Op-amp Circuits: A Simple Multiplication Example and Integrator-Differentiator Experiment, Exams of Engineering

Download Op-amp Circuits: A Simple Multiplication Example and Integrator-Differentiator Experiment and more Exams Engineering in PDF only on Docsity! Electronic Instrumentation ENGR-4300 Spring 2005 Experiment 5 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: Below is a schematic of a typical circuit built with an op-amp. 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 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. K.A. Connor and Susan Bonner 1 of 21 Revised: 12/3/2020 Rensselaer Polytechnic Institute Troy, New York, USA Electronic Instrumentation ENGR-4300 Spring 2005 Experiment 5 [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 below 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. 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: = 0 V 1 - 1 5 V V 2 1 5 V 0 The op-amp chip: Study the chip layout of the 741 op-amp. 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 diagram. The ±VCC connections (7 and 4) MUST be completed for the op-amp to work, although they usually are omitted from simple circuit schematics to improve clarity. K.A. Connor and Susan Bonner 2 of 21 Revised: 12/3/2020 Rensselaer Polytechnic Institute Troy, New York, USA Electronic Instrumentation ENGR-4300 Spring 2005 Experiment 5 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 below in PSpice 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 above on your protoboard. K.A. Connor and Susan Bonner 5 of 21 Revised: 12/3/2020 Rensselaer Polytechnic Institute Troy, New York, USA Electronic Instrumentation ENGR-4300 Spring 2005 Experiment 5 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 below 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  . 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- amp, the voltages at the two input terminals must be the same and no current can enter or leave either K.A. Connor and Susan Bonner 6 of 21 Revised: 12/3/2020 Rensselaer Polytechnic Institute Troy, New York, USA Electronic Instrumentation ENGR-4300 Spring 2005 Experiment 5 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 below in PSpice. 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 on the next page: K.A. Connor and Susan Bonner 7 of 21 Revised: 12/3/2020 Rensselaer Polytechnic Institute Troy, New York, USA Electronic Instrumentation ENGR-4300 Spring 2005 Experiment 5  Build the integration circuit shown below. V1 should have a 100mV amplitude and 1K Hz frequency.  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:  dttvCR tvthen CR if inoutc )( 1 )( 1 2122  o Recall that the integration of sin(t) = (-1/)cos(t). Therefore, the circuit attenuates the integration of the input by a constant equal to -1/R1C2. 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 K.A. Connor and Susan Bonner 10 of 21 Revised: 12/3/2020 Rensselaer Polytechnic Institute Troy, New York, USA Electronic Instrumentation ENGR-4300 Spring 2005 Experiment 5 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 / (RC), where R = R1, C = C2, and =2f. [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 / (R1C2). Note that you need to input the frequency  as 2*pi*Frequency in your PSpice plot. (Pspice recognizes the word “pi” as the value of  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/(2R2C2). 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.  Modify your circuit by replacing the AC source with a DC source and a switch as shown. 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. K.A. Connor and Susan Bonner 11 of 21 Revised: 12/3/2020 Rensselaer Polytechnic Institute Troy, New York, USA Electronic Instrumentation ENGR-4300 Spring 2005 Experiment 5 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. Does it integrate best at lower or higher frequencies? For what range of frequencies does it integrate reasonably well? (This is somewhat subjective.)  Modify the feedback capacitor o Decrease C2 to 0.01F 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 over as wide a range of frequencies?  Create an ideal integrator o The circuit we have been looking at is a Miller integrator. An ideal integrator does not have an extra resistor in the feedback path. What would happen if we changed our circuit to an ideal integrator? o Set the feedback capacitor back to its initial value of 1uF. Remove the resistor from the feedback loop and run your transient analysis again. o You should see that the circuit no longer works. Negate the output voltage again. o Print your output. o What is wrong with the output? The ideal integrator circuit will operate on both the AC and DC inputs. In any real circuit -- no matter how good your equipment is – noise will create a small variable DC offset voltage at the inputs. The problem with this circuit is that there is no DC feedback to keep the DC offset at the input from being integrated. Therefore, the output voltage will continuously increase and, in addition, it will be amplified by the full intrinsic gain of the op-amp. This immediately saturates the op-amp. Building an op-amp integrator and an op-amp differentiator In this part of the experiment, we will build an op-amp integrator and an op-amp differentiator on the protoboard and look at the output for a variety of inputs.  Build the op-amp integrator circuit as shown: K.A. Connor and Susan Bonner 12 of 21 Revised: 12/3/2020 Rensselaer Polytechnic Institute Troy, New York, USA Electronic Instrumentation ENGR-4300 Spring 2005 Experiment 5 Part D – Amplifying the Strain Gauge Signal Background Op-Amp Adders: The figure below shows an adder. Its behavior is governed by the following equation:        2 2 1 1 R V R V RfVout . The gain for each input to the adder depends upon the ratio of the feedback resistance of the circuit to the value of the resistor at that input. The adder is sometimes called a weighted adder because it provides a means of multiplying each of the inputs by a separate constant before adding them all together. It can be used to add any number of inputs and multiply each input by a different constant. This makes it useful in applications like audio mixers. The Differential Amplifier: The circuit below is a differential amplifier, also called a difference amplifier. Its behavior is governed by the following equation:  21 VV Rin Rf Vout  . 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. Experiment PSpice simulation of an amplified bridge circuit In this part of the experiment, we will simulate a circuit which uses an op-amp circuit to amplify the output from our strain gauge bridge.  Build the model of an amplified bridge circuit shown below in PSpice o In the circuit on the next page, the five components: V2, R9, R8, R7 and R6 represent the bridge. There were two legs to the bridge: one consisting of a fixed resistor (R9) and the strain gauge (R8) and the other consisting of two resistors R7 and R6. The final component in the strain gauge bridge is the 5V DC power supply (V2). Since R9, R8, R7 and R6 are all 1K. The bridge in this circuit is always balanced. o The resistor R5 represents the scope impedance. o All the other components (R1, R2, R3, R4, V3, V4 and U1) are the differential amplifier. K.A. Connor and Susan Bonner 15 of 21 Revised: 12/3/2020 Rensselaer Polytechnic Institute Troy, New York, USA Electronic Instrumentation ENGR-4300 Spring 2005 Experiment 5  Add a model for the oscillating signal from the strain gauge o Unfortunately, it is not possible to simulate the operation of the strain gauge directly using PSpice since there is no simple way to make the resistor representing the strain gauge oscillate with time. We can, however, add some components to the bridge to produce the kind of voltages we observed in the actual circuit. o The voltage we want to observe at the node between R9 and R8 has both a DC level of about 2.5 volts (provided already by the bridge) and a small AC signal that oscillates at a frequency of about 20 Hz. We can create this effect by adding an AC source to the circuit, as shown below. (Set the amplitude of V1 to 100mV and the frequency to 20 Hz.)  The source V1 and the resistor R10 represent the function generator. We have also incorporated a DC blocking capacitor, C1, into the circuit. This ensures that the DC voltage at the node between R9 and R8 will not be seen by the function generator. A capacitor is an open circuit at DC (frequency = 0). It will prevent the DC level on one side from causing a change in the DC level on the other side. It is like a gate in a K.A. Connor and Susan Bonner 16 of 21 Revised: 12/3/2020 Rensselaer Polytechnic Institute Troy, New York, USA Electronic Instrumentation ENGR-4300 Spring 2005 Experiment 5 canal lock. The water level exists at two different levels on either side of the gate. If the gate was not there, the water would mix together and reach some intermediate level. If the capacitor was not in the circuit, the DC offset between R8 and R9 would be somewhere between 0 (the DC offset of the source) and the desired offset of 2.5 V at the center of the voltage divider.  Remember there is no function generator in the actual bridge circuit. We are using one to model the oscillation of the beam in this circuit because PSpice has no component for a cantilever beam with a strain gauge on it.  Run the simulation  Set up and run a transient simulation that displays 4 cycles of the output. The input frequency is only 20 hertz, so you will need to think about how much time you will need.  Place voltage markers at the two inputs to the differential amplifier (Vs and Vp). Also place a marker at the output (Vout).  To examine whether the circuit is behaving as it should, add a trace equal to the difference between the two input signals to the amplifier (Vs-Vp).  Mark the amplitude of the output trace and the amplitude of the difference trace.  Print this plot.  What is the theoretical gain of this circuit? What is the gain you found with PSpice? How close are they?  In this circuit, you will notice that R1=R2 and R3=R4. It is necessary that these resistors be the same or the differential amplifier will not work properly. What would happen if we change them? o Change the values of the resistor R2 to 1.5 k. o Rerun the transient simulation. o Print out this plot and describe what happened to the output voltage. That is, how did this voltage change and why.  Add a potentiometer o Now set R2 back to 1k and replace the two 1K resistors in the potentiometer leg with a pot, as shown below. The default resistance of the pot is 1K and the default value of the set parameter is 0.50. This means that if the pot were redrawn as a voltage divider, the top and bottom resistors would both be 500 ohms. K.A. Connor and Susan Bonner 17 of 21 Revised: 12/3/2020 Rensselaer Polytechnic Institute Troy, New York, USA Electronic Instrumentation ENGR-4300 Spring 2005 Experiment 5 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. (1 pt) 2. AC sweep plot of integrator voltage and -Vin/RC with the location of fc and the place where the voltage gets close to ideal indicated. (1 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) (0.5 pt) 5. Agilent Intuilink pictures of your circuit trace (input vs. output) at 10 Hz, 100 Hz and 1K Hz. (3 plots) (1 pt) 6. Agilent Intuilink pictures of your integrator output with sine wave, triangular wave and square wave inputs (input vs. output) (3 plots) (1 pt) 7. Agilent Intuilink picture of your differentiator output with sine wave, triangular wave and square wave inputs (input vs. output) (3 plots) (1 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. (2 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. (1 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 a filter? ...like an integrator? (2 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. (2 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? (1 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? (1 pt) Part D Include the following plots: 1) Transient simulation from PSpice with R1=R2=1K. Indicate the amplitudes of the input and output clearly on the plot. Write on the plot what the gain is and how you found it. (0.5 pt) 2) Transient simulation from PSpice with R1=1K and R2=1.5K. (0.5 pt) 3) Transient simulation from PSpice with R1=R2=1K and a 1K pot. The plot should have the value of set that balanced the bridge written on it. (1 pt) 4) Agilent Intuilink software plot of the scope traces for channel 1 and channel 2 of the circuit you built. Clearly mark which trace corresponds to the signal directly from the strain gauge and the output of the differential amplifier. Show how you calculated the gain. (1 pt) Answer the following questions: 1) What is the theoretical gain of the amplifier you modeled in PSpice (all resistors balanced)? (1 pt) 2) What overall gain did you find for the basic PSpice differential amplifier in plot 1)? Why does the output look correct? (1 pt) 3) Describe the most significant difference between the output of the circuit with the balanced input resistors and the output when they were unbalanced. Explain why you think this has happened using the derivation of the equations for the differential amplifier given in class. (2 pt) K.A. Connor and Susan Bonner 20 of 21 Revised: 12/3/2020 Rensselaer Polytechnic Institute Troy, New York, USA Electronic Instrumentation ENGR-4300 Spring 2005 Experiment 5 4) Redraw the 1k pot in the balanced bridge of plot 3) as two resistors in a voltage divider. Based on the value of your set parameter, provide values for the upper and lower resistances. (1 pt) 5) What overall gain did you find in the circuit that you built on your protoboard? How does the actual gain compare to the theoretical gain of the amplifier you built? ... to the gain in the PSpice simulation? Name two things that can account for the discrepancy. (2 pt) 6) Give an example of a system (electrical, mechanical, chemical or some combination) with negative feedback and an example of a system with positive feedback. (1 pt) Summarize Key Points (1 pt) Mistakes and Problems (1 pt) Member responsibilities (1 pt) Total: 45 points for report Attendance: 3 classes (5 points) 2 classes (3 points) 1 class (0 points) out of 5 possible points No attendance at all = No grade for experiment. K.A. Connor and Susan Bonner 21 of 21 Revised: 12/3/2020 Rensselaer Polytechnic Institute Troy, New York, USA