If you add a 220nF in series with the coil it will be resonant at near 16kHz.
Driving that with a 0-12V, 16kHz square-wave showed a peak inductor current of about 350mA in simulation.
To get more current you could drive the resonant circuit with a push-pull audio bridge circuit such as the TDA2005, which would near double the supply voltage across the load (about 20Vpp with a 12V supply).
That should give a maximum peak current in the resonant circuit (220nF in series with the coil) of over 600mA.

Edit: If you can tolerate some frequency deviation then you could add a series capacitor to form a resonant circuit at about 16kHz and use feedback to form a high current resonant oscillator at this resonant frequency. You can do that with a fairly simple bridge driver circuit, such as below:
The LTspice simulation shows a peak inductor current of a little over 700mA.
Note: R4 needs to be rated for at least 2W.

View attachment 88160

Hi,

Thanks for your valuable feedback.
In the design, for C = 330nF and L = 458uH, the resonant frequency is 12.95KHz.

Thanks & Regards.

 

If you add a 220nF in series with the coil it will be resonant at near 16kHz.
Driving that with a 0-12V, 16kHz square-wave showed a peak inductor current of about 350mA in simulation.
To get more current you could drive the resonant circuit with a push-pull audio bridge circuit such as the TDA2005, which would near double the supply voltage across the load (about 20Vpp with a 12V supply).
That should give a maximum peak current in the resonant circuit (220nF in series with the coil) of over 600mA.

Edit: If you can tolerate some frequency deviation then you could add a series capacitor to form a resonant circuit at about 16kHz and use feedback to form a high current resonant oscillator at this resonant frequency. You can do that with a fairly simple bridge driver circuit, such as below:
The LTspice simulation shows a peak inductor current of a little over 700mA.
Note: R4 needs to be rated for at least 2W.

View attachment 88160

Hi,

Thanks for your valuable feedback.
Would you please confirm the value of R4 resistor?

Thanks & Regards.

 

If you add a 220nF in series with the coil it will be resonant at near 16kHz.
Driving that with a 0-12V, 16kHz square-wave showed a peak inductor current of about 350mA in simulation.
To get more current you could drive the resonant circuit with a push-pull audio bridge circuit such as the TDA2005, which would near double the supply voltage across the load (about 20Vpp with a 12V supply).
That should give a maximum peak current in the resonant circuit (220nF in series with the coil) of over 600mA.

Edit: If you can tolerate some frequency deviation then you could add a series capacitor to form a resonant circuit at about 16kHz and use feedback to form a high current resonant oscillator at this resonant frequency. You can do that with a fairly simple bridge driver circuit, such as below:
The LTspice simulation shows a peak inductor current of a little over 700mA.
Note: R4 needs to be rated for at least 2W.

View attachment 88160

Hi,

Thanks for your valuable feedback.
Would you please explain me the operation of the circuit?

Thanks & Regards.

 

To get 0.5A peak current at 16kHz through a 458uH coil you need roughly 24V peak wave. I you would have to quadruple your 12v supply voltage to get that 48Vpp required.

Hi,

Thanks for your valuable feedback.
Would you please explain this to me?

Thanks & Regards.

 

As AnalogKid said before me, the impedance of that inductor at 16kHz is 47 ohms. V=I*R, so V=47 ohm * 0.5 A, which is roughly 24V.
If you want average current of 0.5A, then the peak current (without using LC resonant tank as mentioned by crutshow) will be 1.414 times higher than the average. And to get that peak current, you also need to have appropriate peak voltage which dictates your supply voltage.

 

As AnalogKid said before me, the impedance of that inductor at 16kHz is 47 ohms. V=I*R, so V=47 ohm * 0.5 A, which is roughly 24V.
If you want average current of 0.5A, then the peak current (without using LC resonant tank as mentioned by crutshow) will be 1.414 times higher than the average. And to get that peak current, you also need to have appropriate peak voltage which dictates your supply voltage.

Hi,

Thanks for your valuable feedback.

With regards.

 

If you add a 220nF in series with the coil it will be resonant at near 16kHz.
Driving that with a 0-12V, 16kHz square-wave showed a peak inductor current of about 350mA in simulation.
To get more current you could drive the resonant circuit with a push-pull audio bridge circuit such as the TDA2005, which would near double the supply voltage across the load (about 20Vpp with a 12V supply).
That should give a maximum peak current in the resonant circuit (220nF in series with the coil) of over 600mA.

Edit: If you can tolerate some frequency deviation then you could add a series capacitor to form a resonant circuit at about 16kHz and use feedback to form a high current resonant oscillator at this resonant frequency. You can do that with a fairly simple bridge driver circuit, such as below:
The LTspice simulation shows a peak inductor current of a little over 700mA.
Note: R4 needs to be rated for at least 2W.

View attachment 88160

Hi,

Thanks for your valuable feedback.
Now, I would like to receive the signal which has been transmitted from the coil of 458uH inductance. At the receiver end, there are two coils are series oppositely connected such that if there is any disturbances in uniformity of the transmitted field, the signal at the receiver end would be detected.
Would you be kind to tell me the value of the capacitors and resistors to receive the signals?

Thanks & Regards.

 

You are missing R4 from +input to ground, google differential amplifier.

 

You are missing R4 from +input to ground, google differential amplifier.

Hi,

Thanks for your valuable feedback.
Would you be kind to tell me the value of the resistors and capacitors to receive the magnetic field generated by the 16KHz sinusoidal signal at the TX-side.

 

Hi,

Thanks for your valuable feedback.

Please see the attachment. It is the receiver circuit.
This circuit will receive a signal of 16KHz in the transmitted EM field which is generated by the 16KHz sinusoidal signal at the Tx side.
The two coils are connected series opposite so that when there is no metal signal would cancelled out and the output of the opamp would be zero. But, if there is any target metal, there would be a non-zero output from the opamp.

Would you please help me to find out the value of the capacitors and resistors.

Thanks & Regards.

 

First you need to say what gain do you want from this circuit. Also, if the coils are already connected out of phase, then you don´t need a differential amplifier as the difference is already done for you by the polarity of the coils, and that might make the circuit a bit simpler.

 

Hi,
Thanks for your valuable feedback.

I want to drive a speaker at the output of the opamp. The speaker will make noise if there is any metal object present in the field between Tx and RX.

Thanks & Regards.

 

First you need to say what gain do you want from this circuit. Also, if the coils are already connected out of phase, then you don´t need a differential amplifier as the difference is already done for you by the polarity of the coils, and that might make the circuit a bit simpler.

Hi,

Thanks for your valuable feedback.
The coils may not be ideal and hence may not cancel the signal perfectly. That is why, the use of differential amplifier would confirm that the signals are being cancelled out for no metal present between Tx and Rx and would make a noise through a speaker if there is any metal between Tx and RX.

Thanks & Regards.

 

Hi,

Thanks for your valuable feedback.
Please see the attachment.
Here, I have used a typical differential amplifier to receive the signal in the EM field which is created by 16KHz signal at the Tx side.
Now, if there is no metal in between Tx and Rx, the output of the opamp would be zero as in that case V2 = V1. But, if there is any metal or conductor present in between Tx & Rx, V2 is not equal to V1 and the difference voltage (V2 - V1) would be amplified.
Would you be kind to tell me if anything change is required in signal amplification to pick up the signal in the EM field of 16KHz which is generated at the TX side.

Thanks & Regards.

 

It seems ok, try it on a breadboard and see how it works.

 

Hi,

Thanks for your valuable feedback.
It seems that the input impedance at the inverting terminal is 1K ohms and that of non-inverting terminal is 101K ohms.
Would you please tell me if I use a 3 opamp instrumentation amplifier, can't it have a great advantages over differential amplifier?

Thanks & Regards.

 

Yes they are different but that really shouldn´t matter much. Instrumentation amp might be better, but also might not make any significant difference.

 

Yes they are different but that really shouldn´t matter much. Instrumentation amp might be better, but also might not make any significant difference.

Hi,

Thanks for your valuable feedback.
I am using LM324 opamp for designing an Instrumentation Amplifier. As it is a quad opamp IC, hence, I can design it with a single IC.
Please let me know if there is any better IC for designing an Instrumentation amplifier in this scenario of metal detection.

Thanks & regards.

 

If you add a 220nF in series with the coil it will be resonant at near 16kHz.
Driving that with a 0-12V, 16kHz square-wave showed a peak inductor current of about 350mA in simulation.
To get more current you could drive the resonant circuit with a push-pull audio bridge circuit such as the TDA2005, which would near double the supply voltage across the load (about 20Vpp with a 12V supply).
That should give a maximum peak current in the resonant circuit (220nF in series with the coil) of over 600mA.

Edit: If you can tolerate some frequency deviation then you could add a series capacitor to form a resonant circuit at about 16kHz and use feedback to form a high current resonant oscillator at this resonant frequency. You can do that with a fairly simple bridge driver circuit, such as below:
The LTspice simulation shows a peak inductor current of a little over 700mA.
Note: R4 needs to be rated for at least 2W.

View attachment 88160

 

If you add a 220nF in series with the coil it will be resonant at near 16kHz.
Driving that with a 0-12V, 16kHz square-wave showed a peak inductor current of about 350mA in simulation.
To get more current you could drive the resonant circuit with a push-pull audio bridge circuit such as the TDA2005, which would near double the supply voltage across the load (about 20Vpp with a 12V supply).
That should give a maximum peak current in the resonant circuit (220nF in series with the coil) of over 600mA.

Edit: If you can tolerate some frequency deviation then you could add a series capacitor to form a resonant circuit at about 16kHz and use feedback to form a high current resonant oscillator at this resonant frequency. You can do that with a fairly simple bridge driver circuit, such as below:
The LTspice simulation shows a peak inductor current of a little over 700mA.
Note: R4 needs to be rated for at least 2W.

View attachment 88160

Hi,

Thanks for your valuable feedback.
Would you be kind to explain me the following point.
1. Now, suppose, i want to reduce the current to 500mA. In that case, which part of the circuit needs modification?
The circuit is attached here.

Thanks & Regards,
Babun Chandra Pal