This WordPress blog is about Australian information and communication technology (ICT) product compliance and certification requirements. It’s owned and run by a professional Compliance Engineering practitioner who participates strongly in several related Australian standards committees (representing industry) and some international (IEC) committees as well, representing Australia. As well he participates on a number of consultative government and regulatory fora in Australia.

Any information posted here is intended only to be of a non-confidential type, as it’s the blogger’s intention not to post anything of a commercially sensitive or confidential nature here (for what should be obvious reasons). Much of the information presented here has come from other sources, which are generally authoritative. I’ll be happy to answer questions as best I can, if I can, if I have the time and inclination! but only on a non-confidential and non-commercially sensitive basis.

Disclaimer:

The postings on this site are my own and don’t necessarily represent IBM’s positions, strategies or opinions. While it is intended information on this site is correct, it’s coming from the perspective of the poster or other people, and won’t necessarily have catered for all other perspectives or all the “if’s and buts” for your own circumstances, so you’ll need to make your own enquiries with your own counsel before you make any business decisions. No responsibility will be accepted for any errors or omissions on this site for whatever reason.

Paul

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Hi Paul, I am designing a product which is going to get certified in accordance with 60950-1. I have a question regarding compounded clearance and creepage across bridging resistors. According to clause 1.5.7.2, we can use a group of resistors to comply with the minimum clearance and creepage. It also directs us to Figure F.13, which I interpreted as compounded clearance/creepage = n x clearance/creepage of single resistor, if the clearance/creepage of a single resistor is larger than ‘X’. (‘X’ determined by Table F.1). Is my understanding correct?

Cheers,

Kay

Hi Kaushayla, Sounds like you’d really need a test lab or consultant to help you decide. The value for X in Figure F.13 and Table F.1 is a minimum value only, but you’d need to evaluate the distances in 2.10.3 (or Annex G) for clearances and 2.10.4 for creepage distances to work out if you’d need larger values for double or reinforced insulation, since that’s what these resistors are bridging, taking into account the working voltage and the pollution degree (usually PD 2), plus the transient voltage (which depends on the overvoltage category). Also, the sum of distances depends on how many resistors in series there are and whether they’re in a straight line or not. If there are only two resistors in the group then the applicable value in Figure F.13 is the smallest distance of either one of the two resistors, since each resistor has to be shorted out one at a time for the evaluation. If there are more than two resistors, then the distance is the smallest sum of distances of n-1 resistors with any single resistor being short circuited. In short, two resistors will be measured as the smallest distance with either other resistor shorted. three resistors in series will be the sum of the smallest distance of any two resistors (taking into account the width of solder pads), and so on. and this also assumes there is no intervening non-connected conductive part in those distances.

Hi Paul, Here I am again 😀

Hope you and your loved ones are keeping safe.

I am going through 60950-1 Clause 5.3.7 and have a question. In 5.3.7 e) it says “Other single faults specified in 1.4.14”. The 1.4.14 is expecting for something like a design FMEA? It is not practical to simulate these faults in a real PCBA. What is your opinion?

Thanks,

Kaushalya