--- title: "Step-up chopper average and rms output" subject: "Power Electronics" chapter: 4 key: "semester_7/power_electronics:step-up-chopper-avg-rms" source: https://www.bctengineer.com/q/semester_7/power_electronics/step-up-chopper-avg-rms site: BCT Engineering --- # Step-up chopper average and rms output Power Electronics · Chapter 4: DC chopper - Asked in: 2074 Asoj, 8 marks - Video: [STEP UP CHOPPERS](https://www.youtube.com/watch?v=fff8j3pGdrE) — Prof. Barapate's Tutorials - 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-avg-rms - Whole chapter: https://www.bctengineer.com/md/semester_7/power_electronics/chapter-4 ## Question Explain the operation of step up chopper. Derive the expression for the average and rms value of output voltage. ## Short answer Inductor stores energy when switch is ON, then releases it in series with input during OFF, raising output voltage. Average output V0 = V_s/(1 - k). RMS of the switched waveform v_x (before filter) is V_s/√(1 - k); with large filter, V0,rms ≈ V0. ## Answer A step-up chopper produces an average output voltage higher than the source. An inductor in series with the source stores energy while the switch is ON and pushes it into the load, in series with $V_s$, while the switch is OFF. #### Circuit and operation ```text L D +---mmm---+---|>|---+------+ | iL | vx | | Vs CH ===C Load V0 | | | | +---------+---------+------+ ``` 1. **CH ON ($0 \le t \le T_{on}$):** node $x$ is shorted to the negative rail ($v_x = 0$). D is reverse biased by $V_0$. $V_s$ appears across L, so $i_L$ rises linearly: $L\,di/dt = V_s$. The capacitor feeds the load. 2. **CH OFF ($T_{on} \le t \le T$):** the inductor EMF reverses and adds to $V_s$; D conducts, $v_x = V_0$, and $i_L$ falls: $L\,di/dt = V_s - V_0$ (negative). ```text vx V0 | ____ ____ | | | | | 0 |___| |____| |___ t iL I2 | /\ /\ | / \ / \ I1 | / \____/ \___ t 0 Ton T ``` #### Average output voltage Rise of current in $T_{on}$ equals the fall in $T_{off}$ (steady state): $$ \begin{aligned} \Delta I &= \frac{V_s}{L}T_{on} = \frac{V_0 - V_s}{L}T_{off} \\ V_0\,T_{off} &= V_s (T_{on} + T_{off}) = V_s T \\ V_0 &= \frac{V_s\,T}{T - T_{on}} = \frac{V_s}{1-k} \end{aligned} $$ Equivalently, the switch-node voltage $v_x$ is 0 for $kT$ and $V_0$ for $(1-k)T$; its average must equal $V_s$ (no average voltage across L): $(1-k)V_0 = V_s$. For $k = 0 \to 1$, $V_0$ varies from $V_s$ to a very large value; e.g. $k = 0.5$ gives $V_0 = 2V_s$. #### RMS value of output voltage - With a large filter capacitor the load voltage is nearly ripple-free, so $V_{0,rms} \approx V_0 = \dfrac{V_s}{1-k}$. - The chopped voltage at the chopper terminals ($v_x$, the voltage the load sees before filtering) is a pulse of height $V_0$ for $(1-k)T$: $$ \begin{aligned} V_{x,rms} &= \left[\frac{1}{T}\int_{kT}^{T}V_0^2\,dt\right]^{1/2} = V_0\sqrt{1-k} \\ &= \frac{V_s}{1-k}\sqrt{1-k} = \frac{V_s}{\sqrt{1-k}} \end{aligned} $$ and its average is $V_{x,avg} = (1-k)V_0 = V_s$. **Example:** $V_s = 100$ V, $k = 0.6$: $V_0 = 100/0.4 = 250$ V; $V_{x,rms} = 100/\sqrt{0.4} = 158.1$ V. **Uses:** regenerative braking of dc motors, battery charging from a lower voltage, boost SMPS. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. During CH ON, what is the inductor equation and what feeds the load? A. L di/dt = Vs; the capacitor feeds the load B. L di/dt = Vs - V0; the source feeds the load C. L di/dt = -V0; the inductor feeds the load D. L di/dt = 0; the diode feeds the load Answer: A. With CH shorting node x, Vs is across L and D is reverse biased. (Key point: Operation when ON) 2. Equating current rise in Ton and fall in Toff gives: A. V0 = k Vs B. V0 = Vs/(1-k) C. V0 = Vs/k D. V0 = Vs(1+k) Answer: B. Vs Ton/L = (V0 - Vs)Toff/L gives V0 = Vs T/Toff = Vs/(1-k). (Key point: Average output derivation) 3. For Vs = 100 V and k = 0.6, V0 is: A. 60 V B. 160 V C. 250 V D. 100 V Answer: C. 100/(1 - 0.6) = 250 V. (Key point: Example) 4. The rms of the switch-node voltage vx (V0 for (1-k)T) is: A. V0 (1-k) B. V0 sqrt(k) C. Vs sqrt(1-k) D. V0 sqrt(1-k) = Vs/sqrt(1-k) Answer: D. sqrt((1/T) V0^2 (1-k)T) = V0 sqrt(1-k); average is (1-k)V0 = Vs. 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