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Download Task 1: Basic Non-Inverting Amplifier
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Operational Amplifier Design Lab By Denny Jung NO lab partner (Missed the lab) All labs done on Multisim TA – Freddy Galindo Jung 1 Task 1: Basic Non-Inverting Amplifier Design Objective: The objective of this task is to amplify the mono audio signal with maximum amplitude of 200mVpp to a maximum of 20Vpp, using a non-inverting operational amplifier. Schematic: Schematic #1… Task 1 Schematic Theory of Operation: Positive and negative charges are connected to the operational amplifier to supply power to it. The bypass capacitors are always used with operational amplifiers to remove any additional sound made. The mono audio signal with amplitude 100 mVp (200 mVpp) goes through the noninverting amplifier to achieve the gain of 100. Using the relationship of a non-inverting operator, the values of the resistors are then set to achieve the gain of 100. Jung 2 Derivations and Analysis: Gain = 100 R1 = 1kΩ, R2 = 99kΩ Experimental Results: **No percent error available because everything was done on Multisim** Capture #1… Capture of Task 1 Discussion of Results / Concluding Thoughts: I first attempted to do this during the lab, however, was not able to finish it due to the faulty wires and complications with the computers. In addition, I had extra wire when setting up the power supply, which only hampered with the progress of the task. When the schematic was set up in the Multisim, as done in the lab, the design worked as expected, yielding almost no experimental error. One mistake that I have made from the start till the end, which forced me to reconfigure all the circuits once again, was the difference between the function generator in VirtualBench and Jung 3 Multisim. In VirtualBench, the system asks for peak-to-peak amplitude, while in Multisim, only the amplitude is asked for. By using 200mVpp in place of 100mVp, the output saturated at +/- 15V. The only way I thought to fix this was to use a higher power source. However, in real life, the operational amplifier type LF412CN only accepts a power source value of 15. Jung 4 Task 2: Difference Amplifier (Karaoke Circuit) Design Objective: The objective of this task is to mix the Left and Right channels, essentially by subtracting the Right channel from the Left channel or vice versa. This in result will remove the identical vocal signal that appears on both the Left and Right channels, while keeping the instrumental signals from the channels, in addition to amplifying the difference to get the desired output amplitude and inverting the Right channel signal. Schematic: Schematic #2… Task 2 Schematic Theory of Operation: From this task on, the bypass capacitors and the power supplies are used for the same reason as stated in task 1. In task 2, two different function generators are used for each channel because each channel consists of vocal and instrumental signals. The two function generators are also put in series with one another so that the signals are summed. In the inverting input, the resistors need a ratio of 100 in order to provide a gain of 100. However, when this is set to 100, the gain of non-inverting input will be 101 due to the relationship of a non-inverting operator. Therefore, two more resistors are added before the non-inverting input to create voltage division. Jung 5 Derivations and Analysis: Gain- = -100, Gain+ = 101 R1 = 1k, R2 = 100k, R3 = 1k, R4 = 100k 𝑅 𝑉𝑥 = 𝑅𝑥 𝑉𝑠 ; Rx is the selected resistor value, RT is the total series circuit resistance 𝑇 Experimental Results: **No percent error available because everything was done on Multisim** Capture #2… Capture of the Right channel and the output Discussion of Results / Concluding Thoughts: When the task was first done, the schematic was missing the Resistor 3 and 4. The output was fairly close to the expected with an experimental error of 1%. Nonetheless, Schematic #2 is the ideal schematic that should be used to get the exact gain of 100 for each channels. In addition to this problem, another problem was that because we were told to measure the input of the Right channel and the output, 440Hz vocal signal was still part of the input. Therefore, it was difficult to spot a single point in which the output had an approximate gain of -100. Jung 6 Task 3: Variable Two Channel Mixer with Unbalanced Inputs Design Objective: The purpose of this task is to mix an unbalanced stereo signal in which the left and right channels have different voltage amplitudes. The circuit requires the left channel to have a fixed gain of -20 while the right channel has a varying gain between -20 and -50. Schematic: Schematic #3a… Task 3 Schematic before short-circuiting a channel Jung 7 Schematic #3b… Task 3 Schematic with right channel short-circuited Schematic #3c… Task 3 Schematic with left channel short-circuited Jung 8 Theory of Operation: Similar to Task 2, this task requires two channels, but both of them are amplified and inverted. This means the non-inverting input is grounded. Left channel, on one hand, will have a fixed gain of -20, while the Right channel has a varying gain between -20 and -50. The Right channel requires a potentiometer, in order to vary the gain. Thus, using a potentiometer, the total resistor value of the Right channel varies from 13.3k to 33.3k, which results in ratio of 50 and 20, respectively. This would result in the Left channel to have a resistor value of 33.3k and the Rf to have a value of 665k. Derivations and Analysis: Gain-Left = -20, Gain-Right = -20 ~ -50 R1 = 1k, R2 = 100k, R3 = 1k, R4 = 100k Experimental Results: **No percent error available because everything was done on Multisim** Capture #3a… Capture showing a gain of -20 from the Left channel (at 0%) Jung 9 Capture #3b… Capture showing a gain of -50 from the Right channel (at 100%) Jung 10 Capture #3c… Capture showing a gain of -20 from the Right channel Discussion of Results / Concluding Thoughts: Without the prior knowledge that we are not allowed to change the resistance value of the potentiometer, I first changed the value of the potentiometer to provide the gain of -20 ~ 50. At this point, no resistor values are unique, as we can easily change the values along with the potentiometer. Using the potentiometer resistance value of 20k, all the resistor values used in this schematic become unique. Moreover, I did not short circuit each side when first measuring; this resulted in a convoluted result that made it impossible to figure out if the operational amplifier was providing the expected gain. Jung 11 Task 4: Level-Shifting Amplifier Design Objective: The purpose of this task is to invert and amplify a mono audio signal that is an output of an electret condenser microphone, while removing the DC offset voltage that is always included in an electret condenser microphone. Schematic: Schematic #4… Task 4 Schematic Theory of Operation: An electret condenser microphone has an output signal, along with a DC voltage offset value of 5V, as provided by the lab handout. In order to amplify and invert the signal, while removing the DC offset from the microphone, a level-shifting amplifier is needed. Using the positive power supply source and the rule of voltage division, I found the ratio of resistors to be 1 to 2. Therefore, the non-inverting input of the operational amplifier is supplied with 5V. From looking at the figure provided in the lab handout, the gain is -100 and by using the relation of inverting operator, I found the values of the resistors. Jung 12 Derivations and Analysis: Gain-Left = -100, Gain-Right = -20 ~ -50 R1 = 1k, R2 = 100k, R3 = 50k, R4 = 1.5k, R5 = 100k 𝑅 𝑅 Vout = − 𝑅2 𝑉𝑎 + 𝑉𝑏 (1 + 𝑅2 ); Va is the input from the inverting side, Vb is the input from the 1 1 non-inverting side Experimental Results: **No percent error available because everything was done on Multisim** Capture #4… Capture of Task 4 Discussion of Results / Concluding Thoughts: In the Theory of Operation section, I purposely left out the reason I included the 1.5kΩ resistor. This is because when I first did the experiment, I did not think this value would prove to have any effect on the circuit. When looking at the oscilloscope captures, that was when the error was spotted: even though the maximum output was approximately 10V, the minimum output was - Jung 13 𝑅 𝑅 5V. By doing the math, this error was resolved. Vout = − 𝑅2 𝑉𝑎 + 𝑉𝑏 (1 + 𝑅2 ), where Va is the 1 1 input signal from the inverting terminal and Vb is from the non-inverting terminal. Therefore, when the 1.5kΩ was not used, it had an “amplified” output – an excess of 5V. Using a 1.5kΩ resistor supplies a lower input voltage at the non-inverting terminal of the operational amplifier but will completely cancel out of the DC offset voltage, without any leftover voltages. Jung 14 Task 5: Variable Level-Shifting Amplifier Design Objective: The objective of this task is to invert and amplify a signal from an electret condenser microphone that has a varying DC offset voltage between 4 and 6. The circuit requires a potentiometer so that it can be adjusted to cancel any DC offset between 4V and 6V. Schematic: Schematic #5a… Task 5 Schematic with 4V DC offset Jung 15 Schematic #5b… Task 5 Schematic with 6V DC offset Theory of Operation: Task 4 and 5 are very similar to each other; the only difference is the potentiometer allowing the circuit to adjust its voltage at the non-inverting input of the operational amplifier. Therefore, a potentiometer arrow is used, in place of where the wire is used for voltage division in Task 4. In doing so, the non-inverting input voltage can be varied between 4 and 6. Using the potentiometer resistance value of 20k and the rule of voltage division, the resistor values needed for offset were found. The 1.5kΩ resistor is put in there for the same reason stated in Task 4. Derivations and Analysis: Gain = -100 R1 = 1k, R2 = 100k, R3 = Rpot = 20k, R4 = 40k, R5 = 90k, R6 = 1.5k 𝑅 𝑅 Vout = − 𝑅2 𝑉𝑎 + 𝑉𝑏 (1 + 𝑅2 ); Va is the input from the inverting side, Vb is the input from the 1 non-inverting side 1 Jung 16 Experimental Results: **No percent error available because everything was done on Multisim** Capture #5a Jung 17 Capture #5b Discussion of Results / Concluding Thoughts: Channel A of the oscilloscope, which measures the input signal, varies between 5.9V and 6.1V and also between 3.9V and 4.1V. The values are centered at 6V and 4V because of the DC offset voltage. Also, they vary by 100mV because of the sine function that is part of the input signal. The expected outcome should be the amplified signal of the sine function, which peaks at +/10V, without any DC offset voltage, that should have been cancelled out by the varying noninverting input. Jung 18 Post-Lab Activities 1. An output signal in the voltage range of 5Vpp ~ 20Vpp actually means 2.5Vp~10Vp. To create a variable-gain amplifier with an output signal as given, we could possibly place a potentiometer in series with the Rf. 100 = 1 + 𝑅𝑓 + 𝑅𝑝𝑜𝑡 𝑅𝑓 + 20𝑘 = 1+ 𝑅𝑠 𝑅𝑠 25 = 1 + 𝑅𝑓 𝑅𝑠 Substituting Rf/Rs, which is 24, in the first equation, we can find the value of Rs, which comes out to 0.26667kΩ and therefore, the Rf value comes out to be 6.4kΩ. Using these values, we can achieve the variable-gain amplifier with the desired output voltage range. 2. If the amplitude of the vocals on the 2 channels that I wanted to cancel out was not identical, we can use voltage division to drop off the difference. For example, if one has 0.5Vp and another has 0.1Vp, we could set up a two-resistor system to drop the voltage on the 0.5Vp signal. One resistor will be connected to the ground, while the other is in series with the signal input. The values of the resistor can be figured out by using the rule of the voltage division. 3. Considering how we are already varying the Right channel gain, the only thing we would need to change in this circuit design is the Left channel. Similar to how the Right channel is designed, we add another potentiometer in series with the previous resistor and change the value of the resistor depending on the value of the gains. 4. The additional complication to wanting a variable signal amplification is having to change the input signal without having to change the gain. If the input was varying in Task 4, which has an input going into the inverting input of the operational amplifier, we would need to use a potentiometer right after the signal input but before the input goes through the op amp. This is because in doing so, we can vary the signal before it goes through the operational amplifier without changing the gain. If the potentiometer is placed on any other spot on the circuit, it will change the gain or the voltage going into the non-inverting input of the operational amplifier. 5. If the DC offset voltage could be either positive or negative, we would need to incorporate both power sources. The problem with this is that in Task 5, the potentiometer is only connected to the positive source, which would never supply any values below 0. I would address this problem by using a switch that will flip between the positive and the negative end of the power sources.