PI Power Output

[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

[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

[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]

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

[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.

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?

[Tinkerer]

True, you can increase the voltage to maintain the same coil current, but then it's no longer equal power, eh? And how practical is going from a 12V battery to a 24V or 48V battery?

One of the chip companies (LTC?) recently sent me an apps booklet on some of their new HV switching supply chips. The only one that caught my eye was one that could supply something like 2-3 amps at 72V (I don't remember the exact numbers). I wondered whether it would make a good coil driver (either for PI or VLF). I have no time to pursue it, so I'll just have to keep wondering!

- Carl

Thanks Carl,

hmmm, 72V pulse... I will look at that.
What I had in mind though, is much more moderate.
The LI-Ion polymer batteries have quite a wide range of voltage from fully charged to low discharge limit.
Like a nominal 11.1V battery has a peak voltage of 12.75V and a low voltage of 8.5V.
To use this voltage straight from the battery for TX makes too much variation. So for a TX voltage of 12V I either need to use a 14.7V battery or boost and regulate the voltage.
At present I use a capacitor switched voltage doubler so I have a few volts left over to play with.

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.
Hmmm... at 72 Volts... it would actually help a lot....

Tinkerer

[Aziz]

Hi all,

the reason for increasing the flyback voltage is as follows:
First, the flyback voltage will be limitted by the mosfet due to avalanche breakdown voltage. Try to make this as high as possible. To stimulate more eddy currents in the target, the magnetic field around the coil must be removed as quickly as possible (Uind = -L*dB/dt).

Remember: The coil is an inductor, resistor and capacitor. The resistor is a series resistor connected to the inductor. The capacitor is a parallel connected capacitance to the coil. It will store or generate energy (on external E/B-field stimulation). Generation is a bad name for this. To be correctly here, it will only convert energy. Target is also an inductor, resistor and capacitor. On the resistive parts (R), energy will be converted into heat, if there is a current flow: P = U*I or P = R*I²

The coil has also a state: empty, partly charged, fully charged and its coupled EMF field state (electric and magnetic fields). Empty state is a steady static state with no stored energy in it.

Lets assume, we have fully charged (energized) the coil and we are going to switch off. We should know, that the coil is charged with the supply voltage on the capacitive elements. The energy of this is: W = 0.5*C*U² (C=capacitance, U=voltage). And of course, it is charged with magnetic field energy: E=0.5*L*I² (L=coil inductance, I=coil current).

The stored energy in the coil should be removed as quickly as possible using the Faraday's Law. You can either recycle this energy (flyback converter are doing this) or convert this into heat on resistive elements (usually on dampening resistor Rd). Lets do the latter one due to its simplicity.

Now we are switching off at time toff. Lets take some figures for calculating the amount of energy: C=300pF (total), U=12 V, I=1A, L=300µH, Rd=400 Ohm.
W = 0.5*300pF*12V*12V = 0.0000000216 Joule (electric field energy)
E = 0.5*300µH*1A*1A = 0.00015 Joule (magnetic field energy)
As you can see, the dominating stored energy is the magnetic field energy. The electric field energy (remember: coil is also a capacitor) will dissipate and be converted into heat during the switch-off time (coil currents reaches zero).

What is happening then? (I will speed up the process a bit).
The coil current reaches zero, a high flyback voltage will be generated and the mosfet is normally clipping this high flyback voltage due to the avalanche breakdown voltage. Lets assume, we reach the maximum possible flyback voltage of 400V. Hey, do not forget the capacitance now! The capacitor will be charged up with much more voltage!
The electric field energy is then:
W = 0.5*300pF*400V*400V = 0.000024 Joule (electric field energy). The electric field energy is becoming very important now. It is 1/6.25 of the magnetic field energy and could affect the flyback process (yes, it does really).

During the avalanche breakdown voltage (and this voltage is nearly constant during this time), the stored magnetic field energy will be converted into heat on resistive elements: U=R*I (R=Rd, Rd=400 Ohm, U=400V), I=U/R = 400V/400Ohm = 1A, P=U*I = 400V*1A = 400 W. The electric field energy won't be dissipated as the avalance breakdown voltage is not changing at the moment. Only the magnetic field energy will be converted into heat with constant and maximum rate. The target stimulation will be energized with maximum rate too.

Now the flyback voltage won't be clipped by the mosfet anymore. It goes below the avalanche breakdown voltage. The stored electric field energy in the coil will be dissipated into heat with the remainder magnetic field energy at the same time now according to the natural decay behaviour of critical damped coil and R/C discharge curve.

Now you are awered of keeping the coil capacitance (+parasitics) as low as possible. It limits the high flyback voltage and delays the damping process. It also delays/slows down the switch-off of the coil.

Now the real reason for the high flyback voltage:

The rate of magnetic field energy conversion into heat can be increased with higher flyback voltages. P = U*I, P = R*I² or P = U²/R.
As the damping resistor Rd is fixed for a coil system configuration, only the current in the damping resistor should be increased. This only can be achieved with more voltage across the damping resistor Rd.
P = U²/Rd. And this behaviour is quadratic. If you double the flyback voltage, you can achive a power dissipation of four times. This will convert the magnetic field energy into heat four times more, in which will generate higher target eddy current. The whole damping process will be more quickly as a result too.

How can the flyback voltage be increased?
- keep the coil capacitance very very low (+ parasitic capacitances: shield, cable, switching mosfet...)
- make the switch-off very very fast (take mosfet drivers and fast mosfets)
- take high voltage rated mosfets (800, 900, 1000 V or more)
- take higher gate voltage to switch the mosfet (10-20 V).
- increase the magnetic field energy by increasing the coil current
- increase the coil drive voltage (to increase the coil current)
- increase the magnetic field energy by increasing the coil inductance (note: the coil capacitance will increase too).
- use split tapped coils to generate higher voltages than the breakdown voltage of the mosfet, which is decoupled from the mosfet. If you use center-tapped coils, you can have differential signal with improved EMI rejection.

Of course, every item has its own advantages and disadvantages. You have to make some comprimise here.

Aziz

[Carl-NC]

Carl, thanks for the numbers.
There is one more way to increase coil current. That is to increase the TX voltage. With higher voltage we can get the same amps quicker into the same coil.
Quicker means more PPS for equal power.
What are the negative effects?

Tinkerer

True, you can increase the voltage to maintain the same coil current, but then it's no longer equal power, eh? And how practical is going from a 12V battery to a 24V or 48V battery?

One of the chip companies (LTC?) recently sent me an apps booklet on some of their new HV switching supply chips. The only one that caught my eye was one that could supply something like 2-3 amps at 72V (I don't remember the exact numbers). I wondered whether it would make a good coil driver (either for PI or VLF). I have no time to pursue it, so I'll just have to keep wondering!

- Carl

[Tinkerer]

Every time you double the PPS you will get a sqrt(2) improvement in noise. But you need to maintain the same coil current as the TX pulse gets shorter, which probably isn't going to happen, unless you were already current limited by a high series R. Instead, the coil current will probably start decreasing, which impacts depth at (roughly) a 6th-power rate. So I think once you reach the practical limit of coil turn-on speed you will be trading a 2nd-power noise improvement for a 6th-power signal decrease.

- Carl

Carl, thanks for the numbers.
There is one more way to increase coil current. That is to increase the TX voltage. With higher voltage we can get the same amps quicker into the same coil.
Quicker means more PPS for equal power.
What are the negative effects?

Tinkerer

[Tinkerer]

Hi B^C,



increase your flyback voltage at any rate. The more, the better in this case (the ultimate brute force method)! Take care to the avalanche breakdown voltage of the mosfet. (limitting the flyback voltage).

Aziz

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.

By the time the peak has been reached, a good amount of the Flyback energy has been dissipated.
As the peak of the Flyback is reached, the second magnetic field #2, the one built by the Flyback, starts collapsing.

Again the collapsing field generates current in the coil, and the current produces magnetic field # 3 but due to the damping resistor the energy is decaying quickly to zero.

Now lets put signs to the fields.
Magnetic field #1 is positive, since it was static for some time, the eddy currents, negative in this case have decayed.
Magnetic field #2 is negative. It generates eddy currents in the target that are positive.
Magnetic field #3, the final one is again positive. It generates negative eddy currents. These eddy currents are weak because the field is much weaker, having only a fraction of the energy of field #2, so it reduces the amplitude of the eddy currents of field #2.

So, if I did not get lost there somewhere under way, what we read, after the Flyback has reached 0V, is the eddy current generated by the magnetic field #2, reduced by Field #3 and decaying since the time the Flyback has reached its peak.

Is that right?
Please correct me if I am wrong.

Tinkerer

[Carl-NC]

Intuitively, using a higher PPS, but with the same power consumption, should produce a better S/N.
Example: 100 PPS with long TX, using 1W total consumption.
Or: 1000 PPS with short TX, using 1W total consumption.
With 1000 PPS you can average 10 times more signal samples. This should better the S/N considerably.
But what is the reality?
What does the Math tell?
Anybody willing to do an analysis?

Every time you double the PPS you will get a sqrt(2) improvement in noise. But you need to maintain the same coil current as the TX pulse gets shorter, which probably isn't going to happen, unless you were already current limited by a high series R. Instead, the coil current will probably start decreasing, which impacts depth at (roughly) a 6th-power rate. So I think once you reach the practical limit of coil turn-on speed you will be trading a 2nd-power noise improvement for a 6th-power signal decrease.

- Carl

[Aziz]

Hi B^C,

Hi All,

What is an "average" control box power output figure to the TX coil?

Eg:
Voltage & Amps

Is there a limit on what this power output should be or is it a case of the more the better type senario?

Is there a preferred Power output figure for some reason?

increase your flyback voltage at any rate. The more, the better in this case (the ultimate brute force method)! Take care to the avalanche breakdown voltage of the mosfet. (limitting the flyback voltage).

Aziz

[bbsailor]

This is a popular misconception. During the time when the current is being established in the TX coil, the target is not being charged. The generation of eddy currents in the target occurs when the "switch" opens. Of course there is some eddy current generation during switch-on, but it is small when compared to switch-off, and plays no part in target detection. The switch-on period needs to be of long enough duration to establish a sufficiently strong magnetic field around the coil. It is the collapsing field that does the work, not the other way round.

Tinkerer,

Look at it this way.

The longer the TX pulse is on, the stronger the magnetic field will be that has the potential to fully stimulate eddy currents in a target that requires a longer pulse such as a silver or copper coin.

The slope of the turn off current is governed by the combined value of the damping resistor (Rd) and the input resistor to the first amplifier (Rin). The resistors Rd and Rin are effectivly in parallel when the clamping dioded are clamping. If Rd is 1K and Rin is also 1K the the effective damping value is 500 ohms. The discharge TC of the coil is now 300uH divided by 500 ohms or 0.6us.

The fastest that a coil could potentially turn off is a perfectly vertical slope. However, since a perfectly vertical slope is not practically achievable, getting the turn off slope as steep as possible is a convenient mental model to contemplate.

Minimizing coil capacitance in the following areas will help achieve this.

1. Coil capacitance between coil turns
2. Coil-to-shield capacitance
3. MOSFET COSS
4. Using a diode between the MOSFET drain and the coil
5. Coax cable capacitance

Theory states that the speed of the current turn off should be 5 times faster than the TC of the smallest target you want to detect.

Based on the above information you can see that PI designs between 100 to 500 PPS will have longer pulses and higher coil current for detecting coins or larger relics.

PI designs 500 pps to about 3000 pps will have shorter TX pulses but use the integration time to help extract the target signal from the noise. The tradeoff is lower PPS frequencies with stronger coil current and longer delays versus faster PPS rates with lower current, that can be damped quicker for shorter delays. There are some intermediate designs that allow the user to select the best of both worlds.

If you want to find very small low conductivity targets, then theory states that you want to use a smaller coil, with more coil turns (still 300uH) with the highest Rd value for a coil discharge TC of between 0.3us to 0.6us or an effective Rd value between 1000 ohms to 500 ohms, or higher effective Rd values, if possible.

Balanced DD coils allow the RX coil to have about 450uH inductance for a little more sensitivity. Also because a balanced DD RX coil does not saturate the opamp as much as a mono coil, this is the way to go for the fastest coil turn-off current and earliest sampling.

The optimium design of a PI circuit starts with defining the TC of the smallest target you wish to detect and work backwards to get critical PPS, pulse width, delay and sample width design parameters.

This should get you moving in the right direction.

bbsailor

[Tinkerer]

Generating Eddy currents

This is a popular misconception. During the time when the current is being established in the TX coil, the target is not being charged. The generation of eddy currents in the target occurs when the "switch" opens. Of course there is some eddy current generation during switch-on, but it is small when compared to switch-off, and plays no part in target detection. The switch-on period needs to be of long enough duration to establish a sufficiently strong magnetic field around the coil. It is the collapsing field that does the work, not the other way round.

I welcome your input. It is time we dispel the misconceptions.
So I suggest you set up a small bread-board and sample during the TX pulse.
And then sample after the TX switches OFF.

Then you show us your results.

I am very curious to see if they are the same as mine.

Now, I know that if you come up with a page full of mathematical formulas and Spice simulations, you can show me that the slow switch ON transient generates much less Eddy currents than the fast switch OFF, but formulas and simulations are only as good as the input.
If the model for the charge or excitation of the target is wrong, the results come out wrong.

So please do the real experiment. You will be very much surprised.

All the best

Tinkerer