Announcement

Collapse
No announcement yet.

NEW !!! Full differential PI front end.

Collapse
X
 
  • Filter
  • Time
  • Show
Clear All
new posts

  • Aziz
    replied
    Hi Paul,

    I finally found sources for good low-noise instrumentation amplifiers. Following IC's should be good enough:
    INA163 (price per piece 4.95 EUR)
    SSM2019 (price per piece 5.72 EUR, more bandwidth, lower noise)

    The SSM2019 is a pin compatible type to a standard 8-DIP instrumentation amplifier. The INA163 offers active feedback bandwidth limitting feature (low pass filter) but has SO-14 surface mount package.

    Other replacements could be:
    INA217 (source not found yet, 8-DIP version available)
    SSM2017 (obsolete product, replacement is INA217 or SSM2019)

    I will order both INA163 and SSM2019 chips soon and will test them. I probably will use the INA128 in the mean time.

    Aziz

    Leave a comment:


  • moodz
    replied
    Well guys ... I 'tuned' the damping on the diff coil and below you can see a scope pic of a flyback decay ( green trace ) of 2us with the output of the diff amp ( white trace ) giving a first sampling point of 3us .... I dont see any point to improve beyond this or there will be no pulse left ..

    100 us tx
    16 turn 25 cm diff coil #22 magnet wire
    12 volt drive
    coil R = 2.0 ohms



    moodz

    Click image for larger version

Name:	diffamp_3us.jpg
Views:	1
Size:	66.9 KB
ID:	322896

    Leave a comment:


  • B^C
    replied
    Moodz,

    Ahha, i get where your coming from, i will read up on the links you posted, thanks for that, it's clearer already.

    Leave a comment:


  • moodz
    replied
    Originally posted by B^C View Post
    Gday Guy's,

    A real dumb question here!

    Could somebody give me the "basic definition" of "Full Differential" in regards to what your trying to achieve.

    Is it just the signal your talking about or ?

    Sorry for the dumb question but i didn't get it when i first read about this with Aziz's posts on another forum & still i don't get it???
    Hi BC ... because there is no 'history' in metal detector design using differential coils / amps ... I will have to fall back to audio.
    www.en.wikipedia.org/wiki/Balanced_audio
    http://en.wikipedia.org/wiki/Differential_pair
    There are countless articles on the virtue of balanced audio designs vs single ended ... you just dont see single ended designs on the top shelf / professional equipment .

    What we are trying to achieve with the differential PI is no different from an audio design problem.

    I will investigate posting comparative scope shots of single ended vs differential for the same conditions. ( coil / target / MOSFET / voltage / damping etc etc ).

    moodz

    Leave a comment:


  • B^C
    replied
    Gday Guy's,

    A real dumb question here!

    Could somebody give me the "basic definition" of "Full Differential" in regards to what your trying to achieve.

    Is it just the signal your talking about or ?

    Sorry for the dumb question but i didn't get it when i first read about this with Aziz's posts on another forum & still i don't get it???

    Leave a comment:


  • Aziz
    replied
    Hi Paul,

    Regarding the dirty gate voltage:

    The parasitic capacitances between gate and drain may cause this. Therefore a low impedance gate drive is necessary.

    Regarding balancing the differential coil ends due to capacitive load of the mosfet and drain diode:

    It is very important, to balance out the differential ends of the coil. So making this totally symmetrically. If you don't use a bipolar tx pulses, the other end of the coil could contain same mosfet (wired for switch off, gate tied to source). You need a drain diode on both mosfets.

    Or use an adjustable capacitor for balancing out the differential coil ends.

    Aziz

    Leave a comment:


  • moodz
    replied
    Thanks for the comments Simon and Aziz ... look forward to your results on the diff amp Aziz. I have ordered some more bits so I can alter the damping more 'dynamically' and adjust for optimum value.
    Below is a scope pic of the output of the first diff amp ... bit under 7 usec on the pulse ... not too bad considering there is more room for improvement yet.
    25 cm differential monocoil
    100 us tx pulse
    12 volts etc etc.
    The white trace is the output from the diff amp.
    The Green trace is the TX MOSFET switch gate drive.
    ... my receive pulses used to look bad ... now my mosfet drive looks worse ... probably due to the 1000 volt p/p now with the very short pulses.

    cheers from moodz

    Click image for larger version

Name:	diiffamp_7us.jpg
Views:	1
Size:	39.7 KB
ID:	322877

    Leave a comment:


  • simonbaker
    replied
    Originally posted by moodz View Post
    Regards moodz
    Thanks for the detailed explaination. It seems reasonable, the difference between motivating the current with an applied voltage versus a magnetic field. I'll have to believe Spice and reality as I'll probably never work the equations to my satisfaction.

    I think the disconnect I still have trouble with is: if the second coil suddenly has the same current as the first coil, you would expect twice the magnetic field compared to the first coil alone -- yet that I don't believe happens. I guess I'll just gnaw on it all for a while.

    Cheers!

    -SB

    Leave a comment:


  • Aziz
    replied
    Hi Paul,

    the balanced coil configuration is presenting itself to drive the transmit pulse alternately on both ends of the coil to achieve a bipolar magnetic field emission. This has the effect of degaussing the ground/target with alternate magnetic field pulses. As the coil driver is then symmetrical, this should have a better common mode rejection.

    We can get some additional information from the ground or target with bipolar transmit pulses (magnetic viscositiy and hysteresis).

    Anyway, the fully differential PI front-end is the way to go. The INA118/128 aren't good enough (according to spice simulation). But the INA163 has enough bandwith to be used as a differential front-end. The spice simulations showing good results which has to be proven by real practice. I will test this at the beginning of next month.

    Keep on good work.
    Aziz

    Leave a comment:


  • moodz
    replied
    Originally posted by simonbaker View Post
    Continuing the train of thought... Yes I agree equal flux at the moment of "TX turn-off", but the flux is all due to coil 1. When MOSFET gate opens (TX turn-off), coil 1 has lots of current, coil 2 does not.
    Well coil 1 can only have a maximum current of V/R where V = supply - diode drops etc and R=coil wiring + mosfet ON resistance. However at TX turn off the coil attempts to maintain that current and it can only do that by greatly increasing the voltage.Like water flowing out of a tap ... if the tap is very suddenly shut off then there will be a pressure spike that is maximum at the tap due to the inertia of the flowing water back up the pipe ... and voltage is analogous to pressure. At this point in time coil 2 only has the induced TX current which is limited by the amount that can flow into the damping resistor ... ie not much. With perfect coupling between the coils and since the coils are 1:1 the induced voltage can only = v_supply so the current will be v_supply/r_damping. Yes the flux is due to coil 1 but coil 2 'sees' it equally.

    The currents cannot change instantaneously, so my thought is that the flux contributed by each coil is quite different and you need to solve for the current in each coil to see how the total flux changes during the period after TX turn-off.
    The current in a coil in response to an applied voltage cannot change instantly .. I agree .... however just at the point of TX off we are talking about two coils connected in series with a flux of field Strength H cutting them. Now the critical thing about flux is that every part ( eg electrons etc ) of the coils 'sees' the flux whereas only the connected part of the coil sees a voltage pulse ( initially ). So if the flux is influencing every electron in the coil ( quantum experts shut eyes now ) then each electron is accelerated by the collapsing flux ie the 'instantaneous' current is proportional to the rate of flux collapse in BOTH coils since both coils see the same flux.
    Imagine the coil as string of beads where each bead is an electron. When I apply a voltage to the first 'bead' it either moves away or is attracted depending on the polarity however when it moves it has to pull or push all the other beads in the coil ... ie the bigger the coil ... the longer the line of beads to push or pull.
    However with a flux .... each bead is pushed or pulled individually so the 'inertia' if you like is not a whole string of beads as in the voltage case but only 1 bead.
    So can you have an instantaneous current ? .... at the timescales that we are working in ... the answer is Yes. ( Quantum experts open eyes now ).

    That is the flux that the target will see. I believe that at the moment of TX turn-off, most of the very large current in coil 1 is absorbed by capacitance since it cannot instantaneously flow through coil 2. At least I think it cannot instantaneously flow through coil 2 -- unless somehow the flux from coil 1 allows it to, but that doesn't feel right.

    You can see from the experimental results that the series connected coils develop a symmetrical voltage across the damping resistor .... intuitively the current must be equal in both coils. I know you don't feel right about the flux ... however remember that the flux was generated by coil 1 it does not 'belong' to coil 1. The diodes in the circuit act as switches and the circuit 'changes' at the TX point turn off.

    In fact, as the current in coil 1 decays, the current in coil 2, I assume, is growing. I would think this might cause the total dphi/dt initially to be diminished because the the total ampere-turns of the two coils are sort of staying constant - their rates-of-change are canceling. Is it possible this reduces the usefulness of the "center-tapped" design?

    I still haven't spiced this so I'm just speculating -- the proof is in your experiments which look good.

    I have spiced the result and it agrees with the experimental. I thought of this walking home from work and only built it up last night at 9pm so I have not done a schematic even.

    Your latest circuit sounds interesting but I don't understand the scope traces. Can you explain each trace in detail?

    Pic 1 is the flyback pulses measured at the input to the diff amp section. The differential coil throws out symmetrical but opposite polarity pulses... just like a centre tap transformer really.

    Pic 2 is a close up of the pulse base showing mild underdamping but demonstrating the symmetry of the signal.

    Pic 3 is just showing the +ve flyback timing in relation to the gate drive.

    Next Pic shows the target response.

    The coil is a 25 cm dia monocoil bifilar wound with 16 effective turns 0.7 mm dia magnet wire.

    Regards,

    -SB
    Regards moodz

    Leave a comment:


  • moodz
    replied
    Originally posted by hobbes_lives View Post
    Are you still using the comparator to drive the MOSFET? If so, you may want to consider using an active-off gate driver using discrete components, or a gate driver IC instead.
    ... yeh I put the comparator in to invert the tx drive from the PIC as the IO line was turning the FET on when the PIC was in reset state. Point taken though ... luckily the hash is not on the side that I need to sample .. but I have got it on the todo list.

    moodz.

    Leave a comment:


  • simonbaker
    replied
    Originally posted by moodz View Post
    Hi Simon .... no I mean equal flux ... the flux is the 'charge' similar to the electric field in a cap ... as the cap charges the field between the plates gets stronger ... so by analogy as the current gets greater during the TX pulse the flux gets stronger ...
    ...
    Regards,

    moodz
    Continuing the train of thought... Yes I agree equal flux at the moment of "TX turn-off", but the flux is all due to coil 1. When MOSFET gate opens (TX turn-off), coil 1 has lots of current, coil 2 does not. The currents cannot change instantaneously, so my thought is that the flux contributed by each coil is quite different and you need to solve for the current in each coil to see how the total flux changes during the period after TX turn-off. That is the flux that the target will see. I believe that at the moment of TX turn-off, most of the very large current in coil 1 is absorbed by capacitance since it cannot instantaneously flow through coil 2. At least I think it cannot instantaneously flow through coil 2 -- unless somehow the flux from coil 1 allows it to, but that doesn't feel right.

    In fact, as the current in coil 1 decays, the current in coil 2, I assume, is growing. I would think this might cause the total dphi/dt initially to be diminished because the the total ampere-turns of the two coils are sort of staying constant - their rates-of-change are canceling. Is it possible this reduces the usefulness of the "center-tapped" design?

    I still haven't spiced this so I'm just speculating -- the proof is in your experiments which look good.

    Your latest circuit sounds interesting but I don't understand the scope traces. Can you explain each trace in detail?

    Regards,

    -SB

    Leave a comment:


  • hobbes_lives
    replied
    Are you still using the comparator to drive the MOSFET? If so, you may want to consider using an active-off gate driver using discrete components, or a gate driver IC instead.

    Leave a comment:


  • moodz
    replied
    ... and now the +ve pulse in relation to the MOSFET gate drive. Note the hash at the base of the rising edge of the flyback that gets into the gate drive ... the circuit is not completely optimised .... I feel it could do better ... but would there be any benefit ??
    ...anyway here is the pic. The flyback scale at 50v/div gate drive at 5v/div.

    Click image for larger version

Name:	diffpulse_gate.jpg
Views:	1
Size:	74.3 KB
ID:	322869

    moodz

    Leave a comment:


  • moodz
    replied
    ... well now I suppose you will say that I'm merely chopping the pulse short with some sort of clamping switch ... I would have made some such remark if I had seen the previous post ... well I am not ... so here is the same close up with copper target.
    ... literally volts of response

    PIC1 .. everything identical to last pulse except for target ...

    Click image for larger version

Name:	diffpulse_target.jpg
Views:	1
Size:	70.0 KB
ID:	322868

    regards ...

    moodz

    Leave a comment:

Working...
X