Minimum distance between two connectors in a high voltage PCB design

Question

My application will have low current (mA) but high voltage lines. High voltage will arriving from external source using banana jacks on the case.

I am looking for an example in which someone uses PCB connectors (i.e., terminal blocks) that carry high voltage (e.g., 5KV). But I'm not having success.

I have not found any connector with more than 2.5kV dielectric strength. But I must put external 5kV signal at one connector and 0V to the other one, if it's possible in the same PCB.

I would like to know how can be calculated the distance between two (separated) connectors that have a 2.5KV breakdown voltage but they need to support a bigger voltage difference between connector 1 and connector 2: 5kV.

• Connector 1: VCC+ (5KV)
• Connector 2: VSS- (0V)

With BV = 2.5KV for me it's clear that bringing the two connectors so close to each other the material will break. Maybe I'm wrong.

I know the creepage and clearance techniques, but if I understood well it's referred to the sheet material and the copper over it, that has its own breakdown voltage/mil. But what about connector material? I thought that E field will be constant between two connectors, but voltage will be decreasing equipotential lines.

1. So, could I find a distance between both connectors in which the VCC+ connector and VSS- connector will be safe from breakdown?

2. Whereas trace and pad spacing literature suggests for certain kind of material to divide nKV/30cm to obtain safety distance between pads or traces with high voltage differences, what can be applied to the gap of connectors? How to get the minimum distance between two connectors with higher voltage level than its breakdown voltage? I need to be sure of minimum distance in which I could solder them in a board without damage.

Take into account that I'm not talking about connector (copper) pads to another printed pad/trace distance. I know how to calculate the minimum clearance and creepage.

I try to say that I will use 2 separated connectors, one for V+ and other one for V-, as far as they can from each other. But how to relate the manufacturer parameter limitation (1kV) with the distance needed for 5kV (or whatever) between them? That's the point of the question. If connector dielectric parameter is negligible, please let me now.

Answer 1

So, could I find a distance between both connectors in which the VCC+ connector and VSS- connector will be safe from breakdown?

Yes, but it's already been done for you in the IPC specs the table below shows this. It might be best to use two separate 1 conductor connectors, instead of finding one that has both (although the connector wouldn't be keyed for polarity). Here is a breakdown of how to calculate spacing between PCB traces:

Source: http://www.smpspowersupply.com/ipc2221pcbclearance.html

Answer 2

If you understand creepage is the ionic breakdown of a good insulator due to the accumulation of dust and humidity, then you will understand why an air gap is good. You may need 2 connectors.

Clean air is about 5kV/mm between smooth parallel surfaces. Plastic can be >10kV/mm.

The highest level of indoor contamination of residential humid dust on the best dielectric reduces to about 0.3kV/mm so 0.2kV/mm became the standard.

There exist various levels of dust& RH vs withstanding voltage in between. Thin Polymer coating such as soldermask only improves a bit. (50%) Thick protection improves a lot, such as conformal coating or potting.