14.8 Active Filters

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14.8 Active Filters

There are three major limitations to the passive filters considered in the previous section. First, they cannot generate gain greater than 1; passive elements cannot add energy to the network. Second, they may require bulky and expensive inductors. Third, they perform poorly at frequencies below the audio frequency range (300 Hz < f < 3,000Hz). Nevertheless, passive filters are useful at high frequencies.

Active filters consist of combinations of resistors, capacitors, and op amps. They offer some adv antages over passive RLC filters. First, they are often smaller and less expensive, because they do not require inductors. This makes feasible the integrated circuit realizations of filters. Second, they can provide amplifier gain in addition to pro viding the same frequenc y response as RLC filters. Third, active filters can be combined with buffer amplifiers (voltage followers) to isolate each stage of the filter from source and load impedance effects. This isolation allows designing the stages independently and then cascading them to realize the desired transfer function. (Bode plots, being log arithmic, may be added when transfer functions are cascaded.) Ho wever, active filters are less reliable and less stable. The practical limit of most active filters is about 100 kHz—most active filters operate well below that frequency.

Filters are often classified according to their order (or number of poles) or their specific design type.

14.8.1 First-Order Low-Pass Filter

One type of first-order filter is shown in Fig. 14.41. The components selected for Zi and Zf determine whether the filter is low-pass or high-pass, but one of the components must be reactive.

Figure 14.42 shows a typical active low-pass filter. For this filter, the transfer function is

$$\mathbf{H}(\omega) = \frac{\mathbf{V}_o}{\mathbf{V}_i} = -\frac{\mathbf{Z}_f}{\mathbf{Z}_i}$$

(14.58)

where Zi = Ri and

$$\mathbf{Z}_f = R_f \left\| \frac{1}{j\omega C_f} = \frac{R_f/j\omega C_f}{R_f + 1/j\omega C_f} = \frac{R_f}{1 + j\omega C_f R_f} \right. \tag{14.59}$$

Therefore,

$$\mathbf{H}(\omega) = -\frac{R_f}{R_i} \frac{1}{1 + j\omega C_f R_f}$$

(14.60)

We notice that Eq. (14.60) is similar to Eq. (14.50), except that there is a low frequency ( ω → 0) gain or dc gain of −Rf∕Ri. Also, the corner frequency is

$$\omega_c = \frac{1}{R_f C_f} \tag{14.61}$$

which does not depend on Ri. This means that several inputs with dif ferent Ri could be summed if required, and the corner frequency would remain the same for each input.

14.8.2 First-Order High-Pass Filter

Figure 14.43 shows a typical high-pass filter. As before,

$$\mathbf{H}(\omega) = \frac{\mathbf{V}_o}{\mathbf{V}_i} = -\frac{\mathbf{Z}_f}{\mathbf{Z}_i}$$

(14.62)

where Zi = Ri + 1∕jωCi and Zf = Rf so that

$$\mathbf{H}(\omega) = -\frac{R_f}{R_i + 1/j\omega C_i} = -\frac{j\omega C_i R_f}{1 + j\omega C_i R_i}$$

(14.63)

This is similar to Eq. (14.52), except that at very high frequencies (ω → ∞), the gain tends to −Rf∕Ri. The corner frequency is

$$\omega_c = \frac{1}{R_i C_i} \tag{14.64}$$

14.8.3 Band-Pass Filter

The circuit in Fig. 14.42 may be combined with that in Fig. 14.43 to form a band-pass filter that will have a gain K over the required range of frequencies. By cascading a unity-gain low-pass filter, a unity-gain

Active first-order high-pass filter.

This way of creating a band-pass filter, not necessarily the best, is perhaps the easiest to understand.

Figure 14.41 A general first-order active filter.

Figure 14.42 Active first-order low-pass filter.

high-pass filter, and an inverter with gain −Rf∕Ri, as shown in the block diagram of Fig. 14.44(a), we can construct a band-pass filter whose frequency response is that in Fig. 14.44(b). The actual construction of the band-pass filter is shown in Fig. 14.45.

Figure 14.44

Active band-pass filter: (a) block diagram, (b) frequency response.

The analysis of the band-pass filter is relatively simple. Its transfer function is obtained by multiplying Eqs. (14.60) and (14.63) with the gain of the inverter; that is,

$$\mathbf{H}(\omega) = \frac{\mathbf{V}_o}{\mathbf{V}_i} = \left(-\frac{1}{1 + j\omega C_1 R}\right) \left(-\frac{j\omega C_2 R}{1 + j\omega C_2 R}\right) \left(-\frac{R_f}{R_i}\right)$$ $$= -\frac{R_f}{R_i} \frac{1}{1 + j\omega C_1 R} \frac{j\omega C_2 R}{1 + j\omega C_2 R}$$

(14.65)

The low-pass section sets the upper corner frequency as

$$\omega_2 = \frac{1}{RC_1} \tag{14.66}$$

while the high-pass section sets the lower corner frequency as

$$\omega_1 = \frac{1}{RC_2} \tag{14.67}$$

With these values of ω1 and ω2, the center frequency, bandwidth, and quality factor are found as follows:

$$\omega_0 = \sqrt{\omega_1 \omega_2} \tag{14.68}$$ $$B = \omega_2 - \omega_1 \tag{14.69}$$ $$Q = \frac{\omega_0}{B} \tag{14.70}$$

To find the passband gain K, we write Eq. (14.65) in the standard form of Eq. (14.15),

To find the passband gain K, we write Eq. (14.65) in the standard
\nn of Eq. (14.15),
\n

$$\mathbf{H}(\omega) = -\frac{R_f}{R_i} \frac{j\omega/\omega_1}{(1 + j\omega/\omega_1)(1 + j\omega/\omega_2)} = -\frac{Rf}{R_i} \frac{j\omega\omega_2}{(\omega_1 + j\omega)(\omega_2 + j\omega)}$$

\n(14.71)

At the center frequency ω0 = √ _____ ω1ω2 , the magnitude of the transfer function is

center frequency

$$\omega_0 = \sqrt{\omega_1 \omega_2}$$

, the magnitude of the transfer
\nn is
\n

$$|\mathbf{H}(\omega_0)| = \left| \frac{R_f}{R_i} \frac{j \omega_0 \omega_2}{(\omega_1 + j \omega_0)(\omega_2 + j \omega_0)} \right| = \frac{R_f}{R_i} \frac{\omega_2}{\omega_1 + \omega_2}$$

(14.72)

Thus, the passband gain is

$$K = \frac{R_f}{R_i} \frac{\omega_2}{\omega_1 + \omega_2} \tag{14.73}$$

14.8.4 Band-Reject (or Notch) Filter

A band-reject filter may be constructed by parallel combination of a lowpass filter and a high-pass filter and a summing amplifier, as shown in the block diagram of Fig. 14.46(a). The circuit is designed such that the lower cutoff frequency ω1 is set by the low-pass filter while the upper cutoff frequency ω2 is set by the high-pass filter. The gap between ω¹ and ω2 is the bandwidth of the filter. As shown in Fig. 14.46(b), the filter passes frequencies below ω1 and above ω2. The block diagram in Fig. 14.46(a) is actually constructed as shown in Fig. 14.47. The transfer function is

$$\mathbf{H}(\omega) = \frac{\mathbf{V}_o}{\mathbf{V}_i} = -\frac{R_f}{R_i} \left( -\frac{1}{1 + j\omega C_1 R} - \frac{j\omega C_2 R}{1 + j\omega C_2 R} \right) \tag{14.74}$$

Figure 14.46 Active band-reject filter: (a) block diagram, (b) frequency response.

Active band-reject filter.

The formulas for calculating the values of ω1, ω2, the center frequency, bandwidth, and quality factor are the same as in Eqs. (14.66) to (14.70).

To determine the passband gain K of the filter, we can write Eq. (14.74) in terms of the upper and lower corner frequencies as

$$\mathbf{H}(\omega) = \frac{R_f}{R_i} \left( \frac{1}{1 + j\omega/\omega_2} + \frac{j\omega/\omega_1}{1 + j\omega/\omega_1} \right)$$

= $\frac{R_f}{R_i} \frac{(1 + j2\omega/\omega_1 + (j\omega)^2/\omega_1\omega_1)}{(1 + j\omega/\omega_2)(1 + j\omega/\omega_1)}$ (14.75)

Comparing this with the standard form in Eq. (14.15) indicates that in the two passbands (ω → 0 and ω → ∞) the gain is

$$K = \frac{R_f}{R_i} \tag{14.76}$$

We can also find the gain at the center frequency by finding the magnitude of the transfer function at ω0 = √ _____ ω1ω2 , writing

and the gain at the center frequency by finding the magni-
\nnsfer function at

$$\omega_0 = \sqrt{\omega_1 \omega_2}$$

, writing
\n

$$H(\omega_0) = \left| \frac{R_f (1 + j2\omega_0/\omega_1 + (j\omega_0)^2/\omega_1 \omega_1)}{R_i (1 + j\omega_0/\omega_2)(1 + j\omega_0/\omega_1)} \right|$$

\n

$$= \frac{R_f}{R_i} \frac{2\omega_1}{\omega_1 + \omega_2}$$

\n(14.77)

Again, the filters treated in this section are only typical. There are many other active filters that are more complex.

Example 14.12 Design a low-pass active filter with a dc gain of 4 and a corner frequency of 500 Hz.

Solution:

From Eq. (14.61), we find

$$\omega_c = 2\pi f_c = 2\pi (500) = \frac{1}{R_f C_f} \tag{14.12.1}$$

The dc gain is

$$H(0) = -\frac{R_f}{R_i} = -4\tag{14.12.2}$$

We have two equations and three unknowns. If we select Cf = 0.2μF, then

tions and three unknowns. If we se

$$R_f = \frac{1}{2\pi (500) 0.2 \times 10^{-6}} = 1.59 \text{ k}\Omega$$

and

$$R_i = \frac{R_f}{4} = 397.5 \ \Omega$$

We use a 1.6-kΩ resistor for Rf and a 400-Ω resistor for Ri . Figure 14.42 shows the filter.

Design a high-pass filter with a high-frequency gain of 5 and a corner Practice Problem 14.12 frequency of 2 kHz. Use a 50-nF capacitor in your design.

Answer: Ri = 1,600 Ω and Rf = 8 kΩ.

Design a band-pass filter in the form of Fig. 14.45 to pass frequencies Example 14.13 between 250 and 3,000 Hz and with K = 10. Select R = 20 kΩ.

Solution:

• 1. Define. The problem is clearly stated and the circuit to be used in the design is specified.
• 2. Present. We are ask ed to use the op amp circuit specified in Fig. 14.45 to design a band-pass filter. We are given the value of R to use (20 kΩ). In addition, the frequency range of the signals to be passed is 250 Hz to 3 kHz.
• 3. Alternative. We will use the equations developed in Section 14.8.3 to obtain a solution. We will then use the resulting transfer function to validate the answer.
• 4. Attempt. Because ω1 = 1∕RC2, we obtain

te the answer.
\napt. Because

$$\omega_1 = 1/RC_2$$

, we obtain
\n

$$C_2 = \frac{1}{R\omega_1} = \frac{1}{2\pi f_1 R} = \frac{1}{2\pi \times 250 \times 20 \times 10^3} = 31.83 \text{ nF}$$

Similarly, since ω2 = 1∕RC1,

$$K\omega_1 = 2\pi f_1 K - 2\pi \times 250 \times 20 \times 10^8$$

arly, since $\omega_2 = 1/RC_1$ ,

$$C_1 = \frac{1}{R\omega_2} = \frac{1}{2\pi f_2 R} = \frac{1}{2\pi \times 3,000 \times 20 \times 10^3} = 2.65 \text{ nF}$$

From Eq. (14.73),

$$\frac{R_f}{R_i} = K \frac{\omega_1 + \omega_2}{\omega_2} = K \frac{f_1 + f_2}{f_2} = \frac{10(3,250)}{3,000} = 10.83$$

If we select Ri = 10 kΩ, then Rf = 10.83Ri ≃ 108.3 kΩ. 5. Evaluate. The output of the first op amp is given by

ct

$$R_i = 10 \text{ k}\Omega$$

, then $R_f = 10.83R_i \approx 108.3 \text{ k}\Omega$ .
The output of the first op amp is given by

$$\frac{V_i - 0}{20 \text{ k}\Omega} + \frac{V_1 - 0}{20 \text{ k}\Omega} + \frac{s2.65 \times 10^{-9} (V_1 - 0)}{1}$$ $$= 0 \rightarrow V_1 = -\frac{V_i}{1 + 5.3 \times 10^{-5} s}$$

The output of the second op amp is given by

$$t \text{ of the second op amp is given by}$$

\n

$$\frac{V_1 - 0}{20 \text{ k}\Omega + \frac{1}{s31.83 \text{ nF}}} + \frac{V_2 - 0}{20 \text{ k}\Omega} = 0 \rightarrow$$

\n

$$V_2 = -\frac{6.366 \times 10^{-4} s V_1}{1 + 6.366 \times 10^{-4} s}$$

\n

$$= \frac{6.366 \times 10^{-4} s V_i}{(1 + 6.366 \times 10^{-4} s)(1 + 5.3 \times 10^{-5} s)}$$

The output of the third op amp is given by

$$\frac{V_2 - 0}{10 \text{ k}\Omega} + \frac{V_o - 0}{108.3 \text{ k}\Omega} = 0 \to V_o = 10.83 V_2 \to j2\pi \times 25^\circ$$ $$V_o = -\frac{6.894 \times 10^{-3} sV_i}{(1 + 6.366 \times 10^{-4} s)(1 + 5.3 \times 10^{-5} s)}$$

Let j2π × 25° and solve for the magnitude of Vo∕Vi.

$$\frac{V_o}{V_i} = \frac{-j10.829}{(1+j1)(1)}$$

∣Vo∕Vi ∣ = (0.7071)10.829, which is the lower corner frequency point. Let s = j2π × 3000 = j18.849 kΩ. We then get

Let

$$s = j2\pi \times 3000 = j18.849 \text{ k}\Omega$$

. We then get
\n

$$\frac{V_o}{V_i} = \frac{-j129.94}{(1+j12)(1+j1)}$$

\n

$$= \frac{129.94/-90^{\circ}}{(12.042/85.24^{\circ})(1.4142/45^{\circ})} = (0.7071)10.791/-18.61^{\circ}$$

Clearly this is the upper corner frequency and the answer checks.

6. Satisfactory? We have satisfactorily designed the circuit and can present the results as a solution to the problem.

Practice Problem 14.13	Design a notch filter based on Fig. 14.47 for ω0 = 20 krad/s, K = 5, and
	Q = 10. Use R = Ri = 10 kΩ.

Answer: C1 = 4.762 nF, C2 = 5.263 nF, and Rf = 50 kΩ.