Signal decay rate??

[wyndham]

Eric and others
Is there a certain decay rate for the eddy current in a target. Is it mass or surface area dependent, is it a quick fall off or gradual. Also is there a max inducion that can be applied to a target where beyond that point is over kill.
Is it correct to see the target as a leaking capacitor that the P.I. induces the current and it decays over fix time.
It seems it should be surface area because electricity travels on the surface, but I need to double check my reasoning.
There seem that there should be a vanishing point where a target is too small to generate a signal, but is this dependent on the target or the reciever's sensitivity to the signal.
I know these may be simple questions for folks that are E engineers, but if I see an equasion with more that a few symbols my eyes glaze over and they carry me out of the room.
Thanks Wyndham

 

[Prospector Al]

Yes Wyndham. there is a decay rate...

The surface area presented to the magnetic flux pulse determines how much voltage is induced in the target.

In response to the induced voltage, an "eddy current" is generated.

The magnitude of the current is determined by the inductance and resistance in the eddy current path.

The decay rate of the current depends on the ratio between the inductance and resistance, and this quantity has been termed "Time Constant". ( T = L/R, where L is the inductance and R is
the resistance.) Sorry about the equation...Are you still there?

The shorter the time constant, the faster the signal decays. T is a time interval, and in four Ts, the signal has vanished altogether.

In practical terms, if you have a disk of foil the size of a quarter, its time constant is very short, compared to the quarter, because the quarter has a much lower resistance in its eddy current path, owing to its thickness.

The conductivity of the target matters also. Thus, a silver quarter has a longer time constant than a "clad" quarter, because of silver's higher conductivity.

The signal from a target is always there, but it may be irretrievably buried in noise, if the target is small.

There is no "overkill" as far as the maximal pulse intensity is concerned--the signal from the target will increase with pulse power, until the target melts.

(By the way, you can't overload a receiver coil either--it's just a piece of wire. The "coil overload" referred to in some User's Manuals actually means that the detector's input stage is
driven into saturation--it has poor "dynamic range" as an engineering type would say.)

The above explains the phenomenon of "invisible gold". Although the size of a nugget may be considerable, the resistance of its eddy current path may be high, owing to its irregular shape. Consequently, its time constant is short, and the signal disappears before the signal gating pulse looks for it...

I hope you're still awake and that you found this useful.

Prospector Al

 

[Prospector Al]

Hi Wyndham,

I wanted to stay away from the highly technical when responding to your question, but lest I be shot down by some EE reading my post, I have to add a little more detail.

The voltage induced in the target is also proportional to the rate-of-change of the coil current, not just its peak value. ( Voltage = dI/dt, as Eric might say.)

To maximally "charge" a target with a short time constant, such as a gold nugget, the coil pulse must return to zero in a time interval that is shorter than the time constant of the target. Otherwise, much of the pulse energy will be wasted, and the target will in fact be
"saturated" at a voltage level determined by how fast the current returns to zero...

Prospector Al

 

[Wirechief]

Hello Al and a quick question, on your example of the clad quarter is the TC going to be very much different if the quarter is standing on edge as compared to laying flat? Thank you and nice to meet you Al. God Bless. Wirechief.

 

[Prospector Al]

Hi Wirechief,

The signal amplitude and the Tc are maximal when the coin lies flat under the search coil.

When the coin is on edge, both decrease by about 50%, but the actual amount depends on coil geometry.

The magnetic lines of force from the search coil are curved, in the shape of donut ( A torus, as Eric would say.) At a certain distance from the coil, the lines of force are parallel to the ground and a coin on edge would be still be hit straight on.

When you sweep across a coin in several directions, at some point you'll get the best signal, regardless of how the coin is oriented.

The curvature of the magnetic field is tighter, the smaller the coil size and with small coils,
you can use another trick: Turn the search coil on edge, to see if the signal improves...

Happy Hunting...

Prospector Al

 

[wyndham]

Thanks Al, I believe the part on the donut as a 3d rep gives me a better insight into the field produced.
I have been trying to understand how small signals behave.
If I understand correctly the smaller the faster decay and the quicker vanishing point.
My point on the math, when I get to a website discussing eddy currents, there are pages of math and the first line or two is ok but, wow.
Geometry I can "see" but symbols without example is hard for me to follow.
Thanks again Wyndham

 

[Wirechief]

Ok Al and thanks again for your input and I figured the field was somewhat the shape of a donut. If I read you right there is a certain amount of the field going into a horizontal plane and is the horizontal field very uniform or are there lobes that extend out further than most of the rest of the field in other words can the field be sort of directional? Thanks a bunch Al. Wirechief

 

[Wirechief]

I have a question, just before the eddy current in a gold nugget is totally disappated is there any kind of spike or is it linear? Thanks. Wirechief.

 

[Prospector Al]

Hi Wirechief,

The signal does not disappear at a linear rate. It's an "exponential decay".

In plain English, if a certain percentage is lost during the first microsecond after the coil pulse, the same percentage OF WHAT IS LEFT, is lost during the next microsecond, etc.

There is no spike at the end--the signal kind of runs into the sand of noise...

Prospector Al

 

[Wirechief]

Ok Al thank you and I am learning all the time. I understand so it is like it has a time constant like a capacitor. Thanks Al. Take care. Wirechief.

 

[Wirechief]

I apologize Al, if I had gone further back in the postings on this subject I would have found the answer to my question. Many thanks again Al. Wirechief.

 

[Eric Foster]

Hi Wirechief, Wyndham and Prospector Al,

It gets even more interesting, and complex, the "deeper" you go into target decay. A sphere is often use for calculations as it is a uniform object with no orientation problems. However, the eddy currents are not a simple exponential until late on in the decay. This is because, just after TX switch off, the currents in the sphere flow on the surface (skin effect). This is to try and maintain the field within the sphere, as physical laws don't like change. These surface currents quickly run out of steam,due to the surface resistivity, and migrate inwards. This migration continues until the currents are flowing in all the sphere, right down to the centre. You can liken it to the skins in an onion, although there are no discrete current layers. The result is that the resulting signal is a sum of many time constants, until the full inward current migration is complete, then it becomes a simple exponential.

A thin ring is the best example of a true single exponential, as there is nowhere for the current to go, other than around the ring. Coins, cylinders and lumps all fall somewhere in between.

Don't ever ask about an iron sphere!

Eric.

 

[Wirechief]

Many thanks Eric for your explanation and I enjoy your sharing the knowledge you have. I am learning so much just by reading your posts and not many people will go to the trouble to do this. Maybe you will expound upon the iron sphere someday. Thank you Eric and CU later. Wirechief.