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modification ATX

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  • #61
    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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    • #62
      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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      • #63
        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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        • #64
          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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