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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.

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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:

The structure in Fig. 6.7 indicates also that the CBJ has a much larger area than the EBJ.

As you said, the collector-base junction (CBJ) has a larger cross sectional area than the emitter-base junction (EBJ). They then continue:

Thus the CB diode \$ D_C \$ has a saturation current \$ I_{SC} \$ that is much larger than the saturation current of the EB diode \$ D_E \$. Tipically, \$ I_{SC} \$ is 10 to 100 times larger than \$ I_{SE} \$.

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