautomation.com\/wp-content\/uploads\/2017\/11\/completecircuit.png\" alt=\"Complete Vd Circuit\" width=\"1174\" height=\"528\" data-srcset=\"https:\/\/bryceautomation.com\/wp-content\/uploads\/2017\/11\/completecircuit.png 1174w, https:\/\/bryceautomation.com\/wp-content\/uploads\/2017\/11\/completecircuit-300x135.png 300w, https:\/\/bryceautomation.com\/wp-content\/uploads\/2017\/11\/completecircuit-768x345.png 768w, https:\/\/bryceautomation.com\/wp-content\/uploads\/2017\/11\/completecircuit-1024x461.png 1024w\" data-sizes=\"(max-width: 1174px) 100vw, 1174px\" src=\"data:image\/svg+xml;base64,PHN2ZyB3aWR0aD0iMSIgaGVpZ2h0PSIxIiB4bWxucz0iaHR0cDovL3d3dy53My5vcmcvMjAwMC9zdmciPjwvc3ZnPg==\" style=\"--smush-placeholder-width: 1174px; --smush-placeholder-aspect-ratio: 1174\/528;\" \/><\/p>\n<h2>Potentiometers<\/h2>\n<p>Potentiometers are also a type of a voltage divider.\u00a0 Internally, you have a resistor.\u00a0 The wiper moves along that resistor to take the voltage at different points.\u00a0 For example, when the wiper is all the way against the left, it&#8217;s very close to ground potential.\u00a0 On the other hand, when the wiper is all the way against the right, it&#8217;s a full potential.\u00a0 \u00a0 If you turn the wiper to the center, there is basically the same amount of resistance on each side of the wiper.\u00a0 Therefore, the wiper is at half of the full voltage.\u00a0 \u00a0Here&#8217;s a diagram I put together on <a href=\"http:\/\/tinkercad.com\">tinkercad<\/a>.<\/p>\n<p><img decoding=\"async\" class=\"aligncenter size-full wp-image-8168 lazyload\" data-src=\"https:\/\/bryceautomation.com\/wp-content\/uploads\/2017\/11\/Screenshot-from-2021-08-26-12-20-09.png\" alt=\"Potentiometer\" width=\"400\" height=\"205\" data-srcset=\"https:\/\/bryceautomation.com\/wp-content\/uploads\/2017\/11\/Screenshot-from-2021-08-26-12-20-09.png 400w, https:\/\/bryceautomation.com\/wp-content\/uploads\/2017\/11\/Screenshot-from-2021-08-26-12-20-09-300x154.png 300w\" data-sizes=\"(max-width: 400px) 100vw, 400px\" src=\"data:image\/svg+xml;base64,PHN2ZyB3aWR0aD0iMSIgaGVpZ2h0PSIxIiB4bWxucz0iaHR0cDovL3d3dy53My5vcmcvMjAwMC9zdmciPjwvc3ZnPg==\" style=\"--smush-placeholder-width: 400px; --smush-placeholder-aspect-ratio: 400\/205;\" \/><\/p>\n<h2>Scaling for Voltage Dividers<\/h2>\n<p>At last, the circuit is finished.\u00a0 \u00a0Keep in mind that the analog input of the arduino has a raw value of 0-1023.\u00a0 \u00a0In other words, when we have the value of 1023 on A0, we know that we have 5 volts.\u00a0 \u00a0 Important to realize, the value of 1023 must be &#8220;Scaled&#8221;. back into volts.\u00a0 \u00a0Keep in mind that when we have 5 volts on A0, this means that we have a true 9 volts on the battery (because of the divider).\u00a0 \u00a0 Therefore the value of 1023 really means that our battery is putting out 9 volts.<\/p>\n<p>Therefore to get the voltage of the battery, we must take the value of A0 * 9 \/ 1023 to get the voltage of the battery.<\/p>\n<h2>Powering Loads<\/h2>\n<p>It&#8217;s important to realize that voltage dividers are good for high impedance inputs.\u00a0 This includes a simple discrete input, or TTL Communication.\u00a0 Additionally, you can use them for analog signals.\u00a0 Please do not use voltage dividers to power variable loads, such as the Arduino processor.\u00a0 As soon as you place a load on a voltage divider circuit, the resistance drops.\u00a0 This causes more current to flow through one of your resistors.\u00a0 Once this happens, the resistor will drop much more voltage.\u00a0 For loads such as the Arduino processor, it&#8217;s better to use a voltage regulator, such as the 7805, or a buck converter.<\/p>\n<p>For other important posts, visit the <a href=\"https:\/\/bryceautomation.com\/index.php\/category\/beginner\/\">beginner&#8217;s category page!<\/a><\/p>\n<p>&#8212; Ricky Bryce<\/p>\n<div id=\"bryce-4146474035\" class=\"bryce-after-content bryce-entity-placement\"><script async src=\"\/\/pagead2.googlesyndication.com\/pagead\/js\/adsbygoogle.js?client=ca-pub-8316758073402323\" crossorigin=\"anonymous\"><\/script><ins class=\"adsbygoogle\" style=\"display:block;\" data-ad-client=\"ca-pub-8316758073402323\" \ndata-ad-slot=\"4667596182\" \ndata-ad-format=\"auto\"><\/ins>\n<script> \n(adsbygoogle = window.adsbygoogle || []).push({}); \n<\/script>\n<\/div>","protected":false},"excerpt":{"rendered":"<p>Introduction to Voltage Dividers Voltage dividers are very important to circuits monitored by microprocessors.\u00a0 \u00a0 Important to realize, microprocessors will usually handle 3.3 volts, or 5 volts.\u00a0 \u00a0 With this in mind, the voltage of the circuits they control can be much higher. In this case, we want to measure the voltage of the 9v <a class=\"moretag btn btn-primary\" href=\"https:\/\/bryceautomation.com\/index.php\/2017\/11\/03\/voltage-dividers\/\">Read More \u00bb<\/a><\/p>\n","protected":false},"author":1,"featured_media":1361,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[5,8],"tags":[86],"class_list":{"0":"post-1357","1":"post","2":"type-post","3":"status-publish","4":"format-standard","5":"has-post-thumbnail","7":"category-arduino-other-microprocessors","8":"category-beginner","9":"tag-voltage-divider","10":"czr-hentry"},"yoast_head":"<!-- This site is optimized with the Yoast SEO plugin v27.3 - https:\/\/yoast.com\/product\/yoast-seo-wordpress\/ -->\n<title>Setting up voltage dividers circuit and scaling the values in the Arduino<\/title>\n<meta name=\"description\" content=\"Here we will set up voltage dividers in a r circuit to measure a battery. 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