# Reverse saturation current in a BJT: active and reverse active modes

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.

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 $I_{SE}$.