Optimizing Target Responses

[green]

Carl, and forum members

I want to propose some design goals to optimize target responses for smaller, lower TC targets.

Find ways to lower the capacitance as seen by the coil to allow a steeper coil discharge slope.

1. Use a low COSS MOSFET
2. Use a series diode between coil and MOSFET.
3. Use a low dielectric coil wire to reduce coil turn to turn capacitance.
4. Use a low coil to shield capacitance construction method.
5. Attempt to reduce coax capacitance by:
5.1 Identifying the coil TX and RX components that can be mounted near the non metallic coil stem to eliminate the bulk of the coax capacitance.
5.2 Identify creative mounting methods to make a very a small active component module that can be mounted near the coil.
5.3 Identify coil building techniques that separate RX and TX coils that allow the highest damping resistor values possible to allow the most vertical coil discharge slope to better stimulate smaller targets.
5.4 Identify the optimum coil size, coil TX frequency and number of integrated signals to improve the sensitivity to low TC targets.

I believe if we as a metal detecting community attempt to achieve the above, that we will minimize coil capacitance, increase damping resistor values and stimulate smaller TC targets better.

Joseph J. Rogowski

Does MOSFET COSS mater if a correct diode(MUR460)or some others is used. Not all diodes work.

[bbsailor]

Carl, and forum members

I want to propose some design goals to optimize target responses for smaller, lower TC targets.

Find ways to lower the capacitance as seen by the coil to allow a steeper coil discharge slope.

1. Use a low COSS MOSFET
2. Use a series diode between coil and MOSFET.
3. Use a low dielectric coil wire to reduce coil turn to turn capacitance.
4. Use a low coil to shield capacitance construction method.
5. Attempt to reduce coax capacitance by:
5.1 Identifying the coil TX and RX components that can be mounted near the non metallic coil stem to eliminate the bulk of the coax capacitance.
5.2 Identify creative mounting methods to make a very a small active component module that can be mounted near the coil.
5.3 Identify coil building techniques that separate RX and TX coils that allow the highest damping resistor values possible to allow the most vertical coil discharge slope to better stimulate smaller targets.
5.4 Identify the optimum coil size, coil TX frequency and number of integrated signals to improve the sensitivity to low TC targets.

I believe if we as a metal detecting community attempt to achieve the above, that we will minimize coil capacitance, increase damping resistor values and stimulate smaller TC targets better.

Joseph J. Rogowski

[Carl-NC]

If target tau is 1us, what formula should be used to calculate coil discharge tau(L/Rd, L/1.4Rd, or L/2Rd)for .2us? Assume separate Tx, Rd only no input resistor. I was thinking L/2Rd(critical damped), maybe not correct.

1/2*L/Rd

Always remember that Rd is ALL damping R, including the input resistor.

[green]

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.
If target tau is 1us, what formula should be used to calculate coil discharge tau(L/Rd, L/1.4Rd, or L/2Rd)for .2us? Assume separate Tx, Rd only no input resistor. I was thinking L/2Rd(critical damped), maybe not correct.

Used spice and Excel to compare(L/Rd, L/1.4Rd and L/2Rd). Don't know if it's an accurate comparison. Looks like using L/R is better.

[green]

If you had an RL circuit with a tau of 318ns it would be 99.3% decayed in 1.59us (5*tau). The turn-off curve above requires 2.25ns to reach 99.3% decayed.

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.
If target tau is 1us, what formula should be used to calculate coil discharge tau(L/Rd, L/1.4Rd, or L/2Rd)for .2us? Assume separate Tx, Rd only no input resistor. I was thinking L/2Rd(critical damped), maybe not correct.

[Carl-NC]

tau Rx is 318ns, what is tau for the usual RL curve? RL and RX the same?

If you had an RL circuit with a tau of 318ns it would be 99.3% decayed in 1.59us (5*tau). The turn-off curve above requires 2.25ns to reach 99.3% decayed.

[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.

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.

Apologize for all the questions. Took to many stupid pills.

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. tau Rx is 318ns, what is tau for the usual RL curve? RL and RX the same?

[green]

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.

Looked at spice simulation again. Maybe only 10% coil energy left when target started to discharge?

Target TC, Coil TC ratio. Need to read it all again.

[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.

[Mr.Jaick]

Thank you so much Carl great lessons

[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.

[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

[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?

[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.

[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.