NAPIER. University School of Engineering. Engineering Applications Module : SE32101 Active Filter Design 2 nd order Butterworth response

Transcription

1 NAPIER. University School of Engineering Engineering Applications Module : SE3101 nd order Butterworth response C1 4.7n 15V + R1 7.04k R 14.09k In C 4.7n OP1 ua R3 10k -15V Out Sallen and key. R4 10k R 83.16k C1 390pF Multi Feedback. + in R k C.34nF R3 8.87k -15V OP1 ua741 out 15V By Klaus Jørgensen. Napier No Teacher. Dr. Mohammad Y Sharif

2 Abstract.: In this paper there is explained how to calculate a nd order low pass Butterworth response by using a Sallen and key or a MFB (Multi Feedback) filter. There are calculations on how to design the s, there is also simulations of the calculated filters using the program Tina, and the filters is furthermore tested in practical, to se how the theoretical and the practical measurement fits together. Page /1

3 Table of contents.: Introduction.:... 4 Specification for the assignment.:... 4 Order of the Butterworth filter... 5 Calculation of the nd order Sallen and key low-pass filter Simulations of nd order Sallen and key filter in Tina Poles in the Butterworth filter using Sallen and key... 9 Practical test of nd order Sallen and key filter Calculation of the nd order MFB (Multi Feedback) low-pass filter Simulations of nd order Multi Feedback filter in Tina Poles in the Butterworth filter using MFB (Multi Feedback) Practical test of nd order Multi Feedback filter Conclusion Reference Appendix Appendix 1. The measurement from the sallen and key filter Appendix. Frequency response for nd order Sallen and key filter Appendix 3. The measurement from the MFB filter Appendix 4. Frequency response for nd order MFB (Multi Feedback) filter... 1 Page 3/1

4 Introduction.: Active filters are used many places, for example in the telecommunication, in the audio frequency range (0 khz to 0 khz), and for modems to the internet connection, and for many more things. There are many types of active filter response, some of them are Butterworth, Bessel, and Chebyshev, and the filter response can be used in Low-pass, High-pass, Band-pass and Band-stop filter setups. Butterworth response is also named the maximally flat filter, because it is maximally flat in the pass-band response, the Butterworth response is perhaps the mote used type of filter response. The Chebyshev can has a ripple in the pass-band, this ripple can be from 0,01 db and up to 3dB ripple, in return for the ripple in the pass-band the Chebyshev response has very good cutoff at the edge of the pass-band as it is showed in figure. [4] Figure 1 shows the poles in Butterworth, Bessel, and Chebyshev filters, there is no zeroes in this filter types. jω Bessel Butterworth Chebyshev σ Figure 1 Figure [4] Specification for the assignment.: Butterworth response. -3dB frequency = 3.4KHz. Stop-band frequency =6KHz. Stop-band attenuation 10dB. Pass-band gain =. Op-amp 741. Page 4/1

5 Order of the Butterworth filter. Before the values off the components in the Butterworth filter can be calculated the order off the filter must be found. H(dώ) Passband Transition -band Stopband S ώ C ώ ώ Figure 3 [] Order off the filter = N [1] 1 Log 1 S N ω * Log ωc ώ C = 3,4KHz. ώ = 6KHz. 10 = 10 = = Log( S ) db S Log 316, 3m. 0* 1 Log 1 1 Log S 1 316,3m 954,4m N = 1,93 ω 6K 493,35m * Log * Log ω 3,4K C The filter is calculated to be a nd order filter, from the calculation above. There will be tow poles in this filter. nd Page 5/1

6 Calculation of the nd order Sallen and key low-pass filter. nd order Sallen and key filter calculations. [1] C1 = C and is chosen to 4,7nF C 1n = C n and is chosen to 1F C C = and C M *ω 1n 1 C C = M *ω n M = = 9,96K C1* ω π * fc * C1 π *3,4K * 4,7n 1 C 1 R1 = * M *9,96K = 7,043KΩ 6,8KΩ E1 R = * M *9,96K = 14,086KΩ 15KΩE1 R3 = R4 and is chosen to be 10KΩ Au = 0 * Log( gain) 0 * Log( ) = 6dB C The diagram there is shown in figure 4 was simulated in the program Tina, with the calculated value, and the outcome is shown in figure 5, the -3dB frequency 3,4KHz and -10,5dB at 6KHz. C1 4.7n 15V + R1 7.04k R 14.09k In OP1 ua Out C 4.7n -15V R3 10k Figure 4 R4 10k Page 6/1

7 Simulations of nd order Sallen and key filter in Tina. Figure 5 shows the gain in db to the frequency, on the output of the filter (figure 4) T a b dB / gain Gain (db) KHz / -3dB 6KHz / -10.5dB k 10.00k k 1.00M Frequency (Hz) Figure 5 Figure 6 shows the gain between the signal In and the signal Out, and the gain is. Input signal : 1KHz / V peek-peek sine wave. Au Uout = peek peek Uinpeek peek 4 = T.00 Signal In Signal Out 1.00 Voltage (V) m 4.00m 6.00m 8.00m 10.00m Time (s) Figure 6 Page 7/1

8 Signal In : 300Hz / V peek-peek Square wave. T Signal In Signal Out 1.00 Voltage (V) m 4.00m 6.00m 8.00m 10.00m 1.00m 14.00m 16.00m 18.00m 0.00m Time (s) Signal In : 3KHz / V peek-peek Square wave. Figure 7 T Signal In Signal Out 1.00 Voltage (V) m.00m 3.00m 4.00m 5.00m Time (s) Figure 8 In figure 7 the output signal is similar to the input signal on 300Hz because the frequency is much lower then the -3dB point at 3,4KHz and the harmonic frequency is getting through the filter. In figure 8 the input signal is a square wave on 3KHz and the harmonic frequency to 3KHz square wave signal is not getting through the filter, and therefore is the signal out of the filter is a sine wave near by 3KHz. Page 8/1

9 Poles in the Butterworth filter using Sallen and key. In a is there no zeros only poles and there is as many poles as the filter order. nd order filter = poles 4 th order filter = 4 poles The position of the poles can be calculated by using a program like MathCAD. 1. The value of the components in the filter has to be defined in MathCAD. R1 := R := R3 := R4 := ( ) ( ) C1 := C := Then the transfer function has to be found, and it can be found by using Tina. W( s) := R4 R3 R4 + ( C1 R3 R1 C R4 R1 C R4 R) s C C1 R4 R R1 s 3. IT is only the part under the division linie of transfer function there has to be used because there are no zeros in a Butterworth filter. The equation from under the linie has to be set equal to And from that MathCAD can calculate the expression for s. R4 + ( C1 R3 R1 C R4 R1 C R4 R) s C C1 R4 R R1 s 0 1 C1 R3 R1 CC1R4RR1 s := 1 C1 R3 R1 CC1R4RR1 C R4 R1 C R4 R1 1 C R4 R ( C1 R3 R1 C1R3R1 C R4 C1R3R1CR4R + C R4 R1 + C R4 R1 R + C R4 R 4CC1R4 + R R1) 1 C R4 R ( C1 R3 R1 C1R3R1 C R4 C1R3R1CR4R + C R4 R1 + C R4 R1 R + C R4 R 4CC1R4 R R1) 5. And from that the imaginary and the real part can be calculated. In Tina a visual plot of the zeros can be simulated by using an ideal op-amp shown in figure 9. T 0.00k s = i i 10 Re = 14,38K + 15,08K Im = 14,38K 15.08K 4 Imaginary part 15.00k 10.00k 5.00k k k k -0.00k -0.00k k k -5.00k k 10.00k Real part Figure 9 Page 9/1

10 Practical test of nd order Sallen and key filter. The circuit on page 6 figure 4 was build on a breadboard, to see how it would work in practices, and the result is shown in figure 10. The practical result is far from the result as the simulation in Tina gave. At about -15dB the curve should have been about 10KHz and NOT about 50KHz as it is in figure 10. The reason for the bump on figure 10 can be several things first of all the resistance there is used in the practical test doesn t have the same, value as the resistance there is used in the simulation in Tina, in the practical test there is used the E1 value of the resistance. Another reason is the capacitors there is used to test with has an accuracy of ±0%. The breadboard there is used in the test is not that good db Frequency. The test setup is showed in figure 11. Frequency Generator Figure10 Oscilloscope Power-supply ±15V Signal In Signal Out Test board Filter. +15V -15V Figure 11 The measurement for figure 10 can be found in appendix 1. Page 10/1

11 Calculation of the nd order MFB (Multi Feedback) low-pass filter. The Multi Feedback filter in figurer 1 has the same specifications as the Sallen and key filter. nd order Multi Feedback filter calculations [4]. C1 is chosen to be 0.5nF chosen to 390pF a1 = 1,414 b1 = 1 C 4* b1* a1 ( 1 Au) 4*1* ( 1 ( ) ) C1* 390 p * =.34nF, nf E6 a1* C R = 1,414*.34n 1,414 [( 4* b1* C1* C* ( 1 Au) ) ( a1 * C )] 4* π * fc * C1* C [( 4*1*390 p *,34n * ( 1 ( ) )) ( 1,414 *,34n )] 4* π *3,4k *390 p *,34n = 83,16kΩ 8kΩ E1 R R R Au 83.16k 1 = = 41,48kΩ 39kΩE1 b1 * C1* C* R 1 *390 p *,34n *83,16k 3 = = 8,17kΩ 7kΩ E1 4* π * fc 4* π *3,4k R 83.16k C1 390pF + in R k C.34nF R3 8.87k -15V OP1 ua741 out Figure 1 15V Page 11/1

12 Simulations of nd order Multi Feedback filter in Tina. T a b dB / gain ,4KHz / -3dB 6KHz / dB Gain (db) k 10k 100k 1M Frequency (Hz) Figure 13 Figure 13 shows the gain in db to the frequency, on the output of the filter. Input signal : 1KHz / V peek-peek sine wave. T.00 Signal In Sinal Out 1.00 Voltage (V) m 5.00m 7.50m 10.00m Time (s) Figure 14 shows the gain between the signal In and the signal Out, and the gain is. Uout peek peek 4 Au = Uin = peek peek Figure 14 Page 1/1

13 Signal In : 300Hz / V peek-peek Square wave. T Signal In Signal Out Voltage (V) m 10.00m 15.00m 0.00m 5.00m 30.00m Time (s) Figure 15 Signal In : 3KHz / V peek-peek Square wave. T Signal In Signal Out m Voltage (V) m u 1.00m 1.50m.00m.50m 3.00m 3.50m 4.00m 4.50m 5.00m Time (s) Figure 16 In figure 15 the output signal is similar to the input signal on 300Hz because the frequency is much lower then the -3dB point at 3,4KHz and the harmonic frequency is getting through the filter. In figure 16 the input signal is a square wave on 3KHz and the harmonic frequency to 3KHz square wave signal is not getting through the filter, and therefore is the signal out of the filter is a sine wave on 3KHz. Page 13/1

14 Poles in the Butterworth filter using MFB (Multi Feedback). The poles is calculated the same way as with the sallen and key filter on page 9 1. The value of the components in the filter has to be defined in MathCAD. R1 := R := R3 := C1 := C := Then the transfer function has to be found, and it can be found by using Tina. w( s) := R1 R + ( R R1 R3 R1 R3 R) C1 s C C1 R3 R R1 s 3. IT is only the part under the division linie of transfer function there has to be used because there are no zeros in a Butterworth filter. The equation from under the linie has to be set equal to 0. R1 + ( R R1 R3 R1 R3 R) C1 s C C1 R3 R R1 s 0 4. And from that MathCAD can calculate the expression for s. 1 C1 CC1R3RR1 s := 1 C1 CC1R3RR1 R R1 R R1 C1 R3 R1 C1 R3 R1 1 C1 R3 R ( C1 R R1 + C1 R R1 R3 + C1 R R1 R3 + C1 R3 R1 + C1 R3 R1 R + C1 R3 R 4CC1R3RR1 + ) 1 C1 R3 R ( C1 R R1 + C1 R R1 R3 + C1 R R1 R3 + C1 R3 R1 + C1 R3 R1 R + C1 R3 R 4CC1R3RR1 ) 5. And from that the imaginary and the real part can be calculated. In Tina a visual plot of the zeros can be simulated by using an ideal op-amp. shown in figure s = i i 10 Re = 15,11K + 15,1 K Im = 15,11K 15,1 K 4 4 T Imaginary part 0.00k 15.00k 10.00k 5.00k k k k -0.00k -0.00k k k -5.00k k 10.00k Real part Figure 17 Page 14/1

15 Practical test of nd order Multi Feedback filter. The circuit shown in figure 1 on pages 11 was build on a breadboard with E1 values. And the result is shown in figure 18 and the result of the practical measurement is close to the Tina simulation in figure 13 on page db Frequency. Figure 18 The test setup is showed in figure 19 Frequency Generator Oscilloscope Power-supply ±15V Signal In Signal Out Test board Filter. +15V -15V Figure 19 The measurement for figure 18 can be found in appendix 3 Page 15/1

16 Conclusion. In the test of the sallen and key filter there is showed on page 10 figure 10 is the cut-off frequency on 3,4KHz not at -3dB and on of the reasons can also be that the capacitor is chosen to be 4,7nF, and the value of the capacitor maybe to large, in setoff 4,7nF the value shout perhaps been chosen to a value on 1nF, that maybe had given another output of the sallen and key circuit. The output of the MFB (Multi Feedback) filter on page 15 figure 18 looks a lot like the simulation from Tina on page 1 figure 13 but it is not quite as good as the simulation but it is acceptable, on of the reasons for this may be that the capacitor was chosen to a value of 1nF, and not 4,7nF as in the sallen and key circuit. If the capacitor in the MFB filter was chosen to a value of 4,7nF in setoff 1nF the test may had looked different. Klaus Jørgensen 19 November 004 Page 16/1

17 Reference. 1. Hand out by Mohammad Y Sharif Klaus Jørgensen. Passive and Active filters, theory and implementations by Wai-kai Chen, University of Illinois at Chicago. ISBN: Handbook by G. S. Moschytz and Petr Horn ISBN: Chapter 16 Techniques. PDF document from Texas Instruments. etype=pdf 5. The Electrical Engineering Handbook. Chapter 9 Active Filters Page 17/1

18 Appendix. Klaus Jørgensen Appendix 1. The measurement from the sallen and key filter. Uout pek pek db = 0 * Log UinPek pek Hz db Uin pek-pek Uout pek-pek Page 18/1

19 Appendix. Klaus Jørgensen Frequency response for nd order Sallen and key filter. Page 19/1

20 Appendix 3. The measurement from the MFB filter. Uout pek pek db = 0 * Log UinPek pek Hz db Uin pek-pek Uout pek-pek Page 0/1

21 Appendix 4. Frequency response for nd order MFB (Multi Feedback) filter. Page 1/1