Ringing in Discontinuous Buck regulator
Erwin
I am building a simple DC/DC Buck converter. I noticed that whenever,
the current goes to Discontinuous mode, there will be a ringing waveform
on the SOURCE pin of the switching Mosfet. This happens when the
inductor current goes to zero. What I am curious is what is the
mechanism that cause this ringing? Where is the Inductance and
Capacitance coming from?
Thanks in advance!
Erwin
Mike Engelhardt
> ...Where is the Inductance and Capacitance coming from?
The capacitance comes the Schottky rectifier, the
MOSFET and some stray wiring capacitance. The
inductance is just the output inductor(the other
end is connected to an AC ground).
--Mike
Bob Wilson
Yes, that is absolutely typical. It you think about it, it even makes
perfect sense.
As the FET turns off, the inductor begins discharging its magnetic energy,
and since the original source of current (i.e. the FET) is now off, it
begins pulling current "up" through the freewheeling diode. As the energy in
the inductor is discharged, current through it (and through the diode)
linearly falls off. The circuit goes "discontinuous" the moment that the
inductor has discharged all its energy. Now, the diode stops conducting. But
the FET is still off as well. Thus, the FET/Diode end of the inductor "sees"
an open circuit (the FET is open and so is the diode since no current flows
through either of them).
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 a
series resonant tank circuit. The output side of the inductor still has the
low impedance of the load connectod to it, and the input end of the inductor
is connected through the series capacitance of the diode to ground. So, this
L/C tank sinply resonates at its own natural resonance frequency.
The resonance is damped, so it eventually dies out after a few cycles, if
the "off" time is long, or it is killed instantly the moment the FET turns
on again ....whichever comes first.
Because of the oscillation you are observing, it is generally not good
practice to design a Buck so it runs discontinuous. It is not damaging or
anything, but the main problem is increased EMI due to the large
amplitude of "ringing" (and also sluggish transient response).
Bob.
Erwin
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 a
> series resonant tank circuit. The output side of the inductor still has the
> low impedance of the load connectod to it, and the input end of the inductor
> is connected through the series capacitance of the diode to ground. So, this
> 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 at
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 take
place?
Am I making sense?
Thanks
Erwin
John Popelish
As long as the inductor current is continuous, it keeps the flyback
diode conducting which keeps a constant voltage across the inductor
(output voltage minus the diode drop). That voltage is also what
drives the inductor current toward zero.
But the moment the inductor current passes through zero, The diode
turns off, and you just have a capacitor (all the stray capacitance of
the diode, transistor inductor, etc.) in series with an inductor. but
there is voltage across the inductor, since the capacitor loaded end
is near zero volts and the other end is at output voltage. So the
inductor current starts to build as that current charges the
capacitance up towards output voltage. The moment the cap gets up to
output voltage, the inductor has no drop to drive its current up, so
its current levels off. But that current keeps driving the cap
voltage up, above the output voltage, creating a growing reverse
voltage across the inductor, which begins to back the inductor current
back toward zero. This continuous until the inductor current again
reaches zero, with the capacitance charged up to almost twice the
output voltage, where the process reverses. Thus is a classic series
LC resonance with the cap pre charged to some offset from the final
equilibrium voltage.
--
John Popelish
Bob Wilson
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
clipped.
Bob.
Erwin
Thanks for the detailed explanation! I understand the ringing mechanism now. :)
Erwin
Søren A.Møller
> Hi all,
>
> Capacitance coming from?
Just FYI:
Maxim makes some DC/DC-converters with circutry to reduce the EMI
caused by this ringing as shown on page 8 in this datasheet:
http://pdfserv.maxim-ic.com/arpdf/MAX1722-MAX1724.pdf
You can probably find more info on this on the Maxim site.
Søren A.Møller