Dear Charles, I'm going to address your question first with an excerpt from one of my previous articles, "Cable Shield Grounding". In high-speed digital applications, a low impedance connection between the shield and the equipment chassis *at both ends* is required in order for the shield to do its job. The shield connection impedance must be low in the frequency range over which you propose for the shield to operate. The measure of shield connection efficacy for a high-speed connector is called the ground transfer impedance, or shield transfer impedance, of the connector, and it is a crucial parameter. In low-speed applications involving high-impedance circuitry, where most of the near-field energy surrounding the conductors is in the electric field mode (as opposed to the magnetic field mode), shields need only be grounded at one end. In this case the shield acts as a Faraday cage surrounding the conductors, prevent the egress (or ingress) of electric fields. In high-speed applications involving low-impedance circuitry, most of the near-field energy surrounding the conductors is in the magnetic field mode, and for that problem, only a magnetic shield will work. That's what the double-grounded shield provides. Grounding both ends of the shield permits high-frequency currents to circulate in the shield, which will counteract the currents flowing in the signal conductors. These counteracting currents create magnetic fields that cancel the magnetic fields emanating from the signal conductors, providing a magnetic shielding effect. For the magnetic shield to operate properly, we must provide means for current to enter (or exit) at both ends of the cable. As a result, a low-impedance connection to the chassis, operative over the frequency range of our digital signals, is required that *both* ends of our shielded cable. (See Henry Ott, "Noise Reduction Techniques in Electronic Systems", 2nd ed., John Wiley & Sons, 1988.) There are exceptions to the "ground coax at both ends rule". These exceptions involve cases where for some reason no currents enter or leave the end-terminal equipment. For example, suppose that at the receiving equipment the coaxial cable feeds the primary winding of an ideal transformer. Let the secondary winding of the transformer feed the receiver circuitry. There is thus no DC connection (no common-mode current connection) through the transformer. Of course, in actual practice you do get current flowing through the parasitic capacitance of the transformer from primary to secondary, and this current then must make its way back to the source, which in your case is one of those big grounding straps between the cabinets. In a high-speed digital product you would therefore likely not pass your radiated emissions tests with that architecture. Another approach useful at lower speeds is to terminate the coaxial cable to itself with 50 ohms (or whatever is appropriate) and then feed only the signal wire into your receiver. A suitably high-impedance receiver input draws very little current, thereby limiting the current that must flow back to the source on the big grounding straps. Once again, at high speeds the current required to drive the parasitic capacitance of the receiver is not insubstantial, and so this architecture won't work. Your old application probably used some combination of either transformer coupling, a high-impedance receiver, or both. (Common-mode chokes can also work). One approach that works well at speeds up to 125 Mbaud on category-5 unshielded twisted pairs is to use a well balanced transformer (sometimes in series with a common-mode choke) to limit the common-mode current flowing into the receiver (or out of the transmitter). Such architectures work and pass FCC/EN radiated emissions tests at speeds up to 125 Mbaud with output signals on the order of 1 V p-p. Best regards, Dr. Howard Johnson, Signal Consulting Inc., tel +1 509-997-0505, howiej@xxxxxxxxxx http:\\sigcon.com -- High-Speed Digital Design articles, books, tools, and seminars -----Original Message----- From: si-list-bounce@xxxxxxxxxxxxx [mailto:si-list-bounce@xxxxxxxxxxxxx]On Behalf Of Charles Sweeney Sent: Thursday, February 20, 2003 10:36 AM To: howiej@xxxxxxxxxx; si-list@xxxxxxxxxxxxx Subject: [SI-LIST] Re: Split planes and ground return wires Howard, This makes me wonder about the situation of connecting together pieces of equipment each with their own separate power supply. I used to design radar systems composed of multiple racks of equipment, and we were concerned about ground loops. One technique we used was to connect all the grounds to a single point (a star configuration) using very thick earth braid, and then all the grounds in the cables (eg shielded ribbon cable) were connected to ground at only one end of each cable. The signals therefore had a reference ground plane very close, but the actual return path was a very long way away. We could have connected the other cable ends to ground through a capacitor to provide the AC return path, but would this have been a "big" improvement? I suppose my question comes down to, is it the close presence of the ground plane that is important, or does the ground plane have to carry the return current. This has implications in PCB's where striplines may have power planes (not ground) on both sides, but the ground plane carrying the return current could be several layers away. This is an increasingly likely scenario in PCB's with different chips having different VCCs and so needing several power planes. Regards Charles -----Original Message----- From: si-list-bounce@xxxxxxxxxxxxx [mailto:si-list-bounce@xxxxxxxxxxxxx] On Behalf Of Dr. Howard Johnson Sent: 20 February 2003 17:37 To: si-list@xxxxxxxxxxxxx Subject: [SI-LIST] Re: Split planes and ground return wires Dear Marc, You have raised what is (to me) a very interesting question, one which deserves a good analytical answer. If we just deal with a group of ten signals surrounding one nearby ground return wire, there are some analytical tools that can help you predict the answer. First, you need to know the mutual inductance between signals in your configuration. Given an aggressive wire at position a, and a victim wire at position v, and one and only one ground wire at position g, all traversing z-axis distance h, if I recall correctly the formula for the mutual inductance shared between the aggressor and victim is: Lm = 5.08 * h * [ln(|v-g|/R) - ln(|v-a|/|g-a|)] where R is the radius of the wire, the dimensions h, |v-g|, |v-a|, |g-a|, and R are inches, the function ln() is the natural logarithm, and the answer Lm appears in nH As you can see the wire radius R appears as a logarithmic term in the equaltion and so changes in the wire radius do not strongly affect the outcome. The height h of the cavity between the boards (the wire length) affects the result directly and is thus your most powerful control variable. Let's do an example with the positions a, v, and g in a straight row with equal spacing s=0.1 inch. |v-a| = .1 |v-g| = .1 |g-a| = .2 R = .01 (radius of 24 AWG wire) h = .2 (from your note, although this sounds awfully tight) Lm = 5.08 * .2 * [ln(.1/.01) - ln(.1/.2)] Lm = 3.04 nH You should re-run this example assuming the only return paths are provided by the so-called "low inductance" ground pins in your backplane connector. Because they are located further away from the signals, the mutual inductance Lm will come out larger -- thus the importance of locating the return paths near the signals. The next issue has to do with your signal risetimes. The risetime (not the operating speed) determines the bandwidth of your signals. Assuming you are working with a modest CMOS process having rise/fall times on the order of Tr = 1 nS, the bandwidth (upper-bounded by approximately 0.5/Tr) works out to f = 500 MHz, at which frequency the impedance Zm represented by the mutual inductor Lm at frequency f equals Zm = 2*pi*f*Lm = 9.55 ohms. The crosstalk you receive has to do with the ratio of Zm to the circuit impedance Z0. If you were using end-terminated transmission lines with Z0=50 ohms, the numbers in this case indicate a crosstalk of approximately 9.55/50 = 19 % FOR EACH AGGRESSOR. Fortunately, in your application the wires are quite short compared to 1 ns and so they are likely not terminated, in which case the current drawn on each wire (and thus the mutual-inductive crosstalk) will be considerably less than 19 %. By the way, if you are using source-terminated lines the current waveform produced on the wire appears as two steps, one positive followed by one negative, with amplitudes equal to (1/2)Vcc/Z0 and separation in time approximately equal to the line delay (depending on where along this line your "connector" appears). If the line length, and thus the separation of the two opposing current waveforms, is small compared to the risetime the two opposing chunks of mutual-inductive crosstalk created in this configuration tend to cancel out. Lastly, the crosstalk that concerns you in this example bears a striking resemblance to the crosstalk that occurs when ordinary vias traverse a pcb. In analyzing those situations one takes into account all the nearby return paths, which tends to diminish the value of Lm quoted above for a single return path. Software for such calculations is provided by, among other companies, Sigrity. Best regards, Dr. Howard Johnson, Signal Consulting Inc., tel +1 509-997-0505, howiej@xxxxxxxxxx http:\\sigcon.com -- High-Speed Digital Design articles, books, tools, and seminars -----Original Message----- From: si-list-bounce@xxxxxxxxxxxxx [mailto:si-list-bounce@xxxxxxxxxxxxx]On Behalf Of Godbout, Marc Sent: Wednesday, February 19, 2003 6:57 AM To: si-list@xxxxxxxxxxxxx Subject: [SI-LIST] Split planes and ground return wires I've got questions concerning ground return wire requirements and hope the list can help. First the configuration. Our system as got boards that are physically two separate PCB's, about 6" X 7", sandwiching, and bonded to, a metal plate. Signals need to traverse from one PCB to another. To accomplish this, we run wires through holes in the plate. These signals are basically microprocessor-type signals, 5V CMOS. The bus speeds are 25MHz or lower. Ground returns are added at about a 10:1 ratio, and placed somewhat near the groups of signals. The ground planes are also connected to each other through the backplane connector through 40 low-inductance pins. So I see this as a not-quite classical split plane problem. The split is crossed by 40 connector wires and maybe 15 z-axis wires, each 0.200" long. We've got some mechanical constraints that are making me look into the minimum requirements of the ground return wires. I feel that the "bigger is better" theory was used to design them in the first place, but I've got to cut back on that right now. So the question is, by how much? So what are the areas or major concerns that I should investigate? How do I go about calculating a minimum wire size? And are there any texts that would help me zero in on my analysis methods? Thanks for any help you could give. -Marc ------------------------------------------------------------ ------ To unsubscribe from si-list: si-list-request@xxxxxxxxxxxxx with 'unsubscribe' in the Subject field or to administer your membership from a web page, go to: //www.freelists.org/webpage/si-list For help: si-list-request@xxxxxxxxxxxxx with 'help' in the Subject field List archives are viewable at: //www.freelists.org/archives/si-list or at our remote archives: http://groups.yahoo.com/group/si-list/messages Old (prior to June 6, 2001) list archives are viewable at: http://www.qsl.net/wb6tpu ------------------------------------------------------------ ------ To unsubscribe from si-list: si-list-request@xxxxxxxxxxxxx with 'unsubscribe' in the Subject field or to administer your membership from a web page, go to: //www.freelists.org/webpage/si-list For help: si-list-request@xxxxxxxxxxxxx with 'help' in the Subject field List archives are viewable at: //www.freelists.org/archives/si-list or at our remote archives: http://groups.yahoo.com/group/si-list/messages Old (prior to June 6, 2001) list archives are viewable at: http://www.qsl.net/wb6tpu ------------------------------------------------------------ ------ To unsubscribe from si-list: si-list-request@xxxxxxxxxxxxx with 'unsubscribe' in the Subject field or to administer your membership from a web page, go to: //www.freelists.org/webpage/si-list For help: si-list-request@xxxxxxxxxxxxx with 'help' in the Subject field List archives are viewable at: //www.freelists.org/archives/si-list or at our remote archives: http://groups.yahoo.com/group/si-list/messages Old (prior to June 6, 2001) list archives are viewable at: http://www.qsl.net/wb6tpu ------------------------------------------------------------------ To unsubscribe from si-list: si-list-request@xxxxxxxxxxxxx with 'unsubscribe' in the Subject field or to administer your membership from a web page, go to: //www.freelists.org/webpage/si-list For help: si-list-request@xxxxxxxxxxxxx with 'help' in the Subject field List archives are viewable at: //www.freelists.org/archives/si-list or at our remote archives: http://groups.yahoo.com/group/si-list/messages Old (prior to June 6, 2001) list archives are viewable at: http://www.qsl.net/wb6tpu