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I understand that in a buck converter dead time happens as both switches are ON simultaneously and this short the power supply. However, I don't see how dead time can exist in boost converter. Could anyone explain it? enter image description here

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I wonder if you have the terms mixed up slightly. Dead time is not the time when both switches are closed (and unwanted currents flow, so called "shoot-through"). It takes a finite amount of time for switches (e.g. MOSFETs) to open. Dead time is the time when both switches are commanded to be open in order to make sure that they aren't partially closed at the same time, to prevent a shoot-through. – Nick Alexeev Feb 20 at 2:51

In your schematic, if S1 and S2 are closed the power supply still cannot be shorted. However, the capacitor C would short. The peak current might be high enough to damage your switching mechanism. So dead time is probably still a good idea.

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Those switches are implemented as transistors in real life. (At least one; a non-synchronous converter can use a diode for one of them). Real transistors cannot switch on and off instantaneously. In addition, remember the diode mention from the sentence before -- real MOSFETs have a body diode as part of their structure.

Observe the Qrr and reverse-recovery parameters from the datasheet -- even when you turn off a switch, there is some time required until current stops flowing due to the reverse-recovery charge that is trapped inside. Most converter controllers have a little bit of deadtime internally -- it doesn't need to be huge. A converter switching at a few hundred kHz may only insert a few ns of deadtime.

So yes, while a controller will ensure it doesn't turn on both switches simultaneously and cause shoot-through, the physical realities of the switching elements can still cause similar behavior.

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