EEE 118 1S2425 Module 4 Study Guide PDF

University of the Philippines EEE118 1S2425 MODULE 4 : OPERATIONAL AMPLIFIERS Learning Outcomes After going through this module, you should be able to: 1. Understand the basics of the operational amplifier. 2. Understand the operation and use of basic...

University of the Philippines EEE118 1S2425 MODULE 4 : OPERATIONAL AMPLIFIERS Learning Outcomes After going through this module, you should be able to: 1. Understand the basics of the operational amplifier. 2. Understand the operation and use of basic operational amplifier circuits: follower, differential, summing, differentiator and integrator, comparator, i-v converter, negative resistance 3. Compare and understand the differences between simulation and breadboard circuit results An Introduction to Operational Amplifiers Operational amplifiers or op amps are voltage amplifying devices designed to be used with components like resistors, and capacitors connected to its terminals. Operational amplifiers had their origins in analog computers, where they were used to perform mathematical operations in many linear, non-linear, and frequency dependent circuits. We can build basic circuits that invert, add, and/or multiply voltages using operational amplifiers Basic I/O of an operational amplifier V+ = Non-Inverting Input V- = Inverting Input Vs+ = Positive Power Supply Pin Vs- = Negative Power Supply Pin Vout = Output The basic operational amplifier construction is of a 3-terminal device, with 2-inputs and 1-output (excluding power connections). 1 University of the Philippines EEE118 1S2425 Operational Amplifier Model As an active element, the op amp must be powered by a voltage supply, typically between 5 to 24V, although power supplies are often not shown in op amp circuit diagrams for simplicity. An operational amplifier follows these main concepts: 1. It has a very high input impedance. The ideal operational amplifier has an infinite input impedance, and no current flows into either of its two inputs. 2. It has a very low output impedance. The ideal operational amplifier has zero output impedance. 3. Op-amps sense the difference between voltage signals applied to their two input terminals and then multiplies it by some predetermined gain called the open loop gain. 4. The quantitative relationship between the input and output of the operational amplifier can be captured by the equation Vout=AOL(V+ - V-) where AOL is the open loop gain. For an ideal operational amplifier, the open loop gain is said to be infinite. 0. Closing the open loop by connecting a resistive or reactive component between the output and one input terminal of the op-amp greatly reduces and controls this open loop gain. Table 1 Typical ranges of op amp parameters Parameter Typical Range Ideal values Open-loop gain, AOL 105 to 108 ∞ Input Resistance 105 to 1013 ∞Ω Output Resistance 10 to 100Ω 0Ω Supply voltage, VS 5 to 24V 2 University of the Philippines EEE118 1S2425 Feedback The concept of feedback is important to understanding op amp circuits. Feedback is achieved when there is a path from the output back to the input. The ratio of the output to the input voltage is then called the closed loop gain. As a result of the connection, it has been observed that the closed loop gain is almost insensitive to the open loop gain of the op amp and is controllable depending on the values of the associated resistances connected to the op amp. Saturation Practical limitations of the op amp is that the magnitude of its output voltage cannot exceed the power supply voltages, VS. In other words, the output voltage is dependent on and is limited by the power supply voltages. Shown in the figure below are the three operating modes depending on the difference of the input voltage, Vd=(V+ - V-) 1. Positive saturation, VO=VS 2. Linear Region, -VS Vin1, and Vout is “-” if Vin2 < Vin1. 8 University of the Philippines EEE118 1S2425 Differential amplifiers are applied as signal noise cancellers. In a twisted two wire system, signal noise introduced in one wire is also experienced by the other wire and their difference is zero and will not be amplified. The Summing Amplifier The Summing Amplifier combines a number of input signals and outputs a single signal which is the weighted sum of the input signals. An example of a Summing Op Amp is shown in Figure 3. F Figure 3: Summing Op Amp: The output voltage is Vin1 multiplied by Gain 1 plus Vin2 multiplied by Gain 2 Its equation is: Vout = (R3/R1) * Vin1 + (R3/R2) * Vin2 And can be extended to: Vout = (Rf/R1) * Vin1 + (Rf/R2) * Vin 2... + (Rf/Ry) * Viny, 9 University of the Philippines EEE118 1S2425 where R3 here is the feedback resistor Rf. One popular use of Summing Op Amps is in Mixer circuits where several tracks (voice, instrument sound, etc.) are combined to produce a single track. The Op Amp Differentiator The Differentiator op amp produces an output signal that is proportional to the differentiation of the input signal. A basic Op Amp Differentiator is shown below: Figure 4: Op Amp Differentiator: The output voltage is the differential of the input voltage With the following output equation: Vout = - RC (dvi/dt) Wherein the output is proportional to the rate of change of the input or the first order derivative of the input. For a ramp input, the output will be a DC voltage since the rate of change of a ramp signal is constant. For a triangular wave, the output is a square wave. For a sine wave input, the output is a sine wave that is 90 degree out phase of the input or a cosine wave. 10 University of the Philippines EEE118 1S2425 Op Amp Differentiators are commonly used for triangular and square wave signals. It is used for wave shaping. It is also useful as high-pass filters as the circuit attenuates low frequency signals. The Op Amp Integrator The Op Amp Integrator produces an output voltage which is proportional to the integral of the input voltage. This is achieved by connecting a capacitor as feedback reactance instead of a resistive one. The feedback is now an RC circuit. As an integrator, the output is determined by the time interval a voltage is present at its input as the current charges and discharges the feedback capacitor. The basic circuit is shown below: Figure 5: Op Amp Integrator: The output voltage is the integral of the input voltage The output equation of the Op Amp Integrator is: 11 University of the Philippines EEE118 1S2425 Vout = - (1/RC) ∫Vin*dt The rate of change of the output voltage is dictated by the RC time constant. Applying a square wave to the input will cause the capacitor to charge and discharge resulting in a triangular output waveform. This type of circuit is also called a Ramp Generator. The Op Amp Integrator acts as a low pass filter as it allows low frequency sine wave signals to pass through. The Op Amp Comparator The Op Amp Comparator compares one input to the other input. One input of the comparator is set as the reference voltage or Vref and it is to Vref that the other input is compared. A basic circuit is shown below: Figure 6: Op Amp Comparator When Vin is less than Vref, Vout is low or zero but when Vin is more than or equal to Vref, Vout is high or near Vs+. In this regard, the Op Amp Comparator converts the changing AC 12 University of the Philippines EEE118 1S2425 input to a digital high-low signal. Notice that the Low output signal is near ground or zero voltage, and the High output signal is near Vs+. When Vref is connected to the inverting input, the output goes High when Vin is equal or higher than Vref. This is what we call a Positive Voltage Comparator. When Vref is connected to the non-inverting input,the output voltage goes Low when Vin is equal or higher than Vref. This is what we call a Negative Voltage Comparator. The I-V Converter Figure 7: I-V Converter 13 University of the Philippines EEE118 1S2425 The I-V Converter, also called a transimpedance amplifier, like a resistor, converts current to voltage. Unlike a resistor, it has a low input impedance and low output impedance, and a very high gain. This circuit uses an op amp and feedback resistor to produce an output voltage that is proportional to the input current. The circuit equation is: Vout = - Rf * Iin A simple application of this circuit is in light intensity measurements using photodiode sensors. The current through a photodiode sensor changes with light intensity, this is converted to voltage, then processed to produce a measurement data. The Negative Impedance Converter A not so common op amp circuit is a Negative Impedance Converter whose basic circuit is shown below: Figure 8: Negative Impedance Converter The equation of the input impedance is: Zin = R1 * (R3/R2) And if R2 = R3 then Zin will just be -R1. The impedance is negative since the input current is 180 degrees out phase with the input voltage. Or simply, the input voltage is V3 to Ground and the input current is negative because it is going towards the positive of the source and not going out of the positive of the source. This circuit can be used to reduce a termination impedance to make it compatible with a source. Another use is reducing a load impedance to increase the drive capacity of a source. 14 University of the Philippines EEE118 1S2425 DISCUSSION ON THE DIFFERENCES BETWEEN SIMULATION RESULTS AND BREADBOARDING RESULTS Figure 9, The left image is a square wave input to an Integrator Op Amp constructed using actual components in a breadboard, the right image is the output. Figure 9: The left image is a square wave input to an Integrator Op Amp simulated in Tinkercad, the right image is the output. Simulations are an important tool in circuit design because it tests the functionality of a circuit design without going through the process of buying parts, waiting for it, assembling. The more significant advantage of simulations is that changes can be made to the circuit without the burden of the aforementioned, buying, waiting, and assembling. However, the accuracy of simulations depend greatly on the modeling of the virtual components and equipment. Breadboarding is also an important tool, as it offers testing the circuit with the affordability of making changes and adjusting component values. However, as breadboarding uses actual components, with inherent parameter tolerances, the results of breadboarding can be different from the results of simulations. The differences cannot be totally eliminated, the designer just needs to be aware of the differences, know where it exists and what causes it. 15 University of the Philippines EEE118 1S2425 Offered here are several other causes of differences between simulations and breadboarding: 1. Resistances: resistances of the wire connections, and intrinsic resistances of the components and equipment, may be different between simulation models and actual circuits 2. Capacitances: capacitances between actual wires, and those intrinsic to the components and equipments, may be different between simulation models and actual circuits 3. The non-ideal outputs of actual power sources and signal generators 4. The non-ideal environment conditions like temperature and humidity LEARNING ACTIVITIES Instructions: Paste all required circuits and displays (pictures) of the activity performed in the laboratory, type the data in the provided answer sheet, include observations and convert the sheet to a pdf file before uploading to the UVLe submission bin. Components List: 100Ω Resistor (1 pc.) 1kΩ Resistor (2 pcs.) 10kΩ Resistor (1 pc.) 1uF Capacitor (1pc.) LM741 Operational Amplifier (1 pc.) Activity 1: Inverting Amplifier (Duration: 45 minutes) Construct the inverting amplifier circuit as shown. Note that the circuit to power up the operational amplifier is not shown for clarity purposes, but must still be powered up as previously taught. Use R =100Ω, R =1kΩ, VS+ = 12V and Vs- = (-12V). The closed loop gain of the i f inverting amplifier is AV = -Rf/Ri. And note that: VOUT = AV * VIN 1. Set Vin at 1kHz sine, 1Vpp and 0 offset. Don’t forget to connect the function generator to the ground. Display the input and output voltages of the inverting amplifier on the oscilloscope. Take a picture of the oscilloscope display and paste it in the answer sheet. 16 University of the Philippines EEE118 1S2425 2. Complete the table below and plot the output peak-to-peak voltage (y-axis) versus the input peak-to-peak voltage (x-axis) between 10mVpp to 2.5Vpp with no offset. Vin peak to peak Vout Av = Vout/Vin 10 mV 50 mV 200 mV 500 mV 1.00 V 2.00 V 2.50 V 3. For an input peak-to-peak voltage of 50mV, complete the table below and plot Av (y-axis) versus frequency (x-axis). Discuss your observations. Frequency Vout Av = Vout/Vin 1kHz 2 kHz 20 kHz 200 kHz 500 kHz 1 MHz 2 MHz 3 MHz 4 MHz 5 MHz 6 MHz 17 University of the Philippines EEE118 1S2425 4. Change your Rf into a 1kΩ Potentiometer. With an input sinusoid with peak-to-peak voltage of 200mV and frequency of 5kHz, observe the effects of varying the potentiometer resistance. Discuss your observations. End of Activity 1 of 2 Activity 2: Summing Op Amp (Duration: 60 minutes) Figure A2: Summing Op Amp Implement the Summing Op Amp shown. Use the LM741 for the Op Amp. Use power supplies for Vs+ at 12V and Vs- at -12V. Use function generators for V3 with an output of -50mV to 50mV sine wave at 60Hz, offset 0, and V4 with an output of -100mV to 100mV sine wave also at 60Hz, offset 0. Set the oscilloscope time per division to show two cycles of the sine waves. 1.a. What waveform do you expect to see at Vout, and at what peak to peak value? Take a picture or grab a screenshot of the following and paste them at the spaces provided in the answer sheet: 1.b. Your circuit 1.c. The Vin1 waveform 1.d. The Vin2 waveform 1.e. The Vout waveform 1.f. What is the actual waveform at Vout, and what is its peak to peak value? 1.g. Discuss your observation of the resulting Vout in relation to Vin1 and Vin2 and explain how this came about. End of Activity 2 of 3 18 University of the Philippines EEE118 1S2425 Activity 3: The Op Amp Differentiator Duration: 60 minutes) Figure A3: Op Amp Differentiator Implement the Op Amp Differentiator shown. Use the LM741 for the Op Amp. Use power supplies for Vs+ at 12V and Vs- at -12V. Use a function generator for V3 with an output of -5V to 5V triangular wave with a frequency of 100Hz. Set the oscilloscope time per division to show two cycles of the sine waves. Please note: the capacitor used here has no polarity. 2.a. What waveform do you expect to see at Vout, and at what peak to peak value? Take a picture or grab a screenshot of the following and paste them at the spaces provided in the answer sheet: 2.b. Your circuit 2.c. The Vin1 waveform 2.d. The Vout waveform 2..e What is the actual waveform at Vout, and what is its peak to peak value? 2.f. Discuss your observation of the resulting Vout in relation to Vin1 and explain how this came about. End of Activity 3 of 3 In closing This module introduced you to different op amp circuits in its most basic forms. As you progress in your courses, enhancements to these basic circuits that will introduce stability, accuracy and precision will lead to better and more reliable results. For now, get to know these basic circuits, ready for use in the future. Do not forget about the differences in the results of simulations and breadboarding. This will be helpful when you do your own designs, as you analyze results. 19 University of the Philippines EEE118 1S2425 Website References and further readings https://www.electrical4u.com/ https://www.electronics-tutorials.ws/ https://www.arrow.com/en https://electricalvoice.com/ https://www.electronicshub.org/ https://www.vedantu.com/ https://www.allaboutcircuits.com/ https://circuitcellar.com/ StackExchange contributors. (2010, November 19). What is the significance of -3dB?. In StackExchange. Retrieved 11:09, September 10, 2021, from What is the significance of -3dB? 20