I've built an LVDT (Linear Variable Differential Transformer) that is a little different than how it is conventionally done with 2 secondary coils positioned adjacent to a primary coil. My design is a primary and secondary coil directly on top of each other and a iron bar that will slide in between them.

There are a lot of resources about transformers that talk about a ratio of number of windings, but this is not the case here since the number of windings is constant, but the voltage changes.

LVDT Homemade

I have a basic understanding of how it works, but I think it is incomplete and I need to relate it to equations.

  1. When I insert the iron core, the inductance of both coils increases.
  2. The first coil that is energized with an AC current, produces changing magnetic flux.
  3. The second coil more strongly experiences the magnetic flux because of the iron core. At the same time, the second coil will produce a higher voltage because the iron core has increased its inductance.

The relevant laws to this, that I have found, are:

  1. Faraday's Law (Change in magnetic environment produces emf)
  2. Bio Savart's Law (Inductance and magnetic field)

Any help would be appreciated!

I've taken some data to relate displacement of the coil to voltage change in the secondary coil. It is mostly a linear relationship, excluding the last couple points.

voltage vs displacement

  • \$\begingroup\$ LVDT? Add it into the question - not in the comments. \$\endgroup\$
    – Transistor
    Mar 13, 2016 at 21:13
  • \$\begingroup\$ @transistor I'm sorry, I don't follow. \$\endgroup\$
    – Klik
    Mar 13, 2016 at 21:14
  • \$\begingroup\$ What's an LVDT? Your question doesn't explain the acronym. \$\endgroup\$
    – Transistor
    Mar 13, 2016 at 21:24
  • \$\begingroup\$ en.wikipedia.org/wiki/Linear_variable_differential_transformer \$\endgroup\$ Mar 13, 2016 at 21:28
  • 1
    \$\begingroup\$ @SpehroPefhany: Actually, more like an NLVT. There's a reason that LVDTs are built the way they are -- most of the nonlinearities cancel out. \$\endgroup\$
    – Dave Tweed
    Mar 13, 2016 at 23:32

1 Answer 1


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:


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...


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