Wireless Power Coupling Coefficient Calculator
Everything about an inductive power link follows from the coupling coefficient, and k collapses as the coils drift apart or sideways. This computes the mutual inductance of two flat spirals directly from the Neumann integral - which is the only honest way to handle misalignment, since no closed form covers it - then turns k and coil Q into the best efficiency the link could ever reach.
Choosing something to solve for makes Coupling coefficient k an editable target and computes the chosen input from it.
Formula
| a, b | turn radii on transmitter and receiver (m) |
| z | vertical gap between coil planes (m) |
| d | lateral misalignment of the coil axes (m) |
| k | coupling coefficient, 0 to 1 |
| Q | unloaded coil quality factor |
What this model assumes, and where it stops
Assumptions
- Both coils are flat spirals of circular turns, evenly spaced between inner and outer radius.
- Air core throughout - no ferrite backing plate and no aluminium shield.
- Coil planes are parallel; misalignment is a pure lateral shift with no tilt.
- Filamentary turns, so conductor cross-section affects only the self-inductance term.
- Q values are supplied rather than derived, and are assumed constant with load.
- Series-series compensation for the capacitor and load figures.
Limitations
- Ferrite backing changes everything. Every practical pad has a ferrite plate to shape the flux, and it typically lifts coupling by half again or more over the bare air-core value this page computes. Treat the result as a floor, not a prediction.
- Aluminium shielding plates behind the ferrite push flux back through the coil and shift both L and k, in the opposite direction to the ferrite.
- Real pads are rarely circular. Rectangular, DD, DDQ and bipolar windings have very different misalignment behaviour, and a circular model will not rank them correctly.
- Tilt and rotation are not modelled, only lateral shift. On a vehicle, suspension travel makes tilt a real term.
- Q is entered rather than computed. It depends on litz construction, ferrite loss and proximity effect, and it falls as the pad heats.
- The efficiency figure is the coil-to-coil ceiling at optimum load. Real systems lose several more points in the inverter, rectifier and control loop, and the optimum load is rarely what the battery presents.
- Nothing here addresses the emissions, foreign-object detection or living-object protection that dominate a real certification effort.
When you need a 3D field solution instead
Closed-form models like the one above hold on idealised geometry. These are the cases where they stop being good enough and a full 3D electromagnetic and thermal solution is the only way to get a trustworthy answer:
- Ferrite backing plates and aluminium shields, which set the real coupling and cannot be captured by an air-core integral.
- Non-circular pad shapes - DD, DDQ, bipolar - where the whole point is the misalignment behaviour a circular model cannot show.
- Stray field at bystander distances, which is what an exposure assessment turns on and is inherently three-dimensional.
- Foreign metal objects in the gap, where induced eddy heating depends on the local field the object actually sees.
- Losses in the vehicle underbody or chassis, which sit right in the flux path.
- Tilt, rotation and the combined misalignment envelope rather than a single lateral offset.
- Coil Q at temperature, where litz proximity effect and ferrite loss both change with operating point.
Common questions
What coupling coefficient should I expect from a wireless power link?
Phone-charger geometry with pads nearly touching reaches k around 0.5 to 0.8. The default automotive-style case here - 200 mm pads with a 150 mm gap - gives k about 0.19, and that is normal: vehicle charging runs happily at k of 0.1 to 0.3 by resonating out the large leakage. If your k is below a few percent, expect the tuning and the electronics, not the coils, to dominate the design effort.
What kQ product do I need for good efficiency?
The coil-pair maximum efficiency depends only on kQ: about 82% at kQ = 10, 90% at 20, and 96.6% at the default case’s kQ of 57. Since k is set by geometry you usually buy efficiency with Q - better litz, ferrite backing - or by simply closing the gap.
How much coupling do I lose to misalignment?
Less than intuition suggests at first: shifting the default 200 mm pads sideways by 50 mm - a quarter of the pad diameter - costs about 6% of k. The loss accelerates sharply beyond that as opposing flux starts threading the receiver. The practical rule: alignment tolerance is bought with pad diameter, which is why vehicle pads are so much larger than the gap would suggest.
References
- Neumann formula - Mutual inductance of two filaments; Grover, Inductance Calculations, Dover 1946
- Maxwell - Coaxial circular filaments via complete elliptic integrals
- SAE J2954 - Wireless power transfer for light-duty plug-in electric vehicles, alignment methodology
- Qi Specification - Wireless Power Consortium, power transmitter and receiver design