I'm currently working on a buck converter topology with a low side switched inductor. The goal is to achieve a constant current in the inductor, therefor only the mean voltage is important and output voltage filtering isn't necessary. The following picture shows a simplified circuit diagram.
Q1 resembles the used switch, L1 the load inductor (which additionally has some serial resistance) and V1 the gate driving voltage source. The diode V1 is a standard silicon diode.
For the following, the starting point is as follows:
It is assumed that a current is already flowing in the free wheeling circuitry through D1 and L1 because of previous switching events of the MOSFET. This current shall be named I1. The MOSFET is currently not conducting.
My understanding of this circuit is as follows:
When the MOSFET is switched on, it has to carry the current I1 and also the reverse recovery current of D1 before the voltage across drain and source of the MOSFET can lower. Then, D1 is reverse biased and the input voltage drops across the load L1 and D1.
My question is as follows:
The datasheet of silicon diodes specifies the reverse recovery time of a diode. When the MOSFET is switched fast, which results in a high di/dt in this case, the peak current (I1 + reverse recovery current of D1) could be reached fast and the voltage drop across the drain-source path can be lowered fast.
If the reverse recovery time of D1 is too long for the application, meaning that it can't block the reverse voltage fast enough, can there be some kind of shoot-through event where the diode still conducts while the voltage over the mosfet decreases resulting in current spikes?
Or did I just not fully understand the circuits behaviour?
Best Answer
Your general understanding of the diode behaviour is correct.
When the FET turns on, the voltage at its drain doesn't begin to fall until the FET's current exceeds the inductor + diode reverse recovery current. Generally the reverse diode current immediately after switching could be more than 2x the original forward current (which was the inductor current).
When switching fast, it is easier to consider the diode's reverse recovery charge. This charge will cause a power loss (Qreverse*Vcc) on every switching cycle.
Depending on your current, voltage and efficiency needs, a Schottky diode might be more suitable than a standard rectifier.
Note your circuit doesn't have an output -- unless your goal is just to have a (triangular waveform) constant current in the inductor.