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Power Over Ethernet Interoperability Guide by Sanjaya Maniktala

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What Happens during the Surge Test
In Fig. 11.4, we monitor the voltage and current associated with the surge. Let us do some simple math around this. If we apply 2 kV across a dead short (say the almost-uncharged caps at the input or output of the PSE), we will get an instantaneous current of I = V/R = 2000/42 ≈ 47 A. Well, this is the simplified upper limit. In reality, we can get much less. There are two related reasons for that. First, the voltage does not reach 2000 V immediately. It takes a few microseconds to get there. And things can change quite dramatically during that brief instant. Which brings us to the second reason: All the surge current must pass through the Y-caps (in series, thus acting as a bottleneck). So if we reduce the ...

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