Experiment 5 for Op-Amp Circuits - Electronic Instrumentation | ENGR 4300, Exams of Engineering

Download Experiment 5 for Op-Amp Circuits - Electronic Instrumentation | ENGR 4300 and more Exams Engineering in PDF only on Docsity! Electronic Instrumentation ENGR-4300 Spring 2006 Experiment 5 K.A. Connor and Susan Bonner 1 of 18 Revised: 1/16/2006 Rensselaer Polytechnic Institute Troy, New York, USA Experiment 5 Op-Amp Circuits Purpose: In this experiment, you will learn about operational amplifiers (or op-amps). Simple circuits containing operational amplifiers can be used to perform mathematical operations, such as addition, subtraction, and multiplication, on signals. They can also be used to take derivatives and integrals. Another important application of an op-amp circuit is the voltage follower, which serves as an isolator between two parts of a circuit. Equipment Required: • HP 34401A Digital Multimeter • HP 33120A 15 MHz Function / Arbitrary Waveform Generator • HP E3631A Power Supply • Protoboard • Some Resistors • 741 op-amp or 1458 dual op-amp Helpful links for this experiment can be found on the links page for this course: http://hibp.ecse.rpi.edu/~connor/education/EILinks.html#Exp5 Part A – Introduction to Op-Amp Circuits Background Elements of an op-amp circuit: Figure A-1 below is a schematic of a typical circuit built with an op-amp. Figure A-1 The circuit performs a mathematical operation on an input signal. This particular op-amp circuit will invert the input signal, V1, and make the amplitude 10 times larger. This is equivalent to multiplying the input by -10. Note that there are two DC voltage sources in addition to the input. These two DC voltages power the op-amp. The circuit needs additional power because the output is bigger than the input. Op-amps always need this additional pair of power sources. The two resistors R3 and R2 determine how much the Electronic Instrumentation ENGR-4300 Spring 2006 Experiment 5 K.A. Connor and Susan Bonner 2 of 18 Revised: 1/16/2006 Rensselaer Polytechnic Institute Troy, New York, USA op-amp will amplify the output. If we change the magnitude of these resistors, we do not change the fact that the circuit multiplies by a negative constant; we only change the factor that it multiplies by. [These components behave much like the parameters of a subroutine.] The load resistor RL is not part of the amplifier. It represents the resistance of the load on the amplifier. Powering the op-amp: The two DC sources, (± VCC), that provide power to the op-amp are typically set to have an equal magnitude but opposite sign with respect to the ground of the circuit. This enables the circuit to handle an input signal which oscillates around 0 volts, like most of the signals we use in this course. (Note the sign on V4 and V5 in the circuit above.) The schematic in figure A-2 shows a standard ± VCC configuration for op-amps. The schematic symbols for a battery are used in this schematic to remind us that these supplies need to be a constant DC voltage. They are not signal sources. Figure A-2 The HP E3631A power supply provides two variable supplies with a common ground (for ± VCC ) and a variable low voltage supply. The power supply jack labeled "COM" between the VCC supplies should be connected to circuit ground. When you supply power to the op-amp, adjust the two voltage levels so that +VCC and - VCC are equal, but opposite in sign, at 15 V. Note that in PSpice, there are two ways to represent a source with a negative sign. Figure A-3 shows the two options: you can either set the voltage source to a negative value, or you can reverse the polarity of the source. = 0 V1 -15V V2 15V 0 Figure A-3 The op-amp chip: Study the chip layout of the 741 op-amp shown in figure A-4. The standard procedure on DIP (dual in-line package) "chips" is to identify pin 1 with a notch in the end of the chip package. The notch always separates pin 1 from the last pin on the chip. In the case of the 741, the notch is between pins 1 and 8. Pin 2 is the inverting input, VN. Pin 3 is the non-inverting input, VP, and the amplifier output, VO, is at pin 6. These three pins are the three terminals that normally appear in an op-amp circuit schematic Electronic Instrumentation ENGR-4300 Spring 2006 Experiment 5 K.A. Connor and Susan Bonner 5 of 18 Revised: 1/16/2006 Rensselaer Polytechnic Institute Troy, New York, USA Experiment The Inverting Amplifier In this part of the experiment, we will wire a very simple op-amp circuit using PSpice and look at its behavior. • Wire the circuit shown in figure A-7 below in PSpice. Figure A-7 o The input should have 100mV amplitude, 1k hertz and no DC offset. o The op-amp is called uA741 and is located in the “EVAL” library. o Be careful to make sure that the + and – inputs are not switched and that the two DC voltage supplies have opposite signs. o Note the location of the input voltage marker. The input to any op-amp circuit goes at Vin which will always be to the left of the input component(s). In this case, R2 is the input resistor, Rin, so the marker goes to its left. • Run a transient simulation of this circuit that displays three cycles. o What does the equation for this type of circuit predict for its behavior? o Use the cursors to mark the amplitudes of the input and output of the circuit. o Calculate the actual gain on the circuit. Is this close to the gain predicted by the equation? o Print out this plot and include it with your report. • Run a transient of the circuit with a much higher input amplitude. o Change the amplitude of the source to 5V and rerun the simulation. o What does the equation predict for the behavior this time? Does the circuit display the output as expected? What happened? o Use the cursors to mark the maximum value of the input and output of the circuit. o What is the magnitude of the output of the circuit at saturation? o Print out this plot and include it with your report. Build an Inverting Amplifier In this part of the circuit, you will build an inverting amplifier. • Build the inverting op-amp circuit in figure A-7 on your protoboard. Electronic Instrumentation ENGR-4300 Spring 2006 Experiment 5 K.A. Connor and Susan Bonner 6 of 18 Revised: 1/16/2006 Rensselaer Polytechnic Institute Troy, New York, USA o Don’t neglect to wire the DC power voltages at pins 4 and 7. o Do not connect pin 4 and 7 to ground. They go through the power supply to ground. o Do not forget to set both the positive and negative values on the DC power supply. One does not automatically set when you set the other. o Do not forget to attach the common ground for the power supply voltages to the ground for the circuit as a whole. • Examine the behavior of your circuit. o Take a picture with the Agilent software of the input and output of the circuit at 1K hertz and 100mV amplitude and include it in your report. o What was the gain of your circuit at this amplitude and frequency? [Use the signals to calculate the gain, not the values of the resistors.] o Increase the amplitude until the op-amp starts to saturate. At about what input amplitude does this happen? What is the magnitude of the output of the circuit at saturation? How does this compare with the saturation voltage found using PSpice? Summary As long as one remains aware of some of their limitations, op-amp circuits can be used to perform many different mathematical operations. That is why collections of op amp circuits have been used in the past to represent dynamic systems in what is called an analog computer. There are some very good pictures of analog computers and other computers through the ages at H. A. Layer’s Mind Machine Web Museum. The link is located on the course links page. Part B – Voltage Followers Background The Voltage Follower: The op-amp configuration in figure B-1 is called a voltage follower or buffer. Note that the circuit above has no resistance in the feedback path. Its behavior is governed by the equation: inout VV = . Figure B-1 If one considers only the equation inout VV = , this circuit would appear to do nothing at all. In circuit design, however, voltage followers are very important and extremely useful. What they allow you to do is completely separate the influence of one part of a circuit from another part. The circuit supplying Vin will see the buffer as a very high impedance, and (as long as the impedance of the input circuit is not very, very high), the buffer will not load down the input. (This is similar to the minimal effect that measuring with the scope has on a circuit.) On the output side, the circuit sees the buffer as an ideal source with no internal resistance. The magnitude and frequency of this source is equal to Vin, but the power is supplied by ± VCC. The voltage follower is a configuration that can serve as an impedance matching device. For an ideal op- Electronic Instrumentation ENGR-4300 Spring 2006 Experiment 5 K.A. Connor and Susan Bonner 7 of 18 Revised: 1/16/2006 Rensselaer Polytechnic Institute Troy, New York, USA amp, the voltages at the two input terminals must be the same and no current can enter or leave either terminal. Thus, the input and output voltages are the same and Zin = Vin/Iin → ∞. In practice Zin is very large which means that the voltage follower does not load down the source. Experiment A Voltage Follower Application In this part, we will investigate the usefulness of a voltage follower using PSpice • Begin by creating the circuit pictured in figure B-2 below in PSpice. Figure B-2 o The source has amplitude of 100mV and a frequency of 1K Hz. o R1 is the impedance of the function generator o R2 and R3 are a voltage divider and R4 is the load on the voltage divider. • Run a simulation that displays three cycles of the input. o Run the simulation, mark the amplitude of the voltages shown, and print the plot for your report. o If we combine R3 and R4 in parallel, we can demonstrate that the amplitude of the output is correct for this circuit. o What if our intention when we built this circuit was to have the input to the 100 ohm resistor be the output of the voltage divider? Ie. We want the voltage across the load (R4) to be ½ of the input voltage. Clearly the relationship between the magnitudes of the 100 ohm resistor and the 1K ohm resistor in the voltage divider will not let this occur. A voltage follower is needed. • Modify the circuit you created by adding an op-amp voltage follower between R3 and R4, as shown in figure B-3 on the next page: Electronic Instrumentation ENGR-4300 Spring 2006 Experiment 5 K.A. Connor and Susan Bonner 10 of 18 Revised: 1/16/2006 Rensselaer Polytechnic Institute Troy, New York, USA Figure C-2 The output of this circuit is the integral of the input INVERTED and amplified by 1/(Rin×Cf). For a sinusoidal input, the magnitude of the gain for this circuit depends on the values of the components and also the input frequency. It is equal to 1/ (ω×Rin×Cf). The circuit will also cause a phase shift of +90 degrees. It is important to remember that there is an inversion in this circuit. For instance, if the input is sin(t), then you would expect the output of an integrator to be -cos(t), a -90 degree phase shift. However, because of the inversion, the output phase shift of this circuit is +90 degrees (-90+180). Also, because the integration of a constant DC offset is a ramp signal and there is no such thing as a real circuit with no DC offset (no matter how small), wiring an ideal integrator will result in an essentially useless circuit. A Miller integrator is an ideal integrator with an additional resistor added in parallel with Cf. It will integrate only at certain frequencies. This distinction is discussed in the power point notes for the course. Experiment Using an op-amp circuit to integrate an AC signal In this section, we will observe the operation of a Miller integrator on a sinusoid. You will examine the way in which the properties of the integrator change both the amplitude and the phase of the input. • Build the integration circuit shown below in figure C-3. V1 should have a 100mV amplitude and 1K Hz frequency. Figure C-3 Electronic Instrumentation ENGR-4300 Spring 2006 Experiment 5 K.A. Connor and Susan Bonner 11 of 18 Revised: 1/16/2006 Rensselaer Polytechnic Institute Troy, New York, USA • Run a transient analysis. o We want to set up the transient to show five cycles, but we also want to display the output starting after the circuit has reached its steady state. Set the run time to 15ms, the start time to 10ms, and the step size to 5us. o Obtain a plot of your results. Just like in mathematical integration, integrators can add a DC offset to the result. Adjust your output so that it is centered around zero by adding a trace that adds or subtracts the appropriate DC value. After you have done this, mark the amplitude of your input and output with the cursors. o Print this plot and include it in your report. • Use the equations for the ideal integrator to verify that the circuit is behaving correctly. o The equation that governs the behavior of this integrator at high frequencies is given by: ∫−≈>> dttvCRtvthenCRif inoutc )( 1)(1 2122 ω o Recall that the integration of sin(ωt) = (-1/ω)cos(ωt). Therefore, the circuit attenuates the integration of the input by a constant equal to -1/ωR1C2. The negative sign means that the output should also be inverted. o What is there about the transient response that tells you that the circuit is working correctly? Is the phase as expected? The amplitude? Above what frequencies should we expect this kind of behavior? • Now we can look at the behavior of the circuit for all frequencies. o Do an AC sweep from 100m to 100K Hz. o Add a second plot and plot the phase of the voltage at Vout. (Either add a p(Vout) trace or add a phase marker at the output of the circuit.) What should the value of the phase be (approximately) if the circuit is working more-or-less like an integrator. Mark the region on the plot where the phase is within +/- 2 degrees of the expected value. o Print this plot o You will mark this sweep with the data from the circuit that you build. • We can also use PSpice to check the magnitude to see when this circuit acts best as an integrator. o Rerun the sweep. Do not add the phase this time. o Using the equation above, we know that at frequencies above fc, Vout = -Vin / (ωRC), where R = R1, C = C2, and ω=2πf. [We plot the negation of the input because the equation for the transfer function of the circuit has an inversion. In a sweep, only the amplitude matters, so the sign is not important.] o Change the plot for the AC sweep of the voltage to show just Vout and -Vin / (ωR1C2). Note that you need to input the frequency ω as 2*pi*Frequency in your PSpice plot. (Pspice recognizes the word “pi” as the value of π and the word “Frequency” as the current input frequency to the circuit. Also note that you must enter numbers for R1 and C2..) o When are these two signals approximately equal? It is at these frequencies that the circuit is acting like an integrator. Mark the point at which the two traces are within 100mV of each other. o Calculate fc=1/(2πR2C2). How close are the amplitudes of the two signals at that frequency? At a frequency much greater than fc, the circuit should start behaving like an integrator. Mark the corner frequency on your plot. o Print this plot. Using an op-amp integrator to integrate a DC signal Another way to demonstrate that integration can be accomplished with this circuit is to replace the AC source with a DC source and a switch. Electronic Instrumentation ENGR-4300 Spring 2006 Experiment 5 K.A. Connor and Susan Bonner 12 of 18 Revised: 1/16/2006 Rensselaer Polytechnic Institute Troy, New York, USA Figure C-4 • Modify your circuit by replacing the AC source with a DC source and a switch as shown in figure C-4. o Note that the switch is set to close at time t=0.01 sec. Use a voltage of 0.1 volts to avoid saturation problems. o The switch is called Sw_tClose and is in the EVAL library. • Analyze the circuit with PSpice. o Do a transient analysis for times from 0 to 50ms with a step of 10us. o Rather than plotting the output voltage (voltage at Vout), plot the negative of the output voltage. You should see that this circuit does seem to integrate reasonably well. o Print this plot. o How close is the output of your circuit to an integration of the input? The integration of a constant should be a ramp signal of slope equal to the constant. The output of an integrating op-amp circuit should be the inversion of the ramp signal multiplied by a constant equal to (1/(R1C2)). Note that since the input is a constant, the output does not depend on the input frequency. o Calculate the approximate slope of the output. Write it on your output plot. Also write the theoretical slope on the plot. For what range of times does it integrate reasonably well? (This is somewhat subjective.) • Modify the feedback capacitor o Decrease C2 to 0.01µF and repeat the simulation. Only run it from 0 to 14ms this time. Don’t forget to plot the negative of the output voltage. o Print your output. o Mark the theoretical slope on the plot. Calculate the theoretical slope of the output. Don’t forget that the constant, (1/(R1C2)), is different because C2 has changed. o Does the circuit integrate -- even approximately -- for any period of time? Can you think of any reason why we might prefer to use a smaller capacitor in the feedback loop, even though the circuit does not integrate as well over as long a period of time? Electronic Instrumentation ENGR-4300 Spring 2006 Experiment 5 K.A. Connor and Susan Bonner 15 of 18 Revised: 1/16/2006 Rensselaer Polytechnic Institute Troy, New York, USA The Differential Amplifier: The circuit in figure D-2 is a differential amplifier, also called a difference amplifier. Its behavior is governed by the following equation: ( )21 VV Rin RfVout −= . Figure D-2 It amplifies the difference between the two input voltages by Rf/Rin, which is the overall gain for the circuit. Note that the ability of this amplifier to effectively take the difference between two signals depends on the fact that it uses two pairs of identical resistances. Also note that the signal that is subtracted goes into the negative input to the op-amp. Be careful with the term “differential”. In spite of its similarity to the term “differentiation”, the differential amplifier does not differentiate its input. Amplifying the output of a bridge circuit: You may recall from Experiment 4, that it was difficult to measure the AC voltage across the output of the bridge circuit because both of the output connections had a finite DC voltage. Without a special probe, the black leads of the scope are always attached to ground. That meant that one could not just connect one of the scope channels across the output, since the scope would short one of the voltages to ground . The differential amplifier allows us to get by this problem, since neither input is grounded. A very large fraction of measurement circuits use some kind of a bridge configuration or are based on some kind of comparison between two voltages. Thus, the operation of the differential amplifier is very important to understand for project 2. Experiment PSpice simulation of an adder • Set up the circuit shown in figure D-3 in PSpice. o Note the polarity of the voltage sources providing +/- Vcc to the op-amp. o Note that the amplitude of the source, V3, is 2V and that the amplitude of the source, V4, is 1V. • Run a simulation that shows 3 cycles of the input. o What is the gain of the adder? What should it do to the two input signals? Is the adder working correctly? o Include this simulation with your report. • Adders are often used as mixers that give different emphasis to each input signal and then combine the inputs together into one signal. What would we have to set R2 to, if we wanted twice as much of the signal from V3 to pass through the adder as the signal from V4. [Note: This does not mean changing Electronic Instrumentation ENGR-4300 Spring 2006 Experiment 5 K.A. Connor and Susan Bonner 16 of 18 Revised: 1/16/2006 Rensselaer Polytechnic Institute Troy, New York, USA nothing because V3 already has twice the amplitude as V4. It means mixing in twice as much of the amplitude of V3 as the amplitude of V4 into the final output signal.] o Modify resistor R2, rerun the simulation, and verify that the output of the signal is as expected. o Include the output of the simulation in your report. Figure D-3 Summary In this experiment we used an adder to add two signals. Then, we modified it so that it would combine the signals with different emphasis, as in an audio mixer. Report and Conclusions The following should be included in your written report. Everything should be clearly labeled and easy to find. Partial credit will be deducted for poor labeling or unclear presentation. Part A (10 points) Include the following plots: 1. PSpice transient of inverting amplifier with input amplitude of 100mV and both traces marked. (1 pt) 2. PSpice transient of inverting amplifier with input amplitude of 5V and both traces marked. (1 pt) 3. Agilent picture of inverting amplifier circuit (1 pt) Answer the following questions: 1. What is the theoretical gain of your inverting amplifier? What gain did you find with PSpice when the input amplitude was 100mV? How close are these? (2 pt) 2. What was the actual gain you got for the inverting amplifier you built? How did this compare to the theoretical gain? to PSpice? (2 pt) 3. What value did you get for the saturation voltage of the 741 op-amp in PSpice? What value did you get for the saturation voltage of the real op-amp in your circuit? How do they compare? (2 pt) 4. At what input voltage did the op-amp in the amplifier you built on the protoboard begin to saturate? (1 pt) Electronic Instrumentation ENGR-4300 Spring 2006 Experiment 5 K.A. Connor and Susan Bonner 17 of 18 Revised: 1/16/2006 Rensselaer Polytechnic Institute Troy, New York, USA Part B (10 points) Include the following plots 1. PSpice transient of voltage divider with 100 ohm load and no voltage follower (1 pt) 2. PSpice transient of voltage divider with 100 ohm load and voltage follower (1 pt) 3. PSpice transient of voltage divider with 1 ohm load and voltage follower (1 pt) 4. PSpice AC sweep of input impedance for the voltage follower. (2 pt) Answer the following questions: 1. Compare the transients of the output with and without the buffer circuit in place. What is the function of the buffer circuit? (2 pt) 2. Why is the follower unable to work properly with a small load resistor? (1 pt) 3. What is the typical value of the input impedance of the voltage follower when it is working properly at low frequencies? (1 pt) 4. Is the magnitude of the input impedance of the voltage follower high enough at high frequencies for it to work effectively? (1 pt) Part C (38 points) Include the following plots: 1. PSpice transient plot of integrator. (1 pt) 2. AC sweep of amplitude (with three experimental points marked) and phase (with three experimental points marked.) The frequency at which the phase gets close to ideal should also be marked. (3 pt) 2. AC sweep plot of integrator voltage and -Vin/ωRC with the location of fc and the place where the voltage gets close to ideal indicated. (2 pt) 3. PSpice plots of integrator with DC source with slope and theoretical slope (if any) indicated on plot. One should be when C2=1uF and the other for C2=0.01uF (2 plots) (2 pt) 4. PSpice plot of ideal integrator (without feedback resistor) (1 pt) 5. Agilent Intuilink pictures of your circuit trace (input vs. output) at 10 Hz, 100 Hz and 1K Hz. (3 plots) (3 pt) 6. Agilent Intuilink pictures of your integrator output with sine wave, triangular wave and square wave inputs (input vs. output) (3 plots) (3 pt) 7. Agilent Intuilink picture of your differentiator output with sine wave, triangular wave and square wave inputs (input vs. output) (3 plots) (3 pt) Answer the following questions: 1. Using the rules for analyzing circuits with op amps, derive the relationship between Vout and Vin for the integrator circuit. (3 pt) 2. Why is the integrator also called a low-pass filter? Take the limits of the transfer function at high and low frequencies to demonstrate this. (3 pt) 3. What are the features of the AC sweep and transient analysis of an integrator that show it is working more-or-less as expected according to the transfer function? For about what range of frequencies does it act like an inverting amplifier? ...like an integrator? (3 pt) 4. Consider the phase shift and the change in amplitude of the output in relation to the input when the circuit is behaving like an integrator. Use the expected change in phase and amplitude (from the ideal equation) to demonstrate that the circuit is actually integrating. (3 pt) 5. Why would we prefer to use the 0.01uF capacitor in the feedback loop even though the circuit does not integrate quite as well over as large a range? (1 pt) 6. What happens when we try to use an ideal integrator? (1 pt) 7. In the hardware implementation, you used a square-wave input to demonstrate that the integrator was working approximately correctly. If it were a perfect integrator, what would the output waveform look like? Is it close? (3 pt) 8. When we built the differentiator, what did the output waveform look like for the square-wave input? What did the differentiator circuit output look like for a triangular wave input? If it were a perfect differentiator, what would the output waveform look like? Is it close? (3 pt)