AMX TX

[IBGold]

Hi Ian ... yep I can send you one. Do you need the programmed CPU chip as well ?

I think I used to have your address ... PM me where you want it sent.

moodz

Will do and yes on the chip you know programming is not one of my strong points.

Regards, Ian.

[IBGold]

PM sent Paul cheers. Regards, Ian.

[moodz]

PM sent Paul cheers. Regards, Ian.

Ok I will spin one up for you in the next coupla days ....

[moodz]

... I have been working on the "Tilt" or Ramp problem ( not that sure that its critical ) .... however energy losses from the CCP coil will cause tilt or ramping of the TX current waveform and there are options with DD coils or similiar that may be worth exploring if the Constant in CCP can actually be kept constant.

So far I have been able to get a control loop working ( simulation ) it seems to be doing the job with a current variation of around 2 mA over 100 microseconds repeatable and a 15 millisecond loop response ( fast enough for coil swing tracking ).

Next step to translate this to real parts to test theory.

The pic shows the simulation response from startup with the ramped coil current ( mono and bipolar pulsing options ) as the loop control voltages stabilises the CCP ramping. The strong target response at top is shown at top.
The loop control voltage obviously responds to changes in losses and thus can be used as a source of target / ground info.

moodz

[Carl-NC]

I've been working on the TX circuit. It needs a way to control the current so this is what I'm leaning toward:



I believe that Paul said he was using a high-side current monitoring chip but it seems to me an opamp will work just as well. The sampling switch is used to avoid the transition points and may be unnecessary if the integrator is slow enough. However, depending on ground effects the integrator may need to be on the fast side. The output of the integrator Vadj provides a bucking voltage to the buck regulator and allows its output to vary from 0V up to some maximum. VTXref controls the flat-top current where the current is 0.1*VTXref/Rsens. That is, if VTXref = 0.5V then the current is 0.5A.

I've also been looking at the math of the circuit, what affects the flat-top slope, and how it might be corrected assuming it needs to be corrected. I've found in simulations that a very short current injection as the transition is ending corrects the slope. This is easy to do in sims but circuitry will take some figurin' out. Also need a way to measure the slope and regulate the amount of injection, which will likely involve a micro.

[Carl-NC]

Paul, you posted while I was typing. Your sims look a lot like mine.

[moodz]

I've been working on the TX circuit. It needs a way to control the current so this is what I'm leaning toward:



I believe that Paul said he was using a high-side current monitoring chip but it seems to me an opamp will work just as well. The sampling switch is used to avoid the transition points and may be unnecessary if the integrator is slow enough. However, depending on ground effects the integrator may need to be on the fast side. The output of the integrator Vadj provides a bucking voltage to the buck regulator and allows its output to vary from 0V up to some maximum. VTXref controls the flat-top current where the current is 0.1*VTXref/Rsens. That is, if VTXref = 0.5V then the current is 0.5A.

I've also been looking at the math of the circuit, what affects the flat-top slope, and how it might be corrected assuming it needs to be corrected. I've found in simulations that a very short current injection as the transition is ending corrects the slope. This is easy to do in sims but circuitry will take some figurin' out. Also need a way to measure the slope and regulate the amount of injection, which will likely involve a micro.

... So this circuit is adjusting the "average" pulse current to keep it constant but does not really adjust the slope ... which makes me realise that my solution adjusts the ramp reasonably well but not the absolute value of the average current ( will have to do some tests ).
The optimal control loop will adjust the pulse current level AND the slope.

moodz

[Carl-NC]

... So this circuit is adjusting the "average" pulse current to keep it constant but does not really adjust the slope ... which makes me realise that my solution adjusts the ramp reasonably well but not the absolute value of the average current ( will have to do some tests ).
The optimal control loop will adjust the pulse current level AND the slope.

moodz

True... my first step was to control the square wave current, not the slope. Controlling the slope will likely require 2 samples, and maybe an ADC, some DSP, and a way to correct. I haven't figured that part out yet, but maybe you have.

[moodz]

True... my first step was to control the square wave current, not the slope. Controlling the slope will likely require 2 samples, and maybe an ADC, some DSP, and a way to correct. I haven't figured that part out yet, but maybe you have.

... I might have ... your post made me think about the average pulse peak current remaining constant ... and I was so focused on the ramp fixing that I neglected that. However fortuitously if you fix the ramping then you get "free" peak current regulation.
Pic below shows a large target being coupled to the TX coil every 10 milliseconds ( ie 10 ms in and 10 ms out ) The loop recovers to the peak current level in 3 or 4 ms. Peak current seems to be the same regardless of target or not.
The target response is shown also.

Clearly you can sample the loop voltage to get a target / loss value indication to add to all the other samplings .....

moodz.

[moodz]

... I cleaned up the schematic somewhat ... Here is the proposed CCPI control loop ... no DSP or fancy current sensing chips I though this could be done also.

It runs off 1.5 volts for 1.2 amp swing / 750 volt peak pulse differential and 2.6 watts consumption.

moodz

[Carl-NC]

Thanks for posting that, I'll dig into when I have time.

[Carl-NC]

I like to mathematically understand the operation of a circuit so I can better understand the various trade-offs available. For the TX circuit, I want to understand how coil parameters affect performance, including power consumption and tilt.

A coil has diameter, number of turns (N), and wire gauge (assuming circular). From that we can measure L, R, and f₀ (resonance frequency). Let's assume a fixed peak current I_p. We can immediately calculate the continuous "DC" power as P_DC = I_p*R. We can also calculate the ampere-turns as N*I_p which gives us a relative TX field strength. From this we can derive an efficiency ratio of the (TX field strength)/(continuous power) = N/R. Since R scales linearly with N, we can see that producing twice the field strength will consume twice the power.

Each transition of the TX driver based on the resonance of the coil. That is, the current transition from +I_p to -I_p is a half-cosine and the flyback voltage produced is a half-sine. The SRF (f₀) tells us the transition time (slew) is 0.5/f₀. From L and f₀ we can calculate the interwinding parasitic C_p = 1/((2πf₀)²*L). The energy in the coil (E_L = 0.5*L*I_p²) gets transferred to the parasitic Cp (E_C = 0.5*C*V_p²) so the peak flyback voltage is



More turns increases L and generally increases C_p but in a complex way and not as fast as L, but generally speaking a coil with more turns will produce more flyback. It is useful to know exactly how much flayback so you can choose the right MOSFETs.

Here is a table of some real 10" coils made with 18awg magnet wire, with all the calculations added (I_p is 0.5A):



Above 1mH the flyback is starting to exceed what's reasonably available in SiC MOSFETs. However, the equation above suggest we can simply add some more capacitance to the coil and reduce the flyback, obviously at the expense of slew rate. This doesn't change the DC power consumption. If C=1000pF is added to all cases then we get this:



So we can control the flyback with capacitance, which is why my schematic has a place for C_extra. The above table adds a VTX column which is the TX drive power supply voltage required to achieve I_p = 0.5A (arbitrarily chosen). VTX is simply R*I_p. Note that the resistances of the MOSFETs have been ignored in both the P_DC and V_TX calculations. I most cases these can be selected to be a fraction of the coil resistance. Another key element missing is the resistance of the coil cable. Normally this is a significant part of the overall resistance (often as much or more than the coil) because in the usual PI designs we are trying to minimize its capacitance. In this design, we can minimize its resistance and easily tolerate the higher capacitance. It's something to keep in mind.

We can also calculate the loss in the current transitions that produce the tilt. The total energy in the coil at the transition is E_L = 0.5*L*I_p²) and we know this transitions from +I_p to -I_p in a half-cosine waveform. This current flows through the coil's parasitic R and dissipates power. The energy lost in the resistance is



where P(t) = i²(t)*R, i(t) = I_p*cos(2πf₀*t), and T = 0.5/f₀. From all this we get



The relative loss is E_R/E_L which is



and from that we can calculate that actual amount of tilt current as I_Loss = I_p*E_Loss. Here are the calculations for the above coils, still assuming C_extra = 1000p:



From this we see that tilt should be fairly independent of coil inductance. However, energy loss is exacerbated by additional capacitance.

So in the end we can use high inductance coils to get higher ampere-turns but they need to be toned down with extra C which slows the transition and worsens tilt. Since C_extra can be placed inside the coil it may make sense to use lower L coils for exceptional speed in, say, hunting for small nuggets and a high-L coil with slower transitions for big deep nuggets.

[moodz]

I like to mathematically understand the operation of a circuit so I can better understand the various trade-offs available. For the TX circuit, I want to understand how coil parameters affect performance, including power consumption and tilt.

A coil has diameter, number of turns (N), and wire gauge (assuming circular). From that we can measure L, R, and f₀ (resonance frequency). Let's assume a fixed peak current I_p. We can immediately calculate the continuous "DC" power as P_DC = I_p*R. We can also calculate the ampere-turns as N*I_p which gives us a relative TX field strength. From this we can derive an efficiency ratio of the (TX field strength)/(continuous power) = N/R. Since R scales linearly with N, we can see that producing twice the field strength will consume twice the power.

Each transition of the TX driver based on the resonance of the coil. That is, the current transition from +I_p to -I_p is a half-cosine and the flyback voltage produced is a half-sine. The SRF (f₀) tells us the transition time (slew) is 0.5/f₀. From L and f₀ we can calculate the interwinding parasitic C_p = 1/((2πf₀)²*L). The energy in the coil (E_L = 0.5*L*I_p²) gets transferred to the parasitic Cp (E_C = 0.5*C*V_p²) so the peak flyback voltage is



More turns increases L and generally increases C_p but in a complex way and not as fast as L, but generally speaking a coil with more turns will produce more flyback. It is useful to know exactly how much flayback so you can choose the right MOSFETs.

Here is a table of some real 10" coils made with 18awg magnet wire, with all the calculations added (I_p is 0.5A):



Above 1mH the flyback is starting to exceed what's reasonably available in SiC MOSFETs. However, the equation above suggest we can simply add some more capacitance to the coil and reduce the flyback, obviously at the expense of slew rate. This doesn't change the DC power consumption. If C=1000pF is added to all cases then we get this:



So we can control the flyback with capacitance, which is why my schematic has a place for C_extra. The above table adds a VTX column which is the TX drive power supply voltage required to achieve I_p = 0.5A (arbitrarily chosen). VTX is simply R*I_p. Note that the resistances of the MOSFETs have been ignored in both the P_DC and V_TX calculations. I most cases these can be selected to be a fraction of the coil resistance. Another key element missing is the resistance of the coil cable. Normally this is a significant part of the overall resistance (often as much or more than the coil) because in the usual PI designs we are trying to minimize its capacitance. In this design, we can minimize its resistance and easily tolerate the higher capacitance. It's something to keep in mind.

We can also calculate the loss in the current transitions that produce the tilt. The total energy in the coil at the transition is E_L = 0.5*L*I_p²) and we know this transitions from +I_p to -I_p in a half-cosine waveform. This current flows through the coil's parasitic R and dissipates power. The energy lost in the resistance is



where P(t) = i²(t)*R, i(t) = I_p*cos(2πf₀*t), and T = 0.5/f₀. From all this we get



The relative loss is E_R/E_L which is



and from that we can calculate that actual amount of tilt current as I_Loss = I_p*E_Loss. Here are the calculations for the above coils, still assuming C_extra = 1000p:



From this we see that tilt should be fairly independent of coil inductance. However, energy loss is exacerbated by additional capacitance.

So in the end we can use high inductance coils to get higher ampere-turns but they need to be toned down with extra C which slows the transition and worsens tilt. Since C_extra can be placed inside the coil it may make sense to use lower L coils for exceptional speed in, say, hunting for small nuggets and a high-L coil with slower transitions for big deep nuggets.

Not disagreeing with you ...

My take on the mathematics is that in an ideal circuit ( coil R = 0 and mosfet switches that are ideal ) intuitively there is no Power Loss ( since L and C cannot dissipate power ).
It would be possible to test this in ltspice using ideal switches and L / C coils but R = 0.

Because there is no power loss with R = 0 there should be no ramping or tilting.

One possible solution ( the one I used ) is to "overcharge" the TX coil causing an "over tilt" error ... target losses will cause a "under tilt" error.. the feedback control loop then causes a current dumper ( ie dummy load ) to be adjusted to soak up the excess power and correct the tilt.
If there are additional losses due to ground or targets then the current dumper backs off but still correcting the tilt error.

Changing the LC does not affect tilt in any substantial way. Only the resistive losses do. So power correction is one possible solution.

Of course it wont correct pulse distortions due to resonances that is an LC issue ( but not a power issue either ).

moodz

[Carl-NC]

Correct, here is a sim of the 500uH with R=0.49Ω and R=0. With R=0.49Ω the current tilt is a little less than the predicted 1.16mA. With R=0 the tilt is not quite zero likely because of the blocking switches.

[Tinkerer]

Correct, here is a sim of the 500uH with R=0.49Ω and R=0. With R=0.49Ω the current tilt is a little less than the predicted 1.16mA. With R=0 the tilt is not quite zero likely because of the blocking switches.

Did you forget to add the simulation?