Question for Dave Johnson

[Anonymous]

As you are our resident VLF expert I was wondering if you could cast some light on the subject of high frequency detectors and their extreme sensitivity to small conductive targets. Case in point: Goldbug II and tiny nuggets. How does the skin effect help when using a 70KHz detector?

 

[Anonymous]

As the frequency goes up, the voltage induced in the inductance of the target goes up in proportion, and hence the current.
This would seem to benefit both large and small targets; however, in the case of a large target, the inductive reactance and not the resistance becomes the factor which limits the current. Also, the need to ground balance a VLF metal detector makes it insensitive to current which is flowing at the reactive phase.
None of the above has anything to do with skin effect. However, skin effect has the effect of increasing the resistance of larger targets, so that you don't lose as much sensitivity on them as otherwise you would.
--Dave J.

 

[Anonymous]

HI Dave and Dave
Have you had a look at the Falcon range of gold prospecting detectors?? 300khz and BFO but extremely sensitive to tiny tiny bits of gold that the gold bugs can't "see". Pitty they are not a bit more stable as their frequency drifts rather badly but for such a simple circuit they don't do too badly.
Cheers
Steve D

 

[Anonymous]

Dave, Thanks for the answer. It's great to be able to ask one's questions to an expert in the field.

 

[Anonymous]

Steve,
It is pretty easy to make a BFO detector which uses slow retuning using a phase locked loop circuit. You can also vary the retune speed for bad ground. Everyone laughs at the old BFO's today but one built into a pocket size case would be great for high grading ore samples.

 

[Anonymous]

that went way over my head! lol
I know its also a little off topic, but I got a new beeper to play with yesterday and already feel its a keeper. Only had time to check out one of the operating modes so far, but if the others perform as well, then Congrats on a super job!!
Tom

 

[Anonymous]

You would think with the technology that we have today that they would be able to produce a stable BFO!!! Also I thought BFO's had fallen out of favor because they couldn't handle mineralization very well. That being said and if it is correct couldn't it actually be mineralzation causing the BFO it be unstable and appear to be drifting? Just a thought on this early 4th of July morning.
HH and Happy 4th of July to one and all <IMG SRC="/forums/images/smile.gif" BORDER=0 ALT="">
Beachcomber

 

[Anonymous]

"in the case of a large target, the inductive reactance and not the resistance becomes the factor which limits the current"
Seems like it would be the other way around, the resistance would be the limiting factor. Also, how does ground balancing affect the sensitivity to reactive current?
- Carl

 

[Anonymous]

In a series L-R circuit, if the inductive reactance is much higher than the resistance, then the current will be determined by the induced voltage and the inductive reactance, not the resistance; and, the phase of the current will lag the induced voltage by nearly 90 degrees.
In an induction balance type metal detector, the phase of the signal induced by ferrite leads by 90 degrees the phase of the signal induced by a resistive current. Therefore, when the demodulator phase is referenced to be orthogonal to ferrite (i.e., iron minerals when ground balanced) it is also orthogonal to the reactive component of the signal from high-conductivity nonferrous targets such as the US silver dollar.
--Dave J.

 

[Anonymous]

Carl
You can describe the current in the target as It = j w M Ic / (R + j w Lt)
Where:
It - is the target current
w - is the frequency
M - is the mutual inductance between the coil and the target
Ic - is the coil current
R - is the target resistance
Lt - is the target inductance
j - is the square root of -1
The numerator is the induced voltage and the denominator is the impedance of the target.
Dave says the induced voltage is proportional to frequency (notice w in the numerator). That assumes that Ic is constant for different frequencies. You could accomplish that by using a drive voltage that is proportional to frequency.
If you want to use the same drive voltage you can use a lower inductance coil in a higher frequency detector. If you pick the coil inductance in inverse proportion to operating frequency you can keep the drive voltage and Ic the same. But M is proportional to the square root of the coil inductance, so in that case you get an induced voltage that is proportional to the square root of the operating frequency.
In a low conductivity target the denominator is dominated by R. In the extreme case you can ignore j w Lt and the current is approximately j w M Ic / R.
In a high conductivity target the denominator is dominated by j w Lt. In the extreme case you can ignore R and the current is approximately M Ic / Lt. Notice that w canceled out.
So increasing frequency favors low conductivity targets over high ones.
This is ignoring skin effect.
Robert

 

[Anonymous]

I didn't consider the frequency dependency of the induced voltage. Thanks guys.
So now I have 2 other questions. If higher frequency helps with low-conductivity targets, and does not appear to hurt high-conductivity response, then this argues for higher frequency. Except, of course, for mineralization. Which is low-conductivity, correct? Such that higher frequency will "light up" the ground. So where is the crossover point, or maybe I should ask, how would you find it?
Second question is: with multifrequency, if the harmonics are smaller than the fundamental (such as the CZ) then the induced voltage of the higher frequencies will be less. Seems like you would have much better performance off the fundamental. Interestingly, I found that the DFX has a stronger harmonic than the fundamental. Maybe this is why?
I really need to get an I/Q demod setup running to look at this stuff, sorta like what you did, Robert.
- Carl

 

[Anonymous]

Carl
I will answer part of the first question.
If you have ever used a 50 kHz or higher detector to look for coins in a park you will know why you don't want to do that. Small pieces of foil hit as hard as copper coins, and all the foil near the surface masks the coins.
The plot below shows the response of a ground balanced detector as a function of phase angle. This is like an antenna gain plot except here the angle refers to phase angle rather than direction. In this plot a ferrite target would fall on the 0 axis. Non-ferrous conductive targets would be in the upper left quadrant. The phase of the detector is adjusted to rotate the pattern so the ground signal falls in the response notch on the right side.
As you increase the frequency all the target signals move toward the 180 axis. There are two bad things about this. High conductivity targets get very close to the notch on the left side, so their signal from the demodulator is weak. Dave J mentioned this in his first response. Also as coin and pulltab signals get scrunched together near 180 it becomes more difficult to discriminate between them.
At frequencies between 3 and 15 kHz conductive signals are well spread out in that quadrant.
At 7 kHz thin foil falls between 90 and 100. Pull tabs are between 110 and 130. Large coins get up to about 160 and you can see how much the response is already down at that angle. You can do the math for 70 kHz and see what happens.
Robert

 

[Anonymous]

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