Objective: Design a one-pole low-pass switched capacitor filter to meet a set of specifications.

Specifications: The circuit with the configuration shown in Figure 15.13(b) is to be designed such that the low-frequency gain is −1 and the cutoff frequency is 1 kHz.

Choices: An ideal op-amp is available and standard-valued capacitors are to be used.

Question

Objective: Design a one-pole low-pass switched capacitor filter to meet a set of specifications.

Specifications: The circuit with the configuration shown in Figure 15.13(b) is to be designed such that the low-frequency gain is −1 and the cutoff frequency is 1 kHz.

Choices: An ideal op-amp is available and standard-valued capacitors are to be used.

Accepted Answer

From Equation (15.32), the low-frequency gain is [latex]−(C_{1}/C_{2})[/latex], and the capacitance ratio must be [latex](C_{1}/C_{2}) = 1[/latex]. From Equation (15.32), the cutoff frequency is

[latex]T(jω) = − \frac{(1/f_{C}C_{2})}{(1/f_{C}C_{1})} \cdot \frac{1}{1 + j \frac{(2π f )C_{F}}{f_{C}C_{2}}} = − \frac{C_{1}}{C_{2}} \cdot \frac{1}{1 + j \frac{f}{f_{3 dB}}} [/latex] (15.32)

[latex]f_{3 dB} = \frac{f_{C}C_{2}}{2π C_{F}} [/latex]
If we set the clock frequency to [latex]f_{C} = 10 kHz[/latex], then

[latex]\frac{C_{2}}{C_{F}} = \frac{2π f_{3 dB}}{f_{C}} = \frac{2π(10^{3})}{10 × 10^{3}} = 0.628[/latex]
Trade-offs: We can use standard-valued capacitors [latex]C_{1} = C_{2} = 75 pF[/latex]. We would need [latex]C_{F} = C_{2}/0.628 = 75/0.628 = 119.4 pF[/latex]. A standard-valued capacitor [latex]C_{F} = 120 pF[/latex] can be used.
Comment: Since the low-frequency gain and cutoff frequency are both functions of capacitor ratios, the absolute capacitor values can be designed for compatibility with IC fabrication.

Question 15.3: Objective: Design a one-pole low-pass switched capacitor fil...

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