Labs / 01
Lab A: Transmission-Line Reflections and Energy Flow
A wave on a line carries voltage and current, in a fixed ratio set by Z0. A backward wave carries them in the opposite ratio: i = −v/Z0 rather than +v/Z0. Which of the two appears to reverse at the far end therefore depends on the sign of ΓL — an open (Γ = +1) returns the voltage unchanged and so doubles it while the currents cancel; a short (Γ = −1) inverts the voltage and so doubles the current instead. That one sign is why the energy has nowhere to go, and why the line rings instead of settling.
Best viewed on a laptop or desktop. These panels are built so you can move a slider and watch several charts answer at once. A phone has no room to put them side by side.
Select a wave with a click or arrow keys. Time pauses at its probe crossing; amber markers link the spatial and probe plots.
Notes
Move Time, select a reflection, or reveal a prediction to follow the same instant here. Energy integrates from before the source edge begins. These are totals for all waves, not energy assigned to one reflection. Source energy is net supplied energy: it can fall when the ideal source absorbs a returning wave.
In introductory electronics, voltage and current are treated as scalar values that exist instantaneously across an entire node. If you close a switch, Ohm's law asserts that current immediately flows through the entire loop. On circuit boards with nanosecond and picosecond edge rates, this lumped-element intuition breaks down completely. Voltage and current do not appear everywhere at once; they travel down copper traces as a coupled electromagnetic wavefront propagating at roughly six inches per nanosecond.
Most textbooks teach transmission line reflections using algebraic bounce diagrams that track voltage alone. This voltage-only framing obscures the fundamental physics: a transmission line cannot carry voltage without carrying current in the exact ratio determined by its characteristic impedance (Z0). When that wave strikes a load boundary, current and voltage do not behave identically — current subtracts where voltage adds.
This interactive lab makes the full wave visible. By stepping through time, you can observe forward and backward waves separating across space, probe the local V/I ratio at any point along the trace, and inspect a real-time energy account that tracks exactly how much energy is stored in fields, absorbed by termination resistors, or trapped bouncing between mismatched boundaries.
Voltage and current in a travelling wave
A transmission line is not a wire with a delay. It is a medium that can only carry energy in one particular combination: a voltage and a current whose ratio is Z0. Launch 833 mV into a 50 Ω line and you have also launched 16.7 mA, whether you wanted to or not. The two are not independent quantities that happen to travel together — they are one wave described two ways.
When a wave meets a boundary it cannot satisfy, it splits. At the load the line insists on V/I = Z0; the resistor insists on V/I = RL. The only way to satisfy both is to launch a second wave going the other way, sized so the totals work out. That is all a reflection coefficient is — the size of the wave you need to add so that Ohm's law and the line's ratio agree at the same point.
Because the backward wave carries its current the other way, the totals behave in opposite directions. At an open end the currents cancel and the voltages add, so you see twice the incident voltage and no current at all. At a short the voltages cancel and the currents add. The same Γ, read through a sign.
The energy account is the part people skip
Open the energy panel and step an unterminated line through its first two round trips. Almost nothing is delivered to the load — an open end absorbs no energy, by definition — and almost nothing is burned in a 10 Ω driver. The energy is simply still on the line, sloshing between the electric field in the dielectric and the magnetic field around the conductor.
That is what ringing is. It is not a control-loop instability or a resonance in the usual sense; it is energy with nowhere to be dissipated, bouncing between two boundaries that both refuse to absorb it. Termination is not a trick to make a waveform look nicer. It is the act of putting something in the circuit that can turn that energy into heat, and the reason a series resistor works is that it makes the driver itself that something.
Go deeper — the lattice, and what the model leaves out
Every wave on this page is one entry in a geometric series. The first forward wave is a divider against the line, not the load:
Each round trip multiplies by ΓLΓs, so wave m leaves the driver at 2mTd with amplitude a0(ΓLΓs)m, and the backward wave leaving the load at (2m+1)Td carries a0ΓL(ΓLΓs)m. Because |ΓLΓs| < 1 for any passive pair, the series converges, and it converges to the plain DC divider VsRL/(Rs+RL) — which is the sanity check worth remembering. However complicated the bouncing looks, the endpoint is the answer you would have got from a resistor divider with no line at all.
What this model deliberately leaves out, and what each omission costs you:
- Loss. The line is lossless, so the ringing decays only through the terminations. A real 9-inch stripline at these edge rates would take the corners off each bounce and shrink the later ones. The arrival times are right; the late-time amplitudes are optimistic.
- Dispersion. Z0 and the propagation velocity are frequency-flat here. In FR-4 both drift with frequency, which rounds edges further and smears the crisp arrivals the lattice draws.
- Discontinuities along the line. There are none — the line is uniform between two boundaries. A via field or a connector in the middle would add a third reflection site, and the lattice would branch at it rather than running edge to edge.
- The return path. Treated as ideal. Every current in the top panels has an equal and opposite current in the reference plane directly beneath it; break that plane and the inductance of the detour becomes a discontinuity the model above cannot represent.
The other half: where the return goes
Everything above treats the line as two terminals and a delay. That is the right model for timing, and it says nothing about where the current goes on its way back — which is the half of the problem that decides crosstalk, emissions, and whether the impedance you designed is the impedance you built.
It is a schematic. The arrows show which way current goes and roughly where it concentrates. They are not streamlines, not field lines, and not the output of any solver — nothing in that figure was computed. Cases 2 and 4 enclose a larger loop than 1 and 3, showing the effect directly, and the drawing will not tell you by how much.
For a number you need a field solver or a measurement. What the figure is for is recognising which of the four you are looking at on a real board, and therefore which question to ask next.
The rest of this lab is computed: the panel above is an exact lattice solution of a uniform line and its model contract says so. This figure is the one place on the page that communicates the idea schematically; streamlines would imply field-solver evidence that this drawing does not provide.
One case that does have a number
Cases 3 and 4 differ only in where the return via sits, and that particular question has an exact closed form — no field solver and no thin-wire approximation. For two parallel round conductors of radius r whose centres are s apart, the external inductance per unit length is:
For a 1.6 mm board with 0.3 mm via barrels, that gives the loop the signal via and its return form:
| Return via at | Loop inductance | and at 1 GHz that is |
|---|---|---|
| 0.5 mm | 0.703 nH | 4.4 Ω |
| 1.2 mm | 1.321 nH | 8.3 Ω |
| 2.0 mm | 1.654 nH | 10.4 Ω |
So moving that one via from half a millimetre to two millimetres costs 0.95 nH — a factor of 2.35, and about 6 Ω of extra series reactance at a gigahertz. That is the quantitative version of the difference between cases 3 and 4, and it is the kind of number worth having before a layout review rather than after one.
Use the arccosh, not the logarithm. At a spacing-to-radius ratio of 2.7 — close barrels, which is exactly the case you care about — the ln form overstates the inductance by 23.3%. It converges: 1.2% at a ratio of 6.7, and under 0.1% by 27. The approximation is fine where the answer does not matter and wrong where it does.
It is an estimate of one contribution, not a bound and not a prediction of the board. This is the external inductance of two uniform round barrels through a board of thickness l. An earlier version of this page called it an upper bound, on the grounds that everything omitted would reduce it. Current shared with other return vias does reduce it. Two other omissions add:
- Internal inductance. A round conductor carries μ₀⁄8π per unit length inside itself at DC. For two 1.6 mm barrels that is 0.160 nH on top of the 1.199 nH external figure at 1 mm spacing — 13.3%. Skin effect expels it as frequency rises, so the formula is close at gigahertz and low at DC.
- The end transitions. A barrel meets a trace, and that junction encloses its own loop. Nothing here accounts for it.
So the error is two-sided, which is why this is not a bound. Compare two placements with it; do not quote either as the loop inductance of your net. It is labelled analytical rather than measured for exactly that reason.
In the real world
The practical question is almost never "is there a reflection" — there always is. It is whether the ringing has settled by the time the receiver looks. A 2-inch net at 170 ps/inch is 340 ps one way and 680 ps per round trip, and with a 60 ps edge it will ring for several nanoseconds at small amplitude. At 100 MHz that interval may have little effect on sampling, while at 3.2 GT/s it occupies the available timing budget.
That is why the rule of thumb is written in terms of edge rate rather than clock rate: if 2Td is shorter than about a third of the rise time, the reflections arrive while the edge is still moving and fold harmlessly into it.
Longer than that and the net is electrically long, which means it has to be analysed as a transmission line rather than as a node — the voltage is no longer one number. It does not automatically follow that a separate terminating component is required. A driver whose output impedance already equals Z0 is series-terminated by construction and needs nothing added; a net whose receiver is reached before the first reflection returns may be fine untermined; an interface with on-die termination has already paid for it. What being electrically long changes is the analysis. Whether it also changes the bill of materials depends on the source impedance, the topology and the timing budget, and the energy panel is the quickest way to see which termination you actually bought.
Where this is explained
This page is the instrument. The mechanisms it lets you change are described on the topic pages below, each one linked for the specific thing it explains rather than as a general reading list.
- Reflections and ringing — the coefficient this lab sweeps, and why the ringing settles at all.
- Characteristic impedance — what the 50 Ω is, and why it is not a resistance a meter can find.
- Termination schemes — the ways to make Γ small, and what each one costs in power and area.
- When a trace becomes a transmission line — whether the round trip matters for your edge rate in the first place.