Wireless power transmission through magnetically coupled inductive coils

Introduction


I think that many of the readers have seen at least one video on popular video services, where electricity is transmitted through empty space using inductive coils.

In this article, we want to address the fundamentals of the process of wireless energy transmission using a magnetic field. Starting with consideration of the simplest inductive coil, and calculating its inductance, we will gradually move on to the theory of electrical circuits, in the framework of which the method of transmitting maximum power will be shown and justified under other equal conditions. So, let's begin.

Magnetic field of a single coil with current


Consider the magnetic field of a single coil with a current. Find the magnetic field of the coil at any point in space. Why is such a review necessary? Because in almost all books, at least in those that the author of the article was able to find, the solution to this problem is limited to finding only one component of the magnetic field and only along the axis of the coil - Bz(z), while we will find the law for the magnetic field in the whole space.

Magnetic field of a single coil with current
Illustration to the law of Bio-Savard-Laplace

To find the magnetic field, we use the law of Bio-Savart-Laplace (see Wikipedia - the Law of Bio-Savard-Laplace ). The figure shows that the center of the coordinate system Ocoincides with the center of the coil. The contour of the circle of the coil is indicated as C, and the radius of the circle - as a. During the coil current flows I.  vecr- is a variable-radius-vector from the origin to an arbitrary point of the coil.  vecr0Is the radius vector to the observation point. We also need a polar angle.  varphi- the angle between the radius vector  vecrand axis OX. The distance from the axis of the turn to the observation point is denoted by  rho. And finally  mathrmd vecr- elementary increment of the radius-vector  vecr.

According to the law of Bio-Savard-Laplace, the circuit element with current  mathrmd vecrcreates an elementary contribution to the magnetic field, which is given by the formula

 mathrmd vecB( vecr0)= frac mu0I4 pi cdot frac[ mathrmd vecr times( vecr0 vecr)]| vecr0 vecr|3


Now we’ll dwell on the variables and expressions included in the formula. Given the axial symmetry of the problem, we can write

 vecr0=( rho cos varphi, rho sin varphi,z) overset varphi=0 rightarrow( rho,0,z)


 vecr=(a cos varphi,a sin varphi,0)


 mathrmd vecr=(a sin varphi,a cos varphi,0) mathrmd varphi


 vecr0 vecr=( rhoa cos varphi,a sin varphi,z)


[\ mathrm {d} \ vec {r} \ times (\ vec {r} _0- \ vec {r})] = \ begin {vmatrix} \ vec {e} _x & \ vec {e} _y & \ vec { e} _z \\ -a \ sin {\ varphi} \, \ mathrm {d} \ varphi & a \ cos {\ varphi} \, \ mathrm {d} \ varphi & 0 \\ \ rho -a \ cos {\ varphi } & -a \ sin {\ varphi} & z \ end {vmatrix} = (az \ cos {\ varphi}, az \ sin {\ varphi}, a ^ 2 -a \ rho \ cos {\ varphi}) \ , \ mathrm {d} \ varphi


| vecr0 vecr|3= left( rho2+a2+z22 rhoa cos varphi right) frac32


In order to find the resultant magnetic field, it is necessary to integrate over the whole loop contour, i.e.

 vecB( vecr0)= intC mathrmd vecB( vecr0)


After the substitution of all expressions and some identical transformations, we obtain expressions for the axial and radial components of the magnetic field, respectively

Bz( rho,z)= frac mu0I4 pi int02 pi frac left(a2 rhoa cos varphi right) mathrmd varphi left( rho2+a2+z22 rhoa cos varphi right) frac32


Br( rho,z)= frac mu0I4 pi int02 pi fracaz mathrmd varphi left( rho2+a2+z22 rhoa cos varphi right) frac32


To find the absolute value of the magnetic field, it is necessary to sum up the components according to the Pythagorean theorem B= sqrtBr2+Bz2.

Let us demonstrate the solution obtained using the example of a radius coil a=0.1(m) and I=1(BUT).

Axial magnetic field amplitude
The amplitude of the axial component of the magnetic field

The amplitude of the radial magnetic field
The amplitude of the radial component of the magnetic field

Absolute magnetic field amplitude
Absolute magnetic field amplitude

Note that for a coil of arbitrary shape, at large distances z ggai.e. much larger than the characteristic size of the coil, the behavior of the magnetic field will tend to the solution found.

Hint...
For such calculations and graphing is convenient to use MathCad 15

Inductor. Magnetically coupled coils


Now that we know the solution for the magnetic field of a single turn, we can find the inductance of the coil, consisting of nturns. By definition, inductance is the coefficient of proportionality between the current in a coil and the magnetic flux through the cross section area. Here we use the ideal coil model, which is dimensionless in the direction of its axis of symmetry. Of course, in practice this does not happen. However, as approximations, the formulas obtained will be good enough. Although coils are considered dimensionless along OZ, you must specify a non-zero radius of the wire cross section. Denote it  deltaand example equal  delta=$0.(mm) Otherwise, when integrating the magnetic flux, the integrand will turn to infinity.

inductively coupled coils
Inductively connected coils

The figure shows two magnetically coupled coils. Let the first coil have a radius a1and contains n1turns, and the second - a2and n2respectively. Then, in order to find the own inductances, it is necessary to calculate the magnetic flux of each coil through its own section.

\ Phi = \ iint_S {\ vec {B} \ cdot \ vec {\, \ mathrm {d} S}} = \ int_0 ^ {2 \ pi} {\ int_0 ^ {a- \ delta} {B_z (\ rho, z) \ rho \, \ mathrm {d} \ rho \, \ mathrm {d} \ varphi} = 2 \ pi \ int_0 ^ {a- \ delta} {B_z (\ rho, z) \ rho \ , \ mathrm {d} \ rho}


Since there are many turns in the coil, we find the quantity called the flux linkage, by multiplying twice the number of turns

 Psi= frac12n2 mu0I int0a delta int02 pi frac left(a2 rhoa cos varphi right) mathrmd varphi left( rho2+a2+z22 rhoa cos varphi right) frac32 rho mathrmd rho


By definition, inductance is a coefficient of proportionality Lin the formula  Psi=LI. Thus, we obtain the own inductances of the coils

L1= frac12n12 mu0 int0a1 delta int02 pi frac left(a12 rhoa1 cos varphi right) mathrmd varphi left( rho2+a122 rhoa1 cos varphi right) frac32 rho mathrmd rho


L2= frac12n22 mu0 int0a2 delta int02 pi frac left(a22 rhoa2 cos varphi right) mathrmd varphi left( rho2+a222 rhoa2 cos varphi right) frac32 rho mathrmd rho


Let the centers of the coils be separated by distance d, lie on the same axis, and their plane of turns is oriented parallel. To find the mutual inductance, it is necessary to calculate the flux linkage formed by one coil through the cross section of another, i.e.

 Psi12= frac12n1n2 mu0I int0a2 delta int02 pi frac left(a12 rhoa1 cos varphi right) mathrmd varphi left( rho2+a12+z22 rhoa1 cos varphi right) frac32 rho mathrmd rho


Then the mutual inductance of the coils is given by

M12= frac12n1n2 mu0 int0a2 delta int02 pi frac left(a12 rhoa1 cos varphi right) mathrmd varphi left( rho2+a12+d22 rhoa1 cos varphi right) frac32 rho mathrmd rho


As far as the author knows, such integrals can be taken only numerically.
Note that as a rule  Psi12= Psi21and M12=M21=M. Coil coupling coefficient is called magnitude

k= fracM sqrtL1L2


We investigate the dependence of the coupling coefficient of the coils on the distance. To do this, consider two identical coils with a radius of turns a1=a2=0.1(m) and the number of turns n1=n2=100. In this case, the inductance of each of the coils will be L1=L2=8.775(mH).

Graph of the coupling coefficient of two identical coils on the distance between them
Coil coupling ratio of the distance between them

The schedule will not change if the number of turns in both coils is equally changed, or the radius of both coils is equally changed. The coupling coefficient is conveniently expressed as a percentage. From the graph it is seen that even with the distance between the coils of 1 (mm), the coupling coefficient is less than 100%. The coefficient drops to 10% at a distance of about 60 (mm), and to 1% at 250 (mm).

Wireless power transmission


So, we know inductance and coupling coefficient. Now let us use the theory of electrical circuits of alternating current to find the optimal parameters at which the transmitted power would be maximum. To understand this paragraph, the reader must be familiar with the concept of electrical impedance, as well as Kirchhoff's laws and Ohm's law. As is known from the theory of circuits, two inductively coupled coils form an air transformer. For the analysis of transformers convenient T-shaped equivalent circuit.

Transformer and its equivalent circuit
Air transformer and its equivalent circuit

The transmitting coil on the left will be called the "trasmitter", and the receiving coil on the right will be called the "receiver". Between coils coupling coefficient k. On the side of the receiver is the consumer represented by the load zL. The load can generally be complex. Input voltage on the transmitter side u1and input current - i1. The voltage transmitted to the receiver - u2and current transmitted i2. The full impedance on the transmitter side is denoted as z1, and the full impedance on the side of the receiver z2.

It is assumed that a sinusoidal voltage is applied to the input of the circuit. u1=u1m sin omegat.

Denote Rcoil1,Rcoil2,Lcoil1,Lcoil2,M- resistance and inductance of the coils (two own and one mutual), respectively. Then, according to transformer theory

z1=Rcoil1+j omega(Lcoil1M)


z2=Rcoil2+j omega(Lcoil2M)+Rload+jXload


On the other hand, according to our designations

z1=r1+jx1


z2=r2+jx2


Where r1,r2- total resistances on the side of the transmitter and receiver, respectively, and x1,x2- full reactive resistances.

Communication impedance is z3=j omegaM=jx3.

Find the input current of the circuit

i1= fracu1z1+z2||z3


where is the sign ||denotes a parallel connection of resistances. Then the voltage transmitted to the receiver

u2=u1i1z1=u1 left(1 fracz1z1+z2||z3 right)


And induced current

i2= fracu2z2= fracu1z2 left(1 fracz1z1+z2||z3 right)


We can find the integrated power transmitted to the receiver

s2=u2i2=p2+jq2


Thus, we have an expression for the complex power

s2=|u1|2z2 left| fracz3z1z2+z1z3+z2z3 right|2


Expression for active power component

p2=|u1|2r2 left| fracz3z1z2+z1z3+z2z3 right|2


Expression for reactive power components

q2=|u1|2x2 left| fracz3z1z2+z1z3+z2z3 right|2


In most practical tasks, the maximum active power is required, therefore

p2 rightarrow mathrmmax Rightarrow left| fracz3z1z2+z1z3+z2z3 right|2 rightarrow mathrmmax


Either that same

 left|z1+z2+ fracz1z2z3 right|2 rightarrow mathrmmin


 left|r1+jx1+r2+jx2+ frac(r1+jx1)(r2+jx2)jx3 right|2 rightarrow mathrmmin


 frac1x32|(r1x3+r2x3+r1x2+r2x1)+j(x1x3+x2x3+x1x2r1r2)|2 rightarrow mathrmmin


For convenience, we introduce the function

f(x1,x2)=(r1x3+r2x3+r1x2+r2x1)+j(x1x3+x2x3+x1x2r1r2)


and examine it for extremes

|f(x1,x2)|2 rightarrow mathrmmin


Where do we get a system of two equations

 frac partial|f|2 partialx1=2 mathbbRe(f)r2+2 mathbbIm(f)(x2+x3)=0


 frac partial|f|2 partialx2=2 mathbbRe(f)r1+2 mathbbIm(f)(x1+x3)=0


This system has five solutions, two of which are nonphysical, since they lead to imaginary values ​​of quantities that are supposed to be valid. Three other physical solutions are given below, together with the corresponding power formulas.
Solution 1

x1=x3, quadx2=x3


Power

p2= frac|u1|2x32r2 left(r1r2+x32 right)2, quadq2= frac|u1|2 ,x33 left(r1r2+x32 right)2


Solution 2 and 3

x1= frac1r2 left( sqrtr1r2 left(x32r1r2 right)r2x3 right), quadx2= frac1r1 left( sqrtr1r2 left(x32r1r2 right)r1x3 right)