# Ethernet - 3000V Isolation

I need some means to increase the Hi-pot rating of an Ethernet port to 3000V to meet regulatory approval. The Ethernet controller module is housed in a large metal enclosure and an 18" patch cord connects the controller to a back panel RJ45 connector. The controller module uses a standard mag jack which has a hi-pot rating of 1500V. Redesigning the controller card is not an option at this point. I was planning on making a small PCB at the back panel with two RJ45 connectors and a LAN transformer in between. This would provide an additional 1500V to 4000V of isolation depending on the transformer chosen. Does anyone see any potential (no pun intended) problems with this approach? Is any biasing required and should the center taps be left floating?

• This sounds like a really, really bad idea. Any reason why you don't simply slap in a media converter and some fiber optic cable? You can get 300kV isolation with that. – Spehro Pefhany Jul 16 '15 at 18:24
• I know there are off the shelf solutions but it is a question of cost. I have seen what looks like a passive transformer based coupler with 5K isolation targeted at medical applications, but it costs over $100. I was hoping for something under$10. – Sean Cahalan Jul 17 '15 at 13:53
• All your measures to save some dollars are worthless, when you fail to increase the isolation rating by missing any nifty detail like creeping distances or corona discharges. If you design your isolation yourself you will have to undergo a certification process to prove your isolation ratings, which will cost some thousands of dollar easily. – Ariser Nov 30 '15 at 13:40

It may be interesting for you:

• Isolation ratings are not added but the minimum is selected instead. For example, if your have 10 kV, 20 kV, and 30 kV rated insulation black boxes connected in series, the resulting insulation level will be min{10,20,30}=10kV instead of 10+20+30=60kV as you presumed.

• Pulse Eng. # HX1224NL Ethernet transformer (for example again) has 4kVrms @ 50/60Hz A.C. rated insulation level. You need to clearly distinct between A.C. and D.C. insulation. Ethernet specifies the A.C. one.

• To protect your scheme properly, you need to ground the center taps of the additional insulation transformer at the side connected to the equipment being protected. It is to eliminate the weak (1.5kVrms) insulation level of the controller module from the result.

P.S. All listed above is on so called basic insulation (provided during 60 secs typ after shock, used for safety) not other(s).

• Do you have any reference for the claim, that isolation ratings cannot be added? Furthermore, grounding of circuitry can break the appliance class of a circuitry as defined in IEC 61140. – Ariser Nov 30 '15 at 13:36
• @Ariser Well... I cannot give you a reference at a glance, instead i can give you some explanation: (rated) isolation level is about the voltage that could be applied between two isolated points of the device during the specified time, causing no damage. It does not specify the equivalent resistance and/or capacitance and theirs tolerances. Therefore when you connect two devices with only the level specified, in series and apply a voltage over them you cannot assume the potential partitioning will be correct --- R/C of each box in the series is not known (specified/controllable). – asndre Dec 10 '15 at 18:44
• In such a case when only ILs are given and no series params are are controllable/specified/toleranced(limited), you can aplly only the minimum voltage to the series assuming guaranteed to not break the weakest element. – asndre Dec 10 '15 at 18:48
• Considering Ethernet, IEC 61140 is not applicable, imo. I think an analysis related to structured cabling systems is one to use. – asndre Dec 10 '15 at 18:51
• This sounds logic. I thought I could deduct the rules from those for creeping distances, where adding is permitted. But here are far more parameters to take care of. – Ariser Dec 10 '15 at 18:54

Isolation voltages in series result in the HIGHEST voltage. The assumption is that when all weaker isolation fails, only the strongest barrier remains, and it must withstand the entire voltage.