You calculate the core flux with the equotion above, and the inductance takes the sum of all fluxes through each turn. The flux through each turn is the same and equal to core flux. The core flux is proportional to N, and the per-turn sum of flux is proportional to \$N^2\$.

Another way to express this dependancy is to say: because of magnetic coupling between turns.

I am not sure if I clearly understand your new sensor, but as far as I did, these are the main physical principles:

- The sensor is a transformer with \$n_1\$ turns in the primary and \$n_2\$ turns on the secondary. This is not an LVDT, because the \$n=n_2/n_1\$ is constant,
- The displacement \$x\$ is measured as the advance of the iron rod. \$x=0\$ mainly out of the coils, \$x=1\$: mainly inside the coils,
- No magnetization/hysteresis effects are relevant (according your OP),
- \$v_1(t)\$ is sinusoidal. We could assume phasors here.

Evidently, because the relative magnetic permeability of the air and the iron are \$\mu^r_{air}=1\$ and \$\mu^r_{iron}=4000\$ aprox. respectively, the magnetic flux \$\phi\$ (in Webers) will be fully passing through the core with \$x=1\$ and almost totally passing through the air with \$x=0\$.

The shape of the flux depend on a 2D Magnetostatic FEM modelation (though the geometry is actually a 3D cylinder, you can simplify it to a rectangular 2D rod). The primary excitation is the Magnetic Field \$H\$ (in A.m) and the rod-air system measurement is the Magnetic Flux Density \$B\$ (in Tesla), or directly the voltage through the coils (depending on your package skills), doing several runs for different \$x\$ values. Any FEM package or program for 2D/3D magnetostatic such as Ansys, Comsol, CST, or any other cheaper/easier alternatives are useful and perhaps better for this case.

If you are more hardware focused, and you don´t want to model anything with FEM, i will not blame you. Because \$\mu^r_{air}<<\mu^r_{iron}\$ the approximation solution can be expressed as:

$$v_2(x)=m(x)+e(x)$$

where \$m(x)\$ is the main linear flux component -the integrated flux density through the rod- and \$e(x)\$ is the remaining error -the integrated flux density out of the rod.
Hence:

- \$m(x)=n \cdot v_1(t) \cdot x\$, that is, no flux amplified with the rod out of the coils and all the flux is amplified with the rod inside the coils. The linear model comes from Maxwell's, as a standard transformer, just in the rod segment. No magnetic nor electrical losses.
- \$e(0)=e_1, e(1)=e_2, e_1>e_2=0\$. \$e_1\$ is the integral of the flux density when the rod is fully out, and \$e_2\$ is zero, because in this case all the flux is explained by the "linear" part. \$e(x)\$ is positive, have its maximum on \$x=0\$, and tends smoothly to zero. This approximation is VERY violent, rough and mostly "empirical" (?). This components ruins the linearity, represents all the integrated flux density over the paired coils, and deviates your sensor from an LVDT, which is much more linear, because on an LVDT, the integrated flux density over the air is much less.

\$v_2(x)\$ indeed should have a maximum for \$x=1\$, a small nonzero value for \$x=0\$ and a minimum with \$x=\pm\inf\$ (with the rod far out), which shows that you make the data only for a piece of the rod inserted. The whole curve should be half bell shaped.

I am really sorry for not including pictures, but my FEM package is not in this computer. But if you really require them, we can start another question with that...

## Best Answer

This is a great question Colin, one which I puzzled over long and hard myself before coming to an intuitive understanding.

Magnetizing inductance and magnetizing current are two aspects of the same departure of a real transformer from the ideal transformer model; in particular, as you mention, the fact that a real transformer core has finite permeability.

The permeability of an ideal transformer core approaches infinity. Interestingly, this means the magnetizing inductance of an ideal transformer is

infinite, not zero. It does not appear in the ideal circuit because it models as anopen circuit, not because it's not present. As a consequence, the flux needed in the ideal core to induce a voltage in the secondary is produced by a vanishingly small current (read zero). So the non-ideality of a real transformer is that it's inductance is less than infinite and therefore requires a current to flow to produce flux.There are three other important factors to note about this:

The change in flux produces a

voltage. It does not by itself produce a current. Essentially, the \$\frac{d\phi}{dt}\$ produces a voltage and the voltage produces a current if it can, based on the impedance of the secondary circuit.For a given permeability and frequency, \$\frac{d\phi}{dt}\$ is proportional to the peak flux. If the peak (or peak-to-peak) flux is lowered, the voltage on the secondary drops. This means that the magnetizing current is the

minimumcurrent that can flow without dropping the output voltage. (Here I'm thinking of magnetizing current as sinusoidal AC. The same principles apply in switching transformers fed with square-wave input voltages but the complicating factors that introduces are perhaps better left until later.)Nothing says the magnetizing current has to be in-phase with the input voltage, in fact, in a well-designed transformer, in general, the magnetizing current will be near 90-degrees out of phase. The real portion of the complex magnetic current gives a measure of the resistive and core losses. The imaginary part is pulsing back and forth, producing flux but not consuming power.

Now, when a load is placed on the secondary, a secondary current flows. This current produces a magnetic field (flux) that

opposesthat produced by the primary. This reduces the effective inductance of the primary which allows more current to flow through it, which produces more flux. This feedback cycle finds equilibrium when the magnetizing flux is restored and the output voltage is maintained. So the magnetizing flux is there at the same level at all times during the transformer's normal (i.e. non overload) operation. There might be a boat-load of other flux happening in both directions as power is transferred in a loaded state, but thenetflux is always at the constant level produced by the magnetizing current.From this you can see that the primary must always produce more flux than is cancelled out by current flow in the secondary and this surplus is provided by the magnetizing current. An open-circuit secondary takes the secondary current out of the picture and allows ready measurement of this magnetizing current.

This might make some sense of the expression "setting up the flux", although I think that's one of those less-than-instructive expressions that only makes sense when you already understand and don't need it any more :)