--- title: "Buck-boost regulator output and ripple derivation" subject: "Advanced Electronics" chapter: 8 key: "semester_4/advanced_electronics:buck-boost-derivation" source: https://www.bctengineer.com/q/semester_4/advanced_electronics/buck-boost-derivation site: BCT Engineering --- # Buck-boost regulator output and ripple derivation Advanced Electronics · Chapter 8: Switched Power Supplies - Asked in: 2072 Chaitra, 7 marks - Video: [Buck Boost Converter Output Voltage Derivation in Power Electronics by Engineering Funda](https://www.youtube.com/watch?v=0eZbpisL7p4) — Engineering Funda - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_4/advanced_electronics/buck-boost-derivation - Whole chapter: https://www.bctengineer.com/md/semester_4/advanced_electronics/chapter-8 ## Question Derive expression for average output voltage, peak to peak ripple current of inductor of Buck-Boost regulator with necessary figure and wave forms. ## Short answer In Buck-Boost converter, switch ON stores energy in inductor; switch OFF transfers it to output with polarity reversal. Average output voltage Va = -Vs·k/(1-k). Peak-to-peak inductor ripple current ΔI = Vs·k/(f·L). ## Answer A **Buck-Boost regulator** gives an output voltage that can be **lower or higher** than the input and is of **opposite polarity**: Va = −k·Vs/(1 − k). #### Circuit ```text Q (switch) Dm +Vs ---/ ---+-----------|<---+------+ | | | L iL C LOAD Va | | (− at top) − --------+----------------+------+ ``` #### Operation **Mode 1 (Q ON, 0 < t < t1 = kT):** Current flows from Vs through Q and L. Inductor current rises linearly from I1 to I2; Dm is reverse biased; C supplies the load. ```text Vs = L·ΔI/t1 → t1 = L·ΔI/Vs ``` **Mode 2 (Q OFF, t1 < t < T):** The inductor keeps the current flowing through C, the load and Dm. Energy stored in L is transferred to the output, which becomes negative. Inductor voltage = Va (negative); current falls from I2 to I1. ```text Va = −L·ΔI/t2 → t2 = −L·ΔI/Va ``` #### Waveforms ```text Q |‾‾‾‾|____|‾‾‾‾|____ 0 kT T vL |+Vs | |+Vs | | |____| |____ Va (negative) iL | /\ /\ I2 |__/ \___/ \___ I1 iD | |\_ |\_ (when Q OFF) ``` #### Average output voltage Rise and fall of inductor current are equal in steady state: ```text Vs·t1/L = −Va·t2/L Vs·kT = −Va(1 − k)T Va = −Vs·k/(1 − k) ``` - k < 0.5: |Va| < Vs (buck) - k = 0.5: |Va| = Vs - k > 0.5: |Va| > Vs (boost) #### Peak-to-peak ripple current of inductor ```text ΔI = Vs·t1/L = Vs·k·T/L ΔI = Vs·k/(f·L) ``` Also, using t1 + t2 = T: ```text T = L·ΔI/Vs − L·ΔI/Va = L·ΔI(Va − Vs)/(Vs·Va) ΔI = Vs·Va/(f·L(Va − Vs)) ``` (Both forms give the same value since Va = −kVs/(1 − k).) **Capacitor ripple** (C supplies Ia during t1): ΔVc = Ia·k/(f·C). Average input current Is = Ia·k/(1 − k) (ideal). The circuit needs only one switch and gives polarity reversal, but the switch and diode carry high peak currents and input current is discontinuous. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. What are the output characteristics of a buck-boost regulator? A. Same polarity, always lower than input B. Opposite polarity, lower or higher than input depending on k C. Same polarity, always higher than input D. Opposite polarity, always lower than input Answer: B. Va = −k·Vs/(1 − k): negative, and |Va| is below Vs for k < 0.5 and above for k > 0.5. (Key point: Polarity and range) 2. What happens in Mode 2 (Q OFF)? A. L is across Vs and iL rises B. The energy in L is transferred through Dm to C and load, making the output negative C. C charges from the source directly D. The inductor current becomes zero immediately Answer: B. The inductor keeps the current flowing through C, load and Dm, with inductor voltage equal to the negative Va. (Key point: Mode 2) 3. Which condition gives the average output voltage? A. Vs·kT = −Va(1 − k)T (equal rise and fall of inductor current) B. Vs(1 − k) = Va·k C. Vs + Va = L·ΔI D. Vs·T = Va·kT Answer: A. Equating rise Vs·t1/L and fall −Va·t2/L gives Va = −Vs·k/(1 − k). (Key point: Output derivation) 4. What is the peak-to-peak inductor ripple current? A. Vs·k/(fL) B. Vs/(8fL) C. Va·k/(fC) D. Vs(1 − k)/(fL) Answer: A. During t1 = kT the full Vs is across L, so ΔI = Vs·t1/L = Vs·k/(fL). 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