New PI technology proposals.

[KingJL]

Do we have comparable Targets with known TC's?

Tinkerer,
I think we'll have to create a chart/table of our own. A good place to start would be to measure the half life time of a target and then divide by 0.693 (ln2) to get the TC. You'll have to make sure that your base line is 0 (or at least compensated for) at your measurement times. You can pick your initial measurement anywhere on the decay curve (as long as you have enough signal left to measure a point at 1/2 the value further down the curve). I would try to minimize the signal distortion caused by excessive amplification. When I get back from the coast, I will be making some measurements after my 10X pre-amp stage. I can't remember whether my signal right off the coil was enough to measure accurately. I would also try to use some 'standard' targets (standard coins, gold wedding band, specific wt. gold nugget, etc.). Positional consistency will also be required to have meaningful data.

It may even be beneficial to start another thread for work on this effort and let this thread follow it's pre-destined path.

Regards,
J. L. King

[Monolith]

NO FLYBACK TEST

Gentlemen, allow me to disagree.
I say it is the expanding magnetic field produced by the coil current, that kicks the target. The Flyback is actually reducing the eddy currents in the target.
The Flyback is too short to "saturate" for want of a better word, the target.
The Flyback induces currents of opposing polarity since the magnetic field of the Flyback expands and then collapses, that is, the field cuts across the target conductor first the one way, the the other way.
The current raising in the coil during TX, generates an expanding magnetic field that cuts across the target \ conductor and induces currents within the target. The first currents are superficial skin effect currents. It takes time for the eddy currents to expand to the core of the target.
How much time? The time needed has been named the TC of the target. High conductive and large size targets have long TC's. Large silver coins can have a TC of hundreds of uS.
If the TX pulse is short, the eddy currents will not have the time to expand to the very core of the target, so they will not reach maximum amplitude.
Now look at the Flyback. It's duration is a matter of nano seconds for the one way and a few micro seconds on the return. Its highest power is at the maximum voltage point that is about even both ways and of a duration of nanoseconds. Just too short to induce much eddy currents. And going both ways, thus canceling the eddy currents induced by the going field, with the field collapsing.
However, the eddy currents induced by the TX coil current persist for some time.
How much time? about the same time it took to generate them The TC of the charge curve equals the TC of the discharge curve.

So, how will I prove my point? I will show you that I can read the eddy currents without having a Flyback.
Better than that, I will show you that, without the Flyback, the amplitude of the eddy currents is much higher.
I will pulse the coil with a usual square wave TX pulse but will modify the circuit to eliminate the Flyback.
Below is the scope picture of the coil current.
The pulse length is about 64us, you can see a slight kink and noise spike where the TX switches OFF. Forgive me for the noise spike, I don't think it can be called a Flyback.
The scope is set at 20us \div and 10mV\div.
Note, this is the coil current, not the TX voltage wave form.

Please allow me some time to setup an RX circuit to read the eddy current. Just the preamp.
I will use US$ 1c, 5c, 10c, 25c as test targets.
Monolith

So here it is, the NO FLYBACK samples.
Just a quick and dirty test to see if it works.

Would somebody be so kind and make a comparative test with a fast switch-off Flyback?
Monolith

[KingJL]

It is all about the TC of the target....What do we know about the Target TC?.....Do we have comparable Targets with known TC's?

Of this whole discussion, this the most important information. This would constitute knowledge, not theory; actual measured datas, not simulation results.

The silence is resounding!

[Tinkerer]

Target TC

G`day Simon

I must say i share the same thoughts as you on this point in regard to the relationship between the slew of the step response and the resulting eddies on targets.I think the amount of time that a target is exposed to a collapsing coil field is to important to be dismissive of it.
Slow targets by there very nature must tell you that they need more time exposed to a collapsing coil field in order to get maximium target stimulation,just like the inductance of a coil when one needs more time to build the current up.
The targets of interest with me are large nuggets from 1 oz to hundreds of ounces,these nuggets are full of surface irregularities and one would think that to some extent the eddies on the surface of these nuggets would cancle or oppose each other to some degree and there for resist the growth of eddies.


"inductance".


Zed

Round and round in circles, but finally back to square one.
It is all about the TC of the target.
There can be no accurate simulation if the Target is not accurately represented.
If the Target signal decay has a TC of 100uS, how long does it take to stimulate this Target to get full response?
What do we know about the Target TC?
Do we have comparable Targets with known TC's?

Tinkerer

[Aziz]

Hi friends,

I can confirm this, that a longer lower target stimulation has some more advantages than a higher and shorter target stimulation.
As we integrate over some period, the total acquired signal is more and has lower noise. There is another disadvantage of higher and shorter target stimulation: we can not process the signal on the coil at this time (high voltage). The recent SPICE simulations showed this elegantly. Unfortunately, they have gone to the hell and I have lost these files yesterday.

Anyway, as soon as my new system is finished, I will remake all the SPICE files and will show you some very interesting results. The SPICE files will be posted here to have also a backup.

Interesting discussion about the energy. We can not create or destroy energy. All what we are doing is converting it. So the total amount of energy isn't lost. It is converted.

Aziz

[ZED]

The point of that analogy is that if we turn off the TX faster, we may stimulate a higher pulse in the target, but for shorter time and perhaps no more energy! The end response may be essentially the same if the target is a "slow responder".

Now for faster targets this could be different -- they tend to follow the pulse more closely. However, the time constant of our RX receive coil may itself have a slower time constant than the target and therefore not respond much differently to a fast current pulse in the target versus a slower one. The target magnetic field in that case looks like the hammer blow to our RX coil!

More food for thought anyway.

Cheers,

-SB

G`day Simon

I must say i share the same thoughts as you on this point in regard to the relationship between the slew of the step response and the resulting eddies on targets.I think the amount of time that a target is exposed to a collapsing coil field is to important to be dismissive of it.
Slow targets by there very nature must tell you that they need more time exposed to a collapsing coil field in order to get maximium target stimulation,just like the inductance of a coil when one needs more time to build the current up.
The targets of interest with me are large nuggets from 1 oz to hundreds of ounces,these nuggets are full of surface irregularities and one would think that to some extent the eddies on the surface of these nuggets would cancle or oppose each other to some degree and there for resist the growth of eddies.


"inductance".


Zed

[simonbaker]

This is a simple limit problem. If you connect the caps with a resistor, the energy is (obviously) lost as heat in the resistor. As you take the limit R ─> 0, you'll find that I ─> ∞ but that the energy loss is constant. So, in this example, 0·∞ = 25 J.

- Carl

And even if you make the wires zero length (in an attempt to get rid of inductance), I'll bet the capacitors become resonant cavities (assuming non lossy dialectric like vacuum) and will still oscillate. So as R -> 0, the energy loss is constant, but the time constant will go to infinity too -- it will take infinitely long time to dissipate the energy (neglecting radiation)!

Let's try it...

Cheers,

-SB

[Carl-NC]

Where did the other 25 Joules of Energy go ????????

This is a simple limit problem. If you connect the caps with a resistor, the energy is (obviously) lost as heat in the resistor. As you take the limit R ─> 0, you'll find that I ─> ∞ but that the energy loss is constant. So, in this example, 0·∞ = 25 J.

- Carl

[simonbaker]

dc-dc converter

Interesting about dc-dc converter problem using only capacitors.

I worked on that myself. While not lossless, you can can do pretty well by putting a bunch of capacitors in series, charge them up, then reconnect them in parallel for a lower voltage. Or the reverse. Called "charge pump".

You can buy chips that do that, albeit much more sophisticated than the design I was trying.

-SB

analogy

Hi simonbaker,

Good Analogy on the fast turn off,

You are right about the oscillations on the capacitors too.

Hi Qiaozhi,

You are right as well, radiation was my point.

do this near a radio and you may get to hear it.


I got on the web and took a look at what I could find on this "classic" problem.

found a bunch of stuff. Google two capacitors energy go


Problem - Charging or discharging a capacitor may cause energy loss even if no dissipative elements are apparent.

Relevance - The problem affects power-loss and efficiency, electromagnetic interference (EMI), contact arcing, power MOSFET drive circuits, snubbers, solid state relays, resonant and duty-cycle switching converters, clocks, and selection of circuit topology.

Solvability - By transferring energy through an inductor or by controlling the charging waveform, a capacitor can be charged without loss of energy (ideally).

Solution - Transfer energy through an inductor or with a resonant or sinusoidal waveform with switching at zero voltage.

Personal - A personal anecdote.

On the Web - Additional information on the Web.

References - One to a few key papers.


Where the energy went.

Loss. The energy lost in directly switching voltage to a capacitor at another potential is lost in parasitic resistance, and if the resistance is too low, in arcing or welding of the switch contacts or in radiation.

Resistance. The easiest loss mechanism to show analytically is the loss in parasitic resistance, such as the capacitor equivalent series resistance (ESR) or wiring resistance. Adding this resistance to the circuit and calculating the power dissipated shows the energy loss. The energy loss is independent of the value of the resistance.

Arcing. Most switches used in power supplies are solid state and arcing is not a problem, but if a capacitor is charged through a contact, arcing may be a problem.

Radiation. High rates of change of voltage or current result in radiation. Directly switching voltage to a capacitor at another potential is a source of radiation (EMI).

The problem in Figure 1 on what happens to the energy when a charged capacitor is switched into another capacitor at a different voltage was a homework problem in my first circuit course so I learned about this early. However, I am constantly amazed at how often such a power electronics "perpetual motion machine" circuit is proposed as an efficient converter.

The most flagrant example I know about was on a three-year government project to make a highly efficient hybrid dc-dc converter using only capacitors and no inductors. When the first paper was given at a respected conference, several from the audience commented on the problem -- suggesting that it was inductance in the banana-plug-wired breadboard that made the circuit work. Undaunted, they reported on circuit improvements the next year, still with their banana plug breadboard. Their plans were to make it a hybrid circuit next year. Next year there was no paper. We speculated that when they finally got rid of wiring inductance, they found the circuit did not work, more than two years after it was obvious to experienced power electronics designers. In all fairness, this was in the mid 1970's and even major power supply companies were having problems with switching-mode power supplies. Concurrently, the sponsoring activity was doing some outstanding work in the field. The lesson-to-be-learned is that when someone questions your work, you should seek a side meeting to explore their concerns - they may know something you need to know. In this case it would have saved going down a blind path.

http://www.smpstech.com/charge.htm

Reflections on the lost energy when two
capacitors are connected, or two masses collide, or
similar problems, suggests a new law of nature:
that stored energy of one kind cannot be
transferred without loss unless it undergoes a
change in energy form in the transfer process.
Many physics and circuit theory textbooks describe
the well-known but curious phenomenon of two
capacitors with a switch between them.

Lossless energy transfer possible, but on one
condition
Note that initial boundary incompatibilities could
also be resolved (and modelling rules complied with)
by adding non-energy-dissipating model elements.
In the two capacitor case, an inductor, for example,
would allow energy to flow from the capacitor into
the inductor, and so into the second capacitor, in a
lossless way, with the energy accounts balanced at
every moment. (Note also the physical justification
for this addition, since all real circuits have
inductance.) This causes a change of energy 'type'
(from capacitive to inductive, or from electric to
magnetic) of precisely the kind referred to in the
Lacy–McCabe law. The only circuit element that
would not resolve the incompatibility would be
another ideal capacitor, which again would
constitute an attempt to violate the Lacy–McCabe
law.

http://www.iop.org/EJ/article/0031-9120/32/2/014/pe7202.pdf?request-id=5a0cc284-df44-490c-8e50-289aa8a9c985

[Aziz]

Hi friends,

I will be offline for some days. My harddisk in my PC got broken one hour ago. I have lost at least 3 weeks of work (also some SPICE files). Fortunately, I made a total backup of my harddisk three weeks ago. So it will delay, until I have a new harddisk and make a fresh new system.
I am using now back-track 2 (Linux) on my USB-stick.

So don't wonder, when I am not available.
Aziz

[simonbaker]

Not necessarily. As I stated before, provided the downward current (field) ramp takes place in 1/10 of the target time constant, then the object is fully stimulated. Designing for a higher flyback voltage and faster switch off will not get you any more signal. Very high flyback voltages have their own problems. i.e very high peak currents in the RX input clipping diodes, and input resistors working outside their voltage ranges. I had an input resistor fail on an early Goldscan for this reason. It didn't fail completely, it just went horribly noisy.

Basic PI theory can be considered by just examining one step in a magnetic field. If a field has been set up for an infinitely long time and then reduced to zero in a infinitely short time, we can see what happens in a conductive object within that field. All the time the field is static, then nothing happens within the object. As soon as the field starts to change, then eddy currents will be generated which try to maintain the field within the object at its pre-switch off value. These eddy currents first flow on the objects surface, and then gradually diffuse inwards, until for times after one time constant, they become exponential. This diffusion is caused by the object's resistance that cause the eddy currents to lose energy (heat loss). Eventually, there is no internal emf and hence no current and the field and object become static again. In this theoretical consideration the initial field could be generated by means other than a coil of wire carrying a current. Something to ponder on during the upcoming holiday period - if a long standing magnetic field (not coil generated) is instantaneously switched off in free space with no conductive objects nearby, is there a flyback voltage?

Eric.

if a long standing magnetic field (not coil generated) is instantaneously switched off in free space with no conductive objects nearby, is there a flyback voltage?

It's called a generator with permanent magnets!

Designing for a higher flyback voltage and faster switch off will not get you any more signal

You said something earlier that was interesting, that if the time constant of the target is slow, then there is a point where turning off the TX current faster doesn't make a difference. It seemed intuitive to me, sort of like removing the force from an object that swings back at its own rate.

I'd like to remind myself better though. It's tempting to think that when the TX coil turn-on phase reaches a steady current, that the target reaches a steady current also. Typically not, of course. As the di/dt of the TX coil diminishes to zero, the current amplitude in the target diminishes to zero.

Now, turning off the TX coil current gives the infamous pulse, which will stimulate a current in the target. Here is where I think it gets complicated.

For long time constant targets, I think it may be useful to think in terms of "delta functions" (infinitely high, zero width, but finite energy). Like striking a pendulum with a hammer -- the blow is over in an instant, but the amplitude of the response depends on the energy of the strike. If the blow is twice as long (still very quick) but half as hard, the pendulum swings the same.

The point of that analogy is that if we turn off the TX faster, we may stimulate a higher pulse in the target, but for shorter time and perhaps no more energy! The end response may be essentially the same if the target is a "slow responder".

Now for faster targets this could be different -- they tend to follow the pulse more closely. However, the time constant of our RX receive coil may itself have a slower time constant than the target and therefore not respond much differently to a fast current pulse in the target versus a slower one. The target magnetic field in that case looks like the hammer blow to our RX coil!

More food for thought anyway.

Cheers,

-SB

[Qiaozhi]

The voltage on a capacitor = Q/C volts

The energy in a capacitor = 1/2 C V^2 Joules

where Q = charge in Colombs

C = capacitance in Farads

V = voltage in Volts

Take two capacitors both with value = 1 Farad

One capacitor in charged to 10 volts

It has a charge of 10 Colombs

The energy in this capacitor is 50 Joules

The other capacitor is uncharged

It has a charge of 0 colombs

The energy in this capacitor is 0 Joules

The leads of the capacitors are shorted together

transfering 1/2 of the charge to the uncharged capactor.

The voltage of each capacitor is now 5 volts

Each capacitor has a charge of 5 Colombs

The energy in each capacitor is 12.5 Joules


If You add the energy of both capacitors you get 25 Joules.

Conservation of Energy says we should have 50 Joules


Where did the other 25 Joules of Energy go ????????

The capacitors are ideal and have no resistance

None of the energy is dissapated as heat in this example

and would be small anyway.

A similar example could be made with inductors/coils but

capacitors are easier for everyone to understand. Usually.

So where did the other 25 Jouldes of Energy go ?????????


(Not a trick question and can be found in Physics books)

This interesting question is commonly known as "the two-capacitor problem".

The usual answer is that the energy is lost in parasitic resistance, ESR of the capacitor, and/or arcing at the switch. I note that you stated the capacitors in your example are ideal and have no resistance, not that the switch or wiring has no resistance. But, if we assume that all these loss mechanisms are negligible, then the only thing left is radiation (EMI), as the two capacitors will act as a loop antenna. However, if the energy is primarily lost in the resistive elements, then this loss is independent of the value of resistance.

[simonbaker]

The voltage on a capacitor = Q/C volts

The energy in a capacitor = 1/2 C V^2 Joules

where Q = charge in Colombs

C = capacitance in Farads

V = voltage in Volts

Take two capacitors both with value = 1 Farad

One capacitor in charged to 10 volts

It has a charge of 10 Colombs

The energy in this capacitor is 50 Joules

The other capacitor is uncharged

It has a charge of 0 colombs

The energy in this capacitor is 0 Joules

The leads of the capacitors are shorted together

transfering 1/2 of the charge to the uncharged capactor.

The voltage of each capacitor is now 5 volts

Each capacitor has a charge of 5 Colombs

The energy in each capacitor is 12.5 Joules


If You add the energy of both capacitors you get 25 Joules.

Conservation of Energy says we should have 50 Joules


Where did the other 25 Joules of Energy go ????????

The capacitors are ideal and have no resistance

None of the energy is dissapated as heat in this example

and would be small anyway.

A similar example could be made with inductors/coils but

capacitors are easier for everyone to understand. Usually.

So where did the other 25 Jouldes of Energy go ?????????


(Not a trick question and can be found in Physics books)

You described a physically impossible circuit. When you connect the two capacitors, if you have no resistance, the current would try to go to infinity. What would happen realistically is that then those "no-resistance" wires would become inductors (even though small inductance, no problem if current trying to go to infinity). So the circuit would oscillate, and the missing energy is in the mag field of the inductor. Probably already answered by someone else.

Radiation would also eventually settle the response down to end state, as Qiaozhi noted.

-SB

Energy

The voltage on a capacitor = Q/C volts

The energy in a capacitor = 1/2 C V^2 Joules

where Q = charge in Colombs

C = capacitance in Farads

V = voltage in Volts

Take two capacitors both with value = 1 Farad

One capacitor in charged to 10 volts

It has a charge of 10 Colombs

The energy in this capacitor is 50 Joules

The other capacitor is uncharged

It has a charge of 0 colombs

The energy in this capacitor is 0 Joules

The leads of the capacitors are shorted together

transfering 1/2 of the charge to the uncharged capactor.

The voltage of each capacitor is now 5 volts

Each capacitor has a charge of 5 Colombs

The energy in each capacitor is 12.5 Joules


If You add the energy of both capacitors you get 25 Joules.

Conservation of Energy says we should have 50 Joules


Where did the other 25 Joules of Energy go ????????

The capacitors are ideal and have no resistance

None of the energy is dissapated as heat in this example

and would be small anyway.

A similar example could be made with inductors/coils but

capacitors are easier for everyone to understand. Usually.

So where did the other 25 Jouldes of Energy go ?????????


(Not a trick question and can be found in Physics books)