SC MOS filter for -20 dB notch band (10–10⁴ rad/s)

Reading the Bode plot. The response is 0 dB up to $\omega = 10$, then falls at −20 dB/decade (a pole at 10), becomes flat at $\omega = 100$ (a zero at 100), stays at −20 dB to $\omega = 1000$, then rises at +20 dB/decade (a zero at 1000) and flattens at $\omega = 10^4$ (a pole at $10^4$). Check: 10 → 100 is one decade, giving −20 dB.

$$\begin{aligned} T(s) &= \frac{(1 + s/100)(1 + s/1000)}{(1 + s/10)(1 + s/10^4)} = \frac{10 \times 10^4}{100 \times 1000}\cdot\frac{(s+100)(s+1000)}{(s+10)(s+10^4)} \\ &= \frac{(s+100)(s+1000)}{(s+10)(s+10^4)} \end{aligned}$$

(gain constant $= 10^5/10^5 = 1$, so 0 dB at DC and at high frequency.)

Building block (first-order SC section). An inverting op-amp stage with input admittance $Y_1 = sC_1 + 1/R_1$ and feedback admittance $Y_2 = sC_2 + 1/R_2$ gives

$$\begin{aligned} T(s) &= -\frac{Y_1}{Y_2} = -\frac{sC_1 + 1/R_1}{sC_2 + 1/R_2} = -\frac{C_1}{C_2}\cdot\frac{s + \frac{1}{R_1C_1}}{s + \frac{1}{R_2C_2}} \end{aligned}$$

Each resistor is replaced by a switched capacitor clocked at $f_c$ (two-phase, non-overlapping clock), for which $R = \frac{1}{f_c C_R}$. Then

$$T(s) = -\frac{sC_1 + f_cC_{R1}}{sC_2 + f_cC_{R2}}, \qquad \omega_z = \frac{f_cC_{R1}}{C_1}, \quad \omega_p = \frac{f_cC_{R2}}{C_2}$$

With $C_1 = C_2 = C$ the high-frequency gain is 1 and $C_{R1} = \frac{C\,\omega_z}{f_c}$, $C_{R2} = \frac{C\,\omega_p}{f_c}$. Two inverting stages in cascade give a positive overall gain.

Splitting into two first-order sections:

$$T(s) = \underbrace{\frac{s+100}{s+10}}_{T_1(s)} \times \underbrace{\frac{s+1000}{s+10^4}}_{T_2(s)}$$

Choice of clock and capacitors. Highest break is $10^4$ rad/s ($\approx 1.6$ kHz); take $f_c = 100$ kHz. Because the breaks span three decades, choose $C = 10$ nF in stage 1 and $C = 100$ pF in stage 2 so that no switched capacitor is below 1 pF. With $C_R = C\omega/f_c$:

$$\begin{aligned} \text{Stage 1: } C_{R1} &= \frac{10\text{ nF}\times 100}{10^5} = 10\text{ pF}, & C_{R2} &= \frac{10\text{ nF}\times 10}{10^5} = 1\text{ pF} \\ \text{Stage 2: } C_{R1} &= \frac{100\text{ pF}\times 1000}{10^5} = 1\text{ pF}, & C_{R2} &= \frac{100\text{ pF}\times 10^4}{10^5} = 10\text{ pF} \end{aligned}$$

Final circuit (two identical-form stages in cascade; each SC resistor is the toggle-switch circuit shown):

                    C2
            +-------||------+
            |               |
            +---[CR2-SC]----+
            |               |
       C1   |   |\          |
 Vi -+--||--+---|-\         |
     |      |   |  >--------+---> Vo
     +[CR1]-+ +-|+/
      (SC)    | |/
             GND

       phi1          phi2
 V1 ----o/o----+----o/o---- V2
               |
              ===  CR
               |
              GND

Clock: $f_c = 100$ kHz, two-phase non-overlapping. Equivalent resistors: stage 1 — 1 MΩ and 10 MΩ; stage 2 — 10 MΩ and 1 MΩ.

Answer: $T(s) = \frac{(s+100)(s+1000)}{(s+10)(s+10^4)}$, realised as two cascaded SC first-order sections (values in the table) at $f_c = 100$ kHz. Only capacitor ratios ($C_R/C$) set the break frequencies, so the response is accurate on a MOS chip. Check of the actual curve: $|T|$ at $\omega = 316$ rad/s is −19.18 dB, close to the −20 dB asymptote.