SaM - 5 Wires LVDT (Linear Variable Differential Transformer)

Remeber:

Now we see a different circuit for an LVDT, with a simple addition of a wire we obtain:

Just by adding a wire the carrier amplifier is no longer needed.

The two mutual inductances $M_{1}$ and $M_{2}$ of the two windings will be (using a linear approximation):

$M_{1} = M_{0} (1 + k x)$ .IMPORTANTE

$M_{2} = M_{0} (1 - k x)$ .IMPORTANTE

$k x$ this time is a little more complicated and it is defined as: $k x = \frac{M _{1} - M _{2}}{2 M _{0}}$
So: $M_{1} - M_{2} = 2 M_{0} k x$

$M_{0}$ is the mutual induction of both $M_{1}$ and $M_{2}$ at rest.

We will obtain as an ouptut two voltages, with different values, so a differential output. This two voltages can be converged into one using a differential amplifier. Here is a graph of the two output voltages for different values of $x$ :

With respect to the previous LVDT sensor, this time we have a differential amplifiers, instead of a carrier amplifier, but I need an additional LPF, and an additional wire, and this could be more complex in terms of connection and cabling.
⇒ Just becouse of cabiling issues, this solution here it's not always used.

Here’s the IC for Read-Out of 5W LVDT. IC: Integrated Ciruit 5W: 5 Wires LVDT: Linear Variable Differential Transformer

Specifically, we have used two operational amplifiers, and two LPF (acting as rectifiers and for noise rejectionNOT_SURE_ABOUT_THIS ).

Finally we can compensate for both $T$ and $E_{1}$ thanks to the differential output, since:
$M_{1} - M_{2} = 2 M_{0} k x$ .
$M_{1} + M_{2} = 2 M_{0}$ .
And both outputs $A$ and $B$ depend on $E_{1}$ .
==So the output will just depend on $k x$ ==.

Memory Card

We just added a reference to ground between the two seondary windings.

Tho just by doing that the carrier amplifier is no longer needed:

We will obtain as an ouptut two voltages, equal and opposed, so a differential output, they can be converged into one using a simple differential amplifier.
So what happens is that I put two different amplifier, and then I need an additional wire, and this could be more complex in terms of connection and cabling.
⇒ Just becouse of cabiling issues, this solution here it’s not always used.

IC for Read-Out of 5W LVDT

IC: Integrated Ciruit 5W: 5 Wires LVDT: Linear Variable Differential Transformer

NOTE: Instead of a simple differential amplifier we have used two Operational Amplifiers, actually 2 Op.Amp plus 2 Rectifier.
After those we have 2 LPF.
This way we also “isolate” the output from the value of $E_{1}$ .

🪴 Quartz 4.0

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SaM - 5 Wires LVDT (Linear Variable Differential Transformer)

Remeber:

Memory Card

IC for Read-Out of 5W LVDT

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