Fundamentals of Microelectronics

It can be shown that the sensitivities of the Tow-Thomas biquad with respect to the component values are equal to 0.5 or 1. An important advantage of this topology over the KHN biquad is accrued in integrated circuit design, where differential integrators obviate the need for the inverting stage in the loop, thus saving one op amp. Illustrated in Fig. 14.38, the idea is to swap

Vin

Figure 14.38 Differential Tow-Thomas filter.

the differential outputs of the second integrator to establish negative feedback.

Example 14.20

Prove that !n and Q of the Tow-Thomas filter can be adjusted (tuned) independently.

Solution

From (14.114), we have

It is therefore possible to adjust !n by R2 or R4 and Q by R3. As expected, if R3 = 1, then Q = 1 and the circuit contains two purely imaginary poles.

Exercise

A Tow-Thomas filter exhibits !n = 2 (10 MHz) and Q = 5. Is it possible to have

R1 = R2 = R3 and C1 = C2?

14.4.3 Biquads Using Simulated Inductors

Recall from Section 14.3.2 that second-order RLC circuits can provide low-pass, high-pass, or band-pass responses, but their usage in integrated circuits is limited because of the difficulty in building high-value, high-quality on-chip inductors. We may therefore ask: is it possible to emulate the behavior of an inductor by means of an active (inductorless) circuit?

Consider the circuit shown in Fig. 14.39(a), where general impedances Z1-Z5 are placed in

A 1

Figure 14.39 General impedance converter.

series and the feedback loops provided by the two (ideal) op amps force V1 , V3 and V3 , V5 to zero:

That is, the op amps establish a current of VX =Z5 through Z5. This current flows through Z4, yielding

The above result suggests that the circuit can “convert” Z5 to a different type of impedance if Z1-Z4 are chosen properly. For example, if Z5 = RX, Z2 = (Cs),1, and Z1 = Z3 = Z4 = RY (Fig. 14.40), we have

Figure 14.40 Example of inductance simulation.

which is an inductor of value RXRY C (why?). We say the circuit converts a resistor to an inductor, i.e., it “simulates” an inductor. 8

Example 14.21

From Eq. (14.125), determine another possible combination of components that yields a simulated inductor.

Solution

It is possible to choose Z4 = (Cs),1 and the remaining passive elements to be resistors: Z1 =

8In today's terminology, we may call this an “emulated” inductor to avoid confusion with circuit simulation programs. But the term “simulated” has been used in this context since the 1960s.

Exercise

Is there yet another possible combination that yields a simulated inductor?

Introduced by Antoniou, the “general impedance converter” (GIC) in Fig. 14.39 and its descendants in Figs. 14.40 and 14.41 prove useful in transforming a passive RLC filter to an active counterpart. For example, as depicted in Fig. 14.42, a high-pass active section is obtained by

Figure 14.42 High-pass filter using a simulated inductor.

replacing L1 with a simulated inductor.

Example 14.22

Prove that node 4 in Fig. 14.42 can also serve as an output.

Sec. 14.4 Active Filters 755

Solution

Thus, V4 is simply an amplified version of Vout. Driven by op amp A1, this port exhibits a lower output impedance than does node 1, and is often utilized as the output of the circuit.

Exercise

Determine the transfer function from Vin to V3.

How is a low-pass filter derived? From the RLC network of Fig. 14.29, we note the need for a floating (rather than grounded) inductor and attempt to create such a device as shown in Fig. 14.43(a). Can this circuit serve as a floating inductor? A simple test is to tie a voltage source to node P and determine the Thevenin equivalent as seen from node Q [Fig. 14.43(b)]. To compute the open-circuit voltage, VThev, recall that the op amps force V5 to be equal to VP (= Vin). Since no current flows through RX ,

To obtain ZThev, we set Vin to zero and apply a voltage to the left terminal of RX [Fig. 14.43(c)]. Since V5 = V1 = 0, IX = VX =RX and hence

Unfortunately, the network operates as a simple resistor rather than a floating inductor! Fortunately, the impedance converter of Fig. 14.39 can overcome this difficulty. Consider the special case illustrated in Fig. 14.44(a), where Z1 = (Cs),1, Z3 = (Cs),1jjRX, and Z2 = Z4 = Z5 = RX. From (14.125), we have

This impedance may be viewed as a “super capacitor” because it is equal to the product of two capacitive components: (Cs),1 and (RX Cs + 1),1.

Now, let us study the circuit depicted in Fig. 14.44(b):

Figure 14.43 (a) General impedance converter considered as a floating impedance, (b) Thevenin equivalent, (c) test circuit for obtaining the output impedance.

This topology thus provides a second-order low-pass response. From Example 14.22, we note that node 4 serves as a better output port.

Example 14.23

Excited by the versatility of the general impedance converter, a student constructs the circuit shown in Fig. 14.45 as an alternative to that in Fig. 14.44. Explain why this topology is less useful.

Solution

Employing Z3 = RX and Z5 = (Cs),1jjRX , this configuration provides the same transfer

(b)

Figure 14.44 (a) General impedance converter producing a “super capacitor,” (b) second-order low-pass filter obtained from (a).

Figure 14.45

function as (14.135). However, V4 is no longer a scaled version of Vout:

Thus, the output can be sensed at only node 1, suffering from a relatively high impedance.

Exercise

Determine the transfer function from Vin to V5.

It can be proved that the sensitivities of the general impedance converter and the resulting filters with respect to component values are equal to 0.5 or 1. Such circuits therefore prove useful in both discrete and integrated design.

14.5 Approximation of Filter Response

How does the design of a filter begin? Based on the expected levels of the desired signal and the interferers, we decide on the required stopband attenuation. Next, depending on how close the interferer frequency is to the desired signal frequency, we choose a slope for the transition band. Finally, depending on the nature of the desired signal (audio, video, etc.), we select the tolerable passband ripple (e.g., 0.5 dB). We thus arrive at a “template” such as that shown in Fig. 14.46 for

H (ω)

Transition

Passband Band

Ripple

Figure 14.46 Frequency response template.

the frequency response of the filter.

With the template in hand, how do we determine the required transfer function? This task is called the “approximation problem,” by which we mean a transfer function is chosen to approximate the response dictated by the template. We prefer to select a transfer function that lends itself to efficient, low-sensitivity circuit realization.

A multitude of approximations with various trade-offs exist that prove useful in practice. Examples include “Butterworth,” “Chebyshev,” “elliptic,” and “Bessel” responses. Most filters suffer from a trade-off between the passband ripple and the transition band slope. We study the first two types here and refer the reader to texts on filter design [1] for others.

14.5.1 Butterworth Response

The Butterworth response completely avoids ripple in the passband (and the stopband) at the cost of the transition band slope. This type of response only stipulates the magnitude of the transfer function as:

where n denotes the order of the filter. 9

9This is called a “maximally-flat” response because the first 2n 1 derivatives of (14.138) vanish at ! = 0.

It is instructive to examine (14.138) carefully and understand its properties. First, we observe that the ,3-dB bandwidth is calculated as:

Interestingly, the ,3-dB bandwidth remains independent of the order of the filter. Second, as n increases, the response assumes a sharper transition band and a greater passband flatness. Third, the response exhibits no ripple (local maxima or minima) because the first derivative of (14.138) with respect to ! vanishes only at ! = 0. Figure 14.47 illustrates these points.

H (ω) n

1

2

2

Example 14.24

A low-pass filter must provide a passband flatness of 0.45 dB for f < f1 = 1 MHz and a stopband attenuation of 9 dB at f2 = 2 MHz. Determine the order of a Butterworth filter satisfying these requirements.

Solution

Figure 14.48 shows the template of the desired response. Noting that jH(f1 = 1 MHz)j = 0:95

20log H (ω)

0.45 dB

0 dB

Figure 14.48

( ,0:45 dB) and jH(f2 = 2 MHz)j = 0:355 ( ,9 dB), we construct two equations with two

760 Chap. 14 Analog Filters

unknowns:

Since f2 = 2f1, the smallest n that satisfies the requirement is 3. With n = 3, we obtain !0 from (14.141):!0 = 2 (1:45 MHz):

Exercise

If the order of the filter must not exceed 2, how much attenuation can be obtained from f1 to f2?

Given filter specifications and hence a template, we can readily choose !0 and n in (14.138) to arrive at an acceptable Butterworth approximation. But how do we translate (14.138) to a transfer function and hence a circuit realization? Equation (14.138) suggests that the corresponding transfer function contains no zeros. To obtain the poles, we make a reverse substitution, ! = s=j, and set the denominator to zero:

but only the roots having a negative real part are acceptable (why?):