Question 6.17: Not all fully differential circuits are free from mirror pol...
Not all fully differential circuits are free from mirror poles. Figure 6.42(a) illustrates an example where current mirrors M_3–M_5 and M_4–M_6 “fold” the signal current. Estimate the low-frequency gain and the transfer function of this circuit.
Neglecting channel-length modulation and using the differential half circuit shown in Fig. 6.42(b), we observe that M_5 multiplies the drain current of M_3 by K, yielding an overall low-frequency voltage gain A_v = g_{m1}K R_D.
To obtain the transfer function, we utilize the equivalent circuit depicted in Fig. 6.42(c), including a source resistance R_S for completeness. To simplify calculations, we assume that R_DC_L is relatively small so that the Miller multiplication of C_{G D5} can be approximated as C_{G D5}(1 + g_{m5}R_D). The circuit thus reduces to that in Fig. 6.42(d), where C_X ≈ C_{GS3} + C_{GS5} + C_{DB3} + C_{G D5}(1 + g_{m5}R_D) + C_{DB1}. The overall transfer function is then equal to V_X /V_{in1} multiplied by V_{out1}/V_X . The former is readily obtained from (6.30) by replacing R_D with 1/g_{m3} and C_{DB} with C_X , while the latter is
\frac{V_{out}}{V_{in}}(s) = \frac{(C_{G D}s − g_m)R_D}{R_S R_Dξ s^2 + [R_S(1 + g_m R_D)C_{G D} + R_SC_{GS} + R_D(C_{G D} + C_{DB})]s + 1} (6.30)
\frac{V_{out1}}{V_X}(s) = −g_{m5}R_D\frac{1}{1 + R_DC_L s} (6.97)
Note that we have neglected the zero due to C_{G D5}.