7
\$\begingroup\$

I am working on my senior project. As part of my project I need to use a metal detector. I am using the following circuit schematic.

The principle is based on the properties of inductor coil:

$$L=N^2\mu \left(\frac{A}{l}\right)$$

Where \$N\$ is number of turns, \$\mu\$ magnetic permeability, \$A\$ is area of the coil, and \$l\$ is the length of the coil (if you think of coil as cylinder, l is its height).

Once a metal object is placed near the coil, it behaves as a core for the inductor, and the \$L\$ changes, it becomes:

$$L = L_{air}\times\mu$$

Hence RLC circuit frequency and the following sound produced by the speaker changes as well. When I set up the circuit, I have used a small inductor about \$1\space\mathrm{cm}\$ in size which was \$10\space\mathrm{mH}\$, as it was specified in circuit (there are different variations of the same circuit with small differences), it was detecting the metal, but the range was very small, at most \$1\space\mathrm{cm}\$.

I researched how I can improve the range (not in academic articles, because I could not find anything specific, would be happy to hear suggestions from you) and read that area of the coil is very important, so I calculated and came up with \$100\space\mathrm{m}\$, \$150\space\mathrm{turns}\$, \$10\space\mathrm{cm}\$ in radius, and \$2\space\mathrm{cm}\$ in length coil.

I bought the wire, and wanted some appropriate round shaped object for that. I went to a place which cuts figures from acrylic, and got this:

coil1 coil1

Now when I connect it to the circuit, sensitivity is even less. I can't get the reason, possibly because of the magnetic properties of acrylic, I tried to look in internet, but the datasheet provides just the electric properties.

Now I am stuck, I have heard that people got \$40\space\mathrm{cm}\$ range, which is more than ok for me. I don't have time to build a different detector. My questions are:

  1. Should I remove acrylic layer?
  2. How can I increase the range? (I have different resistors and capacitors, and a potentiometer, which I can use, but I doubt that they will help with sensitivity, any offers for increasing range are ok)
  3. I am using 9V batteries and they discharge very fast, how do you think what might stand behind it? Sometimes I use potentiometer instead of resistor, and try all values. May it be due to very low resistance?
\$\endgroup\$
4
  • 1
    \$\begingroup\$ (1) No. (2) Use a big block of soft iron as your target (3) Use a 64R speaker. Basically the circuit you are using is not a good design for a sensitive metal detector. \$\endgroup\$ Commented Aug 20, 2015 at 15:46
  • 1
    \$\begingroup\$ The circuit for the metal detector is not too good .Modern metal detectors are very complex .The most basic metal detector is the hetrodyne type which is older than me and would eat your downloaded 555.The 555 timer chip does have some snags that are nentioned on the www.badbeetles.com website.Otherwise you could try a pulse induction type which I did once and it worked fine . \$\endgroup\$
    – Autistic
    Commented Oct 2, 2015 at 10:47
  • \$\begingroup\$ What is the IC in the schematic? \$\endgroup\$ Commented Jul 17, 2017 at 12:33
  • \$\begingroup\$ Should be this en.wikipedia.org/wiki/555_timer_IC as I remember. It is also mentioned in circuit schematic as NE555. \$\endgroup\$
    – UserRR
    Commented Jul 18, 2017 at 13:30

1 Answer 1

2
\$\begingroup\$
  1. No, removing the layer of plastic will not do anything as gravity does not have any particles or anything that must move around to make it work

  2. Increase the amount of energy going to the coil, for example using lower value resistors or potentiometers or increasing the voltage slightly

  3. Try using D cell batteries as they have a higher current capacity then 9V cell batteries

\$\endgroup\$
2
  • \$\begingroup\$ Huh? What does gravity have to do with this? \$\endgroup\$ Commented Jul 17, 2017 at 12:31
  • \$\begingroup\$ Because this discussion was few years ago, I don't remember very well, but I believe he meant that isolation does not have effect on this device. \$\endgroup\$
    – UserRR
    Commented Jul 18, 2017 at 13:32

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.