Square DTC-to-Helmholtz Coil Pair
A flat square planar-spiral (DTC) transmitter drives a wound square receiver. It fits panels and worktops where the transmitter must sit flush, charging an engineered receiver — a tool battery, robot or cart — that can afford some depth.
Worked example
Inputs. Transmitter: 5-turn square planar spiral (DTC), 2 mm pitch, turns at 50, 48, 46, 44, 42 mm half-side. Receiver: 5 co-located square turns at 50 mm half-side. Coaxial, no lateral offset, 1 A at 6.78 MHz, air gaps of 10 mm and 30 mm.
| Air gap | M (µH) | L_tx (µH) | L_rx (µH) | k |
|---|---|---|---|---|
| 10 mm | 2.871 | 5.929 | 8.878 | 0.396 |
| 30 mm | 1.325 | 5.929 | 8.878 | 0.183 |
Going from 10 mm to 30 mm lowers k from 0.396 to 0.183, a 54% drop. The flat panel is the 5.9 µH side, so at a given drive current it needs less voltage across the transmitter than a wound square coil would.
How it was computed. The panel field is the sum of five nested square turns, each built from four finite straight segments; the wound receiver integrates that field over its full 100 mm square, five times. k = M / √(L_tx · L_rx).
When it's the right choice
- Panel or surface-integrated transmitters under a worktop or shelf
- Receivers with room to wind turns, such as tool packs and robot docks
- Industrial docking where the receiver is a fixed, engineered part
Which side should be the flat spiral?
Swapping which side is the flat spiral changes almost nothing about the shared flux, which AirInduct Sim shows directly: this DTC-to-Helmholtz pair gives M = 2.87 µH at a 10 mm gap, versus 2.93 µH for the reverse Helmholtz-to-DTC arrangement, and the coupling coefficients are within 2% (0.396 and 0.403). The small residual comes from the 1 mm grid used for the receiver-area integration, not from the physics — mutual inductance is reciprocal. The real decision is mechanical and electrical. Here the transmitter is the 5.9 µH printed spiral and the receiver the 8.9 µH wound coil, so the panel can sit flush under a worktop, shelf or tray while the receiver’s larger inductance suits a higher-impedance load. Choose this direction for surface-integrated transmitters charging engineered receivers such as tool batteries, carts or robot docks; choose Helmholtz-to-DTC when it is the receiver that must stay thin.
Watch out for
- Receiver depth buys coupling only while its turns stay close to the transmitter plane
- Check current density in the innermost transmitter turns, which carry the same current over a shorter path
Neighbouring topologies
Square Helmholtz Coil Pair
Rectangular test rigs and fixtures. At the worked-example geometry: k = 0.369 at 10 mm.
Swap the receiver to DTCSquare DTC-to-DTC Planar Coil Pair
Tablets, flat modules, PCB-to-PCB links. At the worked-example geometry: k = 0.465 at 10 mm.
Switch to a circular outlineCircular DTC-to-Helmholtz Coil Pair
Earbud, stylus and wearable pads. At the worked-example geometry: k = 0.390 at 10 mm.
All eight pairs side by side, with coupling at both gaps: coil topology comparison.
Check the numbers
Run the Square DTC-to-Helmholtz Coil Pair at your geometry
Frequently asked questions
Should the flat square spiral be the transmitter or the receiver?
Put the spiral on whichever side must stay flush or thin. Coupling barely changes — k = 0.396 for this direction and 0.403 for the reverse at a 10 mm gap — so the choice is driven by mechanics and by which side should carry the larger 8.9 µH winding.
Why are M values for this pair and its reverse not exactly equal?
Physically they must be equal. AirInduct Sim integrates the receiver area on a 1 mm grid, and the nested-square spiral and the solid square are sampled slightly differently, leaving a 2% difference (2.87 vs 2.93 µH). A finer grid narrows it.
How much coupling is left at a 30 mm gap?
For 100 mm square coils with five turns each, k falls from 0.396 at 10 mm to 0.183 at 30 mm. Square coils hold their coupling slightly better with distance than circular coils of the same half-dimension because they enclose more area.