Theory: Accounting for V_be in the CE Amp

Often in CE amps, R_E is small or even zero. Under these circumstances, the effect of $\Delta V_{be}$ can not be neglected in small signal analysis. To account for $\Delta V_{be}$in our expression for voltage gain, we have to express it in terms $\Delta I_c$ as well as known quantities. To do this we go back to our original expression relating V_be and I_c.  

$$I_c=\beta I_s e^{\frac{V_{be}}{V_T}}$$ (33)

Using Taylor series we can obtain the small signal change in I_c that results in a small change in V_be.

$$\Delta I_c = \frac{\partial I_c}{\partial V_{be}} \Delta V_{be}$$ (34)

Performing the differentiation and using 2.19, leads to

$$\Delta I_c = \frac{I_C}{V_T} \vert _{I_C} \Delta V_{be}$$ (35)

Usually, $\frac{\partial I_c}{\partial V_{be}}\vert _{I_C}$ is defined as the small signal transconductance of the BJT, which is designated as g_m. It is important to note that g_m is given for a specific I_C which is determined by the DC bias condition. Thus, the transconductance is given by

$$g_m=\frac{\partial I_c}{\partial V_{be}}\vert _{I_C} =\frac{I_C}{V_T}$$ (36)

and the small signal change in V_be is  

$$\Delta V_{be} = \frac{\Delta I_c}{g_m}$$ (37)

Now, if we substitute $\frac{\Delta I_c}{g_m}$ for $\Delta V_{be}$in Equation (2.14), and take the ratio $\frac{\Delta V_c}{\Delta V_{b}}$, we find that the voltage gain, while including the effect of $\Delta V_{be}$, is:  

$$A_V = \frac{v_{out}}{v_{in}} = \frac{\Delta V_c}{\Delta V_{b}} = \frac{-R_C}{{\frac{1}{g_m}}+R_E}$$ (38)

Now, if R_E were to be totally eliminated from the circuit, then we could just set R_E = 0 in the above equation, and the gain would become:

A_V = -g_mR_C