Cable Shield Transfer Impedance Calculator
Cable Shield Transfer Impedance Calculator
How much voltage a shield current puts onto the inner conductor, per metre — and what a pigtail does to it. A few centimetres of wire between the braid and the backshell adds inductance that overtakes the braid’s own transfer impedance within tens of kilohertz, after which the shield’s quality stops mattering.
Shield transfer impedance and what a pigtail costs
one metre of single-braid coax with a transfer impedance of 15 mΩ/m at DC and 1 nH/m of transfer inductance, carrying 100 mA of shield current at 10 MHz, terminated with a 50 mm pigtail of 1 mm wire
Transfer impedance, and the inductance you add to it
braid: Zt(f) = Rt + jωM solid tube: Zt(f) = Rdc · |kt / sinh kt|, k = (1+j)/δ
pigtail: L = (μ0/2π)[ ℓ asinh(ℓ/g) − √(ℓ²+g²) + g ], g = (d/2)e−1/4
with a pigtail: V = |Ztℓ + jωL| · I crossover at f = Rtℓ / 2πL
braid Rdc = 1 / (σ · C · N · πd²/4 · cosα)
- Zt
- transfer impedance in ohms per metre: the voltage on the inner conductor per ampere on the shield. Lower is better, and it is measured, not calculated from a coverage percentage
- M
- the braid’s transfer inductance, from current leaking through the weave’s holes. A reader input on this page and it stays one: it depends on hole shape, weave angle and carrier count, not just on coverage
- L
- the pigtail’s self-inductance, from the exact Neumann integral with the geometric mean distance of a round section. About 1 nH per mm for ordinary wire, and only weakly dependent on diameter
- g
- the geometric mean distance of a solid round conductor, (d/2)e^(-1/4). Using it puts the internal inductance in correctly; the familiar (μ0ℓ/2π)(ln(2ℓ/a) − 0.75) is this with the small terms dropped
Worked example
one metre of single-braid coax with a transfer impedance of 15 mΩ/m at DC and 1 nH/m of transfer inductance, carrying 100 mA of shield current at 10 MHz, terminated with a 50 mm pigtail of 1 mm wire
At 10 MHz the braid's own transfer impedance is √(0.015² + (2π × 10⁷ × 1 nH)²) = 64.6 mΩ/m, so over one metre a 100 mA shield current puts 6.46 mV on the inner conductor — that is what a 360° backshell would give
The pigtail's inductance, from the exact Neumann integral, is 45.56 nH — about 0.9 nH per millimetre, which is the usual rule of thumb
Its reactance at 10 MHz is 2.863 Ω. The braid's whole-cable transfer impedance is 64.6 mΩ. The pigtail is 44 times larger
Adding them as reactances in series gives 2.926 Ω and a coupled voltage of 292.6 mV instead of 6.46 mV
The pigtail has cost 33.1 dB
That was always going to happen, and early: the crossover — where the pigtail's reactance equals the braid's DC resistance over the whole metre — is at 52.4 kHz. Not megahertz. Tens of kilohertz
Well above the crossover the penalty flattens at 20 log of the inductance ratio, 45.56 nH against 1 nH per metre times one metre, which is 33.2 dB. At 10 MHz it has essentially got there
What the pigtail costs, by length
| Pigtail length | Inductance | Reactance at 10 MHz | Penalty over a 360° bond |
|---|---|---|---|
| 5 mm | 2.322 nH | 145.9 mΩ | 10.2 dB |
| 10 mm | 5.955 nH | 374.2 mΩ | 16.6 dB |
| 25 mm | 19.35 nH | 1.216 Ω | 25.9 dB |
| 50 mm | 45.56 nH | 2.863 Ω | 33.1 dB |
| 100 mm | 104.9 nH | 6.592 Ω | 40.3 dB |
| 200 mm | 237.5 nH | 14.92 Ω | 47.3 dB |
Typical transfer impedance, and why it must be measured
| Shield construction | At DC | Behaviour with frequency |
|---|---|---|
| Aluminium foil with a drain wire | tens to hundreds of mΩ/m | Poor and unpredictable — the drain wire carries most of the current and the foil seam is a slot |
| Single copper braid, around 95% coverage | 10 to 20 mΩ/m | Flat to a few hundred kHz, then rising at 20 dB/decade from the transfer inductance |
| Double braid, or braid over foil | 2 to 10 mΩ/m | Same shape, lower transfer inductance; the gain over a single braid is mostly at HF |
| Solid tube — semi-rigid coax, conduit | a few mΩ/m | FALLS with frequency, about 8.7 dB per skin depth of wall. In a different class above a megahertz |
| Optimised multi-layer braid with a mu-metal or permalloy layer | under 1 mΩ/m | Purpose-built for low-frequency magnetic immunity; expensive and heavy |
Why coverage is not the answer
| What coverage tells you | What it does not |
|---|---|
| Roughly how much metal is in the way, and therefore the DC resistance | The size and shape of the holes, which sets the transfer inductance |
| That 95% is better than 85% | That two braids both at 95% can differ by several times in transfer inductance, because of weave angle and carrier count |
| Something about mechanical robustness and flex life | Anything at all about the termination, which usually dominates |
A pigtail destroys a good shield
A cable shield does not block interference; it gives it somewhere else to go. Current flowing on the outside of the shield produces a small voltage along the inside, and transfer impedance is the constant of proportionality: volts on the inner conductor per ampere on the shield, per metre of cable. Lower is better. It is a measured property, not a calculated one.
A braid is two things at once. At DC it is simply the shield’s resistance. As frequency rises, current begins to leak through the diamond-shaped holes in the weave, and that leakage is inductive — so above a few hundred kilohertz a braid’s transfer impedance climbs at 20 dB per decade. A solid tube has no holes at all, so its transfer impedance does the opposite: the skin effect confines the current to the outer surface and the coupling falls exponentially with wall thickness in skin depths. That is the real difference between semi-rigid coax and a braided cable, and it is far larger than any difference between two braids.
And then the connector throws it away. A pigtail — a few centimetres of wire between the end of the braid and the point where it is bonded — puts its own inductance in series with the whole coupling path. Wire is about a nanohenry per millimetre. A good metre of braid has a transfer inductance of about a nanohenry in total. So fifty millimetres of pigtail is fifty times the braid’s own inductance, and above the crossover frequency the shield’s quality has simply stopped mattering. The crossover, for the numbers above, is tens of kilohertz — not megahertz, which is what most people guess. Everything above it is pigtail.
What to do about it. Terminate the shield through 360 degrees: a conductive backshell that clamps the braid all the way round, an EMI gland, a shield clamp onto the bulkhead. That is the only fix that works, because it is the only one with no length in it. Shortening a pigtail is worth 6 dB per halving and runs out of road quickly; making it fatter is worth under a decibel per doubling, because inductance depends only logarithmically on diameter; using a flat strap instead of round wire is worth a few decibels and no more. The table above shows what each length costs.
Where the shield current comes from. A current probe. It cannot be derived from a schematic — it is set by the common-mode voltage driving the harness, the harness’s impedance to structure, and the parasitic capacitance of everything at both ends, none of which appears on the circuit diagram. This page takes it as an input and says so. Note, though, that the PENALTY the pigtail costs does not depend on the current at all: it is a ratio, and it is the same whatever is flowing.
Why this matters for radiated emissions. A shield that is not properly terminated carries interference straight through the enclosure wall as though the wall were not there, and the harness then radiates it. On most hardware that, and not the box, is what sets the RE102 margin. The cable radiated emissions page turns a common-mode current into a field, and the aperture page covers what leaks through the metalwork itself.
Frequently asked questions
How short does a pigtail have to be?
Shorter than you can make it, and it still will not be short enough. A 5 mm pigtail on a metre of good braid still costs over a dozen decibels at 10 MHz, because 5 mm of wire is about 4 nH against the braid’s 1 nH. The question has no good answer, which is why the answer is a 360° termination instead: a conductive backshell, a shield clamp or an EMI gland has no length in it at all.
Why is my expensive double-braid cable no better than the cheap one?
Almost certainly because both are pigtailed. The penalty a pigtail adds is 20 log of the ratio of its inductance to the shield’s, so a better shield makes the pigtail penalty LARGER while leaving the actual coupled voltage unchanged. The extra money buys nothing until the termination is fixed. Fix the termination first, then the cable.
Where do I get the transfer impedance from?
A measurement — the triaxial method in IEC 62153-4-3 is the usual one — or the manufacturer. Many datasheets do not quote it, which is itself information. Where you cannot get a number, use a pair of bracketing values and see whether your decision changes; if it does, the measurement is worth paying for.
Why will this page not compute the transfer inductance from optical coverage?
Because coverage does not determine it. Transfer inductance comes from the size and shape of the holes in the weave, which depends on the weave angle, the number of carriers and how tightly the braid is laid as well as on how much metal is there. Two braids at the same coverage can differ by a factor of several. The DC transfer RESISTANCE is computable from the geometry and this page does compute it — for a 16 × 7 braid of 0.12 mm wire it returns 15.7 mΩ/m against a published measurement of about 15 mΩ/m for RG-58. The inductance is a different kind of quantity and inventing a model for it would be worse than asking.
Why does a solid shield get better with frequency?
Because coupling through a solid wall happens only by diffusion, and the skin effect confines the shield current to the outer surface as frequency rises. The transfer impedance falls as |kt/sinh kt|, which is about 8.7 dB per skin depth of wall thickness. A braid does the opposite, because its holes let field through directly and that path is inductive. This is the largest single difference between shield constructions.
Does the cable length really just multiply?
While the cable is electrically short, yes — every metre contributes the same voltage in phase. Above about a tenth of a wavelength it stops: contributions from different points arrive with different phase, standing waves develop, and the answer depends on how both ends are terminated. The frequency at which that happens for your length is printed on the page, and above it the number is an order of magnitude rather than a prediction.
Related calculators
References
- Vance EF. Coupling to Shielded Cables. John Wiley & Sons, New York, 1978 (ISBN 0-471-04107-6; reprinted by Krieger, ISBN 0-89874-949-2). The standard treatment of braid transfer impedance, the resistive and inductive contributions, and why the hole inductance is a property of the weave rather than of the coverage.
- Ott HW. Electromagnetic Compatibility Engineering. Wiley, 2009. Chapter 2 Cabling, sections 2.11 Shield Transfer Impedance, 2.15 Shield Terminations and 2.15.1 Pigtails. Section numbers and titles verified against the author’s own published detailed contents listing.
- Akcam N, Karatas MH. Measurement of Transfer Impedance and Screening Attenuation Effects on Cables Using Tri-axial Method. International Journal on Technical and Physical Problems of Engineering, Vol. 4, No. 1, Issue 10, March 2012, pp. 103–107. Reports a DC transfer resistance of about 15 mΩ for RG-58, which is this page’s default, and shows the resistive-to-inductive transition with frequency.
- IEC 62153-4-3, Metallic communication cable test methods — Part 4-3: Electromagnetic compatibility (EMC) — Surface transfer impedance — Triaxial method. The measurement standard. An IEC document, copyrighted; named for the method, with no values reproduced.