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Introduction:

I am designing a board with a standard IEEE 802.3 1000Base-T ethernet interface. The board is a standard 4-layer FR4 board with all components on top and GND on Layer 2. The IEEE 802.3 standard basically requires 1500Vrms isolation from the MDI to the MII part of the interface. IEEE 802.3 references IEC60950-1/IEC62368-1 for test method and compliance criteria. The part of the interface that is relevant here is the magnetics and the connector. Chosing any ethernet transformer will provide the nescessary isolation between the MII and MDI sides. But typically a Bob-Smith termination is included in the MDI side (for common-mode noise supporesion). The capacitor must be a high-voltage capacitor, in order to maintain the 1500Vrms isolation.

My question pertains to the layout of the Bob-Smith termination components: Real estate is expensive on my board so I am forced to place the capacitor in parallel and quite close to one of the 75 Ohm termination resistors. So now I am faced with the question: How far do the capacitor (that has signal GND on one terminal, which is the MII interface) need to be from the neighbouring resistor (which is placed in the MDI part) in order to maintain the 1500Vrms isolation? What I need is basically a Voltage isolation to clearance/creepage distance calculator. Looking online I found http://creepage.com/. Typing in 1500Vrms under creepage gives me the following result:

Creepage distance

So the min. spacing should be 4.1 mm, right? Does anyone know if this is the correct approach to figuring out the required isolation distance?

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1 Answer 1

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I think most pros will use IPC-2221 as their reference for circuit boards. You can find the "A" version as a pdf on-line and here's an extract I sometimes use: -

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Here's the "A" version pdf link: https://www.ipc.org/TOC/IPC-2221A.pdf

If I were to plug-in 1500 volts AC I would choose a peak voltage that is \$\sqrt2\$ higher or 2121 volts. So, for an assembly (A5 column) with a conformal coating, the electrical conductor spacing is 2121-500 volts x 0.00305 mm plus 0.8 mm = 5.74 mm.

This is also the same distance as a bare PCB with bare conductors under a solder mask (polymer coating) i.e. column B4.

If you have tracks that are on internal layers (B1) then 2121 volts requires a distance of 0.25 mm plus 2121-500 volts x 0.0025 mm = 4.3 mm.

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  • \$\begingroup\$ I am aware of IPC-2221, but it seems that this standard is more conservative than the safety standards, i.e. the distances are greater. \$\endgroup\$
    – Lunde
    Mar 30, 2022 at 18:40
  • \$\begingroup\$ You don't have to use it. I do and I have no issues so, it's up to you. \$\endgroup\$
    – Andy aka
    Mar 30, 2022 at 19:01

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