--- title: "Step-up chopper and controllable power transfer" subject: "Power Electronics" chapter: 4 key: "semester_7/power_electronics:step-up-chopper-power-transfer" source: https://www.bctengineer.com/q/semester_7/power_electronics/step-up-chopper-power-transfer site: BCT Engineering --- # Step-up chopper and controllable power transfer Power Electronics · Chapter 4: DC chopper - Asked in: 2078 Bhadra, 8 marks - Video: [41. Step-Up Chopper Operation](https://www.youtube.com/watch?v=vykitDSnOtY) — Electronic Waves - 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-power-transfer - Whole chapter: https://www.bctengineer.com/md/semester_7/power_electronics/chapter-4 ## Question Write down the principle of step up chopper with neat diagram and waveforms. Derive the output value of load voltage and determine the condition for controllable power transfer range. ## Short answer During ON, inductor stores energy; during OFF, it releases energy in series with source, boosting voltage. Output voltage V0 = V_s/(1 - k). For controllable power transfer from a low voltage V_s to a higher fixed E, current must fall during OFF, requiring 0 ≤ V_s ≤ E. ## Answer A step-up (boost) chopper gives an average output voltage higher than the input. It stores energy in an inductor while the switch is ON and releases it, added to the source, into the load while the switch is OFF. #### Principle and circuit ```text L D +---mmm---+---|>|---+------+ | iL | | | Vs CH ===C Load V0 | | | | +---------+---------+------+ ``` - **CH ON ($0 < t < T_{on}$):** D is reverse biased. $V_s$ is across L, so $i_L$ rises linearly from $I_1$ to $I_2$; energy is stored in L. The capacitor supplies the load. - **CH OFF ($T_{on} < t < T$):** the inductor voltage reverses ($L\,di/dt$ adds to $V_s$), D conducts and current flows to the load at voltage $V_0 = V_s + L\,di/dt$. $i_L$ falls from $I_2$ to $I_1$. ```text vCH V0| ____ ____ | | | | | 0|___| |____| |___ t iL I2| /\ /\ | / \ / \ I1| / \____/ \___ t Ton Toff ``` #### Output voltage In steady state, the volt-second balance on L (average inductor voltage = 0): $$ \begin{aligned} V_s T_{on} &= (V_0 - V_s)\,T_{off} \\ V_0 &= V_s\,\frac{T_{on}+T_{off}}{T_{off}} = \frac{V_s}{1-k}, \qquad k = \frac{T_{on}}{T} \end{aligned} $$ The same result follows from energy balance: energy put into L during $T_{on}$ is $V_s I T_{on}$, energy given out during $T_{off}$ is $(V_0 - V_s) I T_{off}$. As $k$ goes from 0 to 1, $V_0$ rises from $V_s$ towards infinity (in practice limited by losses). #### Controllable power transfer (energy from a low voltage to a higher voltage) The same circuit transfers energy from a low-voltage source $V_s$ (e.g. a dc motor's back EMF in regenerative braking) into a fixed higher-voltage dc source $E$ (supply or battery) through D. With resistance $R$ of the source/inductor: $$ \begin{aligned} \text{ON: } & V_s = L\frac{di_1}{dt} + Ri_1 \\ \text{OFF: } & V_s = L\frac{di_2}{dt} + Ri_2 + E \end{aligned} $$ For the current to be controllable: 1. During ON, current must rise: $di_1/dt > 0$, which needs $V_s > 0$. 2. During OFF, current must fall: $di_2/dt < 0$, which needs $V_s < E$ (neglecting $Ri$). So the condition for controllable power transfer is $$ 0 \le V_s \le E $$ Using the boost relation, average values give $V_s = (1-k)E$; as $k$ varies from 1 to 0, $V_s$ can be anywhere from 0 to $E$. If $V_s > E$, current would rise in both intervals and could not be controlled by the switch; if $V_s \le 0$, no energy would flow to $E$. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. What happens while the switch of a step-up chopper is ON? A. Diode is reverse biased, Vs is across L, iL rises and the capacitor feeds the load B. The inductor releases energy to the load C. The diode conducts and load gets Vs + L di/dt D. The load is disconnected and iL falls Answer: A. During ON, the inductor stores energy and D blocks, so the capacitor supplies the load. (Key point: Principle of energy storage) 2. What is the average output voltage of the step-up chopper? A. V0 = k Vs B. V0 = Vs/(1-k) C. V0 = Vs(1-k) D. V0 = Vs/k Answer: B. Volt-second balance Vs Ton = (V0 - Vs) Toff gives V0 = Vs/(1-k). (Key point: Output voltage) 3. What is the condition for controllable power transfer from Vs to a higher source E? A. Vs >= E B. Vs = 2E C. 0 <= Vs <= E D. Vs <= 0 Answer: C. Current must rise when ON (needs Vs > 0) and fall when OFF (needs Vs < E). (Key point: Condition for controllable power transfer) 4. If Vs > E, what happens? A. Current falls in both intervals B. No current flows at all C. Output voltage becomes exactly E D. Current rises in both intervals and cannot be controlled by the switch Answer: D. With Vs > E, di/dt stays positive even in the OFF interval, so the chopper loses control. (Key point: Why Vs > E fails) ## Related in Power Electronics - [Step-up chopper operation and output voltage](https://www.bctengineer.com/question.md?key=semester_7%2Fpower_electronics%3Astep-up-chopper-principle) — Power Electronics, chapter 4, asked 8× - [Step-down chopper: fundamental of load current](https://www.bctengineer.com/question.md?key=semester_7%2Fpower_electronics%3Astep-down-chopper-fundamental-load-current) — Power Electronics, chapter 4 - [Step-up chopper average and rms output](https://www.bctengineer.com/question.md?key=semester_7%2Fpower_electronics%3Astep-up-chopper-avg-rms) — Power Electronics, chapter 4 - [Step-down chopper RLE load, 5 kHz](https://www.bctengineer.com/question.md?key=semester_7%2Fpower_electronics%3Astep-down-chopper-rle-5khz) — Power Electronics, chapter 4 - [Step-down chopper RLE load continuity](https://www.bctengineer.com/question.md?key=semester_7%2Fpower_electronics%3Astep-down-chopper-rle-200v) — Power Electronics, chapter 4 - [Step-down chopper with DC motor load](https://www.bctengineer.com/question.md?key=semester_7%2Fpower_electronics%3Astep-down-chopper-dc-motor) — Power Electronics, chapter 4 ## Related in other subjects - [Chopper classification; step-down chopper RL load](https://www.bctengineer.com/question.md?key=semester_4%2Fadvanced_electronics%3Aclassify-chopper-step-down-rl) — Advanced Electronics, chapter 7 - [Step-down chopper with RL load](https://www.bctengineer.com/question.md?key=semester_4%2Fadvanced_electronics%3Astep-down-chopper-rl) — Advanced Electronics, chapter 7 - [Freewheeling diode; step-up chopper principle](https://www.bctengineer.com/question.md?key=semester_4%2Fadvanced_electronics%3Afreewheeling-diode-step-up-chopper) — Advanced Electronics, chapter 7 - [Chopper classification; step-up converter](https://www.bctengineer.com/question.md?key=semester_4%2Fadvanced_electronics%3Achopper-classification-step-up) — Advanced Electronics, chapter 7