Optimizing Target Responses

[Searchy]

Thank you very much for the introduction series and the very well explanation!
That's exactly what i was looking for.
Hope you go into details about ground balancing, EFE, integration etc. later on aswell.

[Searchy]

Maybe you could also explain the Minelab ZVT technology (https://patents.google.com/patent/US7791345) a bit at this point?

[ZED]

Great thread Carl ! Thank you for spending the time to right it up,this sort of info is usually spread all over the place,it be nice if you could make a sticky out of it,hate to see it disappear into all the other threads.

[waltr]

Making this thread a sticky is a good idea.

Would also be great material to add to the next edition of ITMD book.
I bought the 2nd edition when first study how detectors worked and was a bit disappointed at how little PI info it contained.

[bbsailor]

Carl-NC said: "In another thread Joe/bbsailor suggested you want the turn-off time to be at least 5x faster than the target tau you want to detect. That's pretty fair, as anything better than that is hard fightin' for minimal gain."

I saw an article on Eric Foster's forum about optimizing mine detectors. After World War 2 many mines were still in the ground and very dangerous. In an attempt to optimize the design of mine detectors, they electrically stimulated all the metal parts and tried to fine the optimal mine detector pulse and coil characteristics. One conclusion was that after trying multiple coil discharge TCs that 5 TCs was optimal and going any higher did not increase the target's detectability. I wish Eric Foster could find this article and post it as it would give a practical frame of reference for developing a good mental model about the search coil discharge TC relative to the targets potential detectability.

Here is another thought to add to a good mental model. The coil charge TC follows this current rise graph. Try to visualize how vertical or horizontal the current rise chart would be with the current turn off point at each time constant.
1TC - 63.2% of maximum current
2TC - 86.5%
3TC - 95.1%
4TC - 98.2%
5TC - 99.3%

Most rules of thumb that I have read said that the coil should reach about 3 TCs before the current turns off to not loose some current intensity efficiency. While you are at 3TCs, the current has stopped rising to a great extent and the curve is almost flat and horizontal. If, however you turned the current off at 1TC, the current curve would be more vertical and turning off the current while it is still rising would cancel some already invested current. This might be a good addition to Carl's graphs to make the mental model more complete.

Carl, thanks for the graphs. Good and accurate mental models helps us all to better understand how the many variable interact and ways to optimize a design for a particular range of target TCs.

Joseph J. Rogowski

[Hamid]

Thank you very much Carl

[Old cart]

Will this article continue?

[andreapas79]

Hi everyone, thanks Carl to share your knowledge... it's the real gold for me.. I've a couple of questions:
- with the term "tank cap" you mean the big capacitor in parallel (usually 1000uF)?;
- what change increasing impulse width? because i tried to change it and this is the result. I think the reason is longer pulse, much eddy current stored in the targets... is it correct or I wrong? thanks a lot. I hope I'm not off topic with this last question.

In the picture comparison between pure copper, iron and no target curve with different pulse width, RDump. (NoTarget line in red, target curve in blue)

[Carl-NC]

Will this article continue?

Yes, better late than never!

Part 5: Real Responses

So far we've looked at different responses and applied piecewise-linear solutions to much of it. That is, the PI turn-off was a linear ramp that induced a step EMF in the target. This is a good first-order approximation and a good way to do quick-n-dirty comparisons, like what happens when you try to get a sharper & sharper turn-off slew rate (Part 4).

In reality, the PI coil is an RL circuit during turn-on and an RLC circuit during turn-off. Equations can be derived that accurately describe what they do. I won't derive them here but rather just toss them out. ITMD3 will have full derivations.

The turn-on equation is simple:



The turn-off equation is non-obvious but here it is:



where is the peak current at the turn-off event.

Let's look at a realistic example...

Measured coil parameters:
L = 300uH
Rs = 5 ohms (total series R)
SRF = 500kHz (inc. cable etc)

VTX = 10V

The parasitic C for the coil is



The require damping resistor for critical damping is



Now calculate the taus:





VTX (10V) and Rs (5 ohms) tells us the max flat-top current will be 2A. The TX tau os 60us so we would need a TX pulse width of ~300us to reach the flat-top current. A more normal TX pulse width is 100us which makes the calculated peak current 1.62A at turn-off. This becomes the starting point for the turn-off equation.

Here are both equations plotted in Excel. Note that the turn-on curve spans 100us whereas the turn-off curve spans only 3us.



Also note that in the turn-off side the tau_RX is 318ns. If you are expecting the current to be substantially gone in 5*tau (1.59us) you can see it is not. The 0.67% mark occurs at about 2.25ns which is about 7*tau, about 40% more than the usual RL curve.

But that's not the worst part; let's also look at the flyback voltage. Its equation can also be derived:



We have all the variables from above; the curve looks like this:



This curve is also plotted to only 3us. The peak hits ~560v which can be calculated from the derivative of the above equation. Note that the peak occurs at exactly tau_RX. If you were to assume that any kind of 5*tau settling applies here, then you would take 0.67% of 560v (=3.8v) and see that it occurs at about 2.57us, which is about 8*tau.

But this is the raw coil voltage which is then applied to a preamp with a fairly hefty gain, perhaps 500. It's immediately obvious that 3.8v will seriously overload the preamp. To determine a realistic required settling we need to make an assumption: that the preamp must be, say, within 0.5v of settled for the demods to work well. For a gain of 500, this means the coil voltage must be within 0.5v/500 = 1mv of settled. In the curve above, this happens at about 5.42us, or 17*tau.

You might be thinking that 5.42us is pretty darned good. But this is strictly the coil settling; the diode clamps and the preamp overloading have not been accounted for. And everything above assumes the MOSFET never avalanches.

[boilcoil]

Thank you very much Carl, this is perfect explanation. Мaybe peak of voltage wil be a bit more. The peak is apear immediately after turn-off mosfet (for examp 70-100 ns for IRF840 with good driver). Function of current is smoothly function, so in the first moment peak of voltage could be over of avalanche-breaking limit of transistor (V=I * (Rdamp II Rserial to clamp diodes)) i.e. 1,62A* (750 II 2500)=935V. It will provoke avalanche breakdown and limit voltage to Vbr (IRF840 Vbr=500V). It proceeds for about 0,45us (it aval is harmless for transistor, but make longer dacay of curve).
If transistor is IRF9640, breakdown will proceed 1,9us, and all process to V=0,001V will more then 8us. That is, maybe, reason for fast samples to use either small coil current or trahsistrors with hi breakdown voltage.

[green]

Yes, better late than never!

Part 5: Real Responses

So far we've looked at different responses and applied piecewise-linear solutions to much of it. That is, the PI turn-off was a linear ramp that induced a step EMF in the target. This is a good first-order approximation and a good way to do quick-n-dirty comparisons, like what happens when you try to get a sharper & sharper turn-off slew rate (Part 4).

In reality, the PI coil is an RL circuit during turn-on and an RLC circuit during turn-off. Equations can be derived that accurately describe what they do. I won't derive them here but rather just toss them out. ITMD3 will have full derivations.

The turn-on equation is simple:



The turn-off equation is non-obvious but here it is:



where is the peak current at the turn-off event.

Let's look at a realistic example...

Measured coil parameters:
L = 300uH
Rs = 5 ohms (total series R)
SRF = 500kHz (inc. cable etc)

VTX = 10V

The parasitic C for the coil is



The require damping resistor for critical damping is



Now calculate the taus:





VTX (10V) and Rs (5 ohms) tells us the max flat-top current will be 2A. The TX tau os 60us so we would need a TX pulse width of ~300us to reach the flat-top current. A more normal TX pulse width is 100us which makes the calculated peak current 1.62A at turn-off. This becomes the starting point for the turn-off equation.



[ATTACH]54867[/ATTACH]
[ATTACH]54866[/ATTACH]

An attempt at using spice with your example. Included target and separate Rx response. Change in Tx current charges target(I/T=E/L)E=coil volts-(coil current*coil resistance). Fly back volts charges target. Coil volts during fly back is more than 100 times coil current*coil resistance so can use coil volts for calculation. Coil volts: 300/.0003=1A/us, 30/.0003=.1A/us. Question, is the coil still trying to charge the target at 30V fly back? Simulation suggests, could use a shorter delay with a separate IB Rx coil. Question, was any of what I posted correct?

[toolnuts]

Hi Carl,

That was a load of good information, and I especially appreciated the "nugget".

I am looking forward to the rest of your lesson.

Best Regards,

Paul

[Carl-NC]

An attempt at using spice with your example. Included target and separate Rx response. Change in Tx current charges target(I/T=E/L)E=coil volts-(coil current*coil resistance). Fly back volts charges target. Coil volts during fly back is more than 100 times coil current*coil resistance so can use coil volts for calculation. Coil volts: 300/.0003=1A/us, 30/.0003=.1A/us. Question, is the coil still trying to charge the target at 30V fly back? Simulation suggests, could use a shorter delay with a separate IB Rx coil. Question, was any of what I posted correct?

Sorry, I don't understand.

[Mr.Jaick]

Thank you so much Carl great lessons

[green]

Sorry, I don't understand.

Don't know if I understand. I'm trying to understand when the coil discharge stops charging the target. With spice, it appears not long after coil volts peak and start to decay. Does the energy stored in the coil after coil volts peak charge the target or is it wasted? Been stated, don't let mosfet avalanche. Is it because we don't want fly back to flat top or we don't want avalanche? Peak current is limited with a fast Tx coil if we don't avalanche or snub close to avalanche.

Think I see one of my problems, been looking at log amplitude. Target starts to fall of around 200V Tx not peak.