circuit-designinductanceproximity-sensorstrain-gage
We're doing a school project where we have to measure weight on a fitness machine.
The user can select how many weight plates to use by moving a rod like shown in the picture.
How would you do it?
We've talked about the following solutions:
You need to use the sensor in one-half of a wheatstone bridge circuit. The other half can be made from two reasonably identical resistors forming a potential divider like the strain gauge does.
+Input of instrumentation amp to centre point on resistors and -Input to centre-point of gauges: -
The two 350 ohm resistors that feed the lower input are the two reasonably identical resistors. Your strain gauge is the two resistors to the left feeding the upper input. In fact the inputs can be reversed. Ensure your power rails are sufficient for the AD620 - remember the common mode range that this amp needs is: -
(-Vs + 1.9V) to (+Vs - 1.2V) so if you operating it from a 5V rail and ground it should be OK because the mid-points of the voltage dividers are at ~2.5V give or take a few tens of milli volts.
Black = lower end of the two left resistors, white = upper end and red goes to the IA input. You'll also need to connect the reference input to (possibly) half rail and note that it is usually reasonable to use the junction of the two reasonably matched resistors and maybe add a 10uF cap to ground but check the IA data sheet on this.
Arguably, the easiest way is to just measure it:
+--------+
| |
| [50R]
| |
| [C]
[GEN] |
| +------+
| | |
| [L] [SCOPE]
| | |
+--------+------+
With \$C\$ known, adjust the generator's output frequency for a peak across the scope, then determine the reactance of the capacitor with:
$$X_c=\frac1{2\pi fC}$$
Next, since the reactance of the capacitor and the inductor will be equal at resonance, (the frequency at the maximum amplitude of the peak on the scope) solve for the inductance with:
$$L=\frac{X_c}{2\pi f}$$
Both equations can be combined:
$$L=\frac{1}{(2\pi f)² C}$$
You may need to adjust the value of the \$R\$ in order to get a nice peak.
Best Answer
What not to do
One obvious solution would be to put a load cell in-line with the wire, above the weights. That would probably work well, albeit with some extra peaks and troughs in the signal as the weights move up and down. But: the wire and it's connection to the weights may be safety critical. If the wire parts, it will lash around, and the weights would drop with quite a bang. So don't interfere with the wire unless you're very sure it's remaining strong.
Some Ideas