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SC filter design: 20 dB between 10² and 10³ rad/s

Short answer

A switched-capacitor filter replaces resistors with clock-switched capacitors, enabling precise capacitor-ratio-based time constants. Applications: telecom codecs, tunable filters. For the given 0–20–0 dB plot, fc=20 kHz: section1 C=2 nF, CR1=1 pF, CR2=10 pF; section2 C=20 pF, CR1=10 pF, CR2=1 pF.

What is switched capacitor filter? What are its applications? Design a switched capacitor filter to realize the magnitude response given below: [Figure: |T| dB vs ω rad/s: 0 dB up to ω = 10, rising to 20 dB at ω = 10², flat at 20 dB up to ω = 10³, falling back to 0 dB at ω = 10⁴]

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.

Applications

  • Voice-band filters in telephone PCM codecs and modems.
  • Anti-aliasing and reconstruction filtering in data converters; sigma-delta modulators.
  • Audio equalisers and tone controls; DTMF (touch-tone) receivers.
  • Programmable/tunable filters (cut-off set by the clock) and filter ICs (e.g. MF10 type).
  • Biomedical and instrumentation front ends in mixed-signal ICs.

Transfer function from the plot

Slope changes: +20 dB/dec starts at ω = 10 (zero), flat from 100 (pole), −20 dB/dec from 1000 (pole), flat again from 10⁴ (zero). Check: rise 20log⁡(100/10)20\log(100/10) = 20 dB.

T(s)=(s+10)(s+104)(s+100)(s+1000)=s+10s+100⋅s+104s+1000T(s) = \frac{(s+10)(s+10^4)}{(s+100)(s+1000)} = \frac{s+10}{s+100}\cdot\frac{s+10^4}{s+1000}

DC gain =10×104/(100×1000)=1= 10 \times 10^4/(100 \times 1000) = 1 and gain at high frequency =1= 1 (0 dB).

Building block

A first-order section with an op-amp, input admittance Y1=sC1+1/R1Y_1 = sC_1 + 1/R_1 and feedback admittance Y2=sC2+1/R2Y_2 = sC_2 + 1/R_2:

T(s)=−Y1Y2=−C1C2⋅s+1/(R1C1)s+1/(R2C2)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 C1=C2=CC_1 = C_2 = C the zero is at a=1/(R1C)a = 1/(R_1C) and the pole at b=1/(R2C)b = 1/(R_2C). Each resistor is then replaced by a switched capacitor CR=1/(Rfc)C_R = 1/(Rf_c), i.e.

CR1=aCfc,CR2=bCfcC_{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: fcf_c = 20 kHz (more than 10 times the highest corner, 10⁴ rad/s ≈ 1.6 kHz). Because the corners are spread over three decades, a different CC is used in each section to keep the switched capacitors near 1–10 pF.

SectionZero aaPole bbC1=C2=CC_1 = C_2 = CCR1=aC/fcC_{R1} = aC/f_cCR2=bC/fcC_{R2} = bC/f_c
T1T_1101002 nF1 pF10 pF
T2T_210⁴100020 pF10 pF1 pF

Answer: fcf_c = 20 kHz; section 1: C1=C2C_1 = C_2 = 2 nF, CR1C_{R1} = 1 pF, CR2C_{R2} = 10 pF; section 2: C1=C2C_1 = C_2 = 20 pF, CR1C_{R1} = 10 pF, CR2C_{R2} = 1 pF.

✓ Answer checked by Claude Sonnet 5.5

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