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modification ATX

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  • Detectorist#1
    replied
    Yes, all is OK with this scheme. I forgot that the case with charge transfer between L and C is different and faster than charge of the coil from voltage source. The same case I observe in the truncated half-sine TX MD. The adding of the capacitor in series with the coil haves as result appr. 3 times faster reaching of the same current via the coil. Sorry for my doubts - now all is clear!

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  • Carl-NC
    replied
    During the kickstart phase the coil and the cap form an LC resonant circuit. Dumping the energy from the cap to the inductor creates a 1/4-sine waveform. Again for 300uH, the resonant frequency is 38.8kHz (T=25.75us) so the charge transfer will take 6.44us. Once the cap voltage drops below 5V D1 takes over and you get the flat-top current for the remainder of the pulse. Fig 25-17 is a redrawn version of a measured waveform from an ATX.

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  • Detectorist#1
    replied
    Hi Carl,
    From view point of the energy your equations are right, but In your equations the time not presents. The change of the current Delta L = (U x Delta t)/L. If L=300uH and if U=cte=175V (not true this case) and final current is 5A --> delta t = (delta I x L)/U = ( 5A x 300uH)/175V=1500/175=8.57us. But because the voltage on the capacitor drops down very quick, the time needed for the reaching of 5A current via the inductor will be significantly longer. No chance for the near to square form of the current in only 16us. Maybe the inductor's inductivity is far from 300uH or I have some mistake in the understanding of the process.

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  • Carl-NC
    replied
    Wrong way to look at it. The short pulse is roughly a square wave. The capacitor provides a kick-start voltage to rapidly get the coil current up to 5 amps, after that the 5V supply (via D1) powers the coil. The energy in the cap is 0.5*C*V2 and the (same) energy transferred to the coil is 0.5*L*I2. So you have

    0.5*C*V2 = 0.5*L*I2​
    I2 = V2*C/L

    I don't recall the inductance but let's say it's 300uH. The boost current is

    I = V*sqrt(C/L) = 175*sqrt(56nF/300uH)
    ​I = 2.4A

    This charges the coil as far as it can, then the 5V takes over.

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  • Detectorist#1
    replied
    Hi,
    Something strange haves in this time-diagram (Fig. 25-17) for me. The change of the voltage on a capacitor is "delta U= (I x delta t) / C. Start voltage is 175V. The short pulse is 16us. The capacitance for peek voltage storage capacitor is 56nF. The peek current of the short pulse is 5A. If the form of this current is triangle, average discharge current is 2.5A. As a result: (2.5A x 16us)/56nF= 714V. For 16us the discharge voltage of 56nF capacitor will be 714V but the start voltage is only 175V. Some strange happened. The energy on 56nF capacitor is not enough to generate 5A coil's current in 16us. Or I'm not right in my calculations?

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  • Detectorist#1
    replied
    Hi Carl,
    Interesting, it is too brave solution to use not bipolar TX pulses for mine detector in Recon MD. Thank you for this explaining. It is important to read ITMD3 more slow and with more attention!

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  • Carl-NC
    replied
    Originally posted by Detectorist#1 View Post
    we have to make the conclusion that the two power MOSFETs + the two U3G diodes + one big poly capacitor have to be used for bipolar half-sine or bipolar truncated-sine TX pulses
    No, they are used for energy recycling on the short pulse. Here is the TX circuit (from ITMD3):

    Click image for larger version  Name:	image.png Views:	0 Size:	28.9 KB ID:	444749​
    The TX (current) waveform is a short boosted pulse and a wide normal pulse:

    Click image for larger version  Name:	image.png Views:	0 Size:	22.0 KB ID:	444750​
    Axiom is the same, Recon likely also.

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  • Detectorist#1
    replied
    Hi,
    No posted scheme (or I not found yet) for Garrett ATX but from the pictures of the board we have to make the conclusion that the two power MOSFETs + the two U3G diodes + one big poly capacitor have to be used for bipolar half-sine or bipolar truncated-sine TX pulses (also the power supply solution uses two "banks of 4x alkaline batteries). The civil version of the Garrett's mine detector "Recon Pro AML-1000" have to be very close as idea and case of the original.
    Last edited by Detectorist#1; 01-23-2026, 01:42 PM.

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  • Carl-NC
    replied
    Please post in English, thanks.

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  • fengzhai01
    replied
    这是从ATX里面拆出来的小板子,但是忘记怎么接线了,可以研究一下
    Attached Files

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  • Nexus
    replied
    I realise this thread is a bit old, but I'v been digging recently in old ATX and have some data that could be useful to you guys.
    The stock 12"DD coil has the following parameters;
    TX 270uH, 700mOhms with the cable.
    RX 540 Uh, 12 Ohms with the cable.
    The coil cable is with two pairs individually twisted and screened together. No screen between TX and RX part of the cable. The TX cable wires are about 0.6mm square, the RX cable wires are very thin.
    The RX loop is balanced perfectly. If the RX loop is out of balance the Iron Check does not work properly and probably the discrimination too.

    The 20" x 15" ATX mono is build with a RX pick up coil that is stuck on top of the TX loop. The RX pick up loop is of thin magnet wire. The TX loop is Litz. The electrical parameters are the same as of the 12"DD.
    The RX pick up loop is controlled by a switching smd board (placed in the coil) that takes the TX pulses and trough a switching schematic done with mosfet transistors and no added power supply. The switching board is connecting the RX pick up loop (only when the TX loop is switched off) to analogue ground and the input of the detector like the RX on the 12"DD.
    The reason for this is because the TX of the ATX is a self regenerating type (this is what it looks like to me) or a sort of a pseudo resonance of sorts and it can not function properly if any inductive load is opposing the TX loop.
    The TX loop in the ATX has no reference to analogue (signal) ground or minus of the battery therefore standard Mono coils can never work. The ATX requires a separate RX loop to function properly.

    I never had the time to trace the schematic of this detector. So if anyone else has I will appreciate if they would be willing to share it here. I'm just curious of how this thing works.

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  • am48
    replied
    Try to mak a copy of the pcb inside the atx. Kicad is very useful for that. Then rebuild a complete new atx with a dys coil like the original was.
    Im so excited how this pcp looks like

    excited

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  • USHELEC
    replied
    I realized that something positive.
    With the help of two interpreters.

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  • tifred
    replied
    Thanks Sean_Goddard .
    I sold my ATX two weeks ago.
    USHELEC, I encourage you for you rmodifications

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  • USHELEC
    replied
    I have a photo of waveforms(with oscilloscope foto)for garrett arx coil. Can I publish them?

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