Fundamentals of Microelectronics

Exercise

Calculate the sensitivity of the pole frequency of the circuit in Fig. 14.12(b) with respect to

R1.

14.2 First-Order Filters

As our first step in the analysis of filters, we consider first-order realizations, described by the transfer function

The circuit of Fig. 14.12(b) and its transfer function in (14.4) exemplify this type of filter. Depending on the relative values of z1 and p1, a low-pass or high-pass characteristic results, as illustrated in the plots of Fig. 14.16. Note that the stopband attenuation factor is given by z1=p1.

Figure 14.16 First-order (a) high-pass and (b) low-pass filters.

Let us consider the passive circuit of Fig. 14.12(b) as a candidate for realization of the above transfer function. We note that, since z1 = ,1=(R1C2) and p1 = ,1=[R1(C1 + C2)], the zero always falls above the pole, allowing only the response shown in Fig. 14.16(b).

Example 14.8

Determine the response of the circuit depicted in Fig. 14.17(a).

Solution

We have

The circuit contains a zero at ,1=(R1C1) and a pole at ,[(C1 + C2)R1jjR2],1. Depending on the component values, the zero may lie below or above the pole. Specifically, for the zero frequency to be lower:

Figures 14.17(b) and (c) plot the response for the two cases R2C2 < R1C1 and R2C2 > R1C1, respectively. Note that Vout=Vin = R2=(R1 +R2) at s = 0 because the capacitors act as an open circuit. Similarly, Vout=Vin = C1=(C1 + C2) at s = 1 because the impedance of the capacitors becomes much smaller than R1 and R2 and hence the determining factor.

Exercise

Design the circuit for a high-pass response with a zero frequency of 50 MHz and a pole frequency of 100 MHz. Use capacitors no larger than 10 pF.

Example 14.9

Figure 14.18(a) shows the active counterpart of the filter depicted in Fig. 14.17(a). Compute the response of the circuit. Assume the gain of the op amp is large.

Solution

We have from Chapter 8

As expected, at s = 0, Vout=Vin = ,R2=R1 and at s = 1, Vout=Vin = ,C1=C2. Figures 14.18(b) and (c) plot the response for the two cases R1C1 < R2C2 and R1C1 > R2C3, respectively.

Exercise

Is it possible for the pole frequency to be five times the zero frequency while the passband gain is ten times the stopband gain?

The first-order filters studied above provide only a slope of ,20 dB/dec in the transition band. For a sharper attenuation, we must seek circuits of higher order.

14.3 Second-Order Filters

The general transfer function of second-order filters is given by the “biquadratic” equation:

Unlike the numerator, the denominator is expressed in terms of quantities !n and Q because they signify important aspects of the response. We begin our study by calculating the pole frequencies.

Since most second-order filters incorporate complex poles, we assume (!n=Q)2 , 4!2 < 0,

n

obtaining

Note that as the “quality factor” of the poles, Q, increases, the real part decreases while the imaginary part approaches !n. This behavior is illustrated in Fig. 14.19. In other words, for high Q's, the poles look “very imaginary,” thereby bringing the circuit closer to instability.

j ω

+ωn

Q

σ

− ωn

Figure 14.19 Variation of poles as a function of Q.

14.3.1 Special Cases

It is instructive to consider a few special cases of the biquadratic transfer function that prove useful in practice. First, suppose = = 0 so that the circuit contains only poles4 and operates as a low-pass filter (why?). The magnitude of the transfer function is then obtained by making the substitution s = j! in (14.23) and expressed as

Note that jH(j!)j provides a slope of ,40 dB/dec beyond the passband (i.e., if ! !n). It can be shown (Problem 11) that the response is (a) free from peaking if Q p2=2; and (b) reaches a peak at !np1 , 1=(2Q2) if Q > p2=2 (Fig. 14.20). In the latter case, the peak magnitude

Figure 14.20 Frequency response of second-order system for different values of Q..

normalized to the passband magnitude is equal to Q=p1 , (4Q2),1.

Example 14.10

Suppose a second-order LPF is designed with Q = 3. Estimate the magnitude and frequency of the peak in the frequency response.

4Since H(s) ! 0 at s = 1, we say the circuit exhibits two zeros at infinity.

Solution

Since 2Q2 = 18 1, we observe that the normalized peak magnitude is Q=p1 , 1=(4Q2) Q 3 and the corresponding frequency is !np1 , 1=(2Q2) !n. The behavior is plotted in Fig. 14.21.

γ

ω n2

Figure 14.21

Exercise

Repeat the above example for Q = 1:5.

How does the transfer function in Eq. (14.23) provide a high-pass response? In a manner similar to the first-order realization in Fig. 14.12(b), the zero(s) must fall below the poles. For example, with two zeros at the origin:

we note that H(s) approaches zero as s ! 0 and a constant value, , as s ! 1, thus providing a high-pass behavior (Fig. 14.22). As with the low-pass counterpart, the circuit exhibits a peak if Q > p2=2 with a normalized value of Q=p1 , 1=(4Q2) but at a frequency of

!n=p1 , 1=(2Q2).

Example 14.11

Explain why a high-pass response cannot be obtained if the biquadratic equation contains only one zero.

Solution

Let us express such a case as

Since H(s) ! 0 as s ! 1, the system cannot operate as a high-pass filter.

Figure 14.22 (a) Pole and zero locations and (b) frequency response of a second-order high-pass filter.

Exercise

Calculate the magnitude of H(s).

A second-order system can also provide a band-pass response. Specifically, if

then, the magnitude approaches zero for both s ! 0 and s ! 1, reaching a maximum in between (Fig. 14.23). It can be proved that the peak occurs at ! = !n and has a value of Q=!n.

Figure 14.23 (a) Pole and zero locations and (b) frequency response of a second-order band-pass filter.

Example 14.12

Determine the ,3-dB bandwidth of the response expressed by Eq. (14.28).

Solution

As shown in Fig. 14.24, the response reaches 1=p2 times its peak value at frequencies !1 and

H(ω)

βQ

ωn

1β Q

ωn2

Figure 14.24

!2, exhibiting a bandwidth of !2 ,!1. To calculate !1 and !2, we equate the squared magnitude to ( Q=!n)2(1=p2)2:

obtaining

The total ,3-dB bandwidth spans !1 to !2 and is equal to !o=Q. We say the “normalized” bandwidth is given by 1=Q; i.e., the bandwidth trades with Q.

Exercise

For what value of Q is !2 twice !1?

14.3.2 RLC Realizations

It is possible to implement the second-order transfer function in (14.23) by means of resistors, capacitors, and inductors. Such RLC realizations (a) find practical applications in low-frequency discrete circuits or high-frequency integrated circuits, and (b) prove useful as a procedure for designing active filters. We therefore study their properties here and determine how they can yield low-pass, high-pass, and band-pass responses.

Consider the parallel LC combination (called a “tank”) depicted in Fig. 14.25(a). Writing

Figure 14.25 (a) LC tank, (b) imaginary poles, (c) and frequency response.

we note that the impedance contains a zero at the origin and two imaginary poles at j=pL1C1 [Fig. 14.25(b)]. We also examine the magnitude of the impedance by replacing s with j!:

The magnitude thus begins from zero for ! = 0, goes to infinity at !0 = 1=pL1C1, and returns to zero at ! = 1 [Fig. 14.25(c)]. The infinite impedance at !0 arises simply because the impedances of L1 and C1 cancel each other while operating in parallel.

Example 14.13

Explain intuitively why the impedance of the tank goes to zero at ! = 0 and ! = 1.

Solution

At ! = 0, L1 operates as a short circuit. Similarly, at ! = 1, C1 becomes a short.

Exercise

Explain why the impedance has a zero at the origin.

Now let us turn our attention to the parallel RLC tank depicted in Fig. 14.26(a). We can obtain

The impedance still contains a zero at the origin due to the inductor. To compute the poles, we can factor R1L1C1 from the denominator, thus obtaining a form similar to the denominator of (14.23):

where !n = 1=pL1C1 and Q = R1C1!n = R1pC1=L1. It follows from (14.24) that

or

So long as the excitation of the circuit does not alter its topology,5 the poles are given by (14.39) or (14.43), a point that proves useful in the choice of filter structures.

Before studying different RLC filters, it is instructive to make several observations. Consider the voltage divider shown in Fig. 14.27, where a series impedance ZS and a parallel impedance

Z S

VinVout

Z P

Figure 14.27 Voltage divider using general impedances.

ZP yield

5The “topology” of a circuit is obtained by setting all independent sources to zero.

We note that (a) if, at high frequencies, ZP goes to zero and/or ZS goes to infinity, 6 then the circuit operates as a low-pass filter; (b) if, at low frequencies, ZP goes to zero and/or ZS goes to infinity, then the circuit serves as a high-pass filter; (c) if ZS remains constant but ZP falls to zero at both low and high frequencies then the topology yields a band-pass response. These cases are conceptually illustrated in Fig. 14.28.

Figure 14.28 (a) Low-pass, (b) high-pass, and (c) bandpass responses obtained from the voltage divider of Fig. 14.27.

Low-Pass Filter Following the observation depicted in Fig. 14.28(a), we construct the circuit shown in Fig. 14.29, where

Z S

L1

VinVout

Z P

Figure 14.29 Low-pass filter obtained from Fig. 14.27.

This arrangement provides a low-pass response having the same poles as those given by (14.39) or (14.43) because for Vin = 0, it reduces to the topology of Fig. 14.26. Furthermore, the transition beyond the passband exhibits a second-order roll-off because both ZS ! 1 and ZP ! 0. The reader can show that

Example 14.14

Explain how the transfer function of (14.47) can provide a voltage gain greater than unity.

Solution

If the Q of the network is sufficiently high, the frequency response exhibits “peaking,” i.e., a gain of greater than unity in a certain frequency range. With a constant numerator, the transfer

6We assume ZS and ZP do not go to zero or infinity simultaneously.