In (Sedra; Smith. Microelectronic Circuits), as well as in several other sources, the value of the reverse saturation current (\$I_S\$) is considered the same for the active mode and for the reverse active mode of the BJT:
*all the equations are for an NPN BJT
\$ \alpha_R I_{SC}=\alpha_F I_{SE}=I_S \$
(reciprocity relation)
\$ i_C=I_Se^{v_{BE}/V_T} \$
(in active mode)
\$ i_E=I_Se^{v_{BE}/V_T} \$
(in reverse active mode)
Since it depends on the area of the junction (\$ I_S=\dfrac{AqD_nn_i^2}{N_AW}\$) and – as the primary source itself explained – the area of the BC junction (in forward bias for reverse active mode) is much greater than the area of the BE junction (in forward bias for active mode), I am having trouble understanding how \$I_S\$ does not change from one operation mode to the other, which leads to \$i_{E (reverse)}=i_{C (active)}\$.
I would think that since the only parameter that changes in the equation of \$I_S\$ is \$A\$, maybe this would make more sense to me:
\$ I_{S(active)}=\dfrac{A_EqD_nn_i^2}{N_AW}=\alpha_F I_{SE}\$
\$ I_{S(reverse)}=\dfrac{A_CqD_nn_i^2}{N_AW}=\alpha_R I_{SC}\$
\$\dfrac{\alpha_R I_{SC}}{A_C}=\dfrac{\alpha_F I_{SE}}{A_E}\$
I really appreciate any help. Thank you very much.
Best Answer
The saturation current of a PN junction, as you correctly said, depends on the cross sectional area of the junction itself.
In fact, if you look at a datasheet \$ I_{CBO} \gg I_{EBO} \$, confirming your idea.
Moreover, Sedra/Smith (I'm looking at the 6th edition, page 361) says:
As you said, the collector-base junction (CBJ) has a larger cross sectional area than the emitter-base junction (EBJ). They then continue: