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## Q. 6.17

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.

## Verified Solution

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 $C_{G D5}.$