DEEPER PI DETECTION DEPTH

[Davor]

Remember, when you switch off a coil, which maintains a current flow, this current must be kept to flow further (as long as possible).

Or not - depends on your excitation pulse. In all cases a Tx coil must be terminated with low impedance or even better a short circuit. Current or no current. I'd say we agree on that.

Personally I'd go with step voltage and high L - it can't go any lower Z than that. The equivalent Z of voltage source is a short circuit.

The diagram above is a design that I played with to see the effects of a single coil arrangement, and it is not exactly a ripoff of any existing rig, just my musings. Anyway, pulse duration is just under 20us, cosine shape, and the principle is completely scalable. So, it could have been 100V, and my pulse would again be just under 20us, but with 4A at peak. Funny thing is that Power supply does not suffer any shocks, and power consumption is a few mA. Flyback is under 1us. I'd say it could rock.

...

I think I must review a bit for clarity. I tend to write complicated.
To obtain a transient response you need a voltage pulse, and you need a low Z termination. Depending on initial conditions you will end up with or without residual DC component. In case of single coil arrangement you don't have a choice - you must end up with zero DC. To do that you need a single pulse, which in turn gives not one but two sharp voltage transitions. One is too many because it carries no useful information, in fact it interferes with detection. That is the actual price of single coil operation.

[Aziz]

I=current @ turn-off
Cp = total capacitance
w0 = 1/sqrt(LCp)

This is the classic solution to a critically damped system.

What's happened to the damping resistor Rd?

Aziz

[Aziz]

I=current @ turn-off
Cp = total capacitance
w0 = 1/sqrt(LCp)

This is the classic solution to a critically damped system.

Thanks, it's more clear now.

[Carl-NC]

I=current @ turn-off
Cp = total capacitance
w0 = 1/sqrt(LCp)

This is the classic solution to a critically damped system.

[Aziz]

Hi Carl,

can you elaborate the variables in the formula?
(Just think, I'm stupid.)

Aziz

[Carl-NC]

We will see later, why the flyback period in a PI generates a high voltage pulse. The fact U = -L*dI/dt isn't the real proof. But we aren't so far yet.

L*di/dt is a simplification. The actual equation is

The problem with the traditional PI, is the waiting time to the first sample. This is a place that can be improved.

Depending on what you want to detect, waiting might be the right answer. See above equation.

[Aziz]

Because it must dissipate all the magnetic energy from space around it. That's easy. On the other hand, you do not have to employ high voltages, just go for lower L and higher currents. Less fuss. I think I've found an ideal pulse that comes from the least suspect source. It is just a concept at the moment.

Lo and behold

Hi Davor,

unfortunately not an accurate answer. It doesn't prove the fact however.

It's all about the impedances:
(We are neglecting the parasitic capacitance over the TX coil and the VLF mode at the moment)
The on-time impedance (on-time TX time constant) and the off-time impedance (off-time TX time constant).
Remember, when you switch off a coil, which maintains a current flow, this current must be kept to flow further (as long as possible). Or the magnetic field at switch-off time must be kept as long as possible.

The physics law has no other possibility to react to different impedance.
I=U/Z
Conduct this further and you know, why there is a high flyback voltage in a PI detector.

Aziz

[Tinkerer]

Because it must dissipate all the magnetic energy from space around it. That's easy. On the other hand, you do not have to employ high voltages, just go for lower L and higher currents. Less fuss. I think I've found an ideal pulse that comes from the least suspect source. It is just a concept at the moment.

Lo and behold

If you can get the transient time down to micro seconds instead of milliseconds, you are in business.

Tinkerer

[Tinkerer]

Peak coil current.



Then we need to define a new term "Ampere-turns per us." Still, if a VLF can hit 1A-T/us then a PI can easily hit 20A-T/us.

Agreed, on the "Ampere-turn per us"

The problem with the traditional PI, is the waiting time to the first sample. This is a place that can be improved.

Once we move the first sample closer to switch off, sampling while the coil current is still present and decaying, we find that the reactive signal is still present also.

Tinkerer

[Davor]

Because it must dissipate all the magnetic energy from space around it. That's easy. On the other hand, you do not have to employ high voltages, just go for lower L and higher currents. Less fuss. I think I've found an ideal pulse that comes from the least suspect source. It is just a concept at the moment.

Lo and behold

[Aziz]

BTW,

we have to talk about the (TX) coils reactance X(L) later too:
X(L) = w*L = 2*pi*f*L
And the impedance Z(L) as well.
Z(L) = sqrt(X(L)*X(L) + R(L)*R(L))

We will see later, why the flyback period in a PI generates a high voltage pulse. The fact U = -L*dI/dt isn't the real proof. But we aren't so far yet.

Aziz

[Carl-NC]

Still, the question remains: Is 100mA the power consumption or is this the peak coil current.

Peak coil current.

100mA with 100 turns at 20kHz, brings us back to 1A/us didt.

Then we need to define a new term "Ampere-turns per us." Still, if a VLF can hit 1A-T/us then a PI can easily hit 20A-T/us.

[Davor]

Davor, you keep saying that the RX front end is not good. Can you come up with a better RX front end?

I'm contemplating on it. There are some other things on my mind, like finishing a disertation, so you can only imagine a mess that goes around in my head.
Anyway, from what I've seen so far, there is no need for opamp on frontend. It makes things worse. Instead you need something with hard grip on ground like chopper or Tayloe, or something. Say 1MHz mixer. Next, you must make the most out of the time in which you may detect a signal. To do so you'll need a proper weighting function, in this case some kind of fade-in. Point is that however Tx coil is not exactly emitting power after Tx transistors quench off, the noise power continues to drop for some time, hence, however the target signal is decaying, so is the noise, and you may expect a near constant S/N for some more time. Applying a weighting function gain would make for a much longer duration of signal with good S/N. Hence much longer integration.
Next, discrimination. Zero crossing detection is much easier task with properly weighted signal, so there you have it.

IMHO an opamp at frontend is a mistake. You are not fighting thermal noise there, but a hard slamming switching transistor. A properly balanced analogue switch instead of an opamp with optimum impedance match would be a good start.

About Tx, I just might have a clue on how to make for a nice pi pulse that behaves, and presents itself as low impedance. More to come...

[Aziz]

Hi guys,

thanks for the hot contributions. Everybody is welcome (there is no need to be a "guru").

A clarification to my posting:
The values are chosen somehow arbitrary or reasonable of course. Don't confuse it with power consumption. We assume a comparable configuration of course. And don't focus too much to the irrelevant details. Just focus to the given facts.

Well the term "better" should be interpreted as much gain to the operator as possible:
- could be more depth (well, the most important item I think)
- could be more operating duration (power consumption)
- could be more signal integrity (less noise)
and so on.

Moodz gave you the answer. But I am missing the discussion "why?".

Cheers,
Aziz

PS: Where are the other gurus?

[Tinkerer]

There is no need to measure the current, it's easily calculated. Resonance or not.

Still, the question remains: Is 100mA the power consumption or is this the peak coil current.
100mA with 100 turns at 20kHz, brings us back to 1A/us didt.

Tinkerer