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SC filter applications and resistor simulation

Short answer

A switched-capacitor filter replaces resistors with capacitors and MOS switches clocked at fc, giving Req = 1/(C fc). Applications: PCM codecs, anti-aliasing filters, audio equalizers. Resistor simulation: capacitor C switches between V1 and V2; charge Δq = C(V1–V2) per clock period gives average current I = C fc(V1–V2), so Req = 1/(C fc) = T/C.

What is a switched capacitor filter? What are its applications? How can you simulate a resistor using switched capacitor? Explain with necessary derivations.

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 capacitors, MOS switches and op-amps are all made on one MOS chip.

Applications

  • Voice-band filters in PCM telephone codecs, modems and fax.
  • Anti-aliasing/reconstruction filters and sigma-delta modulators in data converters.
  • Audio equalisers, tone decoders, DTMF receivers.
  • Clock-tunable universal filter ICs and programmable filters.
  • Filters inside mixed-signal ICs (biomedical, sensor interfaces).

Simulating a resistor with a switched capacitor

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

Two non-overlapping clock phases ϕ1\phi_1 and ϕ2\phi_2 at frequency fc=1/Tf_c = 1/T:

  • During ϕ1\phi_1 the capacitor connects to V1V_1: charge q1=CV1q_1 = CV_1.
  • During ϕ2\phi_2 it connects to V2V_2: charge q2=CV2q_2 = CV_2.

Charge moved from V1V_1 to V2V_2 in one clock period: Δq=C(V1−V2)\Delta q = C(V_1 - V_2). Average current:

I=ΔqT=Cfc(V1−V2)⇒Req=V1−V2I=1Cfc=TCI = \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 fcf_c is much higher than the signal frequencies.

Example: CC = 1 pF switched at fcf_c = 100 kHz gives Req=1/(10−12×105)R_{eq} = 1/(10^{-12} \times 10^5) = 10 MΩ, which would need a huge area as a diffused resistor.

Use in an integrator: replacing the input resistor of an RC integrator by C1C_1 gives

VoVi=−1sReqC2=−C1fcC2⋅1s\frac{V_o}{V_i} = -\frac{1}{sR_{eq}C_2} = -\frac{C_1 f_c}{C_2}\cdot\frac{1}{s}

so the time constant C2/(C1fc)C_2/(C_1f_c) depends only on a capacitor ratio and the clock, which is very accurate.

✓ Answer checked by Claude Sonnet 5.5

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