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).
Why in the output characteristic graph of BJTs, within the saturation region, is the collector current positive?
\$\begingroup\$ Are you saying that the graph in your question is wrong? \$\endgroup\$– Andy akaSep 13, 2022 at 13:28
\$\begingroup\$ You should cite where that graphic comes from. I have problems with the 0.7V saturation region and linear breakdown region. \$\endgroup\$– StainlessSteelRatSep 13, 2022 at 14:46
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.