I am trying to understand the reason for having a Schottky diode for protection in a buck switch by reading here.
To summarize, I can understand why two the switches should not be turned on at the same time, and I also can understand why there is a dead time (when both switches are off).
However, I can not understand what can go wrong when both switches are off (if there is no diode). Why is this diode helpful? Thank you.
When switch S1 has been on for a while, a current will have flown through inductor L1, this charges L1. L1 will contain some energy in the form of magnetic flux.
As soon as S1 switches off, but S2 is not on yet, L1 will want that current to continue flowing as inductors resist change in current flow (compare that to a capacitor resisting a voltage change).
If the Schottky diode was not present then the current through the inductor will start to flow through the diode in parallel with S2. Realize that both S1 and S2 are still off.
That diode in parallel with S2 is a Body diode which is part of the MOSFET. Such a body diode is generally not designed to be used like this, I mean, not designed to conduct that current through the inductor. It often can actually be used like that but for many MOSFETs, the properties and limits of that diode are not mentioned in the datasheet so in that case, there is no guarantee. Also, such a body diode is usually not fast enough and since it is a silicon diode, it will drop up to 1 V when the current flows. This wastes energy.
By adding the Schottky diode, the current will flow through the Schottky diode as it has a lower forward voltage (often less than 0.5 V). This takes the strain of the diode inside S2. Also Schottky diodes are generally much faster than body diodes inside a MOSFET. That high speed switching also saves some energy.
Why is there a Schottky diode connected to the low-side switch
The Schottky diode is in parallel with the body diode of the shunt MOSFET S2. The Schottky diode has significantly less forward voltage drop and quicker recovery than the MOSFET body diode. Therefore, current will preferentially flow through the Schottky, lowering power losses, and increasing efficiency.
A diode in either S2 or the diode is fundamental to the operation of a buck converter.
If S2 is switched perfectly the Schottky diode is not needed.
If S2 is mistimed at turn on then the S1-S2 junction will rise in voltage. The Schottky diode conducts if such a timing mismatch exists.
Also, during turn off of S2 and re-turn-on of S1 a degree of non-overlap is necessary to avoid S1-S2 shoot-through. The Schottky diode conducts during this period if needed.
The body diode in the FET performs the same role but it is a very poor diode with high voltage drop and so high loss. The Schottky has much lower losses if it needs to conduct.
