PCB Coil?

[bbsailor]

Bugwhiskers,

Now to answer your other question below.

I tried monitoring the current curve for the TX pulse with a .39 ohms resistor in series but even with a large steel target up close I didn't discern any change.

Monitoring the TX current only shows you the potential decay time of the targets that you want to detect as the faster the current turns off the smaller the targets it will stimulate with eddy currents. Also, The faster the TX current turns off, the higher the flyback voltage will be.

What you want to do is look at the decay time voltage of your target after the first amplifier and see that the TX current turns off (fall time) at least 5 times faster than the target decay time. This mainly becomes an issue with small targets with short decay times. You should see the steel target response when monitoring the voltage output of the first amplifier assuming it has a reasonable gain of about 500. This TX current fall time is important to ensure that you are inducing enough eddy currents into your desired target.

The current rise time is a function of the coil inductance and total resistance of the coil, MOSFET (on resistance) and any series resistor. How long the TX pulse is set to be on will detetmine how high the current rises before it turns off. It takes 3 coil TCs to reach about 95% of the full potential. When you start with designing a coil and TX circuit form the experimenting with the smallest target you want to detect, you need to know that you are putting enough energy into the target to ensure that it can be detected about as well as it potentially can be detected. These techniques help you achieve this. Look at the decay times for a variety of small targets and see if your fall time is at least 5 times faster.

bbsailor

[bugwhiskers]

MOSFET voltage rating vs early sampling

Hi bbsailor,

If the secret to early sampling is to get rid of the back EMF as quickly as possible without ringing or loading down the signal then why do most people use high voltage MOSFETS ie: IRF740 ?

The avalanch diodes inside the MOSFET turn on in nanoseconds and do a very good job of killing off the EMF. Logic would dictate that a remaining EMF of 400 volts is going to decay slower through say a 500 ohm resistor than 200 volts.

If my logic is incorrect, please enlighten me :-))

regards
bugwhiskers

[bugwhiskers]

Eddy currents

The following link contains an interesting description of eddy currents.


http://www.jaycar.com.au/images_uploaded/ferrites.pdf

regards
bugwhiskers

[Carl-NC]

If the secret to early sampling is to get rid of the back EMF as quickly as possible without ringing or loading down the signal then why do most people use high voltage MOSFETS ie: IRF740 ?

The avalanch diodes inside the MOSFET turn on in nanoseconds and do a very good job of killing off the EMF. Logic would dictate that a remaining EMF of 400 volts is going to decay slower through say a 500 ohm resistor than 200 volts.

I thought the same thing, but when I compared different MOSFETs -- the IRF710, 720, and 740 -- the lower breakdown devices had worse recovery times. BBSailor recently sent me some other MOSFETs, and I need to try them out. I'll take photos of the oscope decays and put it all on the forum.

- Carl

[bbsailor]

Bugwhiskers

I will try to answer your questions.

If the secret to early sampling is to get rid of the back EMF as quickly as possible without ringing or loading down the signal then why do most people use high voltage MOSFETS ie: IRF740 ?

Think of the back EMF pulse falling but in three stages to affect it's total fall time.

1. The MOSFET clamps the back EMF spike at the MOSFET voltage rating. But, this clamping looks like a very low rsistance (L divided by a low R equals a longer Time Constant) for a short period of time which tends to extend the fall time until the spike gets below the MOSFET voltage rating then the next two factors take over.

2. The value of the op amp input resistor is effectivley in parallel with the damping resistor (Rd) while the clamping diodes are conducting, until the back EMF pulse and it's oscillations fall below the on-voltage of the clamping diodes.

3. The flyback voltage is then influenced by the value of Rd only when this voltage is below the diodes on-voltage. It is only at this time when Rd is in parallel with the coil by itself and is the only influence on the fall time but that is later in the complete decay cycle.


The avalanch diodes inside the MOSFET turn on in nanoseconds and do a very good job of killing off the EMF. Logic would dictate that a remaining EMF of 400 volts is going to decay slower through say a 500 ohm resistor than 200 volts.

Here is a little experiment you can do to see these effects for yourself.

1. Adjust Rd for a circuit that uses an Rin (op amp input resistor) value of 1K ohms (or what ever you have built already). Then, change Rin to a value about twice as large (2.2K). You will need to readjust Rd as the effective damping load imposed by Rd and Rin has changed. If Rd is 1K, it is in parallel with a 1K input resistor and is effectively 500 ohms but if Rin were 2.2 ohms it would effectivly be 687.5 ohms and Rd would need to be reduced to achieve critical damping with these new values.

2. You can also use a flat-pin type DIP socket modified (cut) to make a 3-pin MOSFET socket. This allows you to plug in many types of MOSFETS with different voltage and capacitance ratings and see the effects on decay time and Rd value. Keep good notes about these values and you will begin to see some patterns evolve that seem, at first, to be counter intutive but will make sense when you consider the answer above.

Post your results on the forum as it would be rather instructive for all the forum members.

I had the same initial thought as you until Eric Foster answered this question before and I did some experiments. The light went on in my head and I can now look at a schematic and see why various components and values are chosen.

Look at the recently posted Goldscan IV schematic and see that there is a 2.2 ohm resistor in series with the coil. Look at the inductance value of the coil and see if you can figure out what that 2.2 ohm resistor does?

I hope this helps?

bbsailor

[bugwhiskers]

Hi bbsailor,

In answer to your question about the 2.2 Ohm resistor, well, it's only going to come into play when the MOSFET is on. It will reduce the ammount of current through the coil. It's value is also Op amp input resistor/Rd.

As for it's relationship to the Inductance of the coil ????

regards

bugwhiskers

[bugwhiskers]

Hi bbsailor,

I am a bit tied up for the next week but will try different values and see what happens. As for different MOSFETS, my circuit is set up for P channel and I only have IRF9640's.


regards
bugwhiskers

[bbsailor]

What The Series Resistor Does

bugwhiskers,

The coil was listed on the schematic as 436 uH at 3.5 ohms. The series resistor of 2.2 ohms preventes the back EMF spike from exceeding the voltage of the MOSFET. Without the resistor, the spike could get about 200 volts higher than the voltage rating of the MOSFET. This voltage limiting helps to sample a little faster and keep the MOSFET cooler. There is still a fair amount of current going through the coil at 450 to 550 PPS.

The relationship to the coil size is that higher inductances have more coil turns and generate a higher back EMF spike.

In Eric Foster's other 3,000 to 10,000 PPS designs, he puts a 27 ohm resistor in series with a 350 uH, 21 turn, 280mm diameter coil that is made with AWG30 wire at 6 ohms. This design provides a current rise time constant (TC) of about 10.6us so that a 50 us wide TX pulse reaches full current in 5 TCs. Even a 30 us pulse achieves about 95 percent of max current.

So if you have a peak back EMF spike the exceeds your MOSFET voltage rating try using a small value series resistor to limit the peak voltage to stay under the MOSFET voltage rating at the max TX pulse width of your design.

bbsailor

[bugwhiskers]

Decay curve

Hi bbsailor,

Could you tell me please just how much target information from the decay curve is never seen because of slow coils/capacitance and because we only get to see the curve after it passes below the clipping of the pre-amps available voltage supply.


regards

bugwhiskers

[bbsailor]

Tough Question

I played with the RC calculator below earlier this morning. Then I had to go out to do some business but had a few hours and some quiet drive time where I could mull this over in my "mental electronic workbench".

http://www.cvs1.uklinux.net/cgi-bin/...time_const.cgi

I looked up the output capacitance (COSS) of your IFR9640 MOSFET and at 10 Volts it is 680pf.

I assumed, for calculation purposes (and you can change the values for your actual values), that your coil capacitance, shield capacitance and coax capacitance adds another 200 pf.

The total capacitance that the coil sees is 880pf.

The Time Constant input would be the equivalent value of the the damping resistor plus the parallel value of the input resistor. I will assume that RD is 1K and Rin is also 1K. The effective parellel resistance is 500 ohms while the diodes are conducting.

The Time Constant is calculated in two stages. Stage one: from 200V down to 0.7V and then stage two: 0.7V to .01V

You want to know how long it will take for the 200V pulse to reach the value of the diode clamping voltage that I assume is .7 Volts.

I will also assume that the starting voltage is 200 volts, the voltage of your MOSFET. 200/0.7 equals 285.7. 1/285.7 equals .0035. 100 percent minus .0035 percent equals 99.9965 percent voltage decay.

Enter 500 ohms into the calculator.
Enter 880pf into the calculator.
Enter 99.9965 percent into the calculator.

The answer is 4.51447us with 880pf and 500 ohms.

At the above time, the voltage will have fallen to 0.7V based on the capacitance of the coil and assuming that Rd damps the inductance/capacitance resonance.

The amplifier will stay in saturation until the input voltage falls below the rail-to rail voltage output that determines the maximum signal level at a gain of 1000. I'll further assume, for ease of calculation, that it is 10V P-P output with a .01 Volt input signal.

Beyond the clamping of the input diodes of 0.7V down to .01V the amplifier is still in saturation but the slope has changed as Rin is no longer in parallel with Rd. That is why Rd is adjusted while looking at the output of the amplifier to accomodate optimizing the damping value to help obtain the fastest sampling time.

Rd is now the only path controlling the RC constant at this point. So back to the calculator we go.

0.7/.01 equals 70

1/70 equals .0142857 which when subtracted from 100 percent equals 99.98572 percent voltage decay or 9.817us for .7V to reach .01V, the point where the amplifier comes out of saturation.

The voltage time of stage 1 is 4.514us plus stage 2 is 9.817us for a total of 14.331us.

While the actual values may not track with what you measure, what is important is the scalling that is necessary to see what circuit design limits are needed to attempt to detect a target with a known total decay time.

Using this technique I was able to come close to something I discovered but never took the time to calculate before today in an attempt to answer your question. I found out that for each 100 pf of capcitance I can eliminate in the TX circuit, I can sample approximately 1us sooner. When I started looking at changing the capacitance values in the calculator I could see how the RC time constants change close to 1us in the scale range typical of PI circuits with coils in the 300 to 400 uH range, 600KHz to 1.2 Mhz self resonance range, Rd 600 ohm to 1K ohm range and Rin 1K ohm to 2.2K ohm range.

Plug in different values into this RC calculator and see how TCs change. It will be very instructive and help you to form a mental model of what changes have the greatest impact on potential sample time. The microsecond range of between 10s to 15us is typically where most sensitive commercial detectors operate. Here you can see how making changes in coil capacitance, cable capacitance, damping and input loading, including DD coils, can impact detector performance.

This has been a very instructive answer for me to think about.

Bugwhiskers, change the capacitance to 780pf and calculate the TC to 0.7V and then from 0.7V to .01V using Rd and Rin of 1K. Then use your values.

bbsailor

[bugwhiskers]

Hi bbsailor,

Thankyou for your encompassing reply.
I suspect that any metal target will have an influence on the decay time at the instant of the cessation of the Tx pulse. Any early signal will therefore be lost in the mud of the parasitic capacitance. I start sampling approx 1 uS before the output of the preamp starts to droop. I have noticed that a target will cause the flat top of the output to be extended and sampling at this point adds to the integration voltage/signal. For me the early signal isn't so important as I need a detector that can discriminate larger gold amidst very ferrous Ironstone. From my experiments I have concluded that it isn't the mass of the target that dictates the strength of signal but rather the area of the face of the target presented to the coil. A 4 inch square piece of Alfoil produces an enormous signal and has a mass of almost nothing. It would seem for my purposes at least the later sample has more information about Gold and Iron as their responses are very similar. Later this week I am visiting a friend who has a collection of > 1 oz nuggets and I will be posting here the results of the discrimination on a huge variety of targets.
The reason for the question was because I needed to know if the early sample was just for fast targets ie small gold in which case my efforts are better directed at producing a coil that goes deep via higher inductance at the expense of higher parasitic capacitance and loss of sensitivity to small targets.

regards
bugwhiskers

[bugwhiskers]

Hi bbsailor,

Further to my previous reply, your calculations are very close. You factored in some shielding capacitance that doesn't exist that would account for the small discrepancy. Your figure of 1uS per 100pF is a nice little gem for tinkerers to anguish over.

regards
bugwhiskers

[bbsailor]

bugwhiskers,

To compare surface area to mass here is some interesting information about a gum wrapper. A silver (color) foil gumwrapper has about a 25us decay time folded evenly 6 or 7 times (as much as you can fold it). Now cut it in half (reduce the mass) and the decay time drops to 20us; cut in half again and it drops to 15us and cut once more and the decay time is 10us. This information is based on a form disussion I had with Eric Foster. He was kind enough to measure the gum wrapper at my request.

I guess that gum wrappers are pretty universally similar so this would make a good standard target to establish some standard time decays.

If you can detect a large piece of gold that seems to have an unusual short time decay, this reaction may be due to the fact that the piece of gold may look like many small gold pieces lumped together.

Try this to simulate the short decay time of some gold. Lay out the gum wrapper foil flat and measure the decay time. Glue the foil to a piece of cardboard using rubber cement. Then using a razor blade, score the alumium foil into successivly smaller squares (down to about 1cm square or a little smaller) and the decay time should get smaller as there are smaller pieces that make up the same surface area.

bbsailor

[bugwhiskers]

Pics of some old gold country.

The attached pics were taken just outside Ararat, VIC Australia. One shows what's on top of the ground and the other what is underneath. There is still gold there, 2 nuggets totalling 1.7grams were found with a GP3500 in 10 minutes, the prospector had other things to do so he went home.

regards

bugwhiskers

[bugwhiskers]

Coil former

Hi bbsailor,

Yes... I am still alive...bad time of year, visiting in-laws and out-laws.
Attached is a pic of the coil former I spoke of earlier. It's PVC stormwater pipe 300mm ID, around 313mm OD. I have yet to put a cut in it to facilitate coil removal. Off to look at your new coil document to get some ideas on wire, #turns etc.

regards
bugwhiskers