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.


1 Answer 1


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} \$.

  • \$\begingroup\$ Although the C-B junction may have a larger area than the B-E junction, the doping differences are the main cause of ISC >> ISE. \$\endgroup\$
    – jp314
    Commented Oct 9, 2021 at 18:42

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.