We will be using the circuit below as the input circuit to the load in order to reduce the effects of changing impedanc

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We will be using the circuit below as the input circuit to theload in order to reduce the effects of changing impedances on the supply voltages to the circuits we willbe building. Note that this is a different op-amp from the 741, but it has the same pinconfiguration. See the handout on the 741 op-amp for the correct pin configuration, if you don’t rememberit; the pin numbers are also shown here (don’t forget the supply voltages — that’s +15 V at Pin7 and −15 V at Pin 4). This circuit will be modified for some measurements

We Will Be Using The Circuit Below As The Input Circuit To The Load In Order To Reduce The Effects Of Changing Impedanc 1 (338.82 KiB) Viewed 55 times

In this lab, we will mostly be measuring RMS voltage with the DMM, not the usual peak-to-peak voltage measured on an oscilloscope. The output of the function generator should be set for a 500 Hz sinusoid, with high load impedance, for all experiments. While the desired input voltage to the circuit is 4 Vp-p, the location at which to measure the voltage changes with some experiments. We will be using the circuit below as the input circuit to the load in order to reduce the effects of changing impedances on the supply voltages to the circuits we will be building. Note that this is a different op-amp from the 741, but it has the same pin configuration. See the handout on the 741 op-amp for the correct pin configuration, if you don't remember it; the pin numbers are also shown here (don't forget the supply voltages - - that's +15 V at Pin 7 and -15 V at Pin 4). This circuit will be modified for some measurements. 500 Hz OPA551 6
Part I. Construct the circuit shown below. Input Circuit A. Adjust the output of the function generator to get a V=4 Vp-p, 500 Hz sine wave, High-Z. 31 ΚΩ B. Measure with the digital multimeter, and use the measured voltage to calculate the wer absorbed by the resistor. Note that the DMM measures rms voltage, not peak- to-peak. V= I= P-R2²- P= R C. Use an ammeter (on the 'AC I' setting) to measure I. Recall that an ammeter is connected in series with the circuit being measured. = P=1²³²R= D. Using the current measured in Part I.C and the voltage from I.B, calculate the power absorbed by the resistor. E. Compare the results of Part I.D with Part I.B and Part I.C. P = VI =
Part II. Construct the circuit shown below. Input Circuit + B. Use an ammeter to measure I 1 ΚΩ A. Keep the function generator output as 4 Vp P-P₂ C. Use an ammeter to measure IR. D. The power absorbed by the resistor is E. Use an ammeter to measure Ic IR 470 nF 500 Hz sine wave, High-Z. I= IR = P=IR²R= Ic= F. Use the measured values of IR and Ic to calculate I. Compare this result with the result from Part II.B. 1 = √ √ 12² + 12² = G. The apparent power absorbed by the load is (use the value of I from Part II.B) VI =
In order to determine the real power dissipated in the circuit, we must determine the phase difference between V and I. To do that, we will insert a resistor in series with the load, and use an oscilloscope to observe the voltages across the resistor Vab and the load simultaneously. Since the voltage across the resistor is in phase with the current, Vab will have the same phase as I. Modify your input circuit, as shown below; the only difference is the 500 resistor. OPA551 6 a 500 Q ww O b + V We can use an oscilloscope to determine the phase difference between V and I, similar to what we did in Lab 1. There is a problem, however. The ground clips on the oscilloscope probes must be connected to the ground node. Since one of the voltages we want to measure is Vab, we need to use one of the Math functions on the scope, specifically subtraction. If we put the Channel 1 probe on a, and the Channel 2 probe on b, we can use the subtraction function to get the voltage across the 500 $2 resistor, which will have the same phase as the total current. We then have Von Channel 2. H. Keep the function generator at V=4 Vp-p, 500 Hz, and High-Z, and determine the phase difference, which we'll call 0, between I and V. K. For a capacitive load, is the power factor leading or lagging? 0= I. Determine the power factor of the load. J. The real power dissipated by the load, P, is given by P= VI cos 0 = Compare this result with Part II.D. cos (=
Additional Report Use the measured value of C and the values of the resistor, voltage, and frequency to calculate IR, IC, I, P, VI, and 0; then, make a table comparing the calculated values with the measured values. I. Let X = 1 WC IR = V R I=V P = V² R 2 P cos 0 = = VI Ic = V X 2 VI = V² 1 1+ 2 ² √ (²₁)² + (²) ²