I'm seeing strange voltage at the switch node of my buck regulator.

There is a decaying sine wave of voltage which seems to happen before the switch is enabled. This is a dual channel regulator and I'm only seeing it on channel 1, the switching waveform for channel 2 is nice and square, so I'm expecting it to be an error in the pcb assembly (done by me, with low skill level). But I can't find anything wrong/missing. Any pointers on where to focus investigation would be most welcome.

Note the bottom side FET is not supported by diode (D3), this is not fitted (in either channel).

Measurement point

Strange waveform at inductor

enter image description here

  • \$\begingroup\$ So you suspect a PCB assembly Problem and ask US to point out what you did wrong without knowing what you even did? \$\endgroup\$
    – PlasmaHH
    Commented May 20, 2018 at 10:34
  • \$\begingroup\$ It would probably be useful to see the gate voltages for both those FETs, as well as knowing the test conditions (input voltage, output voltage, test load impedance). And did you design this regulator to stay in CCM under all load conditions? It looks like M2 or D3 are turning off, putting the inductor in DCM. \$\endgroup\$
    – user16324
    Commented May 20, 2018 at 10:36
  • \$\begingroup\$ @PlasmaHH - I was just wondering if this is a characteristic waveform that anyone recognises from a common error. I've spent a lot of time probing the board but can't find anything wrong. \$\endgroup\$ Commented May 20, 2018 at 10:37
  • \$\begingroup\$ @BrianDrummond Thanks for this, input voltage is 24v, output voltage is 5.7 @ 0.1A dropping to about 5.2 @ 15A, test load is a cheapo electronic load from eBay, not sure what the impedance is. I'll probe the gates, hang on \$\endgroup\$ Commented May 20, 2018 at 10:40
  • \$\begingroup\$ How about not clipping the bottom/sides off the circuit diagram? \$\endgroup\$
    – Andy aka
    Commented May 20, 2018 at 10:41

2 Answers 2


Data sheet extract: -

To select forced continuous operation, tie the MODE/PLLIN pin to a DC voltage below 0.6V (e.g., SGND). To select pulse-skipping mode of operation, tie the MODE/PLLIN pin to INTVCC.

You are allowing the chip to opt-out of continuous conduction mode and this means that both the MOSFETs can turn off and then you get a decaying sinewave due to the drain capacitance of the MOSFETs (being off) and the inductor forming a tuned circuit; the residual energy that can't be used creates a decaying sinewave.

  • \$\begingroup\$ Thanks so much for your help, I get it now. I would like to keep it in pulse-skipping mode if possible. Is this decaying sinewave a problem? And if it is, will increasing the inductance (to, say, 4.7uH) solve it? \$\endgroup\$ Commented May 20, 2018 at 14:58
  • \$\begingroup\$ The decaying sine wave is not regarded as a problem; rather it can be regarded as indicating the purity of your inductor. It's neither a good nor a bad thing if you look at this way - it will ring when you going into skipping mode and when the top MOSFET conducts again there may be a little residual current that adds energy to the energy stored during conduction. It could be that it also subtracts a little energy too. The "uncertainty" means the output voltage regulation becomes noisier. Swings and roundabouts I'm afraid and nothing much you can really do about it. Pointless adding a snubber. \$\endgroup\$
    – Andy aka
    Commented May 20, 2018 at 15:10

Looks like it's in discontinuous conduction mode. This can be fixed by setting the value of the inductor to a greater one. This also means that the controlling loop (most likely) needs to be modified, unless it was already built for continuous conduction mode and you just donwsized the inductor. Here's a basic version in LTspice:


L's value is stepped to be 2\$\mu\$H, then 10\$\mu\$H. V(x) has oscillations when the inductor's current is zero (I(L1), red trace), vs the continuous, green trace. The two V(x) are shifted for better comparison.

DCM can also appear at low loads, e.g. you quote two examples: [email protected] and 5.2V@15A. The first might be a low load that the low peaks of the inductor's current "reach" the ground.

  • \$\begingroup\$ Thanks for thisd, I need to do some more research before I understand! \$\endgroup\$ Commented May 20, 2018 at 14:15

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