0.1uf ± _ ~P-3T_VIRTUAL. , -1\/V R4 2k0 1k0 ~ K = 1 + R3/R4

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1 ELE Electrical Fundamentals Laboratory II ~ Lab 9 - April 1, 2016 Return of the Sallen Key Filter Gabriel Ortiz The purpose of this lab is to experimentally verify the Multisim simulations that you ran in the pre-lab. Specifically we want to compare the performance of the Sall en Key filter to the 1st order filter. Use Multisim to simulate a first order RC filter with R = lk Ohm and C = 0.1 µf. Print out a plot of the gain and the phase shift with the grid on. The frequency range should be the same used in the pre-lab. We will be plotting the measured points on these graphs. You will need three copies of the graphs. Build the following circuit: XFGl Sallen-Key Low Pass Filter C2, !~----, R1 0.1uf 1k0 R = R1, R2 C = C1, C2 ~~U1 ± _ ~P-3T_VIRTUAL 0.1[F R3, -1\V R4 2k0 1k0 ~ K = 1 + R3R4 2k!l R10 1k0 Sallen Key Output R9 and R3 must be equal to keep the DC gain = 1. Run the above circuit for the following cases. Plot the gain and phase shift on the graphs that you just printed out (don't forget to record the actual numbers in your notebook). 1. Let K = 1 (R3 and R9 = 0). Note that f:e gain of the Sallen Key is always lower than the 1st order. Frequency Gain (Vout Vin) (V) Phase Shift (Degrees) loohz Hz ,; 800Hz khz ' 3 khz \ A12.3 7kHz lokhz ~ ~

2 2. Let k = 1.6. This is the Butterworth filter. Is the gain always less than or equal to 1? YES. Frequency Gain (VoutNin) (V) Phase Shift (De2rees) loohz Hz lc Hz 0.954,J k" khz ' \...-" 3kHz kHz lokhz Let K = 2.4. (Note, in the pre-lab we used K=2.0). How does the gain of the Sallen Key act at frequencies less than the cutoff frequency? The gain of the Sallen Key is much higher at frequencies less than the cutoff frequency Find the frequency that has the highest gain khz Compare this frequency to the frequency corresponding to the imaginary part of the pole for K = 2.4. The frequency at K = 2.4 is 1.52 khz. Our values are very close Frequency Gain (VoutNin) (V) Phase Shift (De2rees) loohz Hz Hz ~ 1 khz 1.41.;,2'8'.80 3kHz I kHz lokhz Let K = 3. We can adjust the gain so the pole has no real part. At this point the circuit will oscillate. Start by disconnecting the signal generator. Connect Rl to ground. Now, replace R3 with the Decade Box. Don't worry about the changing R9 in the voltage divider. Set R3 so K = 2.8. There should be no output signal. Increase R3 until you see oscillation. If R3 is too high you will see a square wave. Adjusting R3 can be a little tricky. When you get a nice sine wave measure the frequency. Compare the measured frequency to the fr~quency with the maximum gain in the pre- 1~. ' Calculated Frequency= 1592 Hz ~ Our Measured Frequency= 1610 Hz~ They are pretty close to each other.

3 l5 ELE Electrical Fundamentals Laboratory II Lab 2 - February 1, 2016 DC and AC Bridge Circuits Name: Dylan Heh, China Smith Date: Introduction: This lab explores the use of both DC and AC Bridge Circuits. Part 1: Build the circuit below. _L_Vl -=-12 V T ~"~ lokohm DC l~hm ~ Decade Box j Thermister Adjust the decade resistance box until the voltmeter reads zero volts. The bridge is now balanced, and the decade resistance box has the same resistance as the thermistor at ambient temperature. Now grab the thermistor between your finger and thumb and watch the voltmeter change. The bridge is becoming unbalanced because the thermistor's resistance is changing. What is the maximum voltage change you observe? _31 lm ~ Is the thermistor's resistance increasing or decreasing? Increase Byhowmuch? 30on_ Part 2: Build the circuit given in Part 2 of the Pre-Lab. Put the multimeter into the circuit. Set the multimeter for AC Voltage. Note the oscilloscope will not work in this configuration. Change the resistance in the decade box till the output voltage on the multimeter is equal to 0. V=_OV_lOkn ~ Does this match your expectations? - Yes the resistance would equal the other side to balance the bridge

4 Measure each individual component using the multimeter. Use the measured values to recalculate the given formula. Does this better match the data? yes the formula gives the exact value for the bridge to be stable Change C4 to 0.22 uf adjust the decade box to zero the output. R5 = 21k.Q --- Does the formula give a good measure of the capacitance of C4? yes Use an unknown box for C4 the capacitor is between the black.9d the orange banana plugs. Determine the unknown capacitance. Unknown Box = unknown box #5 Capacitance=.0032uF

5 Part 3: Build the circuit given in Part 3 of the Pre-Lab. Measure the AC Voltage for 10 frequencies between 1 khz and 2 khz. Frequency AC Voltage 1 khz.73 V 1.1 khz.68v 1.2 khz.63 V 1.3 khz.61 V 1.4 khz.59v 1.5 khz.59v khz.60v \..' 1.7 khz.60v 1.8 khz.62v 2kHz.67V Is the measured frequency with the lowest AC Voltage the same as the lowest frequency found in the pre-lab? Yes Change Cl and C2 to 0.01 uf. Find the frequency with the lowest AC Voltage. Frequency= 2.lkHz ~ Is this what you would expect based on the formula No Note the accuracy of the AC part of the Multimeter losses accuracy as the frequency increases.

6 ELE Electrical Fundamentals Laboratory II Lab 10 - April 8, 2016 The Boost Converter Tahsin Tahmid Part 1: Build the circuit below, which is a boost converter. For the lab, I replaced some of the components used in the pre-lab with better components. If there are no IRF530 transistors available, you can substitute a 2N7000. In the circuit below there is no resistor between Ll and Vl as in the pre-lab. This resistor is not needed since Ll already has some resistance. Vl should go from 0-5 V. We can do this by using the CMOS output on the function generator. Measure the voltage across the 1 on resistor with the oscilloscope to monitor the current through the inductor (V = IR). Determine the maximum current through the inductor. Are there any discontinuities in the inductor's current? (Does the current through the inductor ever chang~tly?) Max Current : 310 A= 0.3A. ~ There are no discontinuity. Now measure the output voltage on the other channel. What is the output voltage? Vout = 7.5 V V2 -=-sv ~ - 3 (ct; V1 R1 2kHz 10 n 5V 01 1mH 1N4002GP --~-~ Q1 4 IRF530NS +I C1 10uF 5 Rload 1k0 0 Vout Part 2: We will now measure the voltage, output ripple and efficiency of the boost converter as a function of duty cycle of the input square wave. First replace the 10 ohm resistor with a wire (this will significantly increase the output voltage). Put one oscilloscope probe on the input square wave. Put the second probe on the output. Measure the output voltage, the voltage ripple and calculate the efficiency for duty cycles of.2,.5 and.8. The output ripple is just the variation in the output voltage divided by the mean output voltage. The efficiency is just the power that is dissipated in R1oad (V 2 R) divided by the power into the system from V2 (from the power supply, take the voltage and the current and multiply them). Change R1oad to 2000 ohms. Repeat the measurements. Change R1oad to 470 ohms. Repeat the measurements.

7 Out C cle Out ut 0.2 Vout= Ripple= Efficiency= R=lkO 12.5V 5.76% 63.8% R=2k0 17.4V 8.05% 67.3% R= V 16.1% 65.7% ~ 0.5 I Vout= Ripple= Efficiency= 0.8 Vout= Ripple= Efficiencv= 15V 9.33% 26.0% 13.3 V 10.5% 14.0% 20.2 V 5.94% % 4.37% 17.8% 15.9% Part 3: Make a comment in your report about how a regulated switching power supply can be made. A regulated supply is a power supply that maintains a fixed output voltage ( e.g. 15V) regardless of the load (the amount of current drawn from th~ power supply). We can alter the duty cycle to regulate the power s. I voltage. Important Note: The power rating of Rtoad is V,i Watt. You will be operating the resister at about 1 W. The resister will heat up! Since you are only running for a short period of time, you should be o.k. But don't touch it. L

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