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Chapter 6

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.

6.42

Step-by-Step

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}.