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If its a positive voltage, the maximum voltage will be determined by the rated base current or the power rating of the series resistor, whichever comes first.
If it's a negative voltage, it will be twice (because it is halved by the potential divider) of the transistors BE breakdown voltage.
Be carefull about Emitter Base breakdown, some small signal parts will suffer noise performance changes
and other parameters when E-B experiences breakdown. This can be a permanent change in device per-
Sorry for the confusion. I understand.. I was checking primarily the safe voltage at input.
Suppose I apply 10V at input the divider presents 5V to the Base Emitter junction but since Base Emitter is ON the voltage developed across the base emitter is approximately 0.7V the current is limited by 10k and it will be 1mA base current I guess. And i assume then the power dissipation 1mAx 0.7V + Ic x Vce will be the limiting parameter.
the current flowing into the base is (10 V - V_BE)/(10 k) \[\approx\] 1 mA. Unfortunately the maximum base current is usually not specified in the datasheet. If you perform a websearch regarding the maximum base current you will find different values, here you can find values from a semiconductor manufacturer .