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2 Switchers / Same Vin

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 switchers...is 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!

Reply to
Rich
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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.

John

Reply to
John Larkin

Is the low duty cycle due to a thermal shutdown within the switcher?

Reply to
Rich

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.

John

Reply to
John Larkin

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