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2 Switchers / Same Vin

Discussion in 'Electronic Design' started by Rich, Apr 6, 2006.

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  1. Rich

    Rich Guest

    Hello & thanks for reading...

    I'm analyzing a circuit that has 2 National Semi Simple Switchers running
    from the same Vin (an unregulated +12 Vdc) that are on two separate PCBAs.

    1) Catch Diode: The design guidelines recommend to handle a shored output
    scenario that the diode needs to have an If > Icl of the switcher (1.4A in
    my case). It seems to me if you short the output, very little current will
    flow through the catch diode, did I miss something here? I can see that
    1.3x of the maximum load current making sense for normal operation;
    however, the shorted scenario seems odd!

    2) Both parts are 150 kHz switchers and I'm seeing what I think is a ~450 Hz
    beat frequency on one of the switchers. I surmise in this case that both
    switchers must be operating close to the same frequency as the beat
    frequency is so low. I'm new to this a fairly common
    problem? The appnote suggests to add an inductor that works with Cin on
    one of the two switchers. So this would be an LC filter. It recommends
    solving for L using 1/2pi(LC)^0.5=f where f is 1/10 of 150 kHz. So if I
    target 15kHz resonant frequency then this means at 150 kHz I'd have 40dB of
    attenuation, right? But technically, the other part could be at 110kHz
    (110-173 kHz tolerance). So at 110k I'm at 34.7dB which still seems good
    compared to the 40 target. What's magical about the 40dB target? I know
    in audio each 3dB is a factor of 2x so is this ~13x (40/3) reduction. So I
    could expect a 13mV Vin peak to peak ripple to be 1 mV peak to peak on the
    output? So, if I'm close on this, the LC provides 0dB (no filtering) up to
    ~10 kHz, in the range of 10-20 kHz noise would be amplifed, and higher
    frequencies are highly minimized (attenuated), right? For my 400 uF Cin I
    calculate I need 300 nH for my inductor.

    Thanks again!
  2. John  Larkin

    John Larkin Guest

    If you short the output, the switcher will run at a low duty cycle,
    roughly 5% in your case, and the diode will furnish most of the load
    current, 95% maybe.

  3. Rich

    Rich Guest

    Is the low duty cycle due to a thermal shutdown within the switcher?
  4. John  Larkin

    John Larkin Guest

    Well, the switcher chip is current limiting, which is what it's
    designed to do when you short the output... something inside senses
    the peak current and shuts off for the rest of the clock period.

    Suppose the switcher cuts off at a peak current of 2 amps. Since the
    output voltage (after the inductor) is now zero, to get the 2 amps
    into the short, you need the voltage at the hot end of the inductor to
    average just a little bit positive, just enough to push 2 amps through
    the resistance of the inductor. That condition is obtained when, say,
    the switcher applies +12 to the inductor 5% of the time, and the diode
    applies -0.5 95% of the time. The switcher probably runs cold, and the
    diode gets warm, since it's conducting 2 amps 95% of the time.

    Try it.

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