Why does the BUCK step-down circuit have a strange negative voltage?

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BUCK is a common step-down topology. For the waveform of the BUCK switch node, some articles draw a standard square wave, while others draw a negative pulse waveform?

For example, the two waveforms below have the same high level, but very different negative levels. The first figure has a constant negative level of -0.7V at the switch node, while the second one is more complicated, with a -0.7V pulse negative level and then a 0 level. How are these two waveforms generated?

This starts with the classification of BUCKs. BUCKs are divided into asynchronous BUCKs and synchronous BUCKs. The one using diode D1 in the figure below is an asynchronous BUCK. If the diode is replaced with a switch tube, it becomes a synchronous BUCK.

The working principle of BUCK can be found in previous articles. Please search for "Detailed Principle of BUCK Circuit" . The two waveforms are the same when the voltage level is high, so we will only discuss the situation when the voltage level is low.

In the non-synchronous BUCK shown in the figure below, when the switch S1 is turned off, the energy is discharged to the load through the energy storage inductor L1. The discharge path is L1->load->D1. Therefore, when measured at point A, Vsw will have a constant diode conduction voltage, that is, -0.7V.

For synchronous bucks, if the upper and lower tubes are turned on at the same time, a short circuit will occur. In order to avoid the upper and lower tubes S1 and S2 being turned on at the same time, the dead time needs to be increased .

During the dead time, both the upper and lower tubes are not conducting. At this time, the inductor is discharged through the body diode of MOS. For the principle of MOS body diode, please search the article "Why do MOS tubes need to be connected in parallel with diodes? What is the function?" Since it is discharged through the body diode, there is a negative voltage close to -0.7V on Vsw. After the dead time, the lower tube MOS S2 is turned on, and the discharge path is inductor->load->MOS (the diode is short-circuited by MOS). The conduction impedance of MOS is very small, so the negative voltage of VSW quickly decays from -0.7V to 0V, and then enters the dead time again, and the negative voltage becomes about -0.7V again.