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I am using TPS54260 buck converter to step down 10V to 3.3V. Load = 1A.

While measuring the output voltage and switching frequency using oscilloscope, I am observing this:

  1. During no load condition at the output, no switching at the switching node (before the inductor,) but output voltage = 3.3V.
  2. During loaded condition at the output, switching frequency, duty cycle is observed at the switching node and the output voltage is 3.3V.

How come, during the the 1st case, I am not observing any switching waveforms at the switching node, but I am receiving 3.3V at the output?

Please clarify.

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  • 2
    \$\begingroup\$ Check the time scale of your scope. The converter usually goes into burst mode at no load condition for efficiency reasons. \$\endgroup\$ – Navaro Sep 19 at 8:28
  • \$\begingroup\$ Yes I checked my time scale. The 3.3V Output line is flat at 3.3V Always. Does switching not happen? \$\endgroup\$ – Newbie Sep 19 at 8:35
  • \$\begingroup\$ You missed Navaro’s point. Zoom out in time to observe the pulse skipping. \$\endgroup\$ – winny Sep 19 at 9:11
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Looking at the data sheet the only reference i can find to low power mode is pulse skipping. The switching frequency range is quite wide and all quite high so if you have very little load at the output it could be skipping a lot of switching cycles when in it's 'Eco' mode.

If you think about the logic of the situation, if the datasheet says it doesn't have any other way of providing the output voltage (say an internal LDO which is more efficiency than switching at very light loads) then the the voltage at the output can only be maintained by rail capacitance and/or the regulator switching intermittently.

What switching frequency have you set it to and what value inductor are you using?

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  • \$\begingroup\$ 400kHz and 68uH \$\endgroup\$ – Newbie Sep 19 at 9:47
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How come, during the the 1st case, I am not observing any switching waveforms at the switching node, but I am receiving 3.3V at the output?

Well, you wanted 3.3 volts at the output and you are getting it. Given there is no load current and probably minimal leakage current taken, the chip does not need to try and switch to maintain the output at what is the correct voltage level. If it did switch, it would be injecting a little bit of energy into the output capacitors each cycle and guess what... the output would rise above 3.3 volts and it wouldn't be a very effective regulator would it?

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  • \$\begingroup\$ yes, I wanted 3.3V at the output. But how did I get the 3.3V at the output when I provide 10V at the input, without switching? \$\endgroup\$ – Newbie Sep 19 at 11:54
  • \$\begingroup\$ Because the output capacitor is holding its charge. Holding its charge means that the output voltage remains at 3.3 volts. \$\endgroup\$ – Andy aka Sep 19 at 11:58
  • \$\begingroup\$ Surely the buck did some switching when you turned it on. Try to turn it off, discharge the output capacitor and then monitor the switching node when turning the buck back on. \$\endgroup\$ – TemeV Sep 19 at 17:57
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This is a classical skip mode operation where the feedback voltage (the voltage at the comp pin) is internally monitored by an extra comparator featuring hysteresis. When the feedback is above a certain level (I believe they say 500 mV in the data-sheet), the circuit switches normally with a full-length switching pattern. When the output current reduces, the feedback voltage also does and the duty ratio goes down bringing the peak current down. At some point, the load is such that the feedback voltage passes below the 500-mV threshold. Because of the extra comparator, all switching cycles are interrupted and the power switch is turned off. See the below simplified sketch for illustration:

enter image description here

When the switch turns off, the output is left with the capacitor being charged at your regulation voltage (3.3 V it seems) and the current absorbed by the load and the divider network. As such, the rate at which \$V_{out}\$ falls depends on the time constant \$C_{out}R_{load}\$: if the output current is truly zero amp (no-load) and if the resistive divider is of large ohmic value, then it can take a large amount of time to reach the hysteresis band and reactivate the switching cycles again (as \$V_{out}\$ falls too far from the 3.3-V target, \$v_{FB}(t)\$ rises up again). With your scope, sync in normal mode trigger, not auto, so that you can see the bunches at a low repeating rate. Another option is to slowly reduce the current from full load until it starts skipping pulses. As the current goes lower and lower, the distance between the bunches expands: the converter is now operating in a hysteretic way. If you want to further dig the world of switching converters, you can check this book.

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During no load condition at the output, no switching at the switching node (before the inductor,) but output voltage = 3.3V. This is Pulse-Skipping mode, feature of this IC. More about Pulse-Skipping, you can read this AN from TI (Link)

You maybe need to change Oscilloscope time to see something like this: enter image description here


Update: I'm using TI TINA to simulate your circuit. I downloaded model from here and change Rload to very large. This is result (you see at PH curve): enter image description here Note that the simulation result is little bit different to experiment from TI because affect of output Cap.

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  • \$\begingroup\$ I tried. But I am not observing any pulse skipping waveforms as you have shown \$\endgroup\$ – Newbie Sep 20 at 8:28
  • \$\begingroup\$ Because I don't have your board in hand so I use simulation. I am using TI TINA (it's free) and download model from TI link. Change R load to very small (i.e 1uOhm). \$\endgroup\$ – anhvanthe Sep 20 at 10:53
  • \$\begingroup\$ Sorry for mistake, Change R load to very large. \$\endgroup\$ – anhvanthe Sep 20 at 16:33

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