LPF for removal of radio frequency transmission

[pal114525]

Hi,

I know the basic structure of an LPF.
Would be kind to tell me the following point?

1. How to determine the value of resistor and capacitor for an LPF that would remove radio frequency transmission? This type of filter is required for metal detection.

Thanks & Regards.

 

[jolly jacob]

Hi

Metal detector basically using a tuned LC circuit (parallel Resonance). Inductor will used as detector and this inductor have parallel connection with a capacitor. This tuned circuit will start to oscillate at particular frequency.

f = 1/(2*pi*sqrt(L*C)

if any metal object near to inductor, inductance will very and f also very.

LPF should filter all unwanted frequency other than the LC tuned frequency. Depend on this frequency range you can choose either RC or LC filter.

below link may help you to design RC LPfilter
https://www.electronics-tutorials.ws/filter/filter_2.html

Jolly

 

[LvW]

I know the basic structure of an LPF.
.

...the basic structure? There are at least 10 different "basic" lowpass topologies - each with certain design alternatives.
Most important: Which order?

 

[pal114525]

Hi,

Thanks for your valuable feedback.

Would you be kind to explain me the following point?

In this case, it is a 1st order LPF which consists of a resistor and a capacitor.and the output is taken out from the capacitor terminal.
What should be the value of resistor and capacitor to remove radio transmission frequency (unwanted signal) and filter out ~16KHz frequency signal.

I know that f = 1/(2*pi*R*C). Therefore, if either R or C is known, we can calculate the value of C or R respectively as f is already known.

But, at first, how to select the value of R and depending on that C would be calculated or at first,how to select the value of C and depending on that R would be calculated?

Thanks & Regards.

 

[pal114525]

Hi,

Thanks for your valuable feedback.

Would you be kind to explain me the following point?

In this case, it is a 1st order LPF which consists of a resistor and a capacitor.and the output is taken out from the capacitor terminal.
What should be the value of resistor and capacitor to remove radio transmission frequency (unwanted signal) and filter out ~16KHz frequency signal.

I know that f = 1/(2*pi*R*C). Therefore, if either R or C is known, we can calculate the value of C or R respectively as f is already known.

But, at first, how to select the value of R and depending on that C would be calculated or at first,how to select the value of C and depending on that R would be calculated?

Thanks & Regards.

 

[crutschow]

First you can't "remove" a signal, you can only reduce it.
So the question is, how much do you need to reduce the RF signal so it doesn't interfere with your wanted signal?
Without some estimate of that, you are just guessing as to what you need for a filter.

 

[D.A.(Tony)Stewart]

AM radio can get into the linear amplifiers and due to some non-linearity get rectified and the audio gets amplified instead.

Choose an RC LPF breakpoint is fairly straightforward. In this case choose breakpoint just above the metal detector operating f , well below the AM band above 500kHZ. A ferrite bead and RF cap both help as a second order lossy filter.

The critical part is the impedance ratio of the wanted signal vs the impedance of the filter. such as Zc=1/(2pi f C) for the cap at the Metal detector f.
A filter cannot be designed without knowing the circuit details.

trial and error method
If you can stop the AM interference with your finger in the front end amplifier and it still works, try 100 pF across those points

 

[pal114525]

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.

- - - Updated - - -

Please see the attachment

 

[betwixt]

If the C and R values are for C1/R1 and C2/R2, surely you don't want to filter at that stage as removing any signal will make the circuit less sensitive as a whole. I would consider removing R5 and R6 completely as all they do is reduce the Q of the tuned circuits and placing the RC filter AFTER the op-amp.

You could also consider splitting C4 into two capacitors, one across each coil. You can make one of them variable to null the overall signal when no metal is in range. Make sure the coils are wired in the correct phase, you want their signals to cancel in the absence of metal so they should be wound in the same direction relative to the ground point to allow their difference rather than sum to come from the amplifier.

Brian.

 

[pal114525]

Hi,

Thanks for your valuable feedback.

Please see the attachment.
Would you please help me to find out the values of resistors and capacitors?

Thanks & Regards.

 

[Borber]

Have you told us frequency above which you wish filter to work?

 

[pal114525]

Hi,

Thanks for valuable feedback.
I want to receive an EM field which is created by a 16KHz sinusoidal signal.

Thanks & Regards.

- - - Updated - - -

Hi,

In this regard, there would be no signal at the output of an opamp if the field is received as they would cancel each other. But, there would be a signal at the output of the opamp if there is any deviation in the receiver EM field due to metal or conductor presence.

Thanks & Regards.

 

[betwixt]

The choice of filter frequency is yours of course. I would place it AFTER the op-amp so it didn't attenuate the signal from the pick-up coils and also save two components. Any imbalance between the signals from the coils will still be at 16KHz so you don't want to reduce it at that frequency. There is no 'correct' frequency but if your intention is to reduce signals picked up fom LF broadcast stations which typically start at around 150KHz, I would suggest a frequency in the region of 25 to 50 KHz would be a good compromise.

The actual combination of R value and C value is also your choice, obviously there are infinate combinations which will give the same theoretical result. What you have to take into account is the surrounding circuitry. It would be silly for example to use a very small capacitor value if the capacitance of the surounding circuits 'swamped' it and made it insignificant. Equally, a very low value resistor may cause unwanted high DC currents or force you to use an expensive precision large value capacitor.

If you place the filter after the op-amp and assume the following stage has a relatively high input impedance, you could for example try a 1K resistor and 4.7nF capacitor which would give a 3dB roll off at about 34KHz. Both values are 'middle of the range' selections, not unreasonably big or small and commonly available. If you want a sharper cut-off you could drop the roll off to maybe 25KHz but you would need to employ an LC or active filter.

Brian.

 

[Borber]

Coils loading must be equal for both coils to achieve simetry. Both input impedances of differential amplifier must be equal. C2 is not allowed. C3 and R3 acts as LPF but not equally for both inputs.

 

[pal114525]

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.

 

[Audioguru]

You need a differential amplifier that has inputs with the same impedance. Your differential amplifier has the non-inverting input impedance at 101k ohms and the inverting input impedance is only 1k ohms.
Use an instrumentation amplifier or copy the 3-opamps circuit of an instrumentation amplifier:

 

[pal114525]

Hi,

Thanks for your valuable feedback.
Would you please help me to find out the value of R1, R2 and R3 to receive the signal in an EM filed such that it would give a zero output when there is NO metal and would give some output if there is any metal between Tx and Rx.

Thanks & Regards.

- - - Updated - - -

Hi,

Thanks for your valuable feedback.
Would you please help me how did you calculate the input impedance of inverting and non-inverting terminals?

Thanks & Regards.

 

[Audioguru]

Simply read about Instrumentation Amplifier in google.
If the R1, R2 and R3 resistor values are matched to be the same then the differential amplifier is perfect. A signal on one input perfectly and completely cancels the same signal on the other input.
The resistors cannot be less than about 2k ohms because most opamps cannot drive a load that is lower than 2k ohms.

 

[pal114525]

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.

 

[betwixt]

Personally, I wouldn't say LM324 and 'instrumentation amplifier' in the same breath. It isn't a good choice.

There are many better types, in fact there are few worse ones! TL074 comes to mind.

Brian.