Fundamentals of Microelectronics

function provides this effect if the denominator falls to a local minimum. Writing the squared magnitude of the denominator as

Comparison with (14.41) reveals the poles are complex here. The reader is encouraged to plot the resulting frequency response for different values of R1 and prove that the peak value increases as R1 decreases.

Exercise

Compare the gain at !a with that at 1=pL1C1.

The peaking phenomenon studied in the above example proves undesirable in many applications as it disproportionately amplifies some frequency components of the signal. Viewed as ripple in the passband, peaking must remain below approximately 1 dB (10%) in such cases.

Example 14.15

Consider the low-pass circuit shown in Fig. 14.30 and explain why it is less useful than that of

R1

L1

C1

Figure 14.30

Fig. 14.29.

Solution

This circuit satisfies the conceptual illustration in Fig. 14.28(a) and hence operates as a low-pass

filter. However, at high frequencies, the parallel combination of L1 and R1 is dominated by R1 because L1! ! 1, thereby reducing the circuit to R1 and C1. The filter thus exhibits a roll-off less sharp than the second-order response of the previous design.

Exercise

What type of frequency response is obtained if L1 and C1 are swapped?

High-Pass Filter To obtain a high-pass response, we swap L1 and C1 in Fig. 14.29, arriving at the arrangement depicted in Fig. 14.31. Satisfying the principle illustrated in Fig. 14.28(b), the

C1

Figure 14.31 High-pass filter obtained from Fig. 14.27.

circuit acts as a second-order filter because as s ! 0, C1 approaches an open circuit and L1 a short circuit. The transfer function is given by

The filter therefore contains two zeros at the origin. As with the low-pass counterpart, this circuit can exhibit peaking in its frequency response.

Band-Pass Filter From our observation in Fig. 14.28(c), we postulate that, ZP must contain both a capacitor and an inductor so that it approaches zero as s ! 0 or s ! 1. Depicted in Fig. 14.32 is a candidate. Note that at ! = 1=pL1C1, the parallel combination of L1 and C1 acts as

R1

Figure 14.32 Band-pass filter obtained from Fig. 14.27.

an open circuit, yielding jVout=Vinj = 1. The transfer function is given by

14.4 Active Filters

Our study of second-order systems in the previous section has concentrated on passive RLC realizations. However, passive filters suffer from a number of drawbacks; e.g., they constrain the type of transfer function that can be implemented, and they may require bulky inductors. In this section, we introduce active implementations that provide secondor higher-order responses. Most active filters employ op amps to allow simplifying idealizations and hence a systematic procedure for the design of the circuit. For example, the op-amp-based integrator studied in Chapter 8 and repeated in Fig. 14.10(a) serves as an ideal integrator only when incorporating an ideal op amp, but it still provides a reasonable approximation with a practical op amp. (Thus, the term “integrator” is a simplifying idealization.)

An important concern in the design of active filters stems from the number of op amps required as it determines the power dissipation and even cost of the circuit. We therefore consider secondorder realizations using one, two, or three op amps.

14.4.1 Sallen and Key Filter

The low-pass Sallen and Key (SK) filter employs one op amp to provide a second-order transfer function (Fig. 14.33). Note that the op amp simply serves as a unity-gain buffer, thereby providing

Figure 14.33 Basic Sallen and Key Filter.

maximum bandwidth. Assuming an ideal op amp, we have VX = Vout. Also, since the op amp draws no current, the current flowing through R2 is equal to VX C2 s = VoutC2s, yielding

To obtain a form similar to that in Eq. (14.23), we divide the numerator and the denominator by R1R2C1C2 and define

Example 14.16

The SK topology can provide a passband voltage gain of greater than unity if configured as shown in Fig. 14.34. Assuming an ideal op amp, determine the transfer function of the circuit.

Figure 14.34 Sallen and Key filter with in-band gain.

Solution

Returning to our derivations above, we note that now (1 + R3=R4)VX = Vout, and the current flowing through R2 is given by VX C2s = VoutC2s=(1 + R3=R4). It follows that

Interestingly, the value of !n remains unchanged.

Exercise

Repeat the above analysis if a resistor of value R0 is tied between node Y and ground.

Example 14.17

A common implementation of the SK filter assumes R1 = R2 and C1 = C2. Does such a filter contain complex poles? Consider the general case depicted in Fig. 14.34.

Sec. 14.4 Active Filters 745

Solution

From (14.66), we have

suggesting that Q begins from 1=2 if R3=R4 = 0 (unity-gain feedback) and rises as R3=R4 approaches 2. The poles begin with real, equal values for R3=R4 = 0 and become complex for R3=R4 > 0 (Fig. 14.35).

j ω

R 3

R 4

R 3

R 4 = 0

σ

Coincident

Poles

Figure 14.35

Exercise

Calculate the pole frequencies if R3 = R4.

Sensitivity Analysis With so many components, how is the SK filter designed for a desired frequency response? An important objective in choosing the values is to minimize the sensitivities of the circuit. Considering the topology shown in Fig. 14.34 and defining K = 1 + R3=R4,7 we compute the sensitivity of !n and Q with respect to the resistor and capacitor values.

7In filer literature, the letter K denotes the gain of the filter and should not be confused with the feedback factor (Chapter 12).

The expression in the square brackets is similar to that in (14.74), except for a change in the sign of the second term. Adding and subtracting 2pR2C2=pR1C1 to this expression and substituting for Q from (14.74), we arrive at

Sec. 14.4 Active Filters 747

Example 14.18

Determine the Q sensitivities of the SK filter for the common choice R1 = R2 = R and

C1 = C2 = C.

Solution

From (14.74), we have

The choice of equal component values and K = 1 thus leads to low sensitivities but also a limited Q and hence only a moderate transition slope. Moreover, the circuit provides no voltage gain in the passband.

Exercise

Repeat the above example if R1 = 2R2.

In applications requiring a high Q and/or a high K, one can choose unequal resistors or capacitors so as to maintain reasonable sensitivities. The following example illustrates this point.

Example 14.19

An SK filter must be designed for Q = 2 and K = 2. Determine the choice of filter components for minimum sensitivities.

Solution

748 Chap. 14 Analog Filters

For example, we can choose R1 = 4R2 and C1 = 4C2. But, how about the other sensitivities?

thereby observing that the two sensitivities cannot vanish simultaneously. Moreover, the term

The foregoing observations indicate that some compromise must be made to achieve reason-

The sensitivity to K is quite high and unacceptable in discrete design. In integrated circuits, on the other hand, K (typically the ratio of two resistors) can be controlled very accurately, thus

allowing a large value of SQ.

K

Exercise

Can you choose sensitivities with respect to R1 and C1 such that SQ remains below 2?

K

14.4.2 Integrator-Based Biquads

It is possible to realize the biquadratic transfer function of (14.23) by means of integrators. To this end, let us consider a special case where = = 0:

Cross-multiplying and rearranging the terms, we have

This expression suggests that Vout can be created as the sum of three terms: a scaled version of the input, an integrated version of the output, and a doubly-integrated version of the output. Figure 14.36(a) illustrates how Vout is generated by means of two integrators and a voltage

Figure 14.36 (a) Flow diagram showing the generation of Vout as a weighted sum of three terms, (b) realization of (a), (c) simplified diagram for calculating Vout.

adder. Utilizing the topologies introduced in Chapter 8, we readily arrive at the circuit realization depicted in Fig. 14.36(b). Note that the inherent signal inversion in each integrator necessitates returning VX to the noninverting input of the adder and VY to the inverting input. Since

750 Chap. 14 Analog Filters

It is thus possible to select the component values so as to obtain a desired transfer function. Called the “KHN biquad” after its inventors, Kerwin, Huelsman, and Newcomb, the topology

of Fig. 14.36(b) proves quite versatile. In addition to providing the high-pass transfer function of (14.95), the circuit can also serve as a low-pass and a band-pass filter. Specifically,

which provides a low-pass response.

Perhaps the most important attribute of the KHN biquad is its low sensitivities to component values. It can be shown that the sensitivity of !n with respect to all values is equal to 0.5 and

Interesting, if R3 = R6, then SQ vanishes.

R3;R6

The use of three op amps in the feedback loop of Fig. 14.36(b) raises concern regarding the stability of the circuit because each op amp contributes several poles. Careful simulations are necessary to avoid oscillation.

Another type of biquad developed by Tow and Thomas is shown in Fig. 14.37. Here, the adder and the first integrator are merged, and resistor R3 is introduced to create lossy integration. (Without R3, the loop consisting of two ideal integrators oscillates.) Noting that