As suggested by Andy in these posts proximity sensor technology, range of 150mm, Metal detector "Head sensor" and How can I drive an LC tank circuit to get a nice and clean sine wave?, I built my version with add-ons of graphite sheet shielding inside and with a little different circuit than maijaz99 built.

Andy wrote in one post that the more the current to drive the coils the higher the sensitivity of the detector will be, so I set the current at 7 amperes by adding a transistor and resistor to it.

I tried with Maijaz99's circuit first which was circulating 0.25A in it, but the sensitivity was very limited. Only large pieces of metal were getting detected, and only by a very small shift in phase and amplitude.

My circuit is a basic current control circuit.

  • R calculation : 0.7V / 7A = 0.1 Ohm
  • Resistor wattage calculation = 0.7V X 7A = 5 watt. (0.7V is needed to turn on the transistor, 0.6 ideally)

The problem is when no LC tank is connected and 255 kHz are triggering the MOSFET, at no-load condition my multimeter shows 7 amperes passing through the circuit very easily. When the LC tank is connected as shown in the schematic attached it takes only 100-150 mA. What is going wrong?

There was only one reason for changing the circuit. In that circuit, I would have needed to add a very high wattage resistor to control the current. In my circuit, a smaller wattage resistance can be used. Also as shown in the schematic, the frequency was calculated at 366 kHz, but I achieved resonance at 255 kHz switching frequency.

With my circuit also, I have achieved resonance both in Tx and Rx coils, but as the current is smaller the same thing is happening. Only large pieces of metal are detected.

How can I circulate more current through the coil so that it can detect objects that are in size 1mm or smaller?

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  • \$\begingroup\$ Parallel resonance circuitry impedance gets higher at resonance frequency. Series type is the other way around. Though, suggested methods are good, and though I do not like what it may imply, you may check how the electronic anti tank/vehicle land mines work. \$\endgroup\$
    – jay
    Sep 21, 2021 at 15:48
  • \$\begingroup\$ Third photo shows very little enamel scraped off the inductor leads ... I hope they are making better contact than it looks like. \$\endgroup\$
    – user16324
    Sep 21, 2021 at 17:16
  • \$\begingroup\$ Yes, it makes the connection properly. Otherwise the oscilloscope wouldn’t show the resonance. \$\endgroup\$
    – Rushabh
    Sep 21, 2021 at 17:27
  • \$\begingroup\$ How about my answer in the electronics.stackexchange.com/questions/538642/… having a resonant LC tank? \$\endgroup\$ Sep 21, 2021 at 17:38
  • \$\begingroup\$ @MarkoBuršič Hi, Can you give a bit more practical answer than a theoretical one on this circuit, please? I really don't have the core knowledge of LC circuits. \$\endgroup\$
    – Rushabh
    Sep 22, 2021 at 7:03

3 Answers 3


R calculation : 0.7V / 7A = 0.1 Ohm

You are missing the point here. The DC current that flows through the tuned tank circuit in the drain of the MOSFET should be just a few hundred milliamps. The AC current that circulates in the tank will be several amps. Control the gate voltage through a resistor and potentiometer and drive the gate with a small AC voltage. In fact, close-the-loop and get it to self-oscillate.

255 kHz are triggering the MOSFET

Here's another problem; pharmaceutical metal detectors (due to their quite small coil size/aperture) tend to run at about 1 MHz and will have an oscillator coil inductance of about 400 nH when encased within the metal box they fit in.

I ran all my metal detectors from a 24 volt DC power supply and, as previously mentioned, the standing DC current through the MOSFET (IRFZ44 from memory) was about 250 mA.

but the sensitivity was very limited

The main thing here is a tuned receiver circuit. You have not mentioned anything about how you measured the receive signal but, if you expect to say any movement on the receive waveform with sub mm contaminants using an oscilloscope then forget that. The sort of real signal induced in the coil for a small detectable test sample is less than 100 nV RMS. You'll never see that unless you parallel tune the receive coil (an uplift of 30 ish on voltage to the microvolt level) then a gain of 1000 amplifier to put your detectable signal in the millivolt range. Then you can detect small contaminants.

Maybe you have done all of this but, your post doesn't indicate so.

How can I circulate more current through the coil so that it can detect objects that are in size 1mm or smaller?

  • a 24 volt power supply
  • a bias current of about 250 mA
  • an operating frequency of around 1 MHz
  • the best ceramic NP0 tuning capacitors you can get
  • a tuned receive circuit
  • stable coils that are 3 mm diameter to improve Q factor
  • a good low noise gain of 1000 AC amplifier.
  • a very good synchronous rectification scheme to help detection
  • a great auto-balance circuit that is always trying to actively balance of the offset of the receive coils (except it halts as soon as a signal begins to be detected)
  • really good software filters to give more signal to noise.
  • mechanical stability in coils, coil former and wires to coils.

You appear to be a good way from reaching those targets.

  • \$\begingroup\$ "a great auto-balance circuit that is always trying to actively balance of the offset of the receive coils (except it halts as soon as a signal begins to be detected)" Can you provide some basic info? How to design such a circuit? \$\endgroup\$
    – Rushabh
    Sep 23, 2021 at 11:11
  • \$\begingroup\$ You produce quadrature components of the oscillation waveform and create inverses of them. Then you use something like JFETs to control those four sources to produce an exact replica (but inverted) of the signal you are trying to diminish. The IQ detection circuit then slowly drives the four JFETs in such a way as to create an equal magnitude (but opposite phase) signal that then "sums out" the unwanted offset signal. \$\endgroup\$
    – Andy aka
    Sep 24, 2021 at 7:09

Images taken here

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As you can see, the source current is minimal at resonance. The current through your MOSFET isn't equal to the coil current. You do just cover the losess in the LC tank. Ideally the current would be zero if there would be no resistance.

The capacitor can charge maximally to the source voltage 15V. The energy is tweaking between L and C. So we could say the max energy stored in the capacitor is:

$$W=\dfrac{C\cdot v^2}{2}$$

And it converts into inductor energy:

$$W=\dfrac{L\cdot i^2}{2}$$

Without any losses (R=0), the maximum current is:

$$i_{peak}=\sqrt{\dfrac{C\cdot v^2_{peak}}{L}}$$

The limiting resistor doesn't matter too much, here. Rather the voltage and quality of the LC tank.

  • \$\begingroup\$ Okay, by the quality you mean how stable my capacitors are, right? \$\endgroup\$
    – Rushabh
    Sep 23, 2021 at 7:50
  • \$\begingroup\$ @Rushabh No, the Q factor (quality factor) of the resonant circuit . How much resistance your L and C have, skin effect, ..etc. \$\endgroup\$ Sep 23, 2021 at 8:01

EDIT : This assembly will help you understand why the initial assembly "cannot" work as it is because of Q2, I think.

As stated by @Marko Buršič, after choosing other components values ...

enter image description here

Here another view of current variables. Not yet "steady state". Only "start". enter image description here

Here is a another schematic that can help you ... It is a half-bridge (24 V).Under 12 V, the RMS current will be 4 A. enter image description here

See the voltage across the inductor \$(50uH, @ 250kHz)\$ ... for a current around 8 A rms.

Here an other alternative.

Unless I am wrong : \$L * di/dt = L * 2 * pi * f * Ipeak = +/- 600 V \$ ...

If R3 is higher, voltage and current will be, of course, lower.

  • \$\begingroup\$ I will give it a try. Meanwhile, Can you tell me how to increase the current in my existing circuit? I will give it 24V but as far as the circuit is concerned, Is it suitable for this kinda application? \$\endgroup\$
    – Rushabh
    Sep 22, 2021 at 6:58
  • \$\begingroup\$ @Rushabh Just edited my answer for your schematic. \$\endgroup\$
    – Antonio51
    Sep 22, 2021 at 9:37
  • \$\begingroup\$ Thank you for the update :) \$\endgroup\$
    – Rushabh
    Sep 23, 2021 at 7:49
  • \$\begingroup\$ Can you tell me what is E1 in your schematic? \$\endgroup\$
    – Rushabh
    Oct 3, 2021 at 16:42
  • \$\begingroup\$ @Rushabh - E1 is a "Voltage controlled voltage source" (V of V). Just to control M1 "out of phase" M2. That is why the gain is -1. You can take a driver which can drive a half bridge (up and down). \$\endgroup\$
    – Antonio51
    Oct 4, 2021 at 5:55

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