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# Electronics Sci Instr PHYS 3610

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This 7 page Class Notes was uploaded by Ericka Ryan I on Monday October 26, 2015. The Class Notes belongs to PHYS 3610 at University of Utah taught by Staff in Fall. Since its upload, it has received 23 views. For similar materials see /class/230024/phys-3610-university-of-utah in Physics 2 at University of Utah.

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Date Created: 10/26/15

LECTURE 14 ACTIVE FILTERS There are many occasions when good research demands a frequency selective filter They can be made in three general types Filter Low Pass Doesn39t affect sine waves with a lt 03L Severely attenuates sine waves with m gt m High Pass Doesn39t affect sine waves with m gt m Severely attenuates sine waves with a lt aquot Band Pass Doesn39t affect sine waves with coL lt a lt a Severely attenuates sine waves with a gt aquot or a lt 0 Of course no system is really discontinuous at mL or m One simply approximates the characteristics of a given filter These filters find application in a wide variety of studies in psychology to separate 0 brain waves from Q waves in geology for detection of particular modes of seismic disturbances in radios to tune in a particular station in mechanical engineering to select particular vibrations in anti submarine warfare to select sonar reflections Filtering can be done passively with capacitors resistors and inductors Such an approach often involves expensive bulky components furthermore the signal is often attenuated significantly The use of op amps permits inexpensive small effective filters particularly at low frequencies For a description of a large variety of active filters along with design procedures see Operational Amplifiers by Burr Brown Engineering staff McGraw Hill Inc 1971 In this book as in many engineering texts quotsquot is used for jco A simpler discussion is available in The Active Filter Cookbook by Dan Lancaster Howard Sams 1975 We will illustrate the possibilities with one practical example the active filter with a multiple feedback network 14 2 By setting V O we get the following equations VBVS39IIZI VBI112I322 V0 13ZS V0 VB 1224 V13 1323 Exercise 1 Eliminate VB1112 and 13 from the above equations and solve for VoNs Ans Z Va Zl VS i Z321 Z 22 ZZZ 25 2125 25 Z4 2425 which can be rewritten as V0 ZZZAZS 1 VS 212223 ZIZZZ4 4 le225 ZIZ3Z4 zzzsz4 Low Pass Filter Z1Rl 27 39 J DCz V Z3R3 2 s Z4R4 1 C2 25 l l Before we calculate let us examine the performance qualitatively At very low frequencies C2 and C5 represent infinite impedance For such cases the equivalent circuit is For such a system the gain is R4R1 In the ideal case R3 plays no role since the op amp input neither sources nor sinks current If R3 is not high it can be neglected in actual use At very high frequencies C2 and C5 represent short circuits Thus VB is zero Furthermore the output is connected to V which is at virtual ground Thus the gain is expected to approach 0 as 102 as frequency becomes very large Now let us plug the impedances of Eq 2 into Eg 1 resulting in Eq 3 R E jzmzczcs RR RR R R R V5 3 2 2 R1R3R4 3 quot JOJCZ 3chZ Jmczcs chz jZmZ Multiplying numerator and denominator by RIRBR4 Xi R1R3C2C5 V5 J0 103 jzmz Jw RdCz RJC2 RER4CZC5 RIC2 1 xi R1R3c205 3 V 2 m 1 1 1 1 03 C2 Rl R3 R1 I3RC2C5 Exercise 2 Show that the above equation can be written in the form V Hucuoz f 4 Vs u ndwomo This is the standard form for the pass transfer functionquot Here 12 R l C R R JR R Hu no and on 5 3 4 R1 R3R4C2C5 C2 R4 R3 R1 quotcomplex conj ugate pole pair low 14 4 For mgtgt no VOVSL gtH0mOw2 and that for coltlt no VDVSLgtHOK4R as we predicted Note also that at low frequency V0 and Vs are 1800 out of phase At mmo where the j term is important V0 leads Vs by 90 1010 lid Vs 0 101 9 00501 911 I 03 03 Magnitude response of secondorder lowpass lters for selected at values S 2 High Pass Filter I z c4 1 Joel C1 C3 Z2R2 I vo i i v z 5 S p a 3 ij3 l R2 7 1 z 4 ij4 Z5R5 Once again let us first determine the response qualitatively At low frequencies the capacitive impedances greatly exceed the resistive ones so at point B the input is attenuated to near zero Furthermore the output is at the same potential as V that is near ground Thus the gain goes to zero at low frequencies as 02 At high frequencies the capacitive impedances are much lower than the resistive ones and the equivalent circuit is vs ic 4 1 C3 The gain in this configuration is Z4ZI or 1 imca 3 Vs L C4 chl Now let us calculate the response more carefully Plugging Eqs 5 in Eq 1 Rst Xi J DC4 VS R2 R2 Rst 1 y R2 22 3922 3933 22 J 0 CICS J a CIC4 Jmcl J 0 C1C3C4 J 0 C3C4 J2m3C1 Multiplying numerator and denominator by RZRS szzcl CA Vs J 3 0 j2m11 RSC3 RsC4 RZR5C3C4 R5C3C4 Rearranging nZCl i C V5 2 1 1 Cl 1 m 1 RSC3 RSC4 R5C3C4 RZRSCZC4 Exercise 3 Show that the above equation can be written in the form V Hnu2 Vs u2 0Lmujmbmn2 This is the standard form for the quotcomplexconjugate pole pair high pass transfer functionquot Here 12 11quot EL m0 1 and a g C39 g amp C4 RZRSCSC4 R5 3ch c4 c3 For 3 gtgt no V0VSH gt 110 ClC4 and that for co ltlt 030 IVoVSIH gt Houcu02 These results confirm our expectations Note also that iVOVSH is the same as VDVle with u and a inter changed Thus the graph of IVoVle as a function with mo is the same as IVOVslH if we change mcon to coou One can get a good bandpass filter by placing an appropriately chosen high pass filter in series with an appropriatelychosen low pass filter The reference given at the beginning of this section also gives single op amp bandpass circuits For both the filters we worked out it turns out that good design requires HOO S 100 Design Procedure We must choose the values for the five R s and C39s involved in our filters Since there are only three parameters H0 00 and CL which characterize the filter we have an underdetermined problem Even when we add the condition that the input impedance must be greater than some value we have too much freedom One approach is to start with values for two of the C s which are convenient in terms of our parts kit and capacitor physical sizes We see what the other components must be for our choice of Hmwo and a If these other values are not reasonable too big or too small then we start the prooedure over again There are other procedures For example for the low pass filter we can simplify the situation by requiring R1R3R4R The casL of this choice is that H0 is forced to be 1 The simplified expression for a and we are 123 7 1 1 no R czc5 Our second simplification is to introduce Cand relate C2 andC5 to it by C2 2 3C 0 0LC C5 T This choice is consistent with the on expression above and that 1 a RC 14 7 We pick R by the input impedance desired and the other choices follow from the required mo and a

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