Optional mechanical coding via front socket
Status LED (1-4) red: line fault (lead breakage or short circuit), yellow: state of digital I/O (0/1)Ĭonfiguration LED (AI, AO, DI, DO) white: selected channel mode Mechanical switch with additional resistors (see connection diagram)Ġ.1 % of the signal range at 20 ☌ (68 ☏)ĭiagnostic LED (I) red: module fault, red flashing: communication error, white: fixed parameter set (parameters from com unit are ignored), white flashing: requests parameters from com unit
Manufacturer-specific bus to standard com unitĬan be switched on/off for each channel via configuration tool, configurable via configuration toolįactory setting: > 21 mA Can be parameterized in the range 0 . 22 mAĬan be switched on/off for each channel via configuration tool Release Notes for Pepperl+Fuchs Thin Client Softwareġ2 V DC, only in connection with the power supplies LB9***.Service Level Agreements for ecom instruments.Cybersecurity Information and Reporting.When more than the number of allowed outputs backs up, the universal balancer behaves like a normal balancer, and may not balance properly.Get prepared for the challenges of Industry 4.0-with smart solutions from Pepperl+Fuchs. A throughput limited universal balancer may only have the capacity for a few unused outputs. If a universal balancer is throughput limited, the bottleneck may be in the loops or the balancer itself.
Universal balancers can be throughput limited. These balancers can balance evenly between any inputs and any outputs. Universal balancers solve this issue by having the back-looping built in. Sometimes this can be fixed by looping the unused output back around the balancer and distributing it among the inputs. In essence this means that an n-n balancer is not a functional n-(n-1) balancer. Many balancers fail to balance properly once an output backs up or if an output is not used. This formula is based on the number of nodes in a Beneš network, which is essentially the same as a throughput unlimited balancer - it allows any input to reach any output. For n → n balancers where n is a power of two numbers, n×log 2(n)−n÷2 can be used to calculate how many splitters are needed. This is the case because they use more splitters than the minimum required amount of splitters for a throughput unlimited balancer. The resulting balancer is usually larger than a balancer that was initially designed to be throughput unlimited.
#Input and output factory drawings full#
A guaranteed method to achieve throughput unlimited balancers is to place two balancers back to back that fulfil the first condition for throughput unlimited balancers (100% throughput under full load). However most balancers' bottlenecks can't be solved as easily. This is done by adding two more splitters at the end of the balancer, as it can be seen here: In this particular case, the bottleneck can be fixed by feeding the two middle output belts with more splitters. So, if only one side of that splitter gets input, as can be seen in the gif, it can only output one belt even though the side of the splitter is fed by a splitters which gets two full belts of input. The bottleneck in this balancer is that the two middle belts only get input from one splitter. The gif on the right shows a 4 → 4 balancer being fed by two belts, but only outputting one belt which means that its throughput in that arrangement is 50%.
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