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I believe there is a small misconception here.
As a first order approximation, in the cut off region Ic=Ib=Ie=0.
i.e. there is no current in the emitter branch
If we want a second order analysis, there is a small leakage current in the BJT during cut off, just like the reverse leakage current in the diode in the presence of non-zero voltage accross its terminals.
In the large signal model, we assume there is a leakage current through the emitter-base and collegector base junctions, but these currents are very small to be of practcal interest. (nA->pA's)
If you see very well the curves, that is not true.
Ib is not equal to cero it is called Iss but is very small. The only point where Ic= Ib = Ie = 0 is when Vbe = 0.
If we neglect the leakage currents then there is no current that is flowing from collector to emitter. we know that base has very less width therefore chances for recombination in this region is very less therefore no current flows through base.
But if it has to be consider then current flows due to minority charge carriers.
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