I'm planning the way of connecting 8 devices terminal wires all together within a 6kV point (V+), there will be 8 more wires connected to GND (V-). Those V+ and V- points will be in the metal case. The key is how to join 8 wires in an only point inside the case.

I thought bringing V+ and V- (that comes from external source to the case by 2 jacks) until a PCB with only wire-PCB connectors. No more proposal for this PCB.

I have not any experience at HV design, but I' ve learnt some tips and tricks. I have read all the literature that I can.

Searching components, I have found there are wire-pcb connectors that supports until 2500V (wurth). If I understand correctly, this breakdown voltage is refered to the voltage difference that could be supported between pins or closer connectors before isolation falls.

My suggestions:

  • So if all the connector pins has 6KV --> Vpin1 - Vpin2 = 0V < 2500V will be ok.
  • So if some connector pins are 6KV and other one is 0V --> Vpin1 - Vpin2 = 6KV > 2500V will be a problem and probably it will fail.
  • But, I know that E = V/distance and this has much to see with dielectric strength.

Taking the suggestions into account, guessing they were true, I will launch my two question according to the dielectric subject:

1.- Could I separate them -GND connector from 6KV one- some distance between them in order that voltage doesn' t break connectors isolation?

2.- If answer is "yes"; how to calculate mathematically the air gap/distance between two connectors that will support 6KV knowing that they can support 2.5KV?

enter image description here


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.