PHYS 536 Active Filters

Transcription

1 PHYS 536 Active Filters Introduction Active filters provide a sudden change in signal amplitude for a small change in frequency. Several filters can be used in series to increase the attenuation outside the break frequency. High pass, low pass, band pass, Butterworth, and Chebyshev filters are investigated in this experiment. Two methods of filter design are considered: Convenient time-constants are selected. Then the closed-loop gain (G 0 ) is adjusted to obtain the desired frequency response. The closed-loop gain is selected, and then the time constants are adjusted to obtain the desired frequency response. The Low Pass Filter The low pass circuit is shown in Fig.. Theclosedloopgainforthiscircuitis Figure : The low pass filter

2 G 0 = R 3 + R 4 R 4 () Gain Adjustment Method for the Low Pass Filter To obtain the required component values, set R = R 2 and C = C 2. Next, select a convenient capacitor value. The resistor is then calculated to obtain the desired break frequency. This value is given by R = (2) 2πf n f c C Next, R 4 is selected for convenience and R 3 is calculated to adjust G 0 to the value needed to obtain the desired frequency response (i.e. Bessel, Butterworth, Chebyshev, etc.). R 3 = R 4 (G 0 ) (3) G 0 and f n are obtained from the table of parameters given at the end of the lab instructions. In this table, G 0 is labeled K. For a Butterworth filter, f n =. Multistage Filters Better filter performance can be obtained by including more R-C attenuators. (Each RC attenuator contributes one pole to the filter.) Only two RC attenuatorscanbeusedwithoneop-amp,hence better filters have several op amps in series. For example, an 8-pole filter would use 4 op amps. The individual op amps in the filter do not have the same frequency response as a 2-pole filter of the same type, as shown by the parameters given in the table at the end of the lab. Each op amp in a multistage filter has a different frequency response, but their combined effect is closer to an ideal filter, i.e. no attenuation below the break and no gain above the break. The gain expression for a multistage Butterworth filter is G = G 0 ( f f c ) 2n + (4) where n is the number of poles in the total filter. The gain is approximately ( ) constant (G = G 0 ) below the break frequency (f c ), because the term f f c is very small when n is large and f<f c.thesameterm makes the gain decrease rapidly when f>f c. 2

3 Time Constant Adjustment Method for the Low Pass The gain can be set to any convenient value. In this example, G 0 = is obtained by eliminating R 4 and using a direct connection for R 3. Then the filter acts as a follower below the break frequency. For a 2 pole Butterworth filter we can use and C 2 =2C (5) R = R 2 (6) High Pass Filters R = The high pass filter is shown in Fig πf c C (7) Figure 2: The high pass filter Gain Adjustment Method for the High Pass Gain adjustment method. Set R = R 2 and C = C 2. A convenient capacitor size is selected and the resistor size is calculated to obtain the desired breakfrequency. 3

4 R = The gain is the same as the low pass case. R 3 is given by 2πf n f c C (8) R 3 = R 4 (G 0 ) (9) G 0 and f n are taken from the table at the end of the lab. The fc f term in the Butterworth gain expression is inverted relative to the low pass filter. G = ( fc f G 0 ) 2n + (0) where n is the total number of poles in the filter. In this case, the gain drops rapidly when f is less than the break frequency. Band Pass Filter The band pass filter is shown in Fig. 3. Figure 3: The band pass filter The gain of this circuit is a maximum (G r ) at the resonant frequency f r. 4

5 G r = R 2 R C 2 C + C 2 + G0 A () f 2 r = R + R 3 + G0 A (2π) 2 τ τ 2 C2 C A (2) where τ j = R j C j (3) A is the open loop gain of the op amp at the resonance frequency, f T is the gain frequency product of the op amp. A = f T f r (4) G 0 = R 2 χ c2 =2πf r C 2 R 2 (5) When the open loop gain is large, the two terms containing A can be neglected. f 2 r = R + R 3 (2π) 2 (6) τ τ 2 Therefore, the two time constants set the minimum resonance frequency, but f r can be increased by making R 3 smaller than R. The band-pass (B) is defined as the frequency interval between the high and low break-frequencies (i.e., the frequencies at which gain is decreased by 30%). B = C 2πR 2 C 2 (7) C +C 2 A CA340 op amp will be used for all circuits. The power supply connections (±5 V)should be connected to each circuit. 5

6 Low Pass Filters. Use the values R, C,andC 2 given in Table to calculate the values of R, R 2,andR 3 needed to obtain the specified break frequency f c using the gain adjustment method. 2. Calculate the gain of the filter at 0.5,, 2, 4, and 0 times f c. 3. Build the circuit in Fig.. Measure the break frequency and gain at 0.5, 2, 4, and 0 times the measured break frequency. Use a 0 Vpp sine wave input signal. 4. Use the time constant adjustment method for a two pole, G 0 = Butterworth filter. Use the values of C given and calculate the values of R, R 2,andC 2 needed to obtain the specified f c and Butterworth frequency response. 5. Calculate G at 0.5,, 2, 4, and 0f c. 6. Set up the circuit using the values calculated in step 4 of this section. Measure the break frequency and gain at 0.5,, 2, 4, and 0 times the measured break frequency. Use a 0 Vpp sine wave. 7. Use the gain adjustment method for a four pole Chebyshev (0.5 db) filter. Use the components given in Table. Calculate R, R 2,andR 3 for both stages and total gain. You do not need to build this filter. 2 High Pass Filter. Use the values of R 4, C,andC 2 given to calculate the values of R, R 2, and R 3 needed to obtain the specified break frequency. 2. Calculate the gain of the filter at 2,, 0.5, 0.25, and 0. times f c. 3. Set up the circuit shown in Fig. 2 as designed in the first step of this section. Measure the break frequency and the gain at 2,, 0.5, 0.25, and 0. times the measured break frequency. Use a 5 Vpp sine wave for the input signal. 3 Band Pass Filters. Assume the open loop gain is large enough that A terms may be neglected. Calculate the approximate resonant frequency f r. 2. Use the approximate resonant frequency to calculate f r including the A terms. 6

7 Step R R 2 R 3 R 4 C (nf) C 2 (nf) f c (khz) f n G 0..3 C C C 5.K 2 R.4.6 C C D X C 2.7 C C C 5.K 2 R R C C C 5.K 2 R K 00K X X K 00K 200 X Table : Component values. X = not included, D = direct connection, C = calculated value, R = read from table. f T = 3.7 MHz for a CA Calculate the gain at resonance, G r, and the bandwidth B. 4. Set up the circuit shown in Fig. 3 using the values given in the Table??. Measure f r, G r,andb. B is the interval between the two frequencies when G =0.7G r. The input signal should be small so that v o is not too large at resonance. v i = 20 mvpp is appropriate. Be sure to watch the input voltage as frequency is changed to ensure that v i is constant. 5. The resonance will shift to a higher frequency when R 3 is included. Calculate the change in f r. 6. Add R 3 given in the table and measure f r and G r. The decrease in G r is produced by inaccurate components and a low quality factor ( A G 0 ). Components Required. Op Amps: () CA Resistors: () 200 Ω, () kω, (2) 9. kω, () 3 kω, () 5. kω, (2) 3 kω, () 00 kω 3. Capacitors: (2) 0. μf, (2) nf, (2) 2 nf, () 5 nf, () 0 nf 7

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