For the feedback loop of Fig. 17.29, find the sensitivities of ω0 and Q relative to all the passive components and the op-amp gain. Evaluate these sensitivities for the design considered in the preceding section for which C1 = C2.

Question

For the feedback loop of Fig. 17.29, find the sensitivities of ω₀ and Q relative to all the passive components and the op-amp gain. Evaluate these sensitivities for the design considered in the preceding section for which C₁ = C₂.

Accepted Answer

To find the sensitivities with respect to the passive components, called passive sensitivities, we assume that the op-amp gain is infinite. In this case, ω₀ and Q are given by Eqs. (17.73) and (17.74). Thus for ω₀ we have

[latex]ω_{0} = \frac{1}{\sqrt{C_{1}C_{2}R_{3}R_{4}}}[/latex] (17.73)

[latex]Q = \left[\frac{\sqrt{C_{1}C_{2}R_{3}R_{4}}}{R_{3}} \left(\frac{1}{C_{1}} + \frac{1}{C_{2}} \right) \right]^{-1}[/latex] (17.74)

which can be used together with the sensitivity definition of Eq. (17.82) to obtain

[latex]S_{x}^{y} = \frac{∂ y}{∂ x} \frac{x}{y}[/latex] (17.82)

[latex]S^{ω_{0}}_{C_{1}} = S^{ω_{0}}_{C_{2}} = S^{ω_{0}}_{R_{3}} = S^{ω_{0}}_{R_{4}} = - \frac{1}{2}[/latex]

For Q we have

[latex]Q = \left[\sqrt{C_{1}C_{2}R_{3}R_{4}} \left(\frac{1}{C_{1}} + \frac{1}{C_{2}} \right) \frac{1}{R_{3}}\right]^{-1}[/latex]

to which we apply the sensitivity definition to obtain

[latex]S^{Q}_{C_{1}} = \frac{1}{2} \left(\sqrt{\frac{C_{2}}{C_{1}} - \frac{C_{1}}{C_{2}}}\right) \left(\sqrt{\frac{C_{2}}{C_{1}} + \frac{C_{1}}{C_{2}}}\right)^{-1} [/latex]

For the design with C₁ = C₂ we see that [latex]S^{Q}_{C_{1}}[/latex] = 0. Similarly, we can show that

[latex]S^{Q}_{C_{2}} = 0[/latex], [latex]S^{Q}_{R_{3}} = \frac{1}{2}[/latex], [latex]S^{Q}_{R_{4}} = -\frac{1}{2}[/latex]

It is important to remember that the sensitivity expression should be derived before values corresponding to a particular design are substituted.

Next we consider the sensitivities relative to the amplifier gain. If we assume the op amp to have a finite gain A, the characteristic equation for the loop becomes

1 + At(s) = 0 (17.84)

where t(s) is given in Fig. 17.28(a). To simplify matters we can substitute for the passive components by their design values. This causes no errors in evaluating sensitivities, since we are now finding the sensitivity with respect to the amplifier gain. Using the design values obtained earlier—namely, C₁ = C₂ = C, R₃ = R, R₄ = R/4Q², and CR = 2 Q/ω₀—we get

[latex]t(s) = \frac{s^{2} + s(ω_{0}/Q) + ω_{0}^{2}}{s^{2} + s(ω_{0}/Q) (2 Q^{2} + 1) + ω_{0}^{2}}[/latex] (17.85)

where ω₀ and Q denote the nominal or design values of the pole frequency and Q factor. The actual values are obtained by substituting for t(s) in Eq. (17.84):

[latex]s^{2} + s \frac{ω_{0}}{Q} (2 Q^{2} + 1) + ω_{0}^{2} + A \left(s^{2} + s \frac{ω_{0}}{Q} + ω_{0}^{2}\right) = 0[/latex]

Assuming the gain A to be real and dividing both sides by A + 1, we get

[latex]s^{2} + s \frac{ω_{0}}{Q} \left(1 + \frac{2 Q^{2}}{A + 1}\right) + ω_{0}^{2} = 0[/latex] (17.86)

From this equation we see that the actual pole frequency, ω_0a, and the pole Q, Q_a, are

ω_0a = ω₀ (17.87)

[latex]Q_{a} = \frac{Q}{1 + 2 Q^{2} /(A + 1)}[/latex] (17.88)

Thus

[latex]S^{ω_{0_{a}}}_{A} = 0[/latex]

[latex]S^{Q_{a}}_{A} = \frac{A}{A + 1} \frac{2 Q^{2} / (A + 1)}{1 + 2 Q^{2} / (A + 1)}[/latex]

For A ≫ 2 Q² and A ≫ 1 we obtain

[latex]S^{Q_{a}}_{A} \simeq \frac{2 Q^{2}}{A}[/latex]

It is usual to drop the subscript a in this expression and write

[latex]S^{Q}_{A} \simeq \frac{2 Q^{2}}{A}[/latex] (17.89)

Note that if Q is high (Q ≥ 5), its sensitivity relative to the amplifier gain can be quite high¹⁰.

¹⁰ Because the open-loop gain A of op amps usually has wide tolerance, it is important to keep [latex]S^{ω_{0}}_{A}[/latex] and [latex]S^{Q}_{A}[/latex] very small.

Question 17.3: For the feedback loop of Fig. 17.29, find the sensitivities ...

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