Every fast link is differential and almost nobody asks why. The rejection is a number rather than a property, tight coupling costs decibels that can be computed, and symmetry is lost in ways the schematic cannot show.
The standard list of advantages is well known and partly misleading, so it is worth separating what differential signalling genuinely provides from what it is often credited with.
The two conductors carry equal and opposite currents, so each is the other's return. The loop is small, known and local, and it does not depend on the state of the plane underneath. After deck 02, this is the largest single benefit, and it is the one least often cited.
A disturbance that reaches both conductors equally — a supply shift, a distant aggressor, a ground offset between two boards — cancels at the receiver to the extent that the receiver is balanced. That is genuinely valuable and it is a finite number, computed on slide 06.
The receiver sees the difference, so a given single-ended swing yields twice the decision voltage, worth six decibels of signal-to-noise before anything else is done. This is why differential survives at four hundred millivolts where single-ended would not.
A differential pair rejects an aggressor only to the extent that the aggressor couples equally to both conductors, and a neighbouring trace does not: it is closer to one than the other. Deck 06 computes what is left.
What it costs is twice the conductors, twice the package pins, twice the escape routing and the whole of the symmetry problem this deck is mostly about.
A pair of coupled lines does not have one characteristic impedance, it has two, corresponding to the two ways the pair can be excited. Drive both conductors with the same voltage and the fields between them vanish; that is the even mode, with impedance $Z_{0e}$. Drive them oppositely and the field between them is maximal; that is the odd mode, with impedance $Z_{0o}$.
The numbers the specification quotes follow directly. The differential impedance a receiver terminates is the impedance across the pair, which is twice the odd-mode impedance; the common-mode impedance is the even-mode impedance of the two in parallel, which is half of it:
$$Z_{\text{diff}} = 2 Z_{0o}, \qquad Z_{\text{comm}} = Z_{0e}/2.$$
The coupling coefficient $k = (Z_{0e} - Z_{0o})/(Z_{0e} + Z_{0o})$ measures how different the two modes are, and it is the same quantity as the ratio of mutual to self inductance and, in a uniform dielectric, of mutual to self capacitance. That coincidence is the subject of deck 06 and it is why stripline has no far-end crosstalk.
Cohn solved the edge-coupled stripline exactly in 1955, by conformal mapping, for conductors of zero thickness. That closed form is the right thing to check a numerical field solver against, because it is exact rather than fitted — and the check has to be run at matched assumptions, which means giving the solver the thinnest conductor its grid can represent.
If the closed form is exact, the numerical solver looks redundant. It is not, because Cohn's result covers one cross-section: two identical rectangles of zero thickness, centred between two planes, in a uniform dielectric. It says nothing about a pair beside a guard trace, a pair on an outer layer with air above it, an asymmetric stackup, or copper with a thickness. Those are the cases the rest of this series needs, and the agreement above is what licenses trusting the solver on them.
Every closed-form stripline expression in the literature, Cohn's included, assumes an infinitely thin conductor. Real copper is between twelve and seventy micrometres thick, and the extra metal adds capacitance both to the planes and to the neighbouring trace. The impedance falls, and it falls by more than the ten per cent tolerance a stackup is usually asked to hold.
This is not an error in either calculation; it is a real effect that a zero-thickness formula cannot see. It is also why an impedance coupon is measured rather than calculated, and why a stackup that changes copper weight has to be re-solved rather than scaled.
There is a persistent belief that a differential pair should be coupled as tightly as the technology allows. It is worth testing, and the test has to hold the differential impedance constant, because otherwise it compares geometries that are not alternatives to each other.
Doing that changes the question. Bringing the traces together raises the coupling, which lowers the odd-mode impedance, so the traces must be made narrower to bring it back to a hundred ohms. Narrower traces have a smaller perimeter, and conductor loss goes inversely with perimeter. Tight coupling is therefore paid for in decibels at Nyquist, and the model prices it.
When routing density forces it, which on a modern escape it frequently does; and when the pair has to survive a broken reference plane, because a tightly coupled pair carries more of its own return and cares less about what is underneath. Both are real reasons. Neither is "tighter coupling is better signal integrity", which is what the loss column contradicts.
The textbook statement is that a differential receiver rejects common-mode noise. It rejects it in proportion to how identical the two halves of the system are, and they are never identical. The two traces differ in length and in width, the two terminations differ by their tolerance, and the receiver's two input devices differ by their mismatch.
What survives is the residual, and the residual is what appears in the noise budget alongside crosstalk and receiver noise. The table shows what a fifty-millivolt common-mode disturbance becomes for three levels of balance.
Fifty millivolts of common-mode disturbance is not a pessimistic figure on a board with switching regulators and a shared chassis. Against a differential eye that deck 17 computes at a few tens of millivolts, even a well balanced receiver's residual is a term that has to be carried rather than neglected.
The most important asymmetry is intra-pair skew: a difference in arrival time between the two legs. It comes from length mismatch, from the glass weave of deck 03, from a via on one leg and not the other, and from any bend that makes the inside of the turn shorter than the outside.
Its effect is to move energy out of the differential mode and into the common mode. For a skew $\tau$ the conversion follows $|S_{cd21}| = |\sin(\pi f \tau)|$ and the differential transmission falls as $|\cos(\pi f\tau)|$, so the process is a transfer rather than a loss — the energy is not dissipated, it is moved into a mode the receiver ignores and the chassis radiates.
There is a source of mode conversion that survives perfect length matching. In a uniform dielectric the even and odd modes travel at the same speed, because both are governed by the same permittivity. On an outer layer they do not: the odd mode keeps more of its field in the laminate between the traces and the even mode keeps more of it in the air above, so the two modes have different effective permittivities and different velocities.
A pair that is perfectly symmetric, perfectly length-matched and perfectly fabricated will therefore still convert, purely because it is microstrip. The model computes the conversion for a two per cent velocity difference, which is typical.
Stripline is homogeneous, so its two modes travel at the same speed and this mechanism vanishes identically — the field solver confirms it to five decimal places, in deck 06. Between that, the absence of far-end crosstalk and the shielding from external fields, the case for routing a high-speed pair on an inner layer is close to unanswerable, and the cost is the vias needed to get there, which is deck 04.
A single resistor across the pair terminates the differential mode perfectly and the common mode not at all. Since the previous two slides have established that there will be a common-mode signal, leaving it unterminated means letting it reflect up and down the pair, and a resonant common-mode current on a cable or a long trace is an efficient antenna.
The tee network — two resistors of the odd-mode impedance in series with a shunt resistor to the reference at their midpoint — terminates both, and the shunt element is usually built as a resistor and capacitor in series so that it absorbs the common mode without drawing direct current. It costs a component and a little board area, and it is the difference between an electromagnetic-compatibility result that passes and one that does not.
| Source | Typical size | What it takes to control |
|---|---|---|
| Length mismatch from routing | A few picoseconds; the only term the router reports | Matching within a few thousandths of an inch; cheap and usually already done |
| Bends and serpentines | Sub-picosecond each, but they accumulate | Matching within the bend rather than after it |
| Glass weave | Up to about 5 ps per inch worst case — deck 03 | Routing angle or spread glass; not visible to the router at all |
| Via on one leg only | Tens of picoseconds | Never do it; transition both legs together |
| Package and connector | Vendor-specified, often the largest single term | Budget it; compensate in the receiver if the silicon allows |
| Modal velocity difference | Present on every microstrip pair | Route on inner layers |
Notice the ordering. The term every layout tool measures and every reviewer checks — routed length mismatch — is among the smallest. The terms that dominate are invisible to the tool: the weave is a property of the laminate, the package skew is in a datasheet, and the modal velocity difference is a consequence of which layer the pair is on.
| Quantity | Expression | Worth remembering |
|---|---|---|
| Differential impedance | $Z_{\text{diff}} = 2Z_{0o}$ | What the receiver terminates |
| Common-mode impedance | $Z_{\text{comm}} = Z_{0e}/2$ | What a single resistor across the pair leaves unterminated |
| Coupling coefficient | $k=(Z_{0e}-Z_{0o})/(Z_{0e}+Z_{0o})$ | Equals $L_m/L$ and, in a uniform dielectric, $C_m/C$ |
| Skew conversion | $|S_{cd21}|=|\sin(\pi f\tau)|$ | Complete at half a period of skew |
| Signal advantage | 6 dB | Twice the decision voltage for the same single-ended swing |
| Copper thickness | — | Half-ounce copper lowers a 100 Ω pair by about 10% |
| Coupling cost | — | Holding 100 Ω while tightening from 1.0 to 0.1 mm costs over a decibel in ten inches |
Deck 5 of eleven in Signal Integrity & High-Speed Digital Design. Every figure on this page is computed by si_models/deck05 and embedded as data; nothing is typed in by hand.
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