Lab10_OpAmpFilterInSpice2110-1

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Northeastern University *

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Electrical Engineering

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Apr 3, 2024

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EECE 2510 – Circuits and Signals: Biomedical Applications Lab 10, Spice Analysis of Active Filters Introduction: As discussed in class, Op-amps are useful building blocks in many sensing and measurement applications. To measure the EKG signal, we will be using them to amplify small signals, to reject common-mode signals, and to filter out unwanted low and high- frequency noise and interference. In this lab you will build a model of an Op Amp-based low-pass filter in Spice and test its behavior. 1. In LTSpice, use the 1490 op-amp model in the OpAmps library to build an active low pass filter. Refer back to lab 3 for details. 2. Use R 2 = 100 kΩ, C = 10 nF, and R 1 = 20 kΩ . 3. Use +/- 10V DC power supplies (not shown in diagram below!!!) for the op-amp and connect a voltage source to the input of the filter. Use the advanced button in the source properties to set an 0.1V ac source voltage. Give the filter a 1kΩ load (meaning, connect a 1k resistor from the output of the op-amp to ground). 4. Q1: Document your final circuit with a screen shot. ࠵?(࠵?) = − ࠵? ! ࠵? " ∙ ࠵? # ࠵?࠵? + ࠵? # ࠵? # = 1 ࠵? ! ࠵? |࠵?(࠵?)| = − ࠵? ! ࠵? " ∙ ࠵? # √࠵? ! + ࠵? # !

1. Set up the simulation as follows (using the simulate button and “edit simulation command” on Windows or by going to draft>spice directive on Mac. Set up the simulation for an ac analysis, by decade, 100 points/decade, from 1 Hz to 1,000,000Hz. The command shown on the schematic should be .ac dec 100 1 1000000 2. Run the simulation and create a semilog plot of the response from 1 Hz to 1 MHz. When you run the simulation, you will get a plot window. Next probe the output, and the plot should appear. However, it will probably be in log- log or equivalently decibel (dB) units on the vertical axis so you will have to edit the vertical axis to set it to a linear scale. Note that this is the transfer function, H(ω), except that it is multiplied by 0.1 because the input is 0.1V. Q2: What is the cutoff frequency, f c , (the half power point, or where the voltage is 0.707 times the maximum output voltage)? Is this what you expect, considering the value of w c ? Document the frequency response for your lab report. 3. Q3: What happens if you make your input sinusoid have an amplitude of 0.5 V? Do you get the same transfer function? Why or why not? No screen shot necessary. 4. Time domain view of the filter response: Now a use a pulsed source to produce a 200 Hz square wave with 0.5V amplitude and connect this to the filter input. Note that this procedure is similar to what we did in Lab 4. PULSE(-.25 0.25 0 1e-6 1e-6 .0025 .005 20) As before, use the simulation command on Windows or the spice directive on Mac. The command should be .tran 0 .1 0 1e-6 Use probes to plot both the input and the output voltages . Use the FFT function to display the frequency components for both the input and output waves (you may have to do this in two windows). (note that you should set the vertical axis to a linear scale for the FFT. You can also make the horizontal axis linear and limit the horizontal scale if you like – displaying the FFT up to 8 kHz is about right.

Some notes on Spice source parameters for the square wave: V1 is one of the two voltages, and would be -0.25V for a 0.5V p-p square wave. V2 is the other voltage, + 0.25 for a 0.5V p-p square wave with no DC component. TD is the time delay from the start, which can be set to zero. TR and TF are the rise and fall times of the pulse, which should be significantly smaller than the period, perhaps 1us. If they are too small, PSpice could take a long time to compute the answer, or it might not find a solution. PW is the pulse width, or the time the pulse spends at V2, which should be half of the period for a square wave. PER is the total period of the pulse, which should be double the value of PW for a square wave that has equal times spent in the high and low voltage states. Q4: What do you observe at the input and at the output ? Save/print the output in both the time and frequency domains and try to explain why the output wave looks as it does, rather than just being a square wave. Try thinking in both the time and frequency domains when coming up with your answer. Note that you should have both time and frequency domain plots – the output as a function of time, and the FFT, which gives the magnitude of the frequency components. Hint – it may help you see what is happening if you try inputting square waves with higher and lower frequencies to see what the filter does (10x to 100x higher or lower, maybe). Department of Electrical Engineering, Northeastern University. Last updated: 2/25/20 N. McGruer, 10/24.17 D. Brooks, 10/29/16 D. Brooks,10/24/16, N. McGruer, 80 minutes; 9/30/14 M. Niedre and D. Brooks; 9/30/12, D. Erdogmus and N. McGruer

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