_AA 320 Lab 5

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Operational Amplifiers Kiaan Akbarpour, Hugh Carbrey AA 320: Aerospace Instrumentation, AC William E. Boeing Department of Aeronautics and Astronautics University of Washington 2023-11-11 I. Discussion of Results A: Section 1: In the first part of the lab we were introduced to a basic op amp, which is a device that takes two inputs and outputs the differences between them. We were told to construct a circuit and try to see how V + and V - relate to each other and the voltages that are coming in and out of the circuit. Figure 1: Circuit for Part 1 Thus, we constructed the circuit shown above, where there were 10 kΩ resistors between the potentiometer and the Vs+ and the Vs- on the left side which are both 15 V, respectively. In this, we were told to find the difference between the input and output voltages when V in was close to zero. In doing this the results came up such that the closest value that V in could get to without the circuit “railing” was 0.439 V. For this, we were able to measure the V out value as 14.572 V. V+ (V) V- (V) Vin (V) Vout (V) Gain (V) 0.439 0 0.439 14.572 0.03 Table 1: Basic Op-Amp Circuit Values for Part 1 In this, using an Equation 2: V out = Gain(V+-V-) (2) which would make the expected V out = 0.03(0.439-0) =0.01317 compared to a V out of 14.572, that is an error of -99.9%, which can be expected as looking at the results, it can be seen that the circuit “railed”, considering that the output voltage is a lot higher than zero. Another source of error can be found in the op-amp itself, as the gain was found to be 0.03 rather than a number in the high thousands, which op-amps are supposed to have, suggesting that the op amp itself could be faulty and when V in gets closer to zero, it “rails” too easily and early and it sends a V out based on Vs+ and Vs- rather than V+ and V- based on the

potentiometer being not strong enough to block enough voltage such that the op amp does not “rail”. In this, a small value resistor or the potentiometer being too weak could affect the circuit and cause it to “rail” making finding a value that could work and produce valid results difficult in a lab environment. B: Section 2: In the second part, we were asked to modify the circuit such that a resistor of 10 kΩ between V- and V out is added and removing the ground wire making it such that the amplifier will try to balance the Vin and Vout values rather than take the difference between them. To test this, we changed the potentiometer to let in different voltages and see what the output voltage would be. Vin (V) -5.025 -1.999 0.000 2.002 4.994 Vout (V) -5.025 -2.000 -0.001 1.999 4.995 Table 2: Values for 10 kΩ resistor and differing input voltages From these values , it can be seen that switching to a resistor of 10 kΩ between V- and V out stabilized the difference between the input and output voltages to zero with an average error between the values of 0.044%. This makes sense as the gain in this scenario is 1 and the Vin is therefore not being amplified in any way. The error being greater than zero in this case comes from the fact that the potentiometer and resistor have some uncertainty in blocking voltage but the principle that the circuit is based on limits the amount of error that it could have thus making it close to zero. Resistor Value (kΩ) 10 1 100 Vout (V) 1.001 1.005 1.005 Vin (V) 1.004 1.005 1.008 Table 3: Vin vs Vout with changed resistor values. From the values, when the resistor value was changed the Vout value did not change much when keeping the input voltage the same with an average percent error of 0.2%. This can be explained by the fact that the op-amp functions such that the resistor is independent of the output voltage due to the op amp having a gain of 1 thus changing the resistor should not have any significant effect on the output voltage. Thus, for changing voltages and resistor values, the op-amp can adjust and thus changes in input have no effect on the output voltages. C: Section 3: Using the same setup from section 2 except for an added resistor from to ground, the closed loop gain voltage ( 𝑉 − can be determined. The closed loop gain voltage is the voltage gain from the amplifier with the feedback loop 𝐴 𝑉 ) closed. This is a function of the feedback resistance and input voltage as shown in equation (1) (𝑅 ? ) (𝑅 𝐼 ) (1) 𝐴 ? = 𝑅 ? +𝑅 𝐼 𝑅 𝐼 2

For , and , . This means that the output voltage should be 11 times the input 𝑅 𝐼 = 1𝑘Ω 𝑅 ? = 10𝑘Ω 𝐴 𝑉 = 11 voltage. The input voltage, output voltage, voltage amplification, and percent error are given in table 1. (V) 𝑉 𝑖? (V) 𝑉 ??? 𝐴 𝑉 Percent Error (%) -.944 -11.136 11.8 7.3 -.496 -8.836 17.8 61.8 0 -.025 DNE DNE .494 8.825 17.9 62.7 1.005 11.156 11.1 .909 Table 1: Data from section 3, part A As Table 1 shows, the actual amplification is closer to the ideal amplification as the magnitude of the voltage increases. The values closer to 0, have much higher percent errors. Theoretically, the value of output voltage at 0 should be zero. For part b, the 1k resistor for was replaced with a 100 resistor. This yields an of 101. As before, the input Ω 𝑅 𝐼 Ω 𝐴 𝑉 voltage, output voltage, voltage amplification, and percent error are given in table 2. (V) 𝑉 𝑖? (V) 𝑉 ??? 𝐴 𝑉 Percent Error (%) -.101 -10.303 102 .99 0 .184 DNE DNE .100 9.808 98.08 2.89 Table 2: Data from section 3, part B Table 2 has much lower percent errors across the board. This shows that as the ideal amplification increases, the actual amplification becomes much closer to the ideal. However, the higher amplification yields a higher output voltage at zero input voltage. D: Section 4: For this part of the lab, the input signal was sent to the inverting input of the op amp. The oscilloscope was used to measure the input voltage and the output voltage versus time (figure 2). The trace shows that the input voltage and output voltage are out of phase by 180°. This is due to the input signal being sent to the inverting input op amp, yielding the negative relationship. The amplitude of the output voltage is also smaller, due to the attenuation from the op-amp. 3

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Figure 2: Oscilloscope trace of input voltage (yellow) and output voltage (blue) vs. time Using equation 2 an ideal output voltage can be found when given input voltage and the two resistances. For this case, input voltage ( ) is 15V, the feedback resistor is 10 kΩ, and the input resistor is 20 kΩ. With these values, the 𝑉 𝑖? ideal output voltage can be found: (2) ? = 𝑉 ??? 𝑉 𝑖? = − 𝑅 𝑓 𝑅 𝐼 𝑉 ??? =− 𝑉 𝑖? * 𝑅 𝑓 𝑅 𝐼 = -(15V)* = -7.5V 𝑉 ??? 10 𝑘Ω 20𝑘Ω Compared to the measured value of -6.4V, the percent error is 14.67%. This error can be attributed to many factors, one example being the non-ideal nature of the op-amp. The gain equation used above assumes an ideal op-amp with no input gain error. Practical op-amps on the other hand have open gain error, which means voltage is lost as it moves across the op-amp. This makes the output voltage lower than ideal, which occurred in our data. This error is a function of frequency, so at the low frequencies of the lab the gain error is quite small. Additionally, due to the tolerances of resistors, the resistance is rarely the actual value listed. As equation 1 shows, the output voltage is directly proportional to the ratio of feedback resistance to input resistance. Changes in the resistance ratio yields changes to the output voltage. Additionally from the trace a frequency can be measured. Yielding a measured frequency of 60 ± .0154 Hz. The error can be attributed to the fluctuation of the frequency readout of the oscilloscope. 4

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