8.5. Differential Voltage Gain

Suppose that an input voltage, V_g12 = V_g1 – V_g2, is applied between the inputs. Due to symmetry, the voltage magnitudes at the inputs with respect to ground are V_g1 = V_g12/2 and V_g2 = –V_g12/2, respectively. The noninverting output, V_d2, for this case is, by superposition,

Equation 8.21

The gains for the contributions from V_g1 and V_g2 are (8.19) and (8.20), respectively. This is the case of a pure differential input with resulting gain

Equation 8.22

A similar approach applied to obtain the gain for the output taken as V_d1 produces

Equation 8.23

Assume, for a numerical example, that I_D = 100 μA, R_D1 = 50 KΩ, and g_m = 200 μA/V (V_effn = 0.5 V). In this case the gain is a_vd1 = –10. This would be consistent with V_DD = |V_SS| = 10 V.

The gain for the case of the differential output, a_vd12 = (V_d1 – V_d2)/(V_g1 – V_g2), can be obtained from (V_d1 – V_d2) = –(1/2)g_m1R_D1V_g12 – (1/2)g_m2R_D2V_g12 and is

Equation 8.24

where g_m1 = g_m2 and R_D1 = R_D2 = R_D. Note that (8.22) and (8.23) are the same as (8.15) and (8.17). However, (8.15) and (8.17) are the limiting forms of (8.19) and (8.20) for R_bias → ∞, whereas (8.22) and (8.24) apply for a finite R_bias.