PI Power Output

[Unregistered]

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?

[Qiaozhi]

I have also been getting very interesting results in sampling the ON transient eddy currents. Getting the amps faster into the coil would help with that.

I never said that you wouldn't get any target detection during the coil turn-on time. On the the contrary, there may be some benefits to sampling during that time. However, in a traditional PI (like the HH, for example) the target is detected as a result of eddy current generation due to the collapsing magnetic field, not because the target is somehow "charged" by the initial magnetic field generated during switch-on. Any eddy currents created during the switch-on period are due to the rate-of-change of current in the coil. These eddy currents will of course be smaller than those generated during switch-off. In that case, your second assumption ("Getting the amps faster into the coil would help with that.") is correct.

[Tinkerer]

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?

I am wondering if you refer to my attempt to explain or to Aziz's explanations.
I know that I am not good at explaining things, I am just trying. I am very happy is somebody corrects me and/or comes up with a better explanation.
So please add your comments.

In general we like people to register with a name or handle, so that we know when we are talking to the same person. Trolls tend to be put on the ignore list.
Registering does not cost anything.
It would be OK if you register as "anonymous"

Tinkerer

[porkluvr]

I made a simple comparison between the time it takes for coils connected to IRF640 (200V breakdown) and IRF740 (400V breakdown) to discharge. The TX drive pulse width for each circuit was adjusted so that each coil had the same coil current 2.75A - within a few milliamps.

Now, everybody who reads here knows that I have no honest to god test equipment but do my work in a spice simulator. But, I am working on a machine or two. I compare my situation to the guy who wants a hamburger but is still growing the cow because the local burger joints don't serve Angus. (I'll try not to starve before the Angus is fully grown.)

Anyway, the IRF740 circuit took about 3us to discharge and the IRF640 took about 5us. From what the PI gurus tell us, a 40% increase in discharge speed will equate to more energy directed at the target.

That's too bad because I have about two dozen IRF640 on hand and only a few IRF740, but, facts are facts. I'm too old to waste time lugging around a PI detector running IRF640s now that I know darn well that the IRF740s should give considerably better performance, all other things being the same.

The main advantage I see in using IRF640 is that other components in the system don't need to be rated for the higher voltage, (and also a slight decrease in drive requirement), but I somehow don't think this is a good trade-off.

Perhaps it could be argued that my test is invalid because comparing IRF640 and IRF740 is like comparing apples and oranges, and all I can say to that is take it for what it's worth.

[Tinkerer]

I never said that you wouldn't get any target detection during the coil turn-on time. On the the contrary, there may be some benefits to sampling during that time. However, in a traditional PI (like the HH, for example) the target is detected as a result of eddy current generation due to the collapsing magnetic field, not because the target is somehow "charged" by the initial magnetic field generated during switch-on. Any eddy currents created during the switch-on period are due to the rate-of-change of current in the coil. These eddy currents will of course be smaller than those generated during switch-off. In that case, your second assumption ("Getting the amps faster into the coil would help with that.") is correct.

Qiaozhi,

thanks for your inputs.
I feel we are moving in the right direction. I just have difficulties to put it into the right words. So here is another attempt.

Lets consider the coil has a TC of about 100uS.
Now we pulse it. After 100uS, that is 1TC the coil current has reached about 63% of its maximum possible current.

Now look at a target with a TC of 100uS. It will have been excited to about 63% of its maximum eddy currents.

But a different target with a TC of 300us, will be excited only to about 22% of its maximum eddy currents.

So it does take time to excite or "charge" a target.

It does not matter if this is with the ON transient or with the OFF transient.
Time is time.

OR IS IT????

Does it take the same amount of time to excite a target as it takes for the eddy currents to decay?

Y would think it will be relative to the di/dt.
But what is the di/dt of the switch OFF transient?
At what moment in time?

It is strongest immediately after switch OFF and gets weaker all the time.

But what about the capacitance? The LC tank?
Is it maybe the capacitance that is holding the eddy currents long enough for us to be able to read them?

Tinkerer

[Carl-NC]

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

[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

[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]

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

[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

[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

[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

[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

[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?

[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