Dear Ken Sorry to have taken so long to reply to this. Pressure of work has kept me away from the thread for a week.
I don't know where you get your maths from. The usual formula (commonly available in a number of variants) for the radiated emissions E in V/m at 10 metres due to a common-mode current is: E = 1.26 10^-4 (f.L.ICM) f = frequency in MHz L = length of the radiating cable in metres ICM = common-mode current in the cable in milliamps Using L = 1.5 metres (not a long cable) and f = 100MHz, I find that 92dBuV/m (= 40mV/m) can be created at 10 metres distance by a CM current of just 2mA. 2mA is a far cry from your "significant fraction of an Ampere of common mode rf current". Hence my serious concerns about your math. I also don't agree with a number of other things in your analysis below, especially as you finish by saying: "If you consider that any signal with information content carried by 2 mV is shielded, the issue becomes, once again, a non-problem." I don't think you can make the assumption that cables carrying low level signals are shielded. Remember that this thread began with a discussion of EMC-related safety issues, and where safety is involved one shouldn't make assumptions that everyone else designs equipment as well as you would like them to. If we consider a country such as the US with no mandatory EMC immunity regulations, and a measuring device that uses analogue technology and does not have to meet emissions standards, it is obvious that the lowest cost way to design and market it is to leave out all the shielding and filtering, and I would expect a proportion of manufacturers to do just that. I will spare the emc-pstc a longer email by not responding to the other issues I don't agree with at this time. Maybe other contributors would like to support your analysis below? or not. Regards, Keith Armstrong. PS: It will be another week before I can reply again to postings in this thread. In a message dated 07/01/02 02:46:46 GMT Standard Time, ken.ja...@emccompliance.com writes: > Subj:Re: EMC-related safety issues > Date:07/01/02 02:46:46 GMT Standard Time > From: ken.ja...@emccompliance.com (Ken Javor) > Sender: owner-emc-p...@majordomo.ieee.org > Reply-to: <A > HREF="mailto:ken.ja...@emccompliance.com";>ken.ja...@emccompliance.com</A> > (Ken Javor) > To: cherryclo...@aol.com > CC: emc-p...@majordomo.ieee.org > > > QUOTE: "And I don't think that 92dBuV/m is a high field strength to be > emitted by a PC placed nearby, or for a non-compliant laptop at 10 metres." > > You may not think so, but I am sorry, the numbers just don't add up. 92 > dBuV/m at 10 meters implies an effective radiated power of 5.3 mW. > Consider that the source is not an intentional antenna. It will have no > more directivity than a dipole and its efficiency will be much less since > it isn't matched to the source. If we simply assume no gain (meaning > matching losses just offset directivity) , that means 5.3 mW of rf power > are emitted from the EUT or its attached cables. If one makes the > reasonable assumption that it is common mode rf current which is radiating, > then the potential associated with rf power will be a small number of > millivolts (in the frequency domain). This in turn implies a significant > fraction of an Ampere of common mode rf current. A highly unlikely > situation! Once again, with an impossible conclusion, either the > assumption or the logic must be wrong. You can choose to disbelieve, but > please point out where the logic has gone awry. You have several times > cited Mr. Woodgate for non-constructive criticism. Now I am asking you, > don't give more hearsay: explain where my physics is incorrect. We are > engineers here, not pollsters. > > And if you are saying that specification level compliance at 10 meters can > scale up to 92 dBuV/m nearby, that is either false or misleading depending > on the frequency range. At the low end, say 30 MHz, the area subtended by > position near the offending PC isn't large enough to efficiently radiate or > couple the field (the wavelength is 10 meters, and the other gentleman's > antenna factor calculation assumed a tuned dipole antenna in order to get a > small antenna factor). So the field will not scale up as per your > prediction, and the pickup mechanism will be nowhere near the antenna > factor that gentleman calculated. In fact at 30 MHz your antenna factor > will be on the order of 20 dB or worse (assuming the mutual coupling length > to be 1 m). At the high end (near 1 GHz) you could be in the far field in > close and the field could scale up to a value of 92 dBuV/m, but the antenna > factor of a matched tuned dipole at 1 GHz is 26 dB so the potential from > that perfect antenna is 92 dBuV/m - 26 dB/m = 66 dBuV or 2 mV. If you > consider that any signal with information content carried by 2 mV is > shielded, the issue becomes, once again, a non-problem. > > > on 1/6/02 10:43 AM, cherryclo...@aol.com at cherryclo...@aol.com wrote: > > Snip: And I don't think that 92dBuV/m is a high field strength to be > emitted by a PC placed nearby, or for a non-compliant laptop at 10 metres. >
