Modelling an eddy current sensor

Question

An eddy current sensor is modeled as an air core transformer with shorted secondary:

schematic

^{simulate this circuit – Schematic created using CircuitLab}

According to Kirchoff's laws, the total voltages around the loops are given by:

The equivalent impedance on the primary side is given by the following equation where M is the mutual inductance:

Can someone explain how we arrive at this equation?

Update:

\$\begingroup\$ Try solving it first \$\endgroup\$
– Voltage Spike ♦
Commented Jun 29 at 4:35 — Voltage Spike, Commented Jun 29 at 4:35
\$\begingroup\$ You have made some errors ... i.sstatic.net/8SzVxbTK.png \$\endgroup\$
– Antonio51
Commented Jun 29 at 11:22 — Antonio51, Commented Jun 29 at 11:22

Antonio51 · Accepted Answer · 2024-06-30 10:51:34Z

1

Link to "modelling" ...

With the new update ...
If you solve the two equations now, you should have the results you are searching.
Straightforward.

t1 and t2 are what you are searching for, just rewrite as you want.

Simulated with microcap v12.
Presented Zin = f(frequency) -> Phase (Zin) .vs. abs(Zin) (kc parameter).

answered Jun 29 at 15:37

Antonio51

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\$\begingroup\$ Huh, how does the plot show a complex pole/zero with no capacitors shown? \$\endgroup\$
– Tim Williams
Commented Jun 30 at 11:56
\$\begingroup\$ Parameter kc seems to be the key (?). \$\endgroup\$
– Antonio51
Commented Jun 30 at 12:42
\$\begingroup\$ Well, yeah that's showing the sensitivity, but can you explain why it looks like that? Ohhh, also it's phase, I see, but that's not what I get from simulating the same (I think) thing: i.sstatic.net/GiaWy3QE.png i.sstatic.net/nSuve7vP.png \$\endgroup\$
– Tim Williams
Commented Jun 30 at 15:15
\$\begingroup\$ @TimWilliams ... Is kc varying from 0.9 to 1? It is not "frequency" with axis x. It is the module of "input impedance" ... Although, you are right if it is "frequency" (confirmed). \$\endgroup\$
– Antonio51
Commented Jun 30 at 15:32
\$\begingroup\$ Module? Modulo what--? Oh, magnitude? Ohh, so it's a-- well, it's not a polar graph because it's, not, but it's plotting magnitude against angle. That's... very strange. \$\endgroup\$
– Tim Williams
Commented Jun 30 at 15:35

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