How to Design Analog Filter Circuits.Pdf

a b FIG. 1-TWO LOWPASS FILTERS. Even though the filters use different components, they perform in a similiar fashion. MANNlE HOROWITZ Because almost every analog circuit contains some filters, understandinghow to work with them is important. Here we'll discuss the basics of both active and passive types. THE MAIN PURPOSE OF AN ANALOG FILTER In addition to bandpass and band- age (because inductors can be expensive circuit is to either pass or reject signals rejection filters, circuits can be designed and hard to find); they are generally easier based on their frequency. There are many to only pass frequencies that are either to tune; they can provide gain (and thus types of frequency-selective filter cir- above or below a certain cutoff frequency. they do not necessarily have any insertion cuits; their action can usually be de- If the circuit passes only frequencies that loss); they have a high input impedance, termined from their names. For example, are below the cutoff, the circuit is called a and have a low output impedance. a band-rejection filter will pass all fre- lo~~passfilter, while a circuit that passes A filter can be in a circuit with active quencies except those in a specific band. those frequencies above the cutoff is a devices and still not be an active filter. Consider what happens if a parallel re- higlzpass filter. For example, if a resonant circuit is con- sonant circuit is connected in series with a All of the different filters fall into one . nected in series with two active devices signal source. It lets all frequencies pass of two categories: active or passive. Pas- (such as transistors) it is called a passive freely, with the exception of those in a sive filters are composed of passive com- filter. But, if the same resonant circuit is small band of frequencies around the cir- ponents such as resistors, capacitors, and part of afeedback loop, it is then called an cuit's resonant,freque~zcy.(At its resonant inductors. Inductors, however, are un- active filter. We saw an example of such frequency, the parallel circuit the- desirable at some frequencies because of an active filter when we discussed a tape- oretically acts as an infinite impedance. their size andlor practical performance. player preamplifer during our discussion That impedance is greatly reduced at off- (At low frequencies, a real inductor's of feedback (see the July 1983 issue of resonant frequencies.) properties vary considerably from those Radio-Electronics) . On the other hand, should the parallel of an ideal inductor.)Active filters consist L-C circuit be placed across (in parallel of passive elements along with an active Basic filters with) the signal source, only the narrow element such as a transistor or an A rudimentary passive filter consists of band previously rejected would be integrated-circuit op-amp. Active filters an R-C network or an R-L network. allowed to pass from the source to the are usually designed without using in- Those networks can be found in many amplifier. Thus, we have just described a ductors. Therefore they have the advan- different types of electronic equipment bandpnss filter. tage of being less expensive on the aver- performing different roles. For example, w in power supplies, such circuits are used to filter undesirable ripple. In audio 0 amplifiers, different R-C networks are g (MAXIMUM) - ---I- I used in tone-control, scratch-filter, and I- a3 rumble-filter circuits. I-3 Let's take a look at some actual filter 0 W circuits and determine the effect they will >- I- have on signals that are fed to them. TWO 5 W FIG. 6-A MORE COMPLEX filter circuit. This filter circuits-that perform the same a: filter has more than one corner frequency. function using different components- are shown in Fig. 1. When a low- '018 '014 '012 10 410 810 frequency signal is fed to either one of FREUUENCY-Hz those circuits, the output is identical with FIG. 4-THE OUTPUT OF THE HIGHPASS fllters case it is 2.7K + Ry. Using equation 1, shown in Fig. 3 decreases as the frequency de- we determine that f',, = 11 the input. However, as the frequency is creases. increased, the output is reduced-thus 2s1(2.7K + Ro)C. The frequency at which they are lowpass filters. The output rolls the rolloff starts is j;,, as shown in Fig. 4. off at the rate shown by the curve in Fig. each curve, the slope is the same-6-dB- The next step is to place a short across 2. The important frequency to note is.f,,. per-oc tave. the output and leave e,, open. There is no Any of these filter circuits can be parallel R-C combination in the circuit placed between two amplifier stages. If when that is done; the only frequency - - 7----- 7- that's done, the comer frequency, ,f,,, is indicated in the response calculations of still as predicted by equation 1- the filter is,f,, and its curve in Fig. 4. presuming that the impedance of the cir- Now let's look at a more complex filter cuit feeding the filter is zero and the input circuit-one in which there will be Inore a3 i-3 impedance of the circuit at the output of than one corner frequency. We will use 0 the filter is infinite. Otherwise, resistors the procedure noted above to analyze the LL in series with the input to the filter and in circuit in Fig. 6 and find those comer parallel with the output from the filter, frequencies. One f,, corner frequency is must be considered when determining&,. the beginning of the rolloff characteristic For example, assume that the circuit in curve and a secondj'" corner frequency is 10 8 10 10 2 lo 210 410 a/0 Fig. 3-a is connected between two transis- on a curve with rising characteristics. The FREOUENCY Hr tors, as shown in Fig. 5-a. Figure 5-b is two res~~ltingcurves are then added to FIG. 2-THE OUTPUT OF THE LOWPASS fllters the equivalent circuit involving the filter provide a curve illustrating the overall shown ~n Fig. 1 decreases as the frequency in- and transistors. (The function of coupling response of the circuit as shown in Fig. 7. creases. capacitor C1 is to DC-isolate the transis- First, we short einand note the total resis- tor from the filter. That capacitor is usual- tance across C-R3 in parallel with the ly of such magnitude that it can be treated sum of R1 and R2. Substituting this infor- That's the frequency where the output has as a short for all signal voltages.) It can be mation into equation 1, we find that one of dropped 3 dB from its maximum value simplified by determining the equivalent the corner frequencies, where the high- (0.707 of the maximum). That corner resistance. Ry, of all resistances at the frequency roll off starts, is: ,fr-ecluet~cy(also called the half-power output. The final circuit is shown in Fig. point) can be determined from the values 5-c. Equation 1 applies there also. of the components in the circuit. For the The analysis of this circuit-as well as R-C circuit in Fig. I-a: more complex filter circuits that have We continue by leaving e,, open and more than one corner frequency4an be shorting e,,,,. Then, R2 is across both R3 accomplished in two simple steps. First, and C. By inserting those values into short the input, ei,,, and note the total equation 1 we find for the second corner For the R-L circuit in Fig. 1B: resistance across the capacitor. In this frequency: Figure 3 shows two circuits, either of which can be used to attenuate the 10x1 frequencies (thereby creating a highpass filter). Note that the curve shown in Fig. 4, when compared to Fig. 2, is symmetric about the y-axis. In Fig. 2, the output drops in the linear portion of the curve by 6-dB every time the frequency is doubled. In Fig. 4, the output drops by 6-dB every time the frequency is halved. The important point is that in the linear portion of ul 0 z B 7" W 01; 1 l el"n{t U;1 0 a b 9 FIG. 3-TWO HIGHPASS FILTERS that use different components but perform similiarly. equivalent circuit. .) CURVE WlTH f,, those filters are shown in Figs. 9 and 11 CORNERFREQUENCY To minimize their interrelated effects, The curves add as shown in Fig. 7. The components should be chosen so that the total curve indicates the overall response 4 -10 CORNERFREQUENCY R2-C2 circuit has a minor shunting effect of the filter. The two equations differ olily w w TOTAL EFFECT on R1 and C1 and that the R1-C1 combina- by the resistance factors. Because R21 lR3 -20 OF TWO FILTER tion has only a ininor affect on loading the is smaller than (R1 + R2)l lR3,,f,,, is high- 5 SECTIONS R2-C2 combination. It is best if the two er in frequency than isf,,, . A similar pro- % -30 circuits were completely isolated from cedure may be used to determine the fre- each other by placing each R-C filter be- quency response of other complex tween different transistors in a three-tran- arrangements. fol 2fO1 4fO1 8fO1 32iO164fO1 sistor amplifier circuit. f02 2f02 4fo2 16fo2 32foz Active filters FREQUENCY-Hz FIG. 9-THIS FREQUENCY-RESPONSE CURVE An active filter involving bipolar tran- CURVE WlTH fO2 shows how the rolloff is sharpened by using CORNER FREQUENCY sistors, is shown in Fig.

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