I’ve been puzzling over a question about return currents in transmission lines when the return path is a conducting plane. I was hoping you might shed some light on this.
Let’s assume the case of a microstrip transmission line having one signal trace routed over a conducting referenced plane. You point out in “High Speed Digital Design” that the high-frequency return currents on the reference plane will flow as close as possible to the signal current path to minimize the total loop inductance of the current path. That makes perfect sense to me.
My question is, are there really any high-frequency currents flowing in portions of the transmission line after those portions have been passed by the voltage disturbance moving down the line?
I am assuming the line is driven by a voltage step input with some finite rise time. As the rising voltage wave propagates down a uniform transmission line at constant velocity, a *constant* current is drawn from the supply driving the line. This amplitude of this current equals the amplitude of the voltage wave divided by the characteristic impedance of the line. It seems to me that, in the entire portion of the line between the source and the moving waveform disturbance, both the forward and returning currents must be constant.
However, if the return current in the portion of the line between the source and the moving waveform is constant, then how can this current contain any high-frequency components? It looks like DC current to me. So it seems to me that return currents remote from the actual moving voltage wave front will have to follow the path of least resistance back to the driver, not the path of least loop inductance.
So, if the signal trace takes a few turns (and ignoring the reflections caused by this), I suppose the returning signal current in those portions of the conducting plane that have been passed over by the moving wave front won’t really be following these turns very closely, will it? Won’t it be taking a more direct, low-resistance route back to the source driving the line?
I realize that the current at the wave front itself will indeed have high-frequency components since it is being created by a voltage step, but it seems that when you look backward from the wave front, the return current looks like DC and won’t be following the signal path closely in the return plane.
Do I have the correct picture or am I missing something?
Thanks for your interest in High-Speed Digital Design, and for your excellent question.
You are very close to a complete understanding of how the high and low-frequency portions of a digital current waveform behave on a solid plane.
I am going to answer your question by showing you how to decompose a single step edge into its constituent frequency components.
For a repetitive signal, like a square wave, most electrical engineers know that the signal can be represented as a sum of sinusoidal components, where the sinusoidal frequencies are multiples (harmonics) of the square wave frequency, and where the amplitudes are determined by the exact shape of the square wave (duty cycle, flatness of top and bottom portions, shape and speed of rising edge, etc.). I hope you can see that, above some certain critical frequency, the higher harmonics will all produce return-current waveforms in the reference plane that follow closely the undulations of a serpentine trace, while the harmonics below that same critical frequency produce return-current waveforms that tend to spread out across the plane, flowing back to the source along a wide, straight path. This much you can derive from a sinusoidal analysis, and it sounds like you have no problems with the theory so far.
The difficulty you are having begins with the treatment of a single, isolated edge. In the absence of repetition, harmonic analysis falls completely apart. You cannot express a single edge as a sum of discrete sine waves. You CAN, however, use the following trick to visualize the result.
Suppose I build a 24-channel equalizer. This is the sort of gear you see on fancy stereo systems. It lets you adjust the level of sound in each of 24 individual frequency bands. The instrument works by first passing your signal through a bank of 24 band-pass filters, then applying an independent gain knob to each of the 24 bands, and summing the resulting 24 partial signals to produce a complete result. With the gain in each channel set to unity, the equalizer has no effect.
If you turn up just one of the knobs, but turn down all the others, you can hear (or see, if you are looking at an oscilloscope), the frequency content in just one band at a time. That is the way we are going to view your isolated edge--one band at a time. Figure 1 illustrates the result.
The blue square wave in the figure has a rise/fall time of 200 ps 10-90, with Gaussian edge shaping. The red wave superimposed on top of the blue one shows a low-passed filtered version of this signal. To produce this waveform I used a Gaussian low-pass filter with a time constant of 6400 ps, corresponding to a 3-dB cutoff point of .338/(6400 ps) = 52.8 MHz. The diagram indicates that the low-pass filtered waveform represents the content of the main signal filtered to the band from DC to 6400 ps.
NOTE: The difference between the low-pass filtered version (red) and the main signal (blue) must represent the high-frequency portion of the signal. The high-frequency components, taken together, should fill in the gap (shown in yellow) between the blue and red waveforms. Mathematically, if you sum all the partial band outputs, you get back precisely the main signal waveform.
Moving down the chart, successive signals represent higher and higher frequency bands, corresponding to shorter and shorter scales of time. Down at the bottom, the 100-to-50 ps band does not contain much signal power at all. That makes sense, as the main signal had an intrinsic rise time of 200 ps. On scales of time shorter than that, not much happens.
Now, let's address your question. Each band includes an inevitable degree of "spreading" around the signal edge. Specifically, the band 6400-to-3200 ps band includes significant information over a spread of time comparable to the filter time constant of 6.4 ns. Every time you exercise the signal edge, the band from 52.8-104.6 MHz conveys a significant amount of signal power, and that signal power persists for 6.4 nanoseconds regardless of the signal rise/fall time. Each band has its own persistence time constant.
On a typical pcb board, with the traces about 0.010 in. above the reference planes, using copper traces, the boundary between DC behavior (returning signal current flows straight across the plane taking a wide path) and high-frequency behavior (returning current flows under each trace) occurs somewhere in the range of 1-10 MHz.
Clearly, the band 52.8-104.6 MHz lies well above this high-frequency mode boundary, so we may conclude that the returning signal current associated with that particular band will follow closely the undulations of each pcb trace, and that this current will persist for AT LEAST 6.4 ns.
Your original thinking incorporated only two classes of signal behavior,
- Very high-frequency behavior, with a short time persistence comparable to the edge time, and
- Very low-frequency behavior (DC)
I hope you carry away from this discussion a better understanding of the middle range of frequencies, those high enough to cause interesting return-current behavior (and crosstalk and ground bounce), but still low enough to have a considerable persistence time (many nanoseconds).
Extra for Experts
The band output waveforms may seem small in this example because the bands are narrowly spaced (only one octave each). If you repeat the calculations using a 10x band spacing, the signals in each band rise almost to full level (±0.5 V).
The band outputs in my picture appear perfectly centered around each edge location. I accomplished this using a bank of Gaussian bandpass filters having a perfectly symmetrical (and non-causal) response. Makes a nice picture, but you cannot do this in real life. In a real, physical test, the outputs would be antisymmetric, and they would not start moving until the input edge actually hit the filter.
Dr. Howard Johnson