Switched capacitor filter for (s+200)(s+800)/(s+400)²

Switched-capacitor filter

A switched-capacitor (SC) filter is an active filter in which the resistors of an active-RC filter are replaced by capacitors switched by MOS transistors driven by a clock. Each switched capacitor behaves like a resistor $R = 1/(Cf_c)$, so the time constants depend only on capacitor ratios and the clock frequency.

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

Two non-overlapping clock phases $\phi_1$ and $\phi_2$ at frequency $f_c = 1/T$:

• During $\phi_1$ the capacitor connects to $V_1$: charge $q_1 = CV_1$.
• During $\phi_2$ it connects to $V_2$: charge $q_2 = CV_2$.

Charge moved from $V_1$ to $V_2$ in one clock period: $\Delta q = C(V_1 - V_2)$. Average current:

$$I = \frac{\Delta q}{T} = C f_c (V_1 - V_2) \quad\Rightarrow\quad R_{eq} = \frac{V_1 - V_2}{I} = \frac{1}{C f_c} = \frac{T}{C}$$

This holds when $f_c$ is much higher than the signal frequencies.

Design for $T(s) = \dfrac{(s+200)(s+800)}{(s+400)^2}$

Split into two first-order sections:

$$T(s) = \underbrace{\frac{s+200}{s+400}}_{T_1}\cdot\underbrace{\frac{s+800}{s+400}}_{T_2}$$

DC gain $= 200 \times 800/400^2 = 1$ and high-frequency gain $= 1$.

Building block

A first-order section with an op-amp, input admittance $Y_1 = sC_1 + 1/R_1$ and feedback admittance $Y_2 = sC_2 + 1/R_2$:

$$T(s) = -\frac{Y_1}{Y_2} = -\frac{C_1}{C_2}\cdot\frac{s + 1/(R_1C_1)}{s + 1/(R_2C_2)}$$

With $C_1 = C_2 = C$ the zero is at $a = 1/(R_1C)$ and the pole at $b = 1/(R_2C)$. Each resistor is then replaced by a switched capacitor $C_R = 1/(Rf_c)$, i.e.

$$C_{R1} = \frac{aC}{f_c}, \qquad C_{R2} = \frac{bC}{f_c}$$

            +-------| C2 |-------+
            +----[SC: CR2]-------+
            |                    |
 Vi o--+--| C1 |--+              |
       |          |              |
       +-[SC: CR1]+-----(-)      |
                          A -----+----o Vo
                 GND-----(+)

Choices: clock $f_c$ = 10 kHz (the highest critical frequency, 800 rad/s ≈ 127 Hz, is far below $f_c$), integrating capacitors $C$ = 100 pF.

Each section inverts, so the cascade of two gives $+T(s)$ with the correct sign.

 Vi -->[ T1: C=100p, CR1=2p, CR2=4p ]-->
       [ T2: C=100p, CR1=8p, CR2=4p ]--> Vo
              (fc = 10 kHz)

Answer: $f_c$ = 10 kHz; section 1: $C_1 = C_2$ = 100 pF, $C_{R1}$ = 2 pF, $C_{R2}$ = 4 pF; section 2: $C_1 = C_2$ = 100 pF, $C_{R1}$ = 8 pF, $C_{R2}$ = 4 pF.