AN ACTIVE BANDPASS FILTER

Objective: • Design an active bandpass filter to meet a set of specifications.

Specifications: The center frequency of the bandpass amplifier is to be $f_{o} = 2 kHz$ , the bandwidth is to be Δf = 10 Hz, and the maximum voltage gain is to be $|A_{v}|_{max} = 40$ .

Design Approach: The bandpass amplifier configuration to be designed is shown in Figure 15.53.
Choices: Ideal op-amps are assumed to be available.

Question

AN ACTIVE BANDPASS FILTER

Objective: • Design an active bandpass filter to meet a set of specifications.

Specifications: The center frequency of the bandpass amplifier is to be [latex]f_{o} = 2 kHz[/latex], the bandwidth is to be Δf = 10 Hz, and the maximum voltage gain is to be [latex]|A_{v}|_{max} = 40[/latex].

Design Approach: The bandpass amplifier configuration to be designed is shown in Figure 15.53.
Choices: Ideal op-amps are assumed to be available.

Accepted Answer

(Analysis): Considering the circuit in Figure 15.53, we have
[latex]\frac{v_{o2}}{v_{o}} = − \frac{\frac{1}{s C}}{R_{2}} = \frac{−1}{s R_{2}C} [/latex]

and
[latex]\frac{v_{o3}}{v_{o2}} = −1[/latex]
so
[latex]v_{o3} = \frac{v_{o}}{s R_{2}C} [/latex] (15.128)
Node 1 is at virtual ground. Summing currents at this node, we find
[latex]\frac{v_{i}}{R_{4}} + \frac{v_{o}}{R_{1}} + \frac{v_{o}}{\frac{1}{s C}} + \frac{v_{o3}}{R_{3}} = 0[/latex]
Substituting the expression for [latex] v_{o3} [/latex] from Equation (15.128), we have
[latex]\frac{v_{i}}{R_{4}} = −v_{o} \left(\frac{1}{R_{1}} + s C + \frac{1}{s R_{2} R_{3} C} \right)[/latex]

The overall voltage gain is
[latex]\frac{v_{o}}{v_{i}} = \frac{\frac{−1}{R_{4}}}{\left(\frac{1}{R_{1}} + sC + \frac{1}{s R_{2} R_{3} C} \right)}[/latex]

Setting s = jω to obtain the steady-state frequency response, we obtain
[latex]\frac{v_{o}}{v_{i}} = \frac{\frac{−1}{R_{4}}}{\left[\frac{1}{R_{1}} + j \left(ωC − \frac{1}{ω R_{2} R_{3} C} \right) \right] }[/latex]

The center frequency occurs at the point where the imaginary term in the denominatoris zero, or
[latex]ω_{o} C = \frac{1}{ω_{o} R_{2} R_{3} C} [/latex]
which can be rewritten as
[latex]f_{o} = \frac{1}{2π C \sqrt{R_{2}R_{3}}} [/latex]

The maximum voltage gain occurs at the center frequency, so that
[latex]|A_{v}|_{max} = \frac{R_{1}}{R_{4}} [/latex]
The bandwidth is given by
[latex]BW = \frac{1}{2π R_{1} C} [/latex]
(Design): If we let C = 0.1 µF, then we can find
[latex]R_{1} = \frac{1}{2π (BW)C} = \frac{1}{2π (10) (0.1 × 10^{−6})} = 159 k \Omega[/latex]

From the maximum gain, we determine
[latex]|A_{v}|_{max} = \frac{R_{1}}{R_{4}} ⇒ 40 = \frac{159}{R_{4}} [/latex]
or
[latex]R_{4} = 3.975 k \Omega[/latex]
If we choose [latex]R_{2} = R_{3}[/latex], then from the center frequency
[latex]f_{o} = \frac{1}{2π C \sqrt{R_{2} R_{3}}} [/latex]
we find
[latex] R_{2} = R_{3} = \frac{1}{2π f_{o} C} = \frac{1}{2π(2 × 10^{3})(0.1 × 10^{−6})} [/latex]
or
[latex]R_{2} = R_{3} = 795.8 \Omega[/latex]
(Standard Resistor Values): The closest standard resistor values are [latex]R_{2} = 750 Ω, R_{3} = 820 Ω, R_{1} = 160 kΩ[/latex], and [latex]R_{4} = 3.9 kΩ[/latex]. A capacitor of 0.1 µF is a standard value. Using these circuit elements, we find the center frequency to be
[latex]f_{o} = \frac{1}{2π C \sqrt{R_{2} R_{3}}} = \frac{1}{2π(0.1 × 10^{−6}) \sqrt{(750)(820)}} [/latex]
or
[latex]f_{o} = 2.029 kHz[/latex]
The bandwidth is
[latex]BW = \frac{1}{2π R_{1}C} = \frac{1}{2π(160 × 10^{3})(0.1 × 10^{−6})}[/latex]
or
BW = 9.947 Hz

The maximum voltage gain at the center frequency is
[latex]|A_{v}|_{max} = \frac{R_{1}}{R_{4}} = \frac{160}{3.9} = 41.03 [/latex]

Comment: Using standard resistor values, the center frequency is within 1.5 percent of the design specification, the bandwidth is within 0.53 percent of the design specification, and the maximum gain is within 2.6 percent of the design specification. The circuit elements, of course, have tolerances that will affect the final circuit performance.

Question 15.DA.7: AN ACTIVE BANDPASS FILTER Objective: • Design an active band...

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