ELE302 - Lab 1

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Feb 20, 2024

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Operational Amplifiers Circuit Configurations (Post-Lab and In-Lab) By Justin Pope Course: ELE 302 Section: 17 Instructor: Dr. Venkatesh TA Name: Rene Gilliland-Rocque Date Submitted: Sunday, September 24, 2023

5.0 IN-LAB IMPLEMENTATION & MEASUREMENTS (5 marks in total): (a) Step 1: Connect the circuit as shown in Figure 2.0 . Connect pin (2) of the Op-Amp to common ground. Use a function generator to provide the input voltage, v i (t), by connecting the function generator output to pin (3) of the Op-Amp. Connect Channel (2) of the oscilloscope to display the output voltage v o (t), which is between pin (6) and the common ground. Connect Channel (1) of the oscilloscope to display the input voltage v i (t) provided by the function generator. Press “AutoScale” to capture Channel (1) and Channel (2) signals on the oscilloscope display. Adjust this initial display by setting the oscilloscope “Vertical Controls” as follows: • Channel (1): Coupling => DC, V/div => 2V. • Channel (2): Coupling => DC, V/div => 5V. Setting the oscilloscope “Trigger” and “Horizontal Controls” as follows: • Time base: Trigger Source -> Channel (1), Trigger Type => Edge, Time/div => 2ms. (b) Step 2: Set the controls of the function generator to provide triangular input signal v i (t) of 8V (peak-to- peak) at a frequency of 50Hz. Use Graph 2.0 to plot v i (t) & v o (t) as functions of time. (c) Step 3: Change the oscilloscope setting to the XY-display mode. The oscilloscope are now displaying: [v o (t) vs v i (t)], which is called the voltage-transfer characteristics of the Op-Amp Circuit. Use Graph 3.0 to plot this voltage-transfer [v o (t) vs v i (t)] characteristics of the Op-Amp circuit. (d) Step 4: Change the values of the positive dc-supply voltage to 12V and the negative dc-supply voltage to 9V. Use Graph 3.0 to plot the resulting voltage-transfer characteristics, [v o (t) vs v i (t)].

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(e) Step 5: Change the values of both the positive & negative dc-supply voltages to +15V and -15V, respective, and modify your circuit as shown in Figure 3.0 . Negative feedback is now said to be applied around the Op-Amp, with feedback factor β = R 1 /( R 1 + R 2 ), and the circuit is said to operate as a non- inverting amplifier. Plot the resulting voltage-transfer characteristics on Graph 4.0 . Record feedback factor β in Table 1.0 . Use your graph to record the Dynamic Range and Voltage Gain also in Table 1.0 . (The Dynamic Range is the linear region of the voltage-transfer characteristics .) (f) Step 6: This concludes the first 3-hr lab session. Please demonstrate Step 1 to Step 5 to your TA and submit your answers to the prelab assignments and the results collected from Step 1 to Step 5 to your TA at the end of the lab session. (1 mark)

Table 1.0 (0.5 marks) Circuit Conditions Feedback Factor β Valid Input Range (Dynamic Range) for v i (t) Voltage Gain [v o (t)/v i (t)] Negative feedback [R 1 =10kΩ & R 2 =100kΩ] 0.0909 Min: -1.23 Green Max: 1.38 10.87 Negative feedback [R 1 = 10kΩ & R 2 =47kΩ] 0.175 Min: -2.52 Red Max: 2.50 5.55 Negative feedback [R 1 =inf & R 2 =0] Undefined Min: -4.25 Blue Max: 4.1 0.995 (g) Step 7: Replace R 2 with a 47kΩ resistor on the Figure 3.0 circuit. Use Graph 4.0 to plot the resulting XY-mode voltage transfer curve. Use your plot to fill-in Table 1.0 . (h) Step 8: By replacing the resistor R 1 with an open-circuit and the resistor R 2 with a short-circuit, your circuit is now said to operate as a voltage follower. Use Graph 4.0 to plot the resulting voltage-transfer curve, and fill-in the blanks in Table 1.0 . (i) Step 9: Modify your circuit as shown in Figure 4.0 . The circuit is now said to operate as an inverting amplifier. Use Graph 5.0 to plot the resulting voltage-transfer characteristics and fill-in the blanks in Table 2.0 . (j) Step 10: Replace R 2 with a 47kΩ resistor. Use Graph 5.0 to plot the resulting voltage transfer curve. Use your plot to fill-in the blanks in Table 2.0 . (k) Step 11: Demonstrate Step 7 to Step 10 to your TA. (1 mark)

Table 2.0 (0.5 marks) Circuit Conditions Feedback Factor β Valid Input Range (Dynamic Range) for v i (t) Voltage Gain [v o (t)/v i (t)] Negative feedback [R 1 =10kΩ & R 2 =100kΩ] 0.0909 Min: -1.3 Green Max: 1.475 9.80 Negative feedback [R 1 = 10kΩ & R 2 =47kΩ] 0.175 Min: -2.875 Red Max:2.87 4.72

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6.0 POST-LAB QUESTIONS (2 marks in total, 0.5 marks for each question): (1) By examining your plots on Graph 3.0 , answer the following: a) What are the effects of the dc-supply voltages on the voltage-transfer characteristics? b) Would the solo Op-amp qualify as a practical amplifier circuit? Comment on your answer. a) By examining the plots on graph 3, the graph indicates that the op-amp amplifies the difference in voltage at the input nodes (2 and 3). The op-amp, however, can only amplify the voltage to a certain extent. For example, since the voltage from the power source was +15V and -15V, therefore, the saturation on the graph was seen at those values and the output voltage at node 6 cannot exceed these values. Therefore, this proves that the voltage supplied from the power source sets the boundaries of the saturation levels on the graph. b) The solo op-amp would qualify as a practical amplifier circuit. This is because the device was able to accurately amplify the voltage difference between the two inputs, amplifying the voltage in the range of +15V and -15V. The op-amp also responded to changes in the circuit when different resistors were added, as a result, increasing or decreasing the voltage gain.

(2) By considering your plots on Graphs 3.0, 4.0 & 5.0 , answer the following: a) What does the application of negative feedback (around the Op-amp) have on the amplifier-circuit behavior? b) Does the application of negative feedback have any effect on the output saturation voltages? Comment on your answer. a) The negative feedback resistor helps keep the circuit stabilized, less sensitive to other components, and temperature changes. This resistor also helps to keep the output more linear and minimize distortions. The negative feedback would affect the active region of the graph by creating a deeper or more slanted slope dependent on the value of the feedback resistor. Without the feedback resistor, the active region is a vertical line, indicating an infinite active region(undefined) and an uncontrolled output. b) The negative feedback helps to prevent output saturation to ensure a more linear output and unsaturated range. When feedback is applied, it helps to counteract changes in the output voltage. For example, as the approaches +15V or -15V, the resistor works to bring it back with a more linear and unsaturated range. The larger the feedback resistor value, the more stable the output saturation was.

(3) Thevenin Equivalent Input Resistance: a) Calculate the Thevenin equivalent resistance seen by v i (t) for the non-inverting configuration shown in Figure 3.0 . b) Calculate the Thevenin equivalent resistance seen by v i (t) for the inverting configuration shown in Figure 4.0 . c) Which configuration has higher Thevenin equivalent resistance as seen by v i (t)? a) Voltage Source is Open ࠵? ࠵?࠵? = ࠵?1 || ࠵?2 = 1 1 ࠵? ! + 1 ࠵? 2 = 1 1 10࠵?Ω + 1 100࠵?Ω = 9. 09࠵?Ω b) Voltage Source is Open ࠵? ࠵?࠵? = ࠵? 1 + ࠵? 2 = 10࠵?Ω + 100࠵?Ω = 110࠵?Ω c) As seen by V i (t), the inverting configuration has the larger Thevenin equivalent resistance.

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(4) Thevenin Equivalent Output Resistance: a) Calculate the Thevenin equivalent resistance seen by v o (t) for the non-inverting configuration shown in Figure 3.0. b) Calculate the Thevenin equivalent resistance seen by v o (t) for the inverting configuration shown in Figure 4.0. c) How do the two equivalent resistances compare? a) Recall: ࠵? ࠵? = 75Ω ࠵? ࠵?࠵? = ࠵? ࠵?࠵?࠵? = ࠵? ࠵? ࠵? ࠵? ࠵? ࠵? = ࠵? 1 + ࠵? 2 = ࠵? ࠵? 100࠵?Ω + ࠵? ࠵? ࠵? ࠵? = 100࠵?Ω × ࠵? ࠵? ࠵? ࠵? +100࠵?Ω Note: Given that 100kΩ > R o : ࠵? ࠵?࠵? = 100࠵?Ω×࠵? ࠵? 100࠵?Ω = ࠵? ࠵? = 75Ω b) Recall: ࠵? ࠵? = 75Ω ࠵? ࠵?࠵? = ࠵? ࠵?࠵?࠵? = ࠵? ࠵? ࠵? ࠵? ࠵? ࠵? = ࠵? 1 + ࠵? 2 = ࠵? ࠵? 100࠵?Ω + ࠵? ࠵? ࠵? ࠵? = 100࠵?Ω × ࠵? ࠵? ࠵? ࠵? +100࠵?Ω Note: Given that 100kΩ > R o : ࠵? ࠵?࠵? = 100࠵?Ω×࠵? ࠵? 100࠵?Ω = ࠵? ࠵? = 75Ω c) The Thevenin equivalent resistance in both configurations are the same.