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SC resistor, summing integrator and inverting lossy integrator

Short answer

A switched capacitor filter replaces resistors with capacitors and MOS switches, using a two-phase clock. A resistor is simulated by a capacitor C_R switched at frequency f_c, giving R_eq = 1/(f_c C_R). A summing integrator yields V_o = -(f_c/(s C_F))(C1 V1 + C2 V2). An inverting lossy integrator has transfer function T(s) = -(f_c C1/C_F)/(s + f_c C2/C_F).

What is a switched capacitor filter? How resistor, summing integrator and inverting lossy integrator can be realized using switched capacitor filter? Explain with necessary derivations.

Switched capacitor filter

A switched capacitor (SC) filter is an active filter in which every resistor of an active-RC filter is replaced by a capacitor and MOS switches driven by a two-phase non-overlapping clock. The switched capacitor behaves like a resistor R=1/(fcC)R = 1/(f_cC), so the filter needs only op-amps, capacitors and switches, all easily built on one MOS chip.

Resistor

Consider a capacitor CRC_R with two MOS switches driven by non-overlapping clocks ϕ1\phi_1 and ϕ2\phi_2 of frequency fc=1/Tcf_c = 1/T_c:

       phi1          phi2
 V1 ----o/o----+----o/o---- V2
               |
              ===  CR
               |
              GND
  1. Phase ϕ1\phi_1 (left switch closed): CRC_R charges to V1V_1; stored charge q1=CRV1q_1 = C_RV_1.
  2. Phase ϕ2\phi_2 (right switch closed): CRC_R charges/discharges to V2V_2; stored charge q2=CRV2q_2 = C_RV_2.
  3. Charge moved from node 1 to node 2 in one clock period:
Δq=q1−q2=CR(V1−V2)\Delta q = q_1 - q_2 = C_R(V_1 - V_2)
  1. If fcf_c is much higher than the signal frequency, V1V_1 and V2V_2 are nearly constant over a period, so the average current is
Iavg=ΔqTc=CRfc(V1−V2)Req=V1−V2Iavg=1fcCR=TcCR\begin{aligned} I_{avg} &= \frac{\Delta q}{T_c} = C_Rf_c(V_1 - V_2) \\ R_{eq} &= \frac{V_1 - V_2}{I_{avg}} = \frac{1}{f_cC_R} = \frac{T_c}{C_R} \end{aligned}

Example: CR=1C_R = 1 pF at fc=100f_c = 100 kHz simulates 10 MΩ.

Summing integrator

Summing integrator. Active-RC: Vo=−1sCF(V1R1+V2R2)V_o = -\frac{1}{sC_F}\left(\frac{V_1}{R_1} + \frac{V_2}{R_2}\right). With switched capacitors:

Vo=−fcsCF(C1V1+C2V2)V_o = -\frac{f_c}{sC_F}\left(C_1V_1 + C_2V_2\right)
                    CF
              +-----||-----+
              |            |
 V1 -[C1-SC]--+   |\       |
              +---|-\      |
 V2 -[C2-SC]--+   |  >-----+---> Vo
               +--|+/
               |  |/
              GND

Each input has its own integrator gain fcCi/CFf_cC_i/C_F.

Inverting lossy integrator

Active-RC form: input resistor R1R_1, feedback R2∥CFR_2 \parallel C_F:

T(s)=−1/(R1CF)s+1/(R2CF)T(s) = -\frac{1/(R_1C_F)}{s + 1/(R_2C_F)}

Replacing R1R_1 and R2R_2 by switched capacitors C1C_1 and C2C_2 (1/R=fcC1/R = f_cC):

T(s)=−fcC1/CFs+fcC2/CFT(s) = -\frac{f_cC_1/C_F}{s + f_cC_2/C_F}
                    CF
              +-----||-----+
              |            |
              +--[C2-SC]---+
              |            |
              |   |\       |
 Vi -[C1-SC]--+---|-\      |
                  |  >-----+---> Vo
               +--|+/
               |  |/
              GND

DC gain =−C1/C2= -C_1/C_2 and pole ωp=fcC2/CF\omega_p = f_cC_2/C_F, both set by capacitor ratios. It is the first-order low-pass section used in SC ladders and biquads.

✓ Answer checked by Claude Sonnet 5.5

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