pi power

[bbsailor]

As a general observation, I haven't seen flat-topping make a noticeable difference. It would be good to have a definitive answer.


Tight tolerances are probably only true for professional metal detector manufacturers. Sod's Law says that a certain number of DIYers will make a mistake, and then we get the inevitable post stating that there was a flash/bang event, and the detector is no longer working.
I've also noticed that Eric tends to favour the use of a series resistor. It's saved me on a few occasions.

See this thread: http://www.geotech1.com/forums/showt...-Mine-Detector post#392 where Eric talks about the saw tooth waveform. If the current turns off too soon, meaning while its slope is still rising, then the sudden change in direction at current turn off will cancel out some of the energy just before the turn off and that has the consequence of not allowing the full current field from saturating the mass of the target to have the maximum potential detection response. Then, during the delay time before the RX field is enabled, the target current is decaying. These characteristics are universal but they apply to certain target types more because we have a desire to find them such as small gold nuggets, jewelry and some coins. Here is a good mental picture to think about to better understand this. Compare two gold male wedding bands, one fully formed and conductive and the other one that has a break in the ring. The full ring will be detected at a good distance from the coil because the eddy currents are circulating longer within ring circumference but on the broken ring with eddy currents only circulating within the mass of the metal, the detection distance will be much less.

Each target needs to be fully stimulated to be detected at its potential maximum detection depth. However, target shape, mass, conductivity and orientation all become optimized for detection with different amounts of current intensity and time to fully saturate its mass. Generally, targets do better when the current turns off while the current rise is more horizontal and if the desired target has more mass, the horizontal current on time should be longer to saturate the target deeper rather than just imposing a surface charge that will decay faster.

In some environments where you want to maximize the detection distance for desired targets the ground will not cooperate and that ground becomes energized and becomes a target. Here is where TX and RX parameter adjustment allows for compromise in non optimal situations.

It would be nice if we could cover the same ground only once to find all targets but seeking coins on a beach can easily done with a 10 to 15 inch coil, but you need to change the delay from dry sand to wet sand to adapt. If you now want to find small gold jewelry, you need to change to a smaller coil that is not optimal for finding deep coins and search the area again with different TX and RX settings optimized for smaller gold targets or jewelry.

This may not be a definitive answer but it may be a start!

Joseph J. Rogowski

[ivconic]

Usually i avoid to interfere in debates where i feel inferior comparing to knowledge present at other participants, but this time i can't resist; i must ask for an opinion!
I am offering you here a piece of schematic.
You have "Case A" and "Case B" on it.



"Case B" looks as rather weird solution, but it totally outperforms "Case A", in course of sensitivity, target response and stability.
This is checked and proven in practice.
My question is: how come?
Opinions?

P.S.
People from the other topic will just love me for this!

[Teleno]

"Case B" looks as rather weird solution, but it totally outperforms "Case A", in course of sensitivity, target response and stability.
This is checked and proven in practice.

Two MOSFETs in parallel = half the on-resistance of a single MOSFET.

For the same on-time, half the resistance means twice the coil current.

[Ferric Toes]

Two MOSFETs in parallel = half the on-resistance of a single MOSFET.

For the same on-time, half the resistance means twice the coil current.

I would like to know the inductance and resistance of the coil plus the width of the drive pulse.
Take Rds of the IRF840 as 0.85 ohms.

Eric.

[crane]

Two MOSFETs in parallel = half the on-resistance of a single MOSFET.

For the same on-time, half the resistance means twice the coil current.

It wouldn't result in twice the coil current, it would in fact make no significant difference if it uses a coil with high resistance. The only benefit would be if trying to keep the total coil and switching circuit resistance low, but the circuit would need altering to do this.

As Eric said we need to know the coil and cable resistance, the inductance, pulse width and it would help to know if the 840s avalanche.

[ivconic]

I would like to know the inductance and resistance of the coil plus the width of the drive pulse.
Take Rds of the IRF840 as 0.85 ohms.

Eric.

It was tested with 0.35mH/1.5R (including cable, coill itself is somewhat 1.1R), pulse width 160uS, pps 250.

[Teleno]

It wouldn't result in twice the coil current, it would in fact make no significant difference if it uses a coil with high resistance.

If the coil resistence is dominant then the MOSFET would have little influence. But sinc ivoconic said there IS a difference, I deduced his coil is low Ohm. Tx coils for PI have low resistance of less that 1 Ohm. This is my case.

[Teleno]

It was tested with 0.35mH/1.5R (including cable, coill itself is somewhat 1.1R), pulse width 160uS, pps 250.

With those values your coil current is 3.6 A with one MOSFET and 4A with two in parallel.

[crane]

It was tested with 0.35mH/1.5R (including cable, coill itself is somewhat 1.1R), pulse width 160uS, pps 250.

That's not as low as some but it's still a relatively low resistance coil circuit compared to what we often see here.

Assuming 10v then I get around 2.83A peak coil current with one IRF840 and 3.05A peak current with two 840s in parallel but it becomes harder to turn the FETs on as the resistance drops so it's wise to use a push pull circuit to drive the FET otherwise it spends the early part of the pulse acting as a resistor and this can influence the result.

Looking at the GS5 circuit posted on the forum with a 436uh coil, same pulse length, 3.5 ohm R with a 2.2 ohm series resistor and 0.55R for an IRF740 I get 1.73A, and 1.78A if I add another 740 in parallel.

Most of the projects here use high resistance coil switching circuits, mainly because they are taught this gives the best result when at least one manufacturer has proven it doesn't.

[Ferric Toes]

It was tested with 0.35mH/1.5R (including cable, coill itself is somewhat 1.1R), pulse width 160uS, pps 250.

Working it out very quickly on the parameters you gave -

One IRF840 gives 3.3A at the end of 160uS pulse

Two IRF840 // gives 3.7A at end of 160uS pulse

My conclusion is that it is not enough of a difference to make the significant improvement you suggest. There must be another factor.

Eric.

[Ferric Toes]

I used 12V as the TX supply which would explain why my current is higher than Crane's. See Ivconic's schematic.

Eric

[Teleno]

Working it out very quickly on the parameters you gave -

One IRF840 gives 3.3A at the end of 160uS pulse

Two IRF840 // gives 3.7A at end of 160uS pulse

My conclusion is that it is not enough of a difference to make the significant improvement you suggest. There must be another factor.

Eric.

Taking into account the resistance of the power source, say 1 Ohm, the difference is 3A vs 2.8 A.

That's a 7% increase in current and a 14% increase in power (W = 1/2 L * I^2).

[Qiaozhi]

Working it out very quickly on the parameters you gave -

One IRF840 gives 3.3A at the end of 160uS pulse

Two IRF840 // gives 3.7A at end of 160uS pulse

My conclusion is that it is not enough of a difference to make the significant improvement you suggest. There must be another factor.

Eric.

It could be another factor that is influenced by the increased coil current. For example, the main sample delay used with Q1 alone may be sub-optimal, but with Q2 the circuit reaches an optimised condition. Or maybe the damping resistor needs to be adjusted.
Without seeing the full circuit and more details of the detector settings, we're all just firing arrows into the air.

[crane]

I used 12V as the TX supply which would explain why my current is higher than Crane's. See Ivconic's schematic.

Eric

High Eric,

I didn't see the voltage so 12v gives much the same as your figures, 3.39A and 3.66A. As you say, not enough to make a significant improvement.

Good if that was all there was to it though.

[Ferric Toes]

Taking into account the resistance of the power source, say 1 Ohm, the difference is 3A vs 2.8 A.

That's a 7% increase in current and a 14% increase in power (W = 1/2 L * I^2).

A simple solution to increase the current is to widen the TX pulse to 195uS. This would give 3.7A with the single IRF840, which is the same as the parallel setup at 160uS.

Where do you get the 1 ohm in the power source?

Eric.