Common-mode operation:

In order to derive an expression for the common-mode gain A_c = v_oc/v_c, we view R_EE in Figure as formed of two resistors, each of value 2R_EE, in parallel. The resulting circuit is the half circuit model for common-mode operation and is illustrated in Figure

Analysis follows through inspection as follows :

v_c  = v_π + v_e

v _e  = [(v _π/ r _π  )+ g_m v _π]2R_EE

= 2 R_EE [(1/r _π) + g_m] v_π

v_oc  =- g_m v_π R_C

A  = v_oc / v_c

A_C  = - g_m R_C/1 + 2 R_EE (1+ g_m r_π )/ r_π

Since β = g_m r_π >> 1, this simplifies to

Ac  = - g_m R_C / (1 + 2 g_m R_EE )

 Therefore, by using the value of A_d and definition of CMRR, we get

                                   CMRR = 1 + 2 g_m R_EE

 Since 2 g_m >> 1 and g_m = I_C / V_T =  I _EE /2V_T , at last we get

                           CMRR ≅ I _EE R_EE / V_T

As CMRR must be as large as possible, I_EE may not be increased indefinitely, neither can we use a very large REE. Besides needing a corresponding high voltage power supply for V_EE, a large resistor may not be fabricated by IC technology. Usually This problem is solved by replacing R_EE by a current source, which has a very large output resistance.