# 2 Switchers / Same Vin

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

1. ### RichGuest

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!

2. ### John LarkinGuest

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

3. ### RichGuest

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

4. ### John LarkinGuest

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