In message <[email protected]>, dated Wed, 6 Jan 2010, Doug Smith <[email protected]> writes:
>Yes, I did mean it! But I think there is a misunderstanding. It happens, because when responding to a mailing list message, one hasn't necessarily seen all that has gone before. >If you connect it to a different "ground" then the difference between >the two grounds will show up as a common mode signal to the USB chip. Agreed. >A shielded cable is nearly a perfect transformer meaning that the >voltage impressed across the inductance of the shield is induced into >the center conductor (like a common mode choke). Well, Robert Macy and I have been studying this quite intensively recently, and it's true for sufficiently false (or very carefully defined) values of 'perfect', depending on what is connected to the cable in the real world. Because the shield and the inner conductors are not the same size, their inductances per unit length are different. This can only be true if the magnetic coupling coefficients are significantly less than 1, around 0.7 for typical cables. I hope that we will be producing an accessible paper on this later in the year. >The result (for an ideal cable) is that the signal that arrives at the >end of a shielded cable is just what entered it, in this case at the >mouse end (shield inductive drop cancels the induced voltage in the >center conductor). But, if the shield goes somewhere else than the >signal ground the difference between signal ground and that point will >be added to the output of the shielded cable. Indeed. If you connect the shield to the enclosure, and the signal ground to the enclosure, everything should be OK. What is wrong is to connect the shield to the signal ground on the PC board and the signal ground separately to the enclosure. The inductance of the latter connection lifts the board ground point from 'ground' and across that impedance appears the disturbance voltage that is on the shield. In an unbalanced connection, that voltage is in series with the signal and not remediable unless it's out-of-band; in a balanced connection it is a common-mode signal that challenges the CMRR of the input circuit. -- OOO - Own Opinions Only. Try www.jmwa.demon.co.uk and www.isce.org.uk John Woodgate, J M Woodgate and Associates, Rayleigh, Essex UK I should be disillusioned, but it's not worth the effort. - This message is from the IEEE Product Safety Engineering Society emc-pstc discussion list. To post a message to the list, send your e-mail to <[email protected]> All emc-pstc postings are archived and searchable on the web at: http://www.ieeecommunities.org/emc-pstc Graphics (in well-used formats), large files, etc. can be posted to that URL. Website: http://www.ieee-pses.org/ Instructions: http://listserv.ieee.org/request/user-guide.html List rules: http://www.ieee-pses.org/listrules.html For help, send mail to the list administrators: Scott Douglas <[email protected]> Mike Cantwell <[email protected]> For policy questions, send mail to: Jim Bacher: <[email protected]> David Heald: <[email protected]>
