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?
- Anybody can ask a question
- Anybody can answer
- The best answers are voted up and rise to the top
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.
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.