Confusing case of return current - Page 1
Forum section: electronics / beginners

author: arduino_guy
author_rank: Contributor
timestamp: February 14, 2017, 02:48:03 pm
content: Hi,Recently I started designing boards involving high speed signals and requires proper design to ensure signal integrity and taking good care of return current. I'm still a newbie in this domain. I know that the return current tends to couple with the signal current and keeping them coupled is good for signal integrity, reduce emissions, reducing the impedance of the current path etc. Also return current tends to find the shortest path to ground. Now comes my question. Assume that the PCB is a 2 layer with a solid, unbroken ground plane on bottom layer. Suppose I have a chip spitting out a high freq interface clock, but is placed away from the ground of the power supply(assume power is powered through a barrel connector). On the other hand, the receiver is placed closer to the barrel connector. Obviously, the return current has a lower impedance path back to ground without having to couple with the signal current. So if the return current goes to ground directly without coupling with signal current, we'll have problems like emissions. On the other hand if it tends to couple with the signal current(which is just a wild thought on mine, I know it is impossible since a path with much lesser impedance exists. Just speculating), it is worse from a signal integrity point of view since the signal goes through a higher impedance path. So, in spite of having a solid ground plane, I'm going to have tons of problems(emissions, signal integrity problems etc.). How do I take care of such a situation? Experts chime in your ideas!!!!!!

author: AndyC_772
author_rank: Super Contributor
timestamp: February 14, 2017, 03:33:35 pm
content: Remember that current flows in loops. If you have a high speed device, the loop will be something along the lines of:Decoupling cap +ve terminalTrace from cap to IC VCC pinIC lead frame from VCC pin to dieAcross the die from VCC to output driverIC lead frame from driver to clock output pinExternal tracking and loadGround planeDecoupling cap -ve terminalThat's for driving high. When the output drives low, current will flow from the external tracking and load, backintothe driver output, out to GND via the IC's ground pins.Note to nit-pickers: yes, I've simplified thisThe DC jack and connector have nothing at all to do with the AC current paths. People are sometimes taught that current flows "from the power supply", through the load and "back to the power supply", but that's a gross oversimplification when it comes to high speed design. The "power supply" can be any point on the power distribution net which has low impedance at the frequency of interest.You should really consider using a PCB with at least 4 layers. Above 100 MHz or thereabouts, the only effective capacitance between VCC and GND is the plane-to-plane capacitance in the PCB itself.As a worthwhile academic exercise, work out the inductance of 1mm of PCB trace, and therefore, its impedance at 100 MHz.From this, work out the impedance of the tracks joining your decoupling capacitors to the devices they're nearest to.Then, work out how to minimise this.

author: arduino_guy
author_rank: Contributor
timestamp: February 14, 2017, 04:23:29 pm
content: Thanks Andy for the reply. That makes a lot of sense.

author: AndyC_772
author_rank: Super Contributor
timestamp: February 14, 2017, 04:42:56 pm
content: You're welcomeI should also add something I meant to emphasize, which is that the connector where the mains adapter plugs into the board is rarely, if ever, the "power supply" for the purposes of working out where the current loops flow. It's only the supply at DC, or for very low frequencies where the bulk capacitors in the mains adapter are relevant.At anything from a few MHz upwards, the loops to be considered will start and end on capacitors that are located on, or within, the PCB. On a well designed board, the DC power jack should have a capacitor fitted across the supply right next to it, to short-circuit any high frequency loops that would otherwise involve current flowing up and down the DC cable.

author: dmills
author_rank: Super Contributor
timestamp: February 14, 2017, 05:17:07 pm
content: Also, inductance is everywhere, and there is a subtle gotcha if you make the DC input cap both large and very low ESR, in that the inductance of the planes can ring with this capacitance at a disturbingly low frequency (Think audio band in some cases!), small series resistors (or deliberately lossy ferrites) can be very much your friends.On the subject of ferrite beads, caution is advised as these are not simple components.Another note is that opamps are NOT three terminal devices, no matter what the spice models say. Depending on the current flow in the load the load current can switch quickly between the rails, local decoupling is your friend.Learn to think current loops (And that these loops involve one or both supply pins), you start to see circuits in a whole new light.Regards, Dan.

author: Benta
author_rank: Super Contributor
timestamp: February 14, 2017, 05:27:52 pm
content: I can recommend downloading the Motorola/ON Semi book "MECL System Design Handbook", it has a couple of excellent chapters on impedance matching and transmission line design:http://www.datasheetarchive.com/dl/2b3546ee3d4daa036a6d57b8a41a1d75a07fb4/P/MECL+System+Design+HandbookIt is unfortunately no longer available directly from ON Semi.