How to determine LVDT sensitivity?

Question

I performed an experiment using an LVDT where I used a generator voltage of 4.84V. I determined the secondary voltage at the beginning position, obtaining 2.48V and at the end (displacement os 22.5 cm) 2.12V. Now I have to compute the sensitivity in mV/V/mm. I thought that this was given by $$S=\frac{\Delta V_{sec}}{V_{prim} \Delta x}$$

which leads me to 0.3306 mV/V/mm.

Now apparently this is wrong and I should obtain a value around 4... My guess is that the expression I'm using, even if its dimensions are correct is not right. I read on this: http://www.gatewaycoalition.org/files/bioengg_Instrumentation/site/LinearVariableDifferentialTransformer.html that they basically multiply the output and the input and divide by the excitation voltage times the displacement. But in that case, the units are not right. Can someone help me? Thanks!

Answer 1

Your math looks right, with the numbers shown. I suspect that your measurements, or experimental setup is probably wrong.

Most likely, you are being confused by the AC voltage output measurements. Try using (-2.48) and (+2.12) in your calculations. I then get an answer of 4.22 mV/V/mm.

LVDT's tend to have 2 windings, with the signal output (in VAC) going like:

Distance[mm], Output[Vac]
0 mm,         2.48 Vac
112.5 mm,        0 Vac (right at mid-stroke)
225 mm,       2.12 Vac

So:

dX = 225 mm

and dVac = dVac (1st winding) + dVac (2nd winding), or:

dVac = 2.48 + 2.12 = 4.6 Vac

So, with Vexcitation = 4.84 Vac,

S = 1000*(dVac / Vexc) / dX:

S = 1000 * (4.6 Vac /4.84 Vac ) / 225 mm = 4.22 [mV/V/mm]