When base is open, is the base-emitter junction reverse biased? How does leakage current from collector to emitter flow?
First the \$I_{CBO}\$ current will flow into a &quot;positive&quot; direction in the collector (electrons current in NPN). Thus, \$I_C = I_{C_F} + I_{CBO}\$ where \$I_{C_F}\$ = &quot;forward current&quot;.  But for the base terminal, the \$I_{CBO}\$ current is a &quot;negative current&quot; (holes current) \$I_B = I_{B_F} - I_{CBO} \$
And now after we applied the voltage across the collector and emitter and left the base terminal &quot;open&quot;. We &quot;create&quot;  a &quot;voltage gradient&quot; across the BJT's. Thus, there will be some small voltage drop across the B-E junction, say 0.2V. Therefore the B-E junction will be forward-biased. But the base is left open thus:
\$I_{CEO} =\alpha \: I_{CEO} + I_{CBO} \$
And if we &quot;solve&quot; for  \$I_{CEO}\$
We have this:
\$I_{CEO} = \frac{ I_{CBO}}{ 1 - \:\alpha} = (\beta + 1) I_{CBO}\$