An introduction to the air cored coil

• What is an air core coil?
• Do you need an air coil?
• Single layer coils.
• Multi layer coils.
• Flux density in an air coil.

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See also ...
[ Producing wound components] [A guide to the terminology used in the science of magnetism] [ Power loss in wound components] [The force produced by a magnetic field] [ Faraday's law]

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Do you need an air coil?

• Its inductance is unaffected by the current it carries. This contrasts with the situation with coils using ferromagnetic cores whose inductance tends to reach a peak at moderate field strengths before dropping towards zero as saturation approaches. Sometimes non-linearity in the magnetization curve can be tolerated; for example in switching converters. In circuits such as audio cross over networks in hi-fi speaker systems you must avoid distortion; then you need an air coil. Most radio transmitters rely on air coils to prevent the production of harmonics.
• Air coils are also free of the 'iron losses' which affect ferromagnetic cores. As frequency is increased this advantage becomes progressively more important. You obtain better Q-factor, greater efficiency, greater power handling, and less distortion.
• Lastly, air coils can be designed to perform at frequencies as high as 1 Ghz. Most ferromagnetic cores tend to be rather lossy above 100 MHz.

• Without a high permeability core you must have more and/or larger turns to achieve a given inductance value. More turns means larger coils, lower self-resonance and higher copper loss. At higher frequencies you generally don't need high inductance, so this is then less of a problem.
• Greater stray field radiation and pickup. With the closed magnetic paths used in cored inductors radiation is much less serious. As the diameter increases towards a wavelength (lambda = c / f), loss due to electromagnetic radiation will become significant. Balanis has the gory details. You may be able to reduce this problem by enclosing the coil in a screen, or by mounting it at right angles to other coils it may be coupling with.

  You may be using an air cored coil not because you require a circuit element with a specific inductance per se but because your coil is used as a proximity sensor, loop antenna, induction heater, Tesla coil, electromagnet or deflection yoke etc. Then an external field may be what you want.

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The single layer coil

  L = N2r2 / (228r + 254l)   microhenrys
 
       where  L is inductance in microhenrys
              r is the coil radius in millimetres
              l is the coil length in millimetres (>0.8r)
              N is the number of turns

This formula applies at 'low' frequencies. At frequencies high enough for skin effect to occur a correction of up to about -2% is made.

To construct a self-supporting air cored coil take a length of plain 1 millimetre diameter copper wire and hold one end in a bench vice. Take the other end in a pair of pliers and pull until the wire has stretched slightly - this will straighten it. Using a 5 millimetre diameter drill bit wrap the wire around it until enough turns have been applied. Using 'long nosed' pliers bend the ends of the coil to get them into a radial position.

Small reductions in the inductance obtained can be achieved by pulling the turns apart slightly. This will also reduce self-resonance. Other combinations of wire and coil diameter may be tried but best results are usually obtained when the length of the coil is the same as its diameter.

This property also leads to a disadvantage of the air cored coil: microphony. If you need good frequency stability in the presence of vibration then wind the coil on a support made from a suitable plastic or ceramic former.

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The multi layer coil

This type of winding is very common because it's simple to construct with a winding machine and a mandrel. We'll consider it in some detail. The first question is this: if you have a fixed length of wire then what dimensions of the winding give the greatest inductance? Put another way, what is the most efficient shape?

The ratio of the winding depth to length, which is c/b, needs to be close to unity; so the winding should have a square cross section. This makes sense because only with the square is the average distance between turns at a minimum (a circular cross section would be even better, but that is hard to construct). Keeping the turns close together maintains a high level of magnetic coupling ('flux linkages') between them, and so the general rule that the inductance of a coil increases with the square of the number of turns is maintained.

OK, now what about the mean radius of the turns: dimension a ?

In the Fig. 2 above you see three coils in cross section. Each uses the same length of wire but the diameter of the coil varies. The inductance of any one turn is linearly proportional to its diameter; so you want a large diameter to get the most inductance. Also you need all turns in the winding to be as close as possible to all the others. The coil on the left fulfills these requirements, but it has a problem because, in making the diameter large, you don't have sufficient wire to give it many turns. Since the inductance of the winding as a whole varies as the square of the number of turns the left hand coil won't have high L.

The coil on the right does have a high number of turns, but it isn't optimum for two reasons. The diameter of each turn is small (particularly those near the centre of the coil) which leads to a low inductance per turn. Worse, the distance between turns separated by the diagonals of the winding cross section is large. This leads to weak coupling between them (lower flux linkage) and a failure of the N² relationship with coil inductance.

Can you anticipate a punch line? Brooks, who wrote a paper in 1931, calculated that the ideal value for the mean radius is very close to 3c/2. We call a coil having these dimensions a Brooks coil. It's worth emphasising that the Brooks ratio is not critical. You can have a coil which deviates from it quite significantly before L falls off very much. Also, you may have other considerations than the inductance alone.

If we let S₁ = (c/2a)² then the inductance of any of our three coils can be approximated by

L = 4×10^-7a N²((0.5+S₁/12)ln(8/S₁) -0.84834+0.2041S₁)  henrys

Example: If the mean radius is 2.4 cm, the width is 9mm and N is 350 then L = 6.977 mH.

If we have a proper Brooks coil then the formula above boils down to

L = 1.6994×10^-6aN²  henrys

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Flux density in an air coil

• When the density is important
• The Biot-Savart equation
• Biot-Savart solves a current loop
• The field in a solenoid

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When the density is important

dB = μ₀ I ds sinθ / (4π r²)   teslas

At any point along the circular field line labeled F in figure FCE a tangent to it has a direction fulfilling the requirements just given. Imagine a set of such lines surrounding the current element, spaced in accordance with the expression for dB.

You might complain that the Biot-Savart equation doesn't seem to be buying us very much: an isolated current element is an imposibility in practice and real coils use wires which may be very thick. What saves the day in the case of air coils is the priciple of superposition. Without a ferromagnetic core to spoil the linearity of the system we can simply divide the actual current flow up into any number of smaller regions and then sum, or integrate, the individual fields to obtain the total flux density.

Biot-Savart solves a current loop

In figure FLP we see the loop in cross section and we want the flux density at the point P₁ on the loop axis at a distance r from the conductor. We start by considering the contribution to the field made by a current element at P₂. The figure shows a P₁ which happens to be quite far from the loop. ψ is rather small and the field vector (B_T, shown in blue) is swinging up towards a normal to the axis of the coil. You can resolve B_T into two components: B_N, shown in red and B_A, shown in green. We can disregard the normal component, B_N, because of the symmetry of the loop: it is going to be cancelled by the field generated by the current element on the opposite side of the loop at P₃. We are left with just the axial component (that we need to find): B_A. Its magnitude will be

B_A = B_T sinψ

dB_A = μ₀ I ds sinθ sinψ / (4π r²)

dB_A = μ₀ I ds sinψ / (4π r²)

B_A = a μ₀ I sinψ / (2 r²)

B_A = a μ₀ I sinψ / (2 (a / sinψ)²)

B_A = μ₀ I sin³ψ / (2a)

First we have to make a simplification because the flux lines will not run perfectly true along the axis but will spiral slightly around it (reflecting the helical nature of the coil itself). This effect becomes less if the turns are spaced more closely. In fact this leads on to the concept of a current sheet. If we kept the dimensions of the solenoid the same but used, say, three times the number of turns and at the same time reduced the current to one third then the overall magneto motive force is the same and so the field produced will be much as before.

In the limit we no longer worry about the total number of individual turns and the coil current. Mentally replace them with a thin walled copper tube carrying a total current N × I amps flowing around its circumference and distributed evenly along its length. Don't ask how the current gets there; this is pure hypothesis :-) . Now we can talk of an mmf per unit length of the tube, F_l -

Fl = I × N / b   amps per metre

Equation MLT

F_m = F_l dz

dB = μ₀ F_l dz sin³ψ / (2a)

z = a / tanψ

dz/dψ = -a cosec²ψ

dz = -a dψ cosec²ψ

dB = μ₀ F_l (-a dψ cosec²ψ) sin³ψ / (2a)

dB = -(μ₀ F_l / 2) sinψ dψ

B = -(μ₀ F_l / 2) ∫ sinψ dψ

B = -(μ₀ F_l / 2) _π/2∫^ψ1 sinψ dψ

B = (μ₀ F_l / 2) [cosψ]_π/2^ψ1

B = (μ₀ F_l / 2) cosψ₁

B = μ₀ F_l   teslas

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