NEW !!! Full differential PI front end.

[WM6]

...

11. All of the above is coded inside the FPGA, tested and works and I have only used up 59% of the resources of a cheap ( $20 ) FPGA.

moodz.

Excellent! Thanks moodz.

How to get last source code?

What FPGA periferial circuits to use/build to get usable detector (which frontend etc.)?

[moodz]

... and furthermore .... I'm on a run here .... as some of you may have guessed already ... it is very simple to make an arbitrary waveform generator in the FPGA instead of the pulse generator ... output PWM or play out of the DAC ... to a power amp to drive the coil ... and you can then build any type of metal detector .... VLF, BFO ... triangular waves, sine waves, psuedo random etc etc.
Most analogue circuits can be all done in DSP.

moodz

[Aziz]

Hi Simonbaker,

Continued from prev post... More interesting graph showing sum of currents in coils too. Derivitive of sum gives induced current in target???? (of course this is without MOSFET breakdown).

-SB

if you have only one coil, the current will remain at last current at t-off (switch-off time) and then starts to diminish from this point.

On the bifilar type coil, we have at switch-off period two coupled coils.
The total inductance L will be:
Ltot = L1+L2 + M12 + M21
M12 = M21 = k*sqrt(L1*L2), (mutual inductances)
where k=1, the coupling coefficient (sharing the same coil flux area)
So, Ltot = L1+L2 + 2*sqrt(L1*L2)

The dB/dt applies to both coils L1, L2 at switch-off. From the point of energy balance, the output energy may not be higher than the input energy. The intregral E=U(t)*I(t)*dt (delivered energy from the coil) for the period of switch-off time (until next switch-on time) must be equal to E=0.5*L*(I0)² (emitted magnetic field energy at switch-off time), (I0=switch-off coil current).

TX period: E=0.5*L*(I0)², L=L2 in your example

RX period: E=0.5*Ltot*(Irx)², L1=L2, Ltot=4*L2,

Solving the equation:
0.5*L2*(I0)² = 0.5*4*L2*(Irx)²
(I0)²=4*(Irx)²
(Irx)²=(I0)²/4
Irx = I0/2

So the current in the receive period must be switch-off current I0 /2 and starts then to diminish from this value.
Q.E.D.

Aziz

[moodz]

Excellent! Thanks moodz.

How to get last source code?

What FPGA periferial circuits to use/build to get usable detector (which frontend etc.)?

Thanks ... I am not sure if releasing the source code is a good idea just yet .... It is still changing and needs to live in a proper project page. Releasing source code in a forum is asking for trouble ( never ending questions like ... how do I compile this ? ... and what does this code do ? Im a newby can you explain from beginning ? etc etc etc )
I have been there before . I may just provide it programmed into boards on ebay.

The front ends are the same as your conventional PI's .... I will publish this however .. must finish release v1.000 coding first

moodz.

[moodz]

Hi Don,

I am a bit confused by your naming convention:
square wave and rectangular wave

What do you mean by that?

Aziz

Is Don referring to pulse trains like Minelab PIs.??

[WM6]

Thanks ... I am not sure if releasing the source code is a good idea just yet .... It is still changing and needs to live in a proper project page. Releasing source code in a forum is asking for trouble ( never ending questions like ... how do I compile this ? ... and what does this code do ? Im a newby can you explain from beginning ? etc etc etc )
I have been there before . I may just provide it programmed into boards on ebay.

The front ends are the same as your conventional PI's .... I will publish this however .. must finish release v1.000 coding first

moodz.

I can understand.

I am not sure that eBay can exclude all possible support?

But if you can provided good building guide for complete project, many headaches can be avoided for you and for buyers.

I suppose that whole project will be published here to assess and facilitate the purchase of programed FPGA (and maybe some others part/stages too) decisions?

[Tinkerer]

The Geotech forum definitely need good mathematic equation editor !!!

doncaffe,

thanks for the explanation.
I am a bit confused though. It is one thing to look at the Mosfet drive pulse and another thing to look at the coil magnetic field, which depends on the coil current, not the voltage.
It is the coil current that makes the magnetic field that generates the eddy currents in the target.

So how can I produce a coil current wave form like 5/T with a coil inductance of 300uH?

The TEM TX METHOD is giving a good variety of frequency response as I am showing in the thread:
http://www.geotech1.com/forums/showthread.php?t=16774

I think that Moodz's FPGA will be the ideal platform to implement the TEM TX METHOD and will make it possible to extract a wealth of information about the targets from the response signal.

Tinkerer

[moodz]

Thanks ... I am not sure if releasing the source code is a good idea just yet .... It is still changing and needs to live in a proper project page. Releasing source code in a forum is asking for trouble ( never ending questions like ... how do I compile this ? ... and what does this code do ? Im a newby can you explain from beginning ? etc etc etc )
I have been there before . I may just provide it programmed into boards on ebay.

The front ends are the same as your conventional PI's .... I will publish this however .. must finish release v1.000 coding first

moodz.

OK I have go another cunning plan for releasing the source code.

It goes like this .... lets call it the Contributory Licence Scheme .... the code itself will be GPL or LGPL as some of the cores I use already fall under this licence ... like the CPU core for instance. The contributory licence is a wrapper around the GPL / LGPL and is set out thus ...

The Contributory Licence means you have to contribute to the project in some way. How can you do this ? I dont mean money ( though that is an option BTW) .

Anyone given access to the code for ANY of the following reasons will be deemed an Accredited Developer ( level 1 or 2 ) or Developer Friend ( see clause 3 ).


1. You can contribute by making changes or improvements to the code that in some way are beneficial to functionality or utility of the code. For this clause to be fulfilled you will have to verify in some way to the project maintainers that you at least have the basic development tools as specified by the project maintainers. ( till more people on board .. that is me )

2. In other cases the source code may be used as a learning resource or for using parts of it to do something entirely different. So you contribute by asking questions about the code, providing new ideas maybe suggesting stuff that no-one else thought of. You will still have to fulfill the development requirements of clause 1 above and verify to us the maintainers that you can operate the basic development environment even if you cant understand the code you are looking at. ( well you are supposed to be learning something here OK ? )

3. You can contribute by providing money or material assistance to other developers. Some of the accredited developers have the skills and ideas but lack the means to access equipment or sufficient parts or funds to participate fully in the development.
Or say you want a particular feature modified or added you can make a donation for the effort involved. Even if you use the code commercially you could make a donation in appreciation of the effort that the developers have already 'donated'.
In this case you get access to the source code as an indication that your contribution is actually being used in a productive way.

-------------------------------------------------------------------------------------------------------------------------

To become an Accredited Developer ( level 1 or 2 ) you will have to provide evidence that you have the correct development environment to the project maintainers. So you will need at least one of the supported hardware platforms and have the correct development software installed. ( this will actually avoid alot of questions later on )

The required hardware is Spartan S3E500 starter kit. ( other boards soon )

The required developer environment is .. Webpack ISE 12.1 from Xilinx

Linux or Cygwin environment to run the C compiler for MIPS

Once you have these set up correctly just request by way of email, PM etc for developer accreditation. The project maintainers will send you a file and a relatively simple task. If you can email back the message that appears on the LCD screen of the hardware board then you will be made an Accredited Developer and the maintained source code base is yours..... simple.

Clause 3 is fairly obvious. This will make you a Developer Friend.

moodz.

[Aziz]

Hi Simonbaker,

this is part 2: (I had to shorten my first post due to a visit)

Of course, the coil current I(L2) have to reach the I0/2. But this is more obvious, if transmitter coil is fully decoupled from the bifilar coil (transformer principle, seperate mono TX, bifilar RX).

During the build up of the flyback, there will be some resistive loss and the coil current does not reach fully I0/2.

But the coil current is not relevant. What counts is the first derivate of it (dI/dt).

Aziz

[simonbaker]

Hi Simonbaker,

this is part 2: (I had to shorten my first post due to a visit)

Of course, the coil current I(L2) have to reach the I0/2. But this is more obvious, if transmitter coil is fully decoupled from the bifilar coil (transformer principle, seperate mono TX, bifilar RX).

During the build up of the flyback, there will be some resistive loss and the coil current does not reach fully I0/2.

But the coil current is not relevant. What counts is the first derivate of it (dI/dt).

Aziz

Thanks Aziz, I will study your comments.

First, I was not sure if my LTSpice simulation is physically accurate - I gather you feel it does represent reality, correct?

Second, my purpose with the sim is to identify what we really are gaining with the bifilar / center-tapped coil. Are we gaining any magnetic field advantage, or is it basically the same as a single coil with same amount of wire? I understand (I think) the benefits of the differential design.

Third, some previous discussion suggested that the current in the two coils became instantaneously identical at MOSFET shutoff. However, my LTSpice sim indicates that the current briefly grows in one coil while decreasing in the other, and the difference is made up for by current coming through the diode from V+. When the two coil currents become equal, the sim shows the diode current stops, and both coil currents (really the same current) decaying at a slower exponential rate. I'm not clear whether that initial transition period is physically realistic, or a Spice artifact.

I'll try to look over your equations, probably some answers in there.

Regards,

-SB

[moodz]

Thanks Aziz, I will study your comments.

First, I was not sure if my LTSpice simulation is physically accurate - I gather you feel it does represent reality, correct?

Second, my purpose with the sim is to identify what we really are gaining with the bifilar / center-tapped coil. Are we gaining any magnetic field advantage, or is it basically the same as a single coil with same amount of wire? I understand (I think) the benefits of the differential design.

Third, some previous discussion suggested that the current in the two coils became instantaneously identical at MOSFET shutoff. However, my LTSpice sim indicates that the current briefly grows in one coil while decreasing in the other, and the difference is made up for by current coming through the diode from V+. When the two coil currents become equal, the sim shows the diode current stops, and both coil currents (really the same current) decaying at a slower exponential rate. I'm not clear whether that initial transition period is physically realistic, or a Spice artifact.

I'll try to look over your equations, probably some answers in there.

Regards,

-SB

Simon ... in order to see if there is an actual benefit in diff coils consider the case of the single coil vs the diff coil.

Below I have even reduced the inductance of the single coil greatly yet it cannot match the step response of the differential coil. At the decay / switch off point the current falls much faster than the single mono coil. It does not matter that the current does not actually fall to zero. The flux in the coil decays much faster in a step response.This is ONE of the TWO key features of the diff coil that I have been trying to make you guys see. Critical to remember that only ONE of the coils is carrying substantial current during the TX pulse in a diff coil. Trying to analyse the diff coil on its own with no comparative reference is like trying to find who came second in a one horse race

...and this is backed by physical experimental results not just math ... which is good but needs to be in the framework of experimental data.

moodz.

[simonbaker]

Simon ... in order to see if there is an actual benefit in diff coils consider the case of the single coil vs the diff coil.

Below I have even reduced the inductance of the single coil greatly yet it cannot match the step response of the differential coil. At the decay / switch off point the current falls much faster than the single mono coil. It does not matter that the current does not actually fall to zero. The flux in the coil decays much faster in a step response.This is ONE of the TWO key features of the diff coil that I have been trying to make you guys see. Critical to remember that only ONE of the coils is carrying substantial current during the TX pulse in a diff coil. Trying to analyse the diff coil on its own with no comparative reference is like trying to find who came second in a one horse race

...and this is backed by physical experimental results not just math ... which is good but needs to be in the framework of experimental data.

moodz.

[ATTACH]12608[/ATTACH]

Hi Moodz:

Try graphing I(L1) + I(L2) -- I believe that accounts for the total magnetic field seen by the target, the sum of the ampere-turns from both coils. Even though one falls very fast, the other grows very fast, and the sum does not have that sharp step -- but I'm not sure if that initial step is just a Spice artifact or physical.

-SB

[moodz]

Hi Moodz:

Try graphing I(L1) + I(L2) -- I believe that accounts for the total magnetic field seen by the target, the sum of the ampere-turns from both coils. Even though one falls very fast, the other grows very fast, and the sum does not have that sharp step -- but I'm not sure if that initial step is just a Spice artifact or physical.

-SB

Hi Simon,
yes the graph does produce a curve that is consistant with conservation of energy. This is a standard principle of physics however you are only considering the current and the turns in the coil. How was the energy lost from one coil and transferred to the other so suddenly ?.... the flux transfers the energy .. and what is also in the flux ? ... the target....does the target also experience a rapid current change ?... anyway food for thought.

This problem is very similiar to charging a capacitor then moving the plates apart ... what happens to the voltage across the plates ... even though the charge on the plates has not changed?

moodz.

[simonbaker]

Hi Simon,
yes the graph does produce a curve that is consistant with conservation of energy. This is a standard principle of physics however you are only considering the current and the turns in the coil. How was the energy lost from one coil and transferred to the other so suddenly ?.... the flux transfers the energy .. and what is also in the flux ? ... the target....does the target also experience a rapid current change ?... anyway food for thought.

This problem is very similiar to charging a capacitor then moving the plates apart ... what happens to the voltage across the plates ... even though the charge on the plates has not changed?

moodz.

Hi moodz:

I'm assuming the flux must always be consistent with the currents in the coils (since that is how we calculate the flux, from the currents and coil turns). Both coils are producing flux, and initially I believe the sum of the fluxes is not changing anywhere nearly as fast as that step we see when looking at a single coil. Since the sum of the currents does not have that big step, the target doesn't see any extraordinary dphi / dt, just the typical exponential (my second graph shows the sum of the currents, I(L1) + I(L2), and there is no sudden step).

(I don't know how to compare it to moving capacitor plates apart. You definitely put in physical energy to pull the plates apart, so I'd expect the voltages would rise according.)

However, I think I'm getting the clear advantage of "differential" (bifilar/center-tapped) design from playing more with the sims.

It seems for the same "flyback" voltage across the MOSFET (switch in my sims), you can get more dphi/dt with the center-tapped design, maybe double?

This seems to be due to the fact you get double the voltage across the damping resistor (I guess that shouldn't be an engineering problem) without increasing the MOSFET flyback voltage.

Combined with better noise canceling, it seems like a winner combination.

Regards,

-SB

[moodz]

Hi moodz:

I'm assuming the flux must always be consistent with the currents in the coils (since that is how we calculate the flux, from the currents and coil turns). Both coils are producing flux, and initially I believe the sum of the fluxes is not changing anywhere nearly as fast as that step we see when looking at a single coil. Since the sum of the currents does not have that big step, the target doesn't see any extraordinary dphi / dt, just the typical exponential (my second graph shows the sum of the currents, I(L1) + I(L2), and there is no sudden step).

(I don't know how to compare it to moving capacitor plates apart. You definitely put in physical energy to pull the plates apart, so I'd expect the voltages would rise according.)

However, I think I'm getting the clear advantage of "differential" (bifilar/center-tapped) design from playing more with the sims.

It seems for the same "flyback" voltage across the MOSFET (switch in my sims), you can get more dphi/dt with the center-tapped design, maybe double?

This seems to be due to the fact you get double the voltage across the damping resistor (I guess that shouldn't be an engineering problem) without increasing the MOSFET flyback voltage.

Combined with better noise canceling, it seems like a winner combination.

Regards,

-SB

I guess that if I can get the FPGA project completed ... we will be able to compare some real world results.

moodz