EK307 Active Filters and Steady State Frequency Response

Transcription

1 EK307 Active Filters and Steady State Frequency Response Laboratory Goal: To explore the properties of active signal-processing filters Learning Objectives: Active Filters, Op-Amp Filters, Bode plots Suggested Tools: Function generator, oscilloscope Introduction In this lab assignment, we explore the properties of active filters. In dealing with the concept of each circuit s frequency response, we analyze and test circuits solely in the sinusoidal steady state. As we have learned in class, in the sinusoidal steady state, we can treat resistors, capacitors, and inductors as impedances, wherein resistance comprises but one form of impedance. Using impedances, which often translate to a resistance (or reactance) at a specific frequency, we can utilize all the various circuit techniques from resistive circuit analysis, such as series and parallel combinations, voltage division, and current division. In this context, we can think of a filter as a circuit that operates on each sinusoidal component of the input signal. Prelab: Prelab part one: Draw a schematic of the active low pass filters used in level one. Calculate the Values of R1, R2, and C for the active filter. (Refer to the background material at the end of the lab for the active schematics.) Level 1: Select the appropriate value for R1, R2, and C to result in a desired breakpoint (cutoff) frequency of approximately 1050 Hz and a passband gain of 2. Remember to use practical values for R and C. Practical means not too small or too large. 1kΩ < R < 220kΩ, 100pF <C < 10µF. For VIN frequencies ranging from 10 Hz to 100 khz, observe and record the change in the magnitude ratio VOUT/VIN and phase angle using a function generator and an oscilloscope. Use ten or more points. Be sure some are near the cutoff frequency. Use a sufficiently large voltage amplitude sine wave because your filter will attenuate frequencies above the cutoff. If the amplitude is small to begin with it will quickly disappear into the noise as the frequency goes up. Since we want to know the relative difference between the input (VIN) and output (VOUT), connect one oscilloscope probe to the input of the filter which is the same node as the signal source (function generator) and the other to the output of the OP Amp (VOUT). If the waveforms on your scope are not sinusoidal there is a problem with your setup. You can measure the sine wave magnitude (zero to peak) on each channel of the oscilloscope. Be sure that the scope is in DC coupled mode. You can measure the phase 11/22/2016 1

2 difference by measuring the delay of the zero crossing of VOUT relative to the zero crossing of VIN (Figure 1). Record the data in a spreadsheet and make two plots, one of frequency vs. magnitude (db), the other frequency vs. phase (in degrees). Collect at least 10 different frequency points. One of those frequencies should be at the theoretical breakpoint frequency. You will find the frequency response graph is not very interesting until you are within one decade (factor of 10) of the breakpoint. Therefore it is encouraged to measure more frequency points in the breakpoint region than in the pass band. Use Decibels for your magnitude plot. Note that the decibel scale is defined by db = 20*log10 Vout/Vin, so that db = 0 when Vout/Vin = 1. Phase (degrees) = 360 * Δt / period Or Phase (degrees) = 360 * Δt * Frequency (Hz) Figure 1. Measuring phase difference of two sinusoids of the same frequency using the zero crossing time delay. The phase difference in degrees can be calculated by dividing the time difference by the period then multiplying that quantity times 360. Be mindful of vertical offset between the two waves. Rephrased, the zero (average value) of each sine wave should be the same. Be mindful of units when making this calculation. Level 2: In addition to level 1: Design a second order active low pass filter with the same breakpoint or corner frequency as the low pass filter in level one. It should have a passband gain of 1 and a Q = 1. A Sallen-Key filter is a common implementation of the second order active filter. It is the type listed in the background section of the lab. Here is a link to a good Texas Instruments application note on the filter. Applications notes are often provided by component companies to assist engineers like yourselves in making circuits. They are a great resource that often covers the practical aspects lacking in textbooks and on websites. You may ask why do they provide this information for free? The answer is the application notes usually include the companies components in the designs. For everyone s mutual benefit they want to ensure that you are successful when you use their products. 11/22/2016 2

3 Verify the filter performs by connecting it to the function generator and scope as in level one. Perform the same frequency measurements as in level one. Graph the magnitude and phase results on the same plots as the low pass filters and compare. Background: Passive Filters A passive filter is made solely from resistors, capacitors, and/or inductors. A passive filter can t amplify, nor can it produce power. Rather, it simply allows some frequencies of the input to pass with unchanged amplitude and phase, while attenuating and/or phase-shifting others. The typical passive filter is suitable for driving high-input-impedance or open-circuit loads only. Attaching a low impedance load to the output of a passive filter usually causes the filter s frequency-response properties to change. Consider, for example, the frequency response of the following passive filter circuit. From voltage division in the sinusoidal steady state, If we use this filter to drive a load resistance RL, then the response will change to RP RPC Zin2 where RP = R RL. Similarly, if we attempt to cascade two passive RC filters, as shown below, the net response will not just the simple product of the individual responses, because the input impedance Zin2 of the second filter will load down the first filter: 11/22/2016 3

4 Here, the simple multiplication of the individual responses does not work, because the first filter is loaded by (R2 + 1/jC2) in parallel with C1. Active Filters In contrast to passive filters, an active filter incorporates the properties an op-amp or other integrated circuit capable of driving arbitrary loads. The IC on which the active filter is based is energized by one or more dc sources, thus the active filter can amplify signals and deliver power in addition to filtering the frequency components of its input signal. For example, when an opamp is connected in a negative feedback configuration, its output terminal functions as a dependant voltage source capable of driving low-impedance loads. In this lab, we introduce an entirely new class of op-amp circuit in which capacitors are used as one or more of feedback elements. In such cases, the frequency response of an op-amp activefilter circuit, like its gain Vout/Vin, is independent on op-amp s internal properties. Rather, an active-filter s frequency response depends solely on the values of the R and C components used to make the feedback network. Consider again, for example, the two-filter cascade previously made from R and C only. If we could produce an op-amp circuit having the same frequency response, but capable of driving a load of arbitrary impedance, then it could be used to produce cascades of different op-amp stages. The circuit shown below meets this need. The above equation for Vout/Vin is derived from the formula for an inverting amplifier, but with ZF = ZC ZR2 and ZR1 = R1. This response has precisely the same form as that of the passive RC filter discussed above: 1/(1 + jr2c2). In this case, however, the Vout/Vin ratio is includes multiplication by the op-amp s dc inverting gain R2/R1. If this filter is used to drive a load, the response will be unaffected by the characteristics of the load. Similarly, if we cascade two such filters, as shown below, the response will be given by the simple product of the individual responses: 11/22/2016 4

5 Two-Pole Analog Filters In contrast to simple one-pole filters, for which the response falls off as 1/ away from the breakpoint frequency, the response of a two-pole filter falls off as 1/ 2. Filter Table The op-amp circuits shown on the next page comprise a partial set of two-pole active-filter topologies. 11/22/2016 5

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