{"id":851,"date":"2015-05-21T16:00:15","date_gmt":"2015-05-21T21:00:15","guid":{"rendered":"http:\/\/www.kele.com\/keleblog\/?p=851"},"modified":"2022-03-15T15:08:15","modified_gmt":"2022-03-15T20:08:15","slug":"currently-playing-connecting-a-4-20-ma-constant-current-signal-to-multiple-loads","status":"publish","type":"post","link":"https:\/\/www.kele.com\/content\/blog\/2015\/05\/21\/currently-playing-connecting-a-4-20-ma-constant-current-signal-to-multiple-loads","title":{"rendered":"\u201cCurrently\u201d Playing: Connecting A 4-20 mA Constant-Current Signal To Multiple Loads"},"content":{"rendered":"<p>One of the most popular and long-lived methods for transmitting analog control signals in the HVAC and industrial control worlds is the \u201cconstant-current signal loop.\u201d In this scheme, the value of the current flowing in the circuit is the control variable, rather than any voltages that may appear at different points in the circuit. The constant-current loop is very robust and noise-tolerant and can run for long distances. Even in this high-tech digital age, it remains a very popular analog signaling method.<\/p>\n<p>This article will deal with how to connect multiple loads to a constant-current signal loop. Connecting multiple loads to a constant-current signal loop is totally different than connecting multiple loads to a voltage signal source (which is covered in a previously-published article for those interested).<\/p>\n<p>In the HVAC world, the most common range of current used for constant-current signaling is 4-20 mA (1 mA = 1 milliampere = 0.001 amperes of current). 4 mA represents 0% of the variable\u2019s value and 20 mA represents 100% of the variable\u2019s value. Other current ranges can be used, but today we\u2019ll work exclusively with 4-20 mA current loops.<\/p>\n<p><strong>Current loop load \u201cresistance\u201d or \u201cimpedance\u201d<\/strong><\/p>\n<p>Every load to be connected in the current loop will have an electrical characteristic called \u201cresistance\u201d or \u201cimpedance.\u201d The load\u2019s data sheet might use either term. Don\u2019t worry about whether the data sheet says \u201cresistance\u201d or \u201cimpedance\u201d we\u2019ll take them to be the same thing. In this article we\u2019ll use the term \u201cimpedance.\u201d<\/p>\n<p>Impedance is measured in units called \u201cohms.\u201d If you don\u2019t know what ohms are, don\u2019t worry. All you have to do is use the numbers in some simple calculations later on.<\/p>\n<p><strong>Basic constant current loop configuration<\/strong><\/p>\n<p>Before we discuss connecting multiple loads to a constant current loop, let\u2019s get familiar with the basic constant current loop containing a single load:<\/p>\n<p><a href=\"http:\/\/www.kele.com\/content\/wp-content\/uploads\/2015\/05\/currently-figure-1.png\"><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter size-medium wp-image-857\" src=\"http:\/\/www.kele.com\/content\/wp-content\/uploads\/2015\/05\/currently-figure-1-300x163.png\" alt=\"currently figure 1\" width=\"300\" height=\"163\" \/><\/a><\/p>\n<p>On the left we have a constant-current source that is regulating the current to some desired value (the desired value is determined by a calculation in a controller, measured physical value in a sensor, etc.). The constant current leaves the source, flows through the load, and returns to the source.<\/p>\n<p>Simple, right? What could go wrong? As usual, \u201cthe devil is in the details.\u201d<\/p>\n<p><strong>Two different styles of current sources<\/strong><\/p>\n<p>There are two different styles of current sources you should be aware of. We\u2019ll call the first style the \u201clocally powered\u201d current source. It has its own source of power for the 4-20 mA output and the only other item needed to complete the loop is the load:<\/p>\n<p><a href=\"http:\/\/www.kele.com\/content\/wp-content\/uploads\/2015\/05\/currently-figure-2.png\"><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter size-medium wp-image-858\" src=\"http:\/\/www.kele.com\/content\/wp-content\/uploads\/2015\/05\/currently-figure-2-300x146.png\" alt=\"currently figure 2\" width=\"300\" height=\"146\" \/><\/a><\/p>\n<p>You will usually find this type of current source on the output of a powered controller.<\/p>\n<p>We\u2019ll call the second style the \u201cloop powered\u201d or \u201c2-wire\u201d current source. It does not have its own power source built-in, but rather depends on a separate power source (usually 24VDC) wired into the loop:<\/p>\n<p><a href=\"http:\/\/www.kele.com\/content\/wp-content\/uploads\/2015\/05\/currently-figure-3.png\"><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter size-medium wp-image-859\" src=\"http:\/\/www.kele.com\/content\/wp-content\/uploads\/2015\/05\/currently-figure-3-300x171.png\" alt=\"currently figure 3\" width=\"300\" height=\"171\" \/><\/a><\/p>\n<p>You will find this type of current source in many 4-20 mA loop-powered <a href=\"\/temperature-sensors-and-transmitters-home.aspx\">temperature<\/a>, <a href=\"\/pressure-home.aspx\">pressure<\/a>, and <a href=\"\/humidity-home.aspx\">humidity<\/a> transmitters.<\/p>\n<p><strong>Current sources have a \u201cmaximum load impedance\u201d they can drive<\/strong><\/p>\n<p>Current sources like <strong>small <\/strong>values of load impedance, in fact they can drive their current straight into a short circuit (zero ohms load impedance)!<\/p>\n<p>Current sources run into trouble when the load impedance gets too <strong>large<\/strong>. When the load impedance gets too large, the current source will not be able to make the higher values of current (near the 20 mA end) flow. The current value will plateau at some value less than the desired value.<\/p>\n<p><strong>How do I know the \u201cmaximum load impedance\u201d my current source will support?<\/strong><\/p>\n<p>For a locally-powered current source, you will need to find the maximum load impedance value on the product\u2019s data sheet.<\/p>\n<p>For a loop-powered current source, you may find a parameter called \u201cmaximum load impedance (with 24V supply)\u201d on the data sheet. If you do not, you will have to perform the following steps:<\/p>\n<p>1. Find a parameter called \u201cminimum operating voltage\u201d (or some very similar wording) on the current source data sheet.<\/p>\n<p>2. Calculate maximum load impedance as follows:<\/p>\n<p style=\"padding-left: 30px;\">Maximum load impedance = (Loop power supply \u2013 minimum operating voltage) \/ 0.020<\/p>\n<p style=\"padding-left: 30px;\">For example, suppose we are using a 24V loop power supply and we see on the data sheet that the 2-wire current source needs a minimum of 11V to operate. Then we would calculate:<\/p>\n<p style=\"padding-left: 30px;\">Maximum load impedance = (24V \u2013 11V) \/ 0.020 = 650 ohms<\/p>\n<p>Now that we know something about how a single-load constant-current loop works, let\u2019s see what kind of complications we get into when trying to drive multiple loads!<\/p>\n<p><strong>Connecting multiple loads to the current loop<\/strong><\/p>\n<p><strong>Rule #1: All loads must be connected in series around the loop, never in parallel.<\/strong><\/p>\n<p>Below are sketches of the right way (series) and the wrong way (parallel) to connect loads in a current loop:<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter size-medium wp-image-860\" src=\"http:\/\/www.kele.com\/content\/wp-content\/uploads\/2015\/05\/currently-figure-4-300x141.png\" alt=\"currently figure 4\" width=\"300\" height=\"141\" \/><\/p>\n<p><a href=\"http:\/\/www.kele.com\/content\/wp-content\/uploads\/2015\/05\/currently-figure-5.png\"><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter size-medium wp-image-861\" src=\"http:\/\/www.kele.com\/content\/wp-content\/uploads\/2015\/05\/currently-figure-5-300x150.png\" alt=\"currently figure 5\" width=\"300\" height=\"150\" \/><\/a><\/p>\n<p>At first glance this requirement seems simple enough. That\u2019s because we\u2019ve drawn all the loads as 2-wire loads that don\u2019t require power supplies. A not-so-obvious problem may occur if the loads do require power supplies. The current will always take the \u201cpath of least resistance\u201d back to the current source whatever that path might be. If one of the loads before the last load has a power connection back to the current source, this connection will be the \u201cpath of least resistance\u201d and the 4-20 mA will bypass the rest of the loads in the loop:<\/p>\n<p><a href=\"http:\/\/www.kele.com\/content\/wp-content\/uploads\/2015\/05\/currently-figure-6.png\"><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter size-medium wp-image-862\" src=\"http:\/\/www.kele.com\/content\/wp-content\/uploads\/2015\/05\/currently-figure-6-300x183.png\" alt=\"currently figure 6\" width=\"300\" height=\"183\" \/><\/a><\/p>\n<p>If the powered load is a standalone module, you may be able to use a dedicated \u201cfloating\u201d transformer or small DC power supply to power the load and resolve the issue:<\/p>\n<p><a href=\"http:\/\/www.kele.com\/content\/wp-content\/uploads\/2015\/05\/currently-figure-7.png\"><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter size-medium wp-image-863\" src=\"http:\/\/www.kele.com\/content\/wp-content\/uploads\/2015\/05\/currently-figure-7-300x232.png\" alt=\"currently figure 7\" width=\"300\" height=\"232\" \/><\/a><\/p>\n<p>&nbsp;<\/p>\n<p>If multiple loads in the series connection are powered, each load would need its own power supply that is floating with respect to all the other supplies. For example, if Load 1 and Load 2 were both powered, Load 1 would need a power supply that is floating with respect to Load 2 and the current source, and Load 2 would need a power supply that is floating with respect to Load 1 and the current source. It can get messy quickly if there are several powered loads in the loop!<\/p>\n<p>Consider also that If the powered load is a controller input and the controller is serving other inputs and outputs, it wouldn\u2019t be practical to \u201cfloat\u201d the entire controller subsystem\u2019s power connections.<\/p>\n<p>In cases where a floating power supply for each load is not practical, it\u2019s time to employ a signal <a href=\"\/transducers\/types\/dc-isolator.aspx\">isolator<\/a> module such as the <a href=\"\/transducers\/dt13e.aspx\">DT-13E<\/a> from Kele:<\/p>\n<p><a href=\"http:\/\/www.kele.com\/content\/wp-content\/uploads\/2015\/05\/currently-figure-8.png\"><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter size-medium wp-image-864\" src=\"http:\/\/www.kele.com\/content\/wp-content\/uploads\/2015\/05\/currently-figure-8-300x244.png\" alt=\"currently figure 8\" width=\"300\" height=\"244\" \/><\/a><\/p>\n<p>&nbsp;<\/p>\n<p>The DT-13E signal isolator has its inputs isolated from its outputs, and both inputs and outputs are isolated from the power terminals. So with the DT-13E installed, it just doesn\u2019t matter where Load 2\u2019s power source comes from, it will not corrupt the 4-20 mA current in the loop.<\/p>\n<p><strong>Rule #2: The sum of all the load impedances must be less than the maximum load\u00a0<\/strong><strong>Impedance the current source can support.<\/strong><\/p>\n<p>When the mA loads are wired in series (as they must be) the total load impedance seen by the current source is the sum of the individual load impedances. So if each of the three loads in the drawing below is 250 ohms, the total load impedance seen by the current source would be 250 + 250 + 250 = 750 ohms:<\/p>\n<p><a href=\"http:\/\/www.kele.com\/content\/wp-content\/uploads\/2015\/05\/currently-figure-9.png\"><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter size-medium wp-image-865\" src=\"http:\/\/www.kele.com\/content\/wp-content\/uploads\/2015\/05\/currently-figure-9-300x144.png\" alt=\"currently figure 9\" width=\"300\" height=\"144\" \/><\/a><\/p>\n<p>&nbsp;<\/p>\n<p>Suppose your current source data sheet shows that it can only drive 650 ohms maximum impedance? Houston, we have a problem\u2026<\/p>\n<p>What can you do if your total load impedance adds up to more ohms than the current source can handle? This is another way that the DT-13E signal isolator can sometimes help. Many HVAC 4-20 mA loads have impedances of 250 to 500 ohms, but the DT-13E mA input impedance is only 125 ohms. So by isolating one or more loads using DT-13Es, not only do you remove grounding concerns but you may also lower your total loop impedance to a value the current source can handle:<\/p>\n<p><a href=\"http:\/\/www.kele.com\/content\/wp-content\/uploads\/2015\/05\/currently-figure-10.png\"><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter size-medium wp-image-866\" src=\"http:\/\/www.kele.com\/content\/wp-content\/uploads\/2015\/05\/currently-figure-10-300x203.png\" alt=\"currently figure 10\" width=\"300\" height=\"203\" \/><\/a><\/p>\n<p>&nbsp;<\/p>\n<p>Now the current source only sees 250+125+250 = 625 ohms which is lower than its maximum load impedance, and Rule #2 is satisfied.<\/p>\n<p><strong>Conclusions<\/strong><\/p>\n<p>Multiple loads inserted in a 4-20 mA current loop must be connected in series. If the series-connected loads are powered devices, \u201csneak paths\u201d between the load power supply connections and the current source may misdirect the 4-20 mA so that not all the loads receive the signal. Isolating the series-connected load with a <a href=\"\/transducers\/dt13e.aspx\">DT-13E<\/a> or similar signal isolator will solve this problem.<\/p>\n<p>The sum of the series-connected load impedances may be too high for the current source to handle. The DT-13E signal isolator\u2019s impedance of 125 ohms is less than that of many common HVAC loads. Isolating a load with the DT-13E may lower the total loop impedance to an acceptable value.<\/p>\n","protected":false},"excerpt":{"rendered":"<p>One of the most popular and long-lived methods for transmitting analog control signals in the HVAC and industrial control worlds is the \u201cconstant-current signal loop.\u201d In this scheme, the value of the current flowing in the circuit is the control variable, rather than any voltages that may appear at different points in the circuit. The [&hellip;]<\/p>\n","protected":false},"author":3,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"_acf_changed":false,"_et_pb_use_builder":"","_et_pb_old_content":"","_et_gb_content_width":"","footnotes":""},"categories":[2,7,12,13,14],"tags":[44,49,67,82,89,100,104,112,115,133],"class_list":["post-851","post","type-post","status-publish","format-standard","hentry","category-back-to-basics","category-kele-makes-it-easy","category-technical-reference","category-technically-speaking","category-transducers","tag-current-signal","tag-dt-13e","tag-impedance","tag-loop-powered","tag-multiple-loads","tag-parallel","tag-power-supplies","tag-resistance","tag-series","tag-transducers-2"],"acf":[],"yoast_head":"<!-- This site is optimized with the Yoast SEO plugin v27.4 - https:\/\/yoast.com\/product\/yoast-seo-wordpress\/ -->\r\n<title>\u201cCurrently\u201d Playing: Connecting A 4-20 mA Constant-Current Signal To Multiple Loads - kele.com<\/title>\r\n<meta name=\"robots\" content=\"index, follow, max-snippet:-1, max-image-preview:large, max-video-preview:-1\" \/>\r\n<link rel=\"canonical\" href=\"https:\/\/www.kele.com\/content\/blog\/2015\/05\/21\/currently-playing-connecting-a-4-20-ma-constant-current-signal-to-multiple-loads\" \/>\r\n<meta property=\"og:locale\" content=\"en_US\" \/>\r\n<meta property=\"og:type\" content=\"article\" \/>\r\n<meta property=\"og:title\" content=\"\u201cCurrently\u201d Playing: Connecting A 4-20 mA Constant-Current Signal To Multiple Loads - kele.com\" \/>\r\n<meta property=\"og:description\" content=\"One of the most popular and long-lived methods for transmitting analog control signals in the HVAC and industrial control worlds is the \u201cconstant-current signal loop.\u201d In this scheme, the value of the current flowing in the circuit is the control variable, rather than any voltages that may appear at different points in the circuit. 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