There are many pitfalls in designing overcurrent protection into a system, but why is it so confusing? Where does a design engineer learn that oversizing is necessary?
Maybe we can answer some of these questions by example.
Let’s assume the role of a newly hired, fresh-from-university electrical engineer. Our mentor wants us to determine the overcurrent protection for a control transformer in an electrical control panel for an industrial machine. Our machine has some signal lights, horns, rotating beacons, and a courtesy outlet that all need 120V AC. The machine will be supplied with 480V AC, so we’ll need a control transformer to make 120V AC. For our discussion, let’s say we need 10A of 120V AC.
If you take a quick look at this guide, 10A times 120V gives us 1,200VA. It gives us choices of 1,000VA or 1,500VA. 1,000VA is too small, so 1,500VA it is (which gives us some margin of error).
We’re building an industrial control panel and our mentor gave us a rule book to follow because municipal codes weren’t taught in university. The book we’re following is UL-508A, Underwriters Laboratories Inc. Standard for Safety Industrial Control Panels. To keep this example simple, we’re using the primary-only protection rules on page 84 of UL-508A, dated February, 2010. What current is the primary side of the transformer going to see? 1,500VA divided by 480V gives 3.125A. There’s a fuse-sizing chart at the bottom of the catalogue page for primary-only protection. The chart says 5A is the maximum. Why is there a difference?
Asking the mentor gets us a gruff “go find out about inrush.” Our mentor also reminds us there is a deadline to the project.
Inrush? Oh, at the top of the catalog page. Inrush. 3-10 TIMES steady-state current. That explains it. OK, so we need a 30A fuse. Then our eyes catch sight of the UL-508A book again. Better check that. Page 84, Primary-only protection. Our transformer draws a little over 3A, which from the chart shown means the maximum overcurrent protection can be 167 percent of the primary current. 3.125 times 1.67 gives 5.22A. How is that going to protect from inrush?
We’re a studious engineer; we’ve been reading Design News since being hired because we saw copies of it on the desks of all the senior engineers. We remember several articles about Littelfuse and their willingness to help with selection. A couple of telephone calls later, and we now have a 5A KLDR fuse from Littelfuse to protect our transformer.