--- title: "Buck regulator operation and derivation" subject: "Advanced Electronics" chapter: 8 key: "semester_4/advanced_electronics:buck-regulator-operation" source: https://www.bctengineer.com/q/semester_4/advanced_electronics/buck-regulator-operation site: BCT Engineering --- # Buck regulator operation and derivation Advanced Electronics · Chapter 8: Switched Power Supplies - Asked in: 2073 Shrawan, 7 marks; 2069 Asar, 3+4 marks (asked 2 times) - Video: [Buck Converter Operation and Voltage Equation](https://www.youtube.com/watch?v=H1F4xI2Vgcs) — Engineering with Prof. Kim - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_4/advanced_electronics/buck-regulator-operation - Whole chapter: https://www.bctengineer.com/md/semester_4/advanced_electronics/chapter-8 ## Question Explain the operation of Buck regulator with necessary circuits (diagram) and derivations. ## Short answer The Buck regulator steps down voltage. When switch ON, inductor current rises; when OFF, diode freewheels. Output voltage Va = k·Vs, inductor ripple ΔI = Va(Vs-Va)/(f L Vs), capacitor ripple ΔVc = ΔI/(8 f C). ## Answer A **Buck regulator** is a switching DC-DC converter whose average output voltage Va is **less than** the input Vs: Va = k·Vs. #### Circuit ```text Q (switch) L iL +Vs ---/ ----+------uuuu----+----+------+ | | | | Dm (↑) C LOAD Va | | | | − ---------+--------------+----+------+ control circuit drives Q (PWM, duty k) ``` #### Operation **Mode 1 (Q ON, 0 < t < t1 = kT):** Dm is reverse biased. Inductor current rises linearly from I1 to I2; L stores energy and current flows to C and the load. ```text Vs − Va = L·ΔI/t1 → t1 = L·ΔI/(Vs − Va) ``` **Mode 2 (Q OFF, t1 < t < T):** The inductor current continues through freewheeling diode Dm. Inductor voltage = −Va; current falls from I2 to I1. ```text Va = L·ΔI/t2 → t2 = L·ΔI/Va ``` #### Waveforms ```text vQ gate |‾‾‾|____|‾‾‾|____ 0 kT T iL I2 | /\ /\ I1 | / \__/ \__ vC/Va |~~~~~~~~~~~~~~ small ripple about Va ``` #### Derivations **Output voltage:** equate the two ΔI expressions: ```text (Vs − Va)·t1/L = Va·t2/L (Vs − Va)·kT = Va·(1 − k)T Va = k·Vs ``` **Ripple current of inductor:** ```text T = t1 + t2 = L·ΔI·Vs / (Va(Vs − Va)) ΔI = Va(Vs − Va)/(f·L·Vs) = Vs·k(1 − k)/(f·L) ``` **Ripple voltage of capacitor:** the AC part of iL (triangular) flows into C. Charge added in half a cycle: ```text ΔQ = (1/2)(T/2)(ΔI/2) = ΔI/(8f) ΔVc = ΔQ/C = ΔI/(8fC) ΔVc = Va(Vs − Va)/(8·L·C·f²·Vs) ``` **Critical values** for continuous inductor current and voltage: ```text Lc = (1 − k)R/(2f), Cc = (1 − k)/(16·L·f²) ``` Input current Is = k·Ia (ideal). Advantages: high efficiency (about 90 %), only one switch. Disadvantages: input current is pulsating, and no short-circuit protection is inherent. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. During Mode 1 (Q ON) of the buck regulator, what happens to the inductor current and Dm? A. iL rises linearly and Dm is reverse biased B. iL falls linearly and Dm conducts C. iL stays constant and Dm is reverse biased D. iL rises linearly and Dm conducts Answer: A. With Q on, Vs − Va appears across L so iL rises linearly; Dm is reverse biased. (Key point: Mode 1 operation) 2. What happens when Q turns OFF? A. The load is disconnected and current stops B. The inductor current continues through freewheeling diode Dm and falls from I2 to I1 C. The capacitor charges from the supply directly D. Dm blocks and L voltage becomes +Vs Answer: B. The inductor keeps the current flowing through Dm with inductor voltage −Va, so current falls. (Key point: Mode 2 operation) 3. How is Va = k·Vs obtained? A. By equating the current rise (Vs − Va)kT/L to the current fall Va(1 − k)T/L B. By equating the input and output powers with efficiency 50% C. By setting the capacitor ripple equal to inductor ripple D. By taking the RMS value of the switch voltage Answer: A. In steady state the rise and fall of iL are equal: (Vs − Va)kT = Va(1 − k)T, which gives Va = kVs. (Key point: Output voltage derivation) 4. What is the peak-to-peak ripple voltage of the capacitor? A. ΔI/(8fC) B. ΔI/(2fC) C. ΔI·f/C D. ΔI/(16fC) Answer: A. The AC part of iL is triangular; charge ΔQ = ΔI/(8f), so ΔVc = ΔQ/C = ΔI/(8fC). 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