Why in the output characteristic graph of BJTs, within the saturation region, is the collector current positive?

Question

It is known that in the saturation region of the BJTs the pn junction is reverse biased. The reason for being reversed biased could be explained as follow: Vce=Vcb+Vbe Therefore Vcb=Vce-Vbe We can see that if the Vce<Vbe (saturation region) then Vcb would be negative, which means Vc<Vb and therefore the base-collector junction is forward biased. Since we consider positive direction of collector's current while the base-collector junction is reversed biased, in case of forward-biasing we need to consider the current in opposite direction (negative).

Answer 1

The behavior in the deep saturation region is not correctly shown in that diagram.

Here is a simulation for Ib = 10mA, 0 < Vc < 100mV and a 2N4401:

If the collector voltage is less than about 20mV the net current flow is out of the collector.

Note also that at Vc = 0, the majority of the base current flows out of the collector rather than the emitter. This is a hint as to how you can usually identify collector from emitter on an unknown transistor.