VEBO parameters in transistors

I know that the VEBO is the emmitter to base voltage limit on transistor specifications, but is this the reverse breakdown limit on a bjt? i do not understand why this is stated. Since the potential between the base and the emmitter will be 0.6V.
Does anyone have the answer?
Does anyone know where i can get help explaining the transistor datasheets and what each parameter is for?

THANkX

ROVER

Hi,

Vebo is the emitter to base voltage. It is the max reverse voltage (or breakdown voltage), typicaly comprised between 3v and 5v. As i was student, it was also named "reverse Vbe".

Vbeo is the direct base to emitter voltage, typicaly comprised between 0.6v and 0.7v. Note than Vbeo max is about 1.2v

One further item is the o at the end means that the remaining terminal (collector in this case) is left floating (open).

VBEO ( Base Emitter Open Collector Breakdown Voltage ) is a limit for every bipolar transistor. It can be up to 2V and beyond this limit , B-E junction of the transistor may be broken.
It depends on semicondictor technology and doping concentration in junctions.
Rgrds