PI Power Output

[Aziz]

Hi ZED,

it is only a matter of exposed energy and how you control this at damping process. The target has some capacity to store and consume energy (due to inductivity and resistance and has some tiny capacitance too). It cannot produce more energy than it is available. Then you have two options:
- slow damping, longer sampling with lower target signal response or
- fast damping, shorter and earlier sampling at higher target signal response

So there is really no free lunch. This is a fact.

Sure, making the system faster has some disadvantages: higher noise susceptibility due to shorter detection window and reduced bandwidth of the front-end (slow response). On the other side, slow response needs really more time to sample at low signal levels. If I consider, then you are right, that the slow system has more obvious advantages compared to a fast response system.

So using an integrator for sampling is very convenient, as the target response is a time-domain response (distributed over time).

Now the question is: What is better? Using higher target signals for short sampling periods or using lower target signals for longer sampling periods. I think, the latter one is obvious as the ground itself has not much energy storing capability and will (should) decay faster than the targets.

From the point of processing (circuit), whether which option is better, should be analysed further.

Aziz

[ZED]

But from the point of target stimulation, there is definitely no other way. The more, the better. At the end: The higher the flyback voltage, the more the target stimulation and target eddy currents, as this gives the highest dB/dt damping. It doesn't matter, how you get the high flyback voltage: either with strong short pulses or weaker long pulses.



Regards,
Aziz

Aziz i beg to differ,your not considering the inductance of targets,targets need time to build up eddy currents,your suggesting to reduce this time in favour of increasing the speed of the coil field,it is true that the faster a field moves the stronger the eddy currents but what about the inductance of targets that resist the growth of eddy currents,increasing the speed and reducing the window for which targets are exposed to a moving field in my opinion produces a negative improvement or zero improvement for targets with high inductance.
Speed of the collapsing field and the exposure time to this field are both important criteria.

Zed

[Aziz]

Hi B^C,

I think, there are benefits from other side using shorter TX pulses. The CMRR, PSRR will be better, if your power supply won't be stressed much. There will be less ripples and regulatation in the supply part. Also less EMI during TX switching. This will improve the analog part of the front-end and SNR of course.

But from the point of target stimulation, there is definitely no other way. The more, the better. At the end: The higher the flyback voltage, the more the target stimulation and target eddy currents, as this gives the highest dB/dt damping. It doesn't matter, how you get the high flyback voltage: either with strong short pulses or weaker long pulses.

A common misinterpretation is the ground: Ground should also be seen as a weak target: It has conductivity, inductivity, capacitance, permeability, permittivity, coupling to the transmitter coil, parasitic capacitances to the transmitter coil, ...
It is obvious, that the ground will react to the transmitter stimulation. And it is also obvious, that all parameters will interact with each other, which makes the discrimination and ground balancing difficult. So varying the transmitter pulse width gives some help to get some of the parameters (time constant of the target for instance). It is also obvious, that high excited pulse energy will give more ground signals. And it is only a question of SNR (signal compared to noise level) and dynamic range of the measuring system.

I think, there is still a common misinterpretation of target eddy current stimulation in a PI device. We had here an interesting contraversal discussion last year (or was it this year?) . As we all know, the transmit on pulse can be neglected due to less stimulation on MONO coil systems. I think, not all people have really understood the importance of the damping process.

Regards,
Aziz

[B^C]

Tinkerer

I understand that ampere turns is important but only if we can take full advantage of it, it's hard to explain. Let me test something out & i'll try to answer this differently. I tested some things & found that sometimes a little less is more so to speak. It's like have a 500hp engine but you can only get half throttle, i found something interesting & putting it words is not easy.


Yes short TX is the only way to go.

I guess the figures above are calculated from your operating frequency which is important.

What calculation are you using for your current rise vs time vs voltage?

Hi Tinker,

I have been following this thread and find it very interesting. I believe some of the questions you are asking are due to inductive coupling vs true pulse induction. Many of the great answers from the others are explaining what happens when the current is switched off and there after. Your observations are during switch on while the current is building. From what I have read you are using an induction balanced coil. It can be looked at as an air core transformer that has very low coupling. When a resistive target is placed into the changing magnetic flux, it creates an impeadance to the flow. This in turn upsets the balance in your transfomer and causes coupling. Ractive targets, in simplistic terms inhance the flow of flux and cause a disbalance in the transformer and coupling. A true PI is measures the decay of eddy currents induced into the target after the coil current collaspes. If you see how standard TR detectors operate it is the same as what you are doing during switch on. You are using a step function vs a sine wave. Time domain vs Frequency domain remember T=1/f and F=1/t. Keep up the interesting experiments

Regards
Mark

[Tinkerer]

You guys have been busy with some good replies.

Gday Aziz, i agree with your explanation with increasing power to the coil.

Tinkerer, we talk the same language--plain ol english!

Yes higher voltages do work & it is easier & faster for higher voltages to travel through things rather than trying to force Amps through something.
I have been working on a system that has 90v & 600mA hence the original question to get a conversation going about it.
I have had to refine it & detune it a bit as it is capable of much higher output, it's all but finished & i will test again in the next few days hopefully.
I have been working in terms of total power to the coil where a more std setup has around 15v & 1.2A=18W, mine has 90v & 600mA= 54W.

I have had near 360v & 1.6A from the same setup but of course battery life suffers & personal life expectancy lowers dramatially especially on wet days. Also i didn't want to have a big rock on the coil to stop it bouncing hahaha!


Battery size is not an issue at this stage, the minelab batteries last from 8 to 10hrs from full charge & there not heavy at all. If these style of batteries only lasted a couple of hrs before recharge it would suit me if the detectors were stronger.

Tx on time?
The thing to remember is that the TX on time makes NO difference as the Target Decay signal is the same regardless, something to do with the time constant maybe?.
In Fact Long TX on times is an absolute no no as it induces the ground more & then things become troublesome.
Short sharp pulses are in order here as the ground is not induced as much & we get longer stronger target signals to read.

TX signal shape?
Does it really matter what the TX signal shape is?
Ok in the past with low power we have seen the TX switch off very fast for flyback reasons i guess.
My question is that with a higher output TX signal do we need to be as fast with the switch off?
For a given TX power output what time frame is sufficient, you would think for low output figures that very fast would be correct, but what about with higher output, how slow is to slow in microseconds?

Having a strong magnetic field from a coil is only part of the equation, the thing we really need is a strong magnetic field that has a lot of Push associated with it. So i guess a fast switch off is still necessary, but how fast?

Hi B^C,

higher voltage will get the same amps faster into the coil. However, for the ground and target penetration the important part is the magnetic field strength.
What counts is the Ampere Turns.
Now, I have heard that it is an advantage in the aggressive Australian soil to use very short TX pulses. I believe ML uses short and long pulses.
So there may be a real advantage in using high voltage to reduce the TX time.
In a coil with 325uH inductance and 2.8 Ohm DC resistance, it takes 100uS to bring the coil current to 2.5A, at 12 V
With 120 V it takes only 7uS to reach 2.5A coil current.
It also looks like a saving of about 30% power.
Tinkerer

[Tinkerer]

To me it is confusing to think in terms of separate magnetic fields (although your analogy may be perfectly useful). I just see a magnetic field that was growing and leveled off and then suddenly starts shrinking. Just a continual change. The target sees varying dphi/dt, which it responds to accordingly. And I agree the dphi/dt changes direction at discharge because di/dt changes direction -- even though the current is still going in the same direction through the TX coil. So the current in the target I assume changes direction too.

So I'm quibbling over description there -- more importantly, maybe there is something to your observations we could explain and make use of:

Suppose we can create a more powerful target "response" (measured by our RX coil) by actually reversing the current in the target rather than just stimulating it in one direction.

As Carl explained - maximum target-stimulation from the TX turn-on happens at t=0. As you let the field build up, it slows down in rate and so your target current dwindles too -- but you get a nice fat magnetic field ready for the big turn-off pulse.

But is there a possible trade-off? What if we turn off the current while the TX is still building fairly fast? Could we get a bigger (or longer) di/dt in the target, and thus more pulse in our RX coil?

-SB

There is nothing wrong to look at the same problem from different angles or to describe it in different words. It is good to have the technical description from Carl for reference.
I will add one more piece of the puzzle. I re-post two scope pictures from the coil thread, for the other pictures please go to that thread.

The TX is about 42uS. The coil TC is about 116uS.
Now look closely at the end of the TX pulse at the right. The coil current is at about 1.3A. The gain is 40.
Look at its position at the 0V line with no target.
See how it goes down or negative with the silver coin at the coil.
See how it goes positive with the ferrite.
The top trace is set at 1V div. and 1uS div.
The bottom trace shows the sample pulse and is set at 5V div. and 1uS div.
For the ON time sample this is the point of the highest signal amplitude.
The highest amplitude is at the time of the highest coil current when the TX pulse is up to one TC.

When I add a resistor in series with the coil, to shorten the coil TC, the highest signal amplitude is at about 1 coil TC.
But, the highest signal amplitude changes in time with the TC of the target.

I have noticed that the time of highest signal amplitude is also influenced by the capacitor that I add in the feedback of the preamp.

So it is a puzzle of many pieces. How does it all fit together?

Tinkerer

[B^C]

You guys have been busy with some good replies.

Gday Aziz, i agree with your explanation with increasing power to the coil.

Tinkerer, we talk the same language--plain ol english!

Yes higher voltages do work & it is easier & faster for higher voltages to travel through things rather than trying to force Amps through something.
I have been working on a system that has 90v & 600mA hence the original question to get a conversation going about it.
I have had to refine it & detune it a bit as it is capable of much higher output, it's all but finished & i will test again in the next few days hopefully.
I have been working in terms of total power to the coil where a more std setup has around 15v & 1.2A=18W, mine has 90v & 600mA= 54W.

I have had near 360v & 1.6A from the same setup but of course battery life suffers & personal life expectancy lowers dramatially especially on wet days. Also i didn't want to have a big rock on the coil to stop it bouncing hahaha!


Battery size is not an issue at this stage, the minelab batteries last from 8 to 10hrs from full charge & there not heavy at all. If these style of batteries only lasted a couple of hrs before recharge it would suit me if the detectors were stronger.

Tx on time?
The thing to remember is that the TX on time makes NO difference as the Target Decay signal is the same regardless, something to do with the time constant maybe?.
In Fact Long TX on times is an absolute no no as it induces the ground more & then things become troublesome.
Short sharp pulses are in order here as the ground is not induced as much & we get longer stronger target signals to read.

TX signal shape?
Does it really matter what the TX signal shape is?
Ok in the past with low power we have seen the TX switch off very fast for flyback reasons i guess.
My question is that with a higher output TX signal do we need to be as fast with the switch off?
For a given TX power output what time frame is sufficient, you would think for low output figures that very fast would be correct, but what about with higher output, how slow is to slow in microseconds?

Having a strong magnetic field from a coil is only part of the equation, the thing we really need is a strong magnetic field that has a lot of Push associated with it. So i guess a fast switch off is still necessary, but how fast?

[simonbaker]

The coil turn-on current follows a rising exponential with a tau of L/Rs, where Rs is the total series resistance. The maximum di/dt occurs right at turn-on (t=0), and is Imax*Rs/L. As the coil current exponentially rises to Imax, the di/dt exponentially decays to zero.

dB/dt follows di/dt, so the peak excitation for a target also occurs at t=0. Even though dB/dt is decaying thereafter, it is still providing a continuous excitation to the target, and the target response will be an integration of exponentials, which is an exponential.

The same thing happens when the switch turns off. The only difference is the turn-off tau is now L/Rd, where Rd is the damping resistor. The max di/dt occurs right at turn-off (t=toff) and is Io*Rd/L. Io is the peak current achieved right at toff, and could be as high as Imax, but maybe not.

What all this means is that the target sees 2 different dB/dt's. In both cases, the target is "hit" with an initial maximum dB/dt, and it is all downhill from there. The only "charging" of the target occurs right at turn-on and turn-off; after that, it is discharging.

What I have argued is that -- regardless of the tau of the target and how far it has discharged before toff occurs -- the turn-on dB/dt is very weak compared to the turn-off dB/dt. If you compare the peak ratio (at t=0 and t=toff) it is Rd/Rs... typically this will be 200 or more. But the reality is the ratio is even higher, because at t=toff the target has probably decayed out quite a bit. If you have a long enough coil turn-on time the target will be completely decayed, and the ratio is infinity. But even if you turn the coil off while the target still has turn-on eddy currents running around, those eddy current are relatively very small compared to the eddies generated by the turn-off dB/dt. I believe they just don't matter much.

- Carl

What I have argued is that -- regardless of the tau of the target and how far it has discharged before toff occurs -- the turn-on dB/dt is very weak compared to the turn-off dB/dt.

A question about the above statement -- if you model a target as two separate inductances in series , only one of which is coupled to the magnetic field (not saying this is real), then it would seem you could "charge" the target, because even a small dphi/dt would "integrate" in the second inductor. Is it possible that eddy currents could do something like that?


-SB

[simonbaker]

Are you taking the change in polarity into account?

Tinkerer

To me it is confusing to think in terms of separate magnetic fields (although your analogy may be perfectly useful). I just see a magnetic field that was growing and leveled off and then suddenly starts shrinking. Just a continual change. The target sees varying dphi/dt, which it responds to accordingly. And I agree the dphi/dt changes direction at discharge because di/dt changes direction -- even though the current is still going in the same direction through the TX coil. So the current in the target I assume changes direction too.

So I'm quibbling over description there -- more importantly, maybe there is something to your observations we could explain and make use of:

Suppose we can create a more powerful target "response" (measured by our RX coil) by actually reversing the current in the target rather than just stimulating it in one direction.

As Carl explained - maximum target-stimulation from the TX turn-on happens at t=0. As you let the field build up, it slows down in rate and so your target current dwindles too -- but you get a nice fat magnetic field ready for the big turn-off pulse.

But is there a possible trade-off? What if we turn off the current while the TX is still building fairly fast? Could we get a bigger (or longer) di/dt in the target, and thus more pulse in our RX coil?

-SB

[Tinkerer]

The coil turn-on current follows a rising exponential with a tau of L/Rs, where Rs is the total series resistance. The maximum di/dt occurs right at turn-on (t=0), and is Imax*Rs/L. As the coil current exponentially rises to Imax, the di/dt exponentially decays to zero.

dB/dt follows di/dt, so the peak excitation for a target also occurs at t=0. Even though dB/dt is decaying thereafter, it is still providing a continuous excitation to the target, and the target response will be an integration of exponentials, which is an exponential.

The same thing happens when the switch turns off. The only difference is the turn-off tau is now L/Rd, where Rd is the damping resistor. The max di/dt occurs right at turn-off (t=toff) and is Io*Rd/L. Io is the peak current achieved right at toff, and could be as high as Imax, but maybe not.

What all this means is that the target sees 2 different dB/dt's. In both cases, the target is "hit" with an initial maximum dB/dt, and it is all downhill from there. The only "charging" of the target occurs right at turn-on and turn-off; after that, it is discharging.

What I have argued is that -- regardless of the tau of the target and how far it has discharged before toff occurs -- the turn-on dB/dt is very weak compared to the turn-off dB/dt. If you compare the peak ratio (at t=0 and t=toff) it is Rd/Rs... typically this will be 200 or more. But the reality is the ratio is even higher, because at t=toff the target has probably decayed out quite a bit. If you have a long enough coil turn-on time the target will be completely decayed, and the ratio is infinity. But even if you turn the coil off while the target still has turn-on eddy currents running around, those eddy current are relatively very small compared to the eddies generated by the turn-off dB/dt. I believe they just don't matter much.

- Carl

Carl, thanks again for the correction.
So now we have the proper scientific explanation.
And I admit, it makes very much sense to me.
BUT, and this is the purpose of me starting this discussion, there is a problem:
When I sample DURING TX, I get more response from the target.
The signal is of higher amplitude.
Now, I have made hundreds of tests and it is always like that.
Other people have found the same result.

My explanation for this phenomena is that I take the sample at the very moment the response is happening, not some time later.
This implies that the decay or losses, during the "some time" is very considerable.

With the 3# magnetic field I tried to find a reason for these losses.

But really what I would like to see, is more other people doing the experiment, see the results and then help figuring out the reason.

Tinkerer

[Tinkerer]

I think the above statement over-simplifies and causes confusion -- when we switch off, the field does not just collapse -- it can only collapse as fast as we can reduce the current. So I don't see a "second magnetic field" -- just the first one hanging around trying to die; and we know it resists that vigorously by creating voltage to try to keep current going and itself alive. But we want it to die, and quickly. So as Aziz outlined, one way is really high voltage through high resistance so that we get lot of power with very small current in short time, and let energy laws do the rest.

Nothing new I'm saying, just trying to clarify.

Cheers,

-SB

Are you taking the change in polarity into account?

Tinkerer

[simonbaker]

OK, lets look at the Flyback.
So we have a TX pulse for X amount of time, lets say time does not matter, only the final peak coil current matters.
The first magnetic #1, field has reached its maximum intensity or field strength.
When we switch OFF, this field collapses, the magnetic field lines cut across the target (conductor) at very high speed, generating eddy currents.
The collapsing field also cuts across the coil itself and generates a high voltage. The Flyback.
The coil also has a damping resistor across its terminals. This damping resistor dissipates the Flyback energy.
Current is flowing in the coil. This current generates a new magnetic field, the second #2, magnetic field.
After a very short time the Flyback has reached its peak, the damping resistor regulated this speed and the hight of the peak.

When we switch OFF, this field collapses, the magnetic field lines cut across the target (conductor) at very high speed, generating eddy currents.

I think the above statement over-simplifies and causes confusion -- when we switch off, the field does not just collapse -- it can only collapse as fast as we can reduce the current. So I don't see a "second magnetic field" -- just the first one hanging around trying to die; and we know it resists that vigorously by creating voltage to try to keep current going and itself alive. But we want it to die, and quickly. So as Aziz outlined, one way is really high voltage through high resistance so that we get lot of power with very small current in short time, and let energy laws do the rest.

Nothing new I'm saying, just trying to clarify.

Cheers,

-SB

[Qiaozhi]

-- the turn-on dB/dt is very weak compared to the turn-off dB/dt.

Exactly my point.

And ... if you were to charge a capacitor, and then discharge it into the coil, you would still generate eddy currents in the target. However, in this case, where is the fictitious "charging" of the target at turn-on?

As I said previously, this co-called "charging" is a common misconception.

[simonbaker]

Some parts of the synopsis are correct but others are dead wrong. Does throwing a larger rock into a pond allow you to see the ripples more clearly?

Yes it does, if wind has created small ripples (noise) on the water...

To be fair, it depends on where the noise comes from. If the "noise" is actually ground mineralization patterns or trash, which are repeatable and not time-varying, then your analogy is probably correct -- more power won't help.

Cheers,

-SB