From: firstname.lastname@example.org (Bob Wilson)
Subject: Re: Ringing in Discontinuous Buck regulator
Date: Wed, 25 Sep 2002 04:12:52 -0000
Organization: Your Organization
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References: <3D8F348B.D8BBCEA7@singnet.com.sg> <3D90941F.7A4AB009@singnet.com.sg>
In article <3D90941F.7A4AB009@singnet.com.sg>, email@example.com says...
>Regarding your answer....
>> But the non-conducting diode has capacitance to ground (and also there is
>> other stray capacitance as well). So once it goes discontinuous, you have
>> series resonant tank circuit. The output side of the inductor still has
>> low impedance of the load connectod to it, and the input end of the
>> is connected through the series capacitance of the diode to ground. So,
>> L/C tank sinply resonates at its own natural resonance frequency.
>At the instant that the current goes to zero in the inductor, will it stop
>zero or continue to go to negative? If it goes to zero exactly, how does it
>charge the stray capacitance? Don't you need some energy for oscillation to
>Am I making sense?
Actually, not really :) You need very little energy to excite an LC tank
into resonance, and if it has a decently high Q, it can ring freely for many
cycles. Typically what happens is that when the inductor is still
discharging its energy (and therefore the diode is still conducting), the
voltage at the diode end of the inductor is one diode drop below ground. The
moment the inductor "runs out of gas" and the diode stops conducting, the
voltage at the diode/inductor node jumps up to the same voltage as at the
OTHER end of the inductor (namely the output voltage).
But this is in an ideal situation. In reality, it does exactly as you saw it
doing (which makes me wonder why you are asking). That is, it does a rapid
step up to the same voltage as the output, then it keeps on going up in an
undamped positive-going half-sinewave. This rapid jump from just under
zero Volts up to the output voltage value, is what starts the oscillation
going. What follows is an somewhat damped sinusoidal decay of several
cycles, whereupon the sinewave eventually decays to a straight line whose
voltage is exactly the same as that at the OTHER end of the inductor.
In some cases, after the first positive-going half sine, the peak of the
negative half sine that follows can go slightly negaive and thus turns on
the diode again. when this happens, the tip of the negative half-sine is