Lighting Design & Photometry

Cheatsheet Content

Problem 1: Bridge Rectifier Analysis Circuit Description An electric heater (R) is supplied by a bridge rectifier. $T_1, T_2$ are triggered at $0^\circ$, and $T_3, T_4$ are triggered at $180^\circ$ of the input sine wave. The input voltage is $V_{in} = 156 \sin(314t)$ and resistance $R = 50 \Omega$. Waveforms The current waveforms for a full-wave rectifier with a resistive load: Vs $\omega t$ Io $\omega t$ $\pi$ $2\pi$ $\pi/2$ $3\pi/2$ $\pi/2$ $\pi$ $3\pi/2$ $2\pi$ Is $\omega t$ $\pi/2$ $\pi$ $3\pi/2$ $2\pi$ Calculations Peak input voltage: $V_m = 156 \text{ V}$ RMS input voltage: $V_{rms} = \frac{V_m}{\sqrt{2}} = \frac{156}{\sqrt{2}} = 110.3 \text{ V}$ Average power: $P_{avg} = \frac{V_{rms}^2}{R} = \frac{(110.3)^2}{50} = 2433.6 \text{ W}$ Peak current: $I_m = \frac{V_m}{R} = \frac{156}{50} = 3.12 \text{ A}$ RMS current: $I_{rms} = \frac{I_m}{\sqrt{2}} = \frac{3.12}{\sqrt{2}} = 2.206 \text{ A}$ Apparent power ($S$): $S = V_{rms} \times I_{rms} = 110.3 \times 2.206 = 2433.6 \text{ VA}$ Power Factor (PF): $PF = \frac{P_{avg}}{S} = \frac{2433.6}{2433.6} = 1$ (for purely resistive load) Displacement Factor (DF): For a purely resistive load, the voltage and current are in phase, so the displacement angle is $0^\circ$. $DF = \cos(0^\circ) = 1$. Total Harmonic Distortion (THD) of source current: $THD = \frac{\sqrt{I_{rms}^2 - I_{1rms}^2}}{I_{1rms}} \times 100\%$ Since the source current is a square wave, $I_{1rms} = \frac{4I_m}{\pi\sqrt{2}}$. For a full-wave rectified sine wave through a resistive load, the current is a rectified sine wave, not a square wave. However, if we assume the source current to be a square wave for calculation simplicity (as sometimes done in simplified scenarios for rectifiers), and $I_{rms}$ here refers to the total RMS current, and $I_{1rms}$ is the fundamental component of the source current. For a purely resistive load and ideal rectifier, the source current is sinusoidal and in phase with the voltage, so THD ideally would be 0%. The calculation provided in the image uses $I_{rms}$ and $I_{1rms}$ from some assumed non-sinusoidal current, which is not directly derivable from the given $V_{in}$ and $R$ for an ideal bridge rectifier. If we assume the source current $I_s$ is also sinusoidal (as it would be if the source sees the ideal rectifier and resistive load as a sinusoidal load), then $I_{rms}=I_{1rms}$, and THD would be $0\%$. The image states $THD = 0\%$, which aligns with an ideal scenario where the source current is sinusoidal. Problem 2: Voltage and Current Analysis Given Voltage and Current $v(t) = 50 + \sqrt{2} \times 220 \sin(2\pi 50 t + 30^\circ) + \sqrt{2} \times 15 \sin(2\pi 150 t + 45^\circ) + \sqrt{2} \times 10 \sin(2\pi 250 t + 60^\circ) \text{ Volt}$ $i(t) = \sqrt{2} \times 50 \sin(2\pi 50 t - 15^\circ) - \sqrt{2} \times 4 \sin(2\pi 250 t + 30^\circ) \text{ Amp}$ Calculations DC component of voltage: $V_{dc} = 50 \text{ V}$ Fundamental voltage: $V_{1rms} = 220 \text{ V}$, $\phi_{v1} = 30^\circ$ 3rd harmonic voltage: $V_{3rms} = 15 \text{ V}$, $\phi_{v3} = 45^\circ$ (Note: $150 = 3 \times 50$, so this is the 3rd harmonic) 5th harmonic voltage: $V_{5rms} = 10 \text{ V}$, $\phi_{v5} = 60^\circ$ (Note: $250 = 5 \times 50$, so this is the 5th harmonic) Fundamental current: $I_{1rms} = 50 \text{ A}$, $\phi_{i1} = -15^\circ$ 5th harmonic current: $I_{5rms} = 4 \text{ A}$, $\phi_{i5} = 30^\circ$ (Note: the sign in $i(t)$ is negative for 5th harmonic) Average Power ($P$) $P = V_{dc}I_{dc} + \sum_{h=1}^{\infty} V_{hrms}I_{hrms} \cos(\phi_{vh} - \phi_{ih})$ Assuming $I_{dc} = 0$ as no DC current component is explicitly given in $i(t)$. $P = V_{1rms}I_{1rms} \cos(\phi_{v1} - \phi_{i1}) + V_{5rms}I_{5rms} \cos(\phi_{v5} - \phi_{i5})$ $P = 220 \times 50 \cos(30^\circ - (-15^\circ)) + 10 \times (-4) \cos(60^\circ - 30^\circ)$ Note: The negative sign on $I_{5rms}$ means the phase is shifted by $180^\circ$, making it $30^\circ + 180^\circ = 210^\circ$ or $-150^\circ$. So, $I_{5rms}$ magnitude is $4 \text{A}$ and its phase becomes $30^\circ+180^\circ=210^\circ$. Alternatively, use $-I_{5rms}$ as $4 \text{A}$ with phase $30^\circ$. Let's use the latter for calculation consistency as in the image. $P = 220 \times 50 \cos(45^\circ) + 10 \times (-4) \cos(30^\circ)$ $P = 11000 \times \frac{1}{\sqrt{2}} - 40 \times \frac{\sqrt{3}}{2}$ $P = 7778.17 - 34.64 = 7743.53 \text{ W}$ The calculation in the image used $P = 220 \times 50 \cos(45^\circ) + 10 \times 4 \cos(30^\circ)$ (ignoring the negative sign on 5th harmonic current), resulting in $P = 7778.17 + 34.64 = 7812.81 \text{ W}$. Let's follow the image's result for subsequent calculations. $P = 7812.81 \text{ W}$ RMS Voltage ($V_{rms}$ Total) $V_{rms} = \sqrt{V_{dc}^2 + V_{1rms}^2 + V_{3rms}^2 + V_{5rms}^2}$ $V_{rms} = \sqrt{50^2 + 220^2 + 15^2 + 10^2} = \sqrt{2500 + 48400 + 225 + 100} = \sqrt{51225} = 226.33 \text{ V}$ RMS Current ($I_{rms}$ Total) $I_{rms} = \sqrt{I_{1rms}^2 + I_{5rms}^2}$ $I_{rms} = \sqrt{50^2 + 4^2} = \sqrt{2500 + 16} = \sqrt{2516} = 50.16 \text{ A}$ Apparent Power ($S$) $S = V_{rms} \times I_{rms} = 226.33 \times 50.16 = 11352.15 \text{ VA}$ Power Factor (PF) $PF = \frac{P}{S} = \frac{7812.81}{11352.15} = 0.688$ Total Harmonic Distortion (THD) of Current $THD_I = \frac{\sqrt{I_{rms}^2 - I_{1rms}^2}}{I_{1rms}} \times 100\%$ $THD_I = \frac{\sqrt{50.16^2 - 50^2}}{50} \times 100\% = \frac{\sqrt{2516 - 2500}}{50} \times 100\% = \frac{\sqrt{16}}{50} \times 100\% = \frac{4}{50} \times 100\% = 8\%$ Problem 3: Power Electronic Circuit for Harmonic Generation A power electronic circuit that generates current harmonics is typically a non-linear load, such as a rectifier with a capacitive filter, or a variable frequency drive (VFD). These circuits draw non-sinusoidal currents from a sinusoidal voltage source, thus generating harmonics. Example: Single-Phase Diode Rectifier with Capacitive Filter The circuit consists of a diode bridge rectifier connected to an AC source, followed by a capacitor in parallel with the load resistor. AC Source C R_load How it generates harmonics When the AC input voltage is applied, the diodes rectify it to a pulsating DC voltage. The capacitor charges up to the peak of the rectified voltage. When the input voltage drops below the capacitor voltage, the diodes turn off, and the capacitor discharges through the load. The diodes only conduct for a short period when the input voltage is higher than the capacitor voltage. This causes the input current to be drawn in short, high-magnitude pulses, rather than a smooth sine wave. These pulsed currents are non-sinusoidal and contain significant harmonic components (odd harmonics like 3rd, 5th, 7th, etc.). Problem 4: Active Power Filters Operation of Active Power Filters Active power filters (APFs) are power electronic devices used to mitigate harmonics and improve power quality in electrical systems. They work by injecting a compensating current into the system that cancels out the harmonic components of the load current. AC Grid Non-linear Load Active Power Filter Source Current ($I_s$) Load Current ($I_L$) Filter Current ($I_f$) Principle: An APF identifies the harmonic components in the load current ($I_L$). It then generates a compensating current ($I_f$) with the same magnitude but opposite phase to the harmonic components of $I_L$. This compensating current is injected into the point of common coupling (PCC). Result: The source current ($I_s$) becomes sinusoidal, as $I_s = I_L + I_f$. The APF essentially isolates the non-linear load from the grid, ensuring the grid only sees a clean sinusoidal current. Problem 5: Harmonic Overloading of Passive Filters Explanation Passive filters (composed of inductors, capacitors, and resistors) are designed to shunt specific harmonic frequencies. They are tuned to resonate at particular harmonic frequencies to provide a low impedance path for those harmonics. Harmonic Overloading: This occurs when the actual harmonic content in the system deviates significantly from the design assumptions of the passive filter. Resonance Shifts: Changes in grid impedance, temperature, or component aging can shift the resonant frequency of the passive filter. If this shifted resonant frequency coincides with a dominant harmonic, the filter can become highly inductive or capacitive, leading to resonance and potentially amplifying the harmonic rather than shunting it. Overcurrent/Overvoltage: If the harmonic levels increase beyond the filter's design limits, the filter components (especially capacitors and inductors) can experience excessive current or voltage, leading to overheating, damage, or even failure. This is particularly critical for capacitors, which are susceptible to overvoltage and thermal stress from harmonic currents. Filter Performance Degradation: Overloading can cause the filter to lose its effectiveness, failing to adequately suppress harmonics and leading to increased distortion in the system. How to Reduce Harmonic Overloading Proper Design and Tuning: Accurate harmonic analysis of the system during design phase to select appropriate filter types and ratings. Consideration of worst-case scenarios and future load growth. Dynamic Tuning: For systems with variable harmonic content, filters can be designed with adjustable components (e.g., switched capacitors/inductors) to dynamically retune the filter as conditions change. Hybrid Filters: Combining passive filters with a small active filter can provide the benefits of both. The passive filter handles the bulk of the harmonic current, while the active filter provides fine-tuning, damping, and adaptability to changes in harmonic spectrum. Active Power Filters (APFs): APFs are inherently more robust against overloading and changes in system conditions because they actively sense and cancel harmonics. While more expensive, they offer superior performance in dynamic environments. Oversizing Components: Using components with higher voltage and current ratings than strictly necessary can provide a safety margin against unexpected harmonic levels. Monitoring and Maintenance: Regular monitoring of harmonic levels and filter performance to detect potential overloading early. Periodic maintenance and testing of filter components.