How does the MOSFET intrinsic body diode cause short-circuit in an 3-phase inverter?

[powersys]

Can the MOSFET intrinsic body diode cause short-circuit in an 3-phase inverter?

Found the following article:
Why MOSFETs fail in Solid State TC duty
https://www.richieburnett.co.uk/mosfail.html

The following comments were copied from the article:

Shoot-through, (cross conduction.)
If the control signals to two opposing MOSFETs overlap, then a situation can occur where both MOSFETs are switched on together. This effectively short-circuits the supply and is known as a shoot-through condition. If this occurs, the supply decoupling capacitor is discharged rapidly through both devices every time a switching transition occurs! This results in very short but incredibly intense current pulses through both switching devices.

Slow reverse recovery of MOSFET body diode
...MOSFET body diodes generally have a long reverse recovery time compared to the performance of the MOSFET itself. If the body diode of one MOSFET is conducting when the opposing device is switched on, then a "short circuit" occurs similar to the shoot-through condition described above... This problem is usually eased by the addition of two diodes surrounding each MOSFET. Firstly, a Schottky diode is connected in series with the MOSFET source. The schottky diode prevents the MOSFET body diode from ever being forward biased by the free-wheeling current. Secondly, a high speed (fast recovery) diode is connected in parallel to the MOSFET/Schottky pair so that the freewheeling current bypasses the MOSFET and Schottky completely.

Can the slow reverse recovery body diode of MOSFET cause "short-circuit" in a 3-phase inverter shown below (only one phase leg is shown)?

 

[BradtheRad]

Along with time lags due to internal capacitance being cited as another reason, it's why a certain period of 'dead time' is recommended (both mosfets off).

The aim is to eliminate all opportunities for shoot-through to occur.

 

[powersys]

Another article below showing that "an effective short circuit across the supply" may occur in an inverter due to diode reverse recovery. I wish to understand how the so called "effective short circuit across the supply" can happen in an inverter.

The article below states that "... voltage drop across the diode does not change appreciably from its steady state value till the diode current reaches reverse recovery level...". First, what does "reverse recovery level" mean in the Fig. 2.13? The article also states that the voltage drop across the diode remains ~0.7V (voltage forward drop of a diode) during the early stage of reverse recovery.

Now, let's refer to the phase-A inverter leg shown in my first post above. Let's say the Rds_on of the MOSFET is 0.5Ω and the VDC is 50V. Assume the upper and lower MOSFETs are OFF. Assume current flows from drain to source via the lower MOSFET's body diode (the lower MOSFET is off) to phase-A ohmic-inductive load. When current is still flowing through the lower MOSFET's body diode, the upper MOSFET is turned on. So, the lower MOSFET's body diode will go into reverse recovery. At the early stage of the reverse recovery process, the voltage drop across the diode or the lower MOSFET is 0.7V (assume the forward voltage drop of the MOSFET's body diode is similar to that of a typical diode, i.e. 0.7V). The current that flows through the upper MOSFET (which is ON now) is expected to be (50-0.7)/0.5 = 98.6A. Will this huge current flow through the lower MOSFET's body diode?