Minelab GPZ 7000

[Monolith]

If all of the energy stored in the coil is transferred to the parasitic capacitance at the peak of the flyback, then how do you explain that 65% of the initial current is still flowing in the coil at that point?

All is probably not the correct word. Intuitively I would say that the impedance of the coil is higher, so more energy is transferred to the capacitor initially. The amount of energy of course depends on the actual total capacitance of Mosfet, cable, coil inter-wire etc. In fact, specific coils have been built such that ALL the energy was transferred to the total parasitic capacitance.
For usual PI coils your number 35% is probably correct.

[Qiaozhi]

Teleno, I agree with you here. I will share the simulation with you by email.

If all of the energy stored in the coil is transferred to the parasitic capacitance at the peak of the flyback, then how do you explain that 65% of the initial current is still flowing in the coil at that point?

[Monolith]

At the peak of the flyback, all the energy that was previously stored in the coil gets transferred to the capacitance. Because the latter has such a small value, all that energy implies a very high voltage, as per your formula: => . For the same energy, the lower the capacitance the higher the voltage.

Teleno, I agree with you here. I will share the simulation with you by email.

[Qiaozhi]

At the peak of the flyback, all the energy that was previously stored in the coil gets transferred to the capacitance. Because the latter has such a small value, all that energy implies a very high voltage, as per your formula: => . For the same energy, the lower the capacitance the higher the voltage.

That would only be true in the absence of resistive losses, and with no damping resistor present.

When the MOSFET is switched off, the self-capacitance of the coil plus the output capacitance of the MOSFET look like a temporary short-circuit, and a relatively small amount of energy is transferred from the coil to the capacitance. If you examine the coil current decay after MOSFET switch-off, you will initially see a small drop [kink] in the decay curve. This is caused by the transfer of energy to the parasitic capacitances. Once the flyback voltage reaches its peak, the current in the aforementioned capacitors passes through zero. However, the coil current is still significant (about 65%) at this point, and the coil current is being dissipated only by the damping resistor. After the flyback peak has passed, the parasitic capacitors give back their stored energy (which is still relatively small). So the energy stored in the parasitic capacitance is not the dominant factor, and it is not the flyback voltage that kicks the target.

[Teleno]

The inductance dominates over the capacitance. Considering the very small capacitance values involved, coupled with the fact that the voltage across the coil during charging is also very small, it is clear that the energy stored in the aforementioned capacitors is miniscule in comparison to the energy stored in the inductor.

At the peak of the flyback, all the energy that was previously stored in the coil gets transferred to the capacitance. Because the latter has such a small value, all that energy implies a very high voltage, as per your formula: => . For the same energy, the lower the capacitance the higher the voltage.

[Qiaozhi]

I think what you point to, is that our TX coil circuit has inductance as well as capacitance. So, could we review the statement above?
Perfect inductor L = current source
Perfect capacitor C = voltage source

TX coil = L+C =?

The inductance dominates over the capacitance. Considering the very small capacitance values involved, coupled with the fact that the voltage across the coil during charging is also very small, it is clear that the energy stored in the aforementioned capacitors is miniscule in comparison to the energy stored in the inductor.

[Teleno]

I think what you point to, is that our TX coil circuit has inductance as well as capacitance. So, could we review the statement above?
Perfect inductor L = current source
Perfect capacitor C = voltage source

TX coil = L+C =?

Now as I said before, discharge the Tx coil with a current sink of I/2 and see what happens to resonance.

[Monolith]

From your two equations, when driving a coil L of capacitance Cc with current I and a MOSFET whose output capacitance is Co, the maximum flyback voltage is:

I think what you point to, is that our TX coil circuit has inductance as well as capacitance. So, could we review the statement above?
Perfect inductor L = current source
Perfect capacitor C = voltage source

TX coil = L+C =?

[Qiaozhi]

From your two equations, when driving a coil L of capacitance Cc with current I and a MOSFET whose output capacitance is Co, the maximum flyback voltage is:

Nice.
Someone else who can use LaTeX.

[Teleno]

The energy stored in the coil is

The amount of energy is proportional to the square of the current. When the MOSFET is switched off, the coil acts as a current source , not a voltage source. As a consequence, the flyback is an obvious [unwanted] artifact of the collapsing magnetic field. It is a side effect, not the cause.

The energy stored in a capacitor is

In this case the amount of energy is proportional to the square of the voltage. And, when the capacitor is discharged, the current produced is a consequence of the voltage developed across the load. In other words, the capacitor acts like a voltage source.

From your two equations, when driving a coil L of capacitance Cc with current I and a MOSFET whose output capacitance is Co, the maximum flyback voltage is:

[Qiaozhi]

It is an old discussion. One could say it depends on how you look at the Flyback and the "kick".

The energy stored in the coil is

The amount of energy is proportional to the square of the current. When the MOSFET is switched off, the coil acts as a current source , not a voltage source. As a consequence, the flyback is an obvious [unwanted] artifact of the collapsing magnetic field. It is a side effect, not the cause.

The energy stored in a capacitor is

In this case the amount of energy is proportional to the square of the voltage. And, when the capacitor is discharged, the current produced is a consequence of the voltage developed across the load. In other words, the capacitor acts like a voltage source.

[Teleno]

Try different damping resistors, from low Ohm up to the point of critical damping and measure the decay time.

Instead of a resistor, try a constant current sink equal to I/2 and see what happens (I = current in the coil before switch off).

The device has some operating modes. High Yield/General/Extra Deep. The scope shots show frequency and duration of tx.

That square current pulse (assuming your scope screenshots show the current) is an interesting challenge to produce. First you need to charge an auxilliary inductor to a current higher than he current you want for the Tx coil, then apply its flyback to the Tx coil (to achieve the high rise time), then maintain that level of current at the Tx coil (short circuit it), then switch off the coil. A while back I've designed a circuit that can just do that.

[Monolith]

This statement is not correct. The flyback voltage is an artifact, and not relevant to eddy current generation in the target.
For example:
Let's say we have a current established in the TX coil. When the MOSFET turns off, the current in the coil starts to decay. Since the effective resistance across the coil is now much higher, the [flyback] voltage across the coil will also increase. The negative amplitude of the flyback is not due to the current changing polarity (as some people have assumed in the past), since the current remains positive. What's happening is that the coil is no longer acting as a load, but has become a generator (or source), which means that it has a negative impedance. It is the collapsing magnetic field that kicks the target and creates the eddy currents. In other words, it's the coil current decay that is important, not the voltage. The flyback voltage itself doesn't even appear in the equations. The high flyback voltage is not "pushing" any current through the coil. The current is already established, and after switch-off it starts to rapidly decay. The high negative flyback voltage is just an artifact.

It is an old discussion. One could say it depends on how you look at the Flyback and the "kick".
Here is a simple experiment to show the importance of the high Flyback voltage:

Charge your TX coil.
Short it to ground.
What is the Flyback voltage?
How long does it take for the collapsing magnetic field to decay to 0V?

Try different damping resistors, from low Ohm up to the point of critical damping and measure the decay time.

Now tell us a different way to produce the high di/dt we need for a good "kick" without the high Flyback voltage.

[Qiaozhi]

It is the high voltage that pushes a high current through the coil.

This statement is not correct. The flyback voltage is an artifact, and not relevant to eddy current generation in the target.
For example:
Let's say we have a current established in the TX coil. When the MOSFET turns off, the current in the coil starts to decay. Since the effective resistance across the coil is now much higher, the [flyback] voltage across the coil will also increase. The negative amplitude of the flyback is not due to the current changing polarity (as some people have assumed in the past), since the current remains positive. What's happening is that the coil is no longer acting as a load, but has become a generator (or source), which means that it has a negative impedance. It is the collapsing magnetic field that kicks the target and creates the eddy currents. In other words, it's the coil current decay that is important, not the voltage. The flyback voltage itself doesn't even appear in the equations. The high flyback voltage is not "pushing" any current through the coil. The current is already established, and after switch-off it starts to rapidly decay. The high negative flyback voltage is just an artifact.

[Monolith]

Indeed, the Flyback voltage does not need to appear in the expression. However, it does have a significant role in the process. It is the high voltage that pushes a high current through the coil.

We charge the TX coil.
Then we discharge the TX coil. If we have only a few mOhm resistance in the discharge path, it takes a very long time to discharge the joules in the coil.
With a large damping resistor the Flyback voltage goes high and the discharge time is short.
At present, the practical Mosfet blocking voltage is the limiting factor in this process.

It is interesting that the basic principle of the "innovative TX method" has been in the public domain since about 10 years and was discussed at length on this forum, about 6-7 years ago.