--- title: "Step-up chopper operation and output voltage" subject: "Power Electronics" chapter: 4 key: "semester_7/power_electronics:step-up-chopper-principle" source: https://www.bctengineer.com/q/semester_7/power_electronics/step-up-chopper-principle site: BCT Engineering --- # Step-up chopper operation and output voltage Power Electronics · Chapter 4: DC chopper - Asked in: 2082 Baishakh, 8 marks; 2081 Baishakh, 8 marks; 2080 Bhadra, 8 marks; 2076 Chaitra, 8 marks; 2076 Asoj, 8 marks; 2074 Chaitra, 8 marks; 2072 Kartik, 8 marks; 2072 Chaitra, 8 marks (asked 8 times) - Video: [Step Up Chopper Part 1 : Operation & Derivation of Output Voltage](https://www.youtube.com/watch?v=8eXc8c__ImA) — Learning Electronics - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_7/power_electronics/step-up-chopper-principle - Whole chapter: https://www.bctengineer.com/md/semester_7/power_electronics/chapter-4 ## Question Explain the operation of a step-up DC chopper and derive the expression for average output voltage in terms of input voltage and duty cycle. ## Short answer A step-up chopper uses an inductor to store energy when the switch is on and deliver it with the source to the load when the switch is off. The average output voltage is V_o = V_s / (1 - D), where D is the duty cycle. ## Answer A **step-up (boost) chopper** is a dc–dc converter whose average output voltage is **greater than** the input voltage. It uses an inductor to store energy while the switch is ON and releases it, added to the source voltage, to the load while the switch is OFF. ```text L D +---^^^^^---+---->|---+--------+ | iL -> | | | | | | | (+) Vs S === C Load Vo | (chopper switch) | | | | | | +-----------+---------+--------+ ``` #### Operation Let the switch S (thyristor with commutation circuit, MOSFET, IGBT, etc.) be ON for $T_{on}$ and OFF for $T_{off}$; period $T = T_{on} + T_{off}$, duty cycle $D = T_{on}/T$. **Mode 1: S ON ($0 < t < T_{on}$)** - The inductor is connected directly across the source; diode D is reverse biased (blocked by the capacitor voltage). - $v_L = V_s$, so $i_L$ rises linearly from $I_1$ to $I_2$; energy is stored in L. - The capacitor supplies the load. **Mode 2: S OFF ($T_{on} < t < T$)** - The inductor current cannot change suddenly, so the inductor voltage reverses and adds to $V_s$; D becomes forward biased. - Current flows from source + inductor to the load and capacitor: $v_L = V_s - V_o$ (negative), so $i_L$ falls from $I_2$ to $I_1$. - Energy from both the source and the inductor is delivered to the load, so $V_o > V_s$. ```text vL +Vs +------+ +------+ | | | | 0 ---+ | | | +------+ +--- -(Vo-Vs) iL /\ /\ I2 (max) / \ / \ / \ / \ I1 (min) ||| ``` #### Derivation of average output voltage In steady state the average voltage across an inductor over one period is zero (volt-second balance), i.e. the rise in current in Mode 1 equals the fall in Mode 2: $$ \begin{aligned} \Delta I &= \frac{V_s}{L}T_{on} = \frac{V_o - V_s}{L}T_{off} \\ V_sT_{on} &= (V_o - V_s)T_{off} \\ V_o T_{off} &= V_s(T_{on} + T_{off}) = V_sT \end{aligned} $$ $$ V_o = V_s\frac{T}{T_{off}} = V_s\frac{T}{T - T_{on}} = \frac{V_s}{1 - D} $$ Alternative (energy method): energy taken by L in Mode 1, $V_sIT_{on}$, equals energy given in Mode 2, $(V_o - V_s)IT_{off}$, which leads to the same result (assuming ripple-free $I$). #### Discussion - As $D$ varies from 0 to 1, $V_o$ varies from $V_s$ to (theoretically) infinity; in practice losses limit the gain, and D is kept below about 0.8–0.9. - Input current $I_s = I_o/(1-D)$ for a lossless converter, since $V_sI_s = V_oI_o$. - Example: $V_s = 100$ V, $D = 0.6$: $V_o = 100/(1 - 0.6) = 250$ V. - Applications: regenerative braking of dc motors (motor emf pumping energy back into a higher-voltage supply), battery-powered systems, PFC front ends and solar PV converters. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. When the switch S of a step-up chopper is ON, what happens? A. The inductor is across the source, iL rises linearly and diode D is reverse biased B. The inductor discharges into the load through D C. The source is disconnected from the inductor D. The capacitor charges from the inductor Answer: A. With S ON, v_L = Vs, so iL rises and energy is stored in L while the capacitor supplies the load. (Key point: Mode 1 operation) 2. When S turns OFF, why can the load voltage exceed the source voltage? A. The capacitor doubles the voltage B. The inductor voltage reverses and adds to Vs while D conducts C. The diode amplifies voltage D. The load resistance rises Answer: B. The inductor current cannot change suddenly, so its voltage reverses and adds to Vs, delivering both source and inductor energy to the load. (Key point: Why output exceeds input) 3. Equating the current rise Vs Ton / L to the fall (Vo - Vs) Toff / L gives: A. Vo = Vs D B. Vo = Vs / (1 - D) C. Vo = Vs / D D. Vo = Vs (1 - D) Answer: B. Vs Ton = (Vo - Vs) Toff gives Vo Toff = Vs T, so Vo = Vs T / Toff = Vs / (1 - D). (Key point: Output voltage expression) 4. For Vs = 100 V and D = 0.6, the output voltage is: A. 60 V B. 160 V C. 250 V D. 40 V Answer: C. Vo = 100 / (1 - 0.6) = 250 V. 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