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SC summer, inverting and non-inverting integrators

Short answer

Switched-capacitor filters replace resistors with capacitors and MOS switches, giving Req=1/(Cfc). Using a four-switch branch, inverting integrator: H(s)≈ –(C1 fc/C2)/s; non-inverting integrator: H(s)≈ (C1 fc/C2)/s by swapping input phases. Summer: multiple branches feeding an op-amp with feedback capacitor CF; with reset across CF, output is –(C1/CF V1 + C2/CF V2).

What is switched capacitor filter? How summer, inverting integrator and non-inverting integrator can be realized using switched capacitor? Explain with necessary diagrams and transfer function.

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/(Cfc)R = 1/(Cf_c), so the time constants depend only on capacitor ratios and the clock frequency.

The basic SC resistor (charge C1(V1−V2)C_1(V_1 - V_2) moved each clock period) gives Req=1/(C1fc)R_{eq} = 1/(C_1f_c). The building blocks below use the parasitic-insensitive four-switch branch: capacitor C1C_1 with two switches on each plate.

        S1        C1        S3
 Vi o---/ ---+---| |---+---/ ---o to (-) of op-amp
             |         |        (virtual ground)
             / S2      / S4
             |         |
            GND       GND

Inverting integrator

Switching: ϕ1\phi_1: S2 and S4 closed (C1 discharged). ϕ2\phi_2: S1 and S3 closed: C1C_1 charges to ViV_i through the virtual ground, so charge C1ViC_1V_i is pushed onto the feedback capacitor C2C_2 with negative sign.

H(z)=−C1C2⋅11−z−1≈H(s)=−C1fcC2⋅1sH(z) = -\frac{C_1}{C_2}\cdot\frac{1}{1 - z^{-1}} \qquad\approx\qquad H(s) = -\frac{C_1 f_c}{C_2}\cdot\frac{1}{s}

Non-inverting integrator

Change the phases of the left switches: ϕ1\phi_1: S1 and S4 closed (C1C_1 charges to ViV_i with its right plate grounded). ϕ2\phi_2: S2 and S3 closed (left plate grounded, right plate to the virtual ground). The charge is now delivered with the opposite polarity, one half-period later:

H(z)=+C1C2⋅z−11−z−1≈H(s)=+C1fcC2⋅1sH(z) = +\frac{C_1}{C_2}\cdot\frac{z^{-1}}{1 - z^{-1}} \qquad\approx\qquad H(s) = +\frac{C_1 f_c}{C_2}\cdot\frac{1}{s}
SwitchInverting integratorNon-inverting integrator
S1 (input side)ϕ2\phi_2ϕ1\phi_1
S2 (left plate to ground)ϕ1\phi_1ϕ2\phi_2
S3 (right plate to op-amp)ϕ2\phi_2ϕ2\phi_2
S4 (right plate to ground)ϕ1\phi_1ϕ1\phi_1

Circuit for both integrators (the same box, with the switch phases of the table above):

                    +-----| C2 |-----+
                    |                |
 Vi o--[SC branch C1]--+---(-)       |
                       |      A -----+--o Vo
                 GND--(+)

Summer (and summing integrator)

Connect several SC branches C1,C2,…C_1, C_2, \dots (one per input) to the same virtual-ground node. The charges add:

              +------------| CF |------------+
              |        (reset switch across CF|
              |         for the summer)       |
              |            |\                 |
 V1 o--[SC C1]+------------|-\                |
              |            |  >---------------+--o Vo
 V2 o--[SC C2]+       +----|+/
                      |    |/
                     GND

Each box [SC Ci] is the four-switch branch shown above.

  • With a feedback capacitor CFC_F only: summing integrator Vo=−fcs(C1CFV1+C2CFV2)V_o = -\dfrac{f_c}{s}\left(\dfrac{C_1}{C_F}V_1 + \dfrac{C_2}{C_F}V_2\right).
  • With CFC_F reset each period (a switch across CFC_F), or with an SC resistor in feedback: summer (amplifier) Vo=−(C1CFV1+C2CFV2)V_o = -\left(\dfrac{C_1}{C_F}V_1 + \dfrac{C_2}{C_F}V_2\right).

A branch wired with non-inverting phasing gives a + sign for that input, so sums and differences are both possible with one op-amp. All gains are capacitor ratios, hence accurate.

✓ Answer checked by Claude Sonnet 5.5

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