How a Boost Converter Steps Up DC Voltage | Power Electronics Ch. 2

How can a circuit make a DC voltage higher than the battery that feeds it, without a transformer? In this second chapter of the Power Electronics series we rearrange the parts of the buck converter into a boost converter and follow one switching cycle: the inductor charges from the source, then adds its voltage to the source and pushes current into the output. Volt-second balance gives U₂ = U₁/(1 − δ), and a worked example puts numbers on currents, ripple and power. Model: ideal switch and diode, periodic steady state, continuous inductor current. The values in the worked example (12 V, 50 kHz, δ = 0.6, 72 µH, 15 Ω, 100 µF) are illustrative. Chapters 0:00 The problem: raising a DC voltage 0:59 Recap: inductor and capacitor laws 1:39 The boost circuit and notation 2:24 Assumptions and the duty ratio 2:56 Switch on: the inductor charges 3:33 Switch off: source and inductor in series 4:22 Waveforms over two periods 5:06 Volt-second balance: U2 = U1/(1 - delta) 6:14 Input and output currents 6:53 Worked example: setup 7:41 Worked example: inductor ripple 8:28 Output capacitor ripple 9:15 Practical limits 10:07 Recap Previous chapter: Ch. 1 Buck converter,    • How a Buck Converter Steps Down DC Voltage...   Animations made with Manim Community #PowerElectronics #BoostConverter #ElectricalEngineering