Power Factors Correction in Circuit Analysis II Laboratory | EE 304, Lab Reports of Microelectronic Circuits

Download Power Factors Correction in Circuit Analysis II Laboratory | EE 304 and more Lab Reports Microelectronic Circuits in PDF only on Docsity! Name: ___________________________ Lab Instructor: _________________ Date Performed: ________________ Date Due: _______________________ Lab Partner(s): ______________________________________________________ © 2008 Simon J. Tritschler. All Rights Reserved. EE 304 Laboratory III Power Factor Correction Up until now, we have been concerned with measuring voltages and currents in AC circuits and verifying circuit laws pertaining to such. Now we focus on how circuits behave when attempting to deliver power into various loads. The important point to consider with AC circuits such as these is that their reactance causes a disparity between real power (actual power dissipated, measured in watts) and apparent power (complex load voltage times complex load current, measured in volt-amperes). Therefore, in order to quantify and interpret this disparity, we use the concept of power factor, which is expressed as the ratio of real power to apparent power. Consider the following circuit: + This is a classic case of a resistive source attempting to deliver power into an inductive load. The problem is that inductive reactance in the load causes its apparent power to be artificially high, which is undesirable because it is strictly real power that does useful work. What we will attempt to do is correct for this inductive load reactance by shunting it with a capacitor. LLOAD 10 mH RSOURCE 620 Ω VSOURCE 3.54 VRMS @ 10 kHz SINE RLOAD 620 Ω © 2008 Simon J. Tritschler. All Rights Reserved. 1. Build the circuit as shown. Measure the phasor VLOAD, giving your voltage in RMS form. + VLOAD = _____ ∟ _____ VRMS Now calculate the load current, ILOAD, by finding the voltage dropped across RSOURCE and dividing by its resistance, converting your units into mARMS: ILOAD = (VSOURCE − VLOAD) / RSOURCE ILOAD = (3.54 ∟ 0° − _____ ∟ _____ ) / 0.620 kΩ ILOAD = _____ ∟ _____ mARMS Now we can calculate the apparent power of the circuit: S = VLOAD * ILOAD = _____ ∟ _____ * _____ ∟ _____ S = _____ ∟ _____ mVA LLOAD 10 mH RSOURCE 620 Ω VSOURCE 3.54 VRMS @ 10 kHz SINE RLOAD 620 Ω + VR − VLOAD ILOAD − + © 2008 Simon J. Tritschler. All Rights Reserved. 3. Measure the compensated load voltage magnitude and convert it to its RMS value, remembering that the phase angle has been adjusted to zero: VLOAD(COMP) = _____ ∟ 0° VRMS Now calculate the compensated load current: ILOAD(COMP) = (VSOURCE − VLOAD(COMP)) / RSOURCE ILOAD(COMP) = (3.54 ∟ 0° − _____ ∟ 0° ) / 0.620 kΩ ILOAD(COMP) = _____ ∟ 0° mARMS Apparent power in the compensated load can now be found: S(COMP) = VLOAD(COMP) * ILOAD(COMP) = _____ ∟ 0° * _____ ∟ 0° S(COMP) = _____ ∟ 0° mVA Measure the RMS voltage across the resistive portion of the load and calculate the resulting real power: VR(COMP) = __________ VRMS P(COMP) = VR(COMP)2 / RLOAD = _____ 2 / 0.620 kΩ = __________ mW © 2008 Simon J. Tritschler. All Rights Reserved. Finally, the moment of truth…calculate the corrected power factor by dividing the real power by the magnitude of the apparent power: pf(CORRECTED) = P(COMP) / |S(COMP)| = __________ mW / __________ mVA pf(CORRECTED) = __________ Post Lab: 1. Calculate the power factor of the original circuit. 2. Analytically determine the value of shunt capacitance necessary for complete power factor correction. 3. Why, in practice, might it be very difficult to achieve perfect power factor correction in real-life circuits?