--- title: "Half-wave diode battery charger, 10 A" subject: "Power Electronics" chapter: 2 key: "semester_7/power_electronics:half-wave-battery-charger-24v" source: https://www.bctengineer.com/q/semester_7/power_electronics/half-wave-battery-charger-24v site: BCT Engineering --- # Half-wave diode battery charger, 10 A Power Electronics · Chapter 2: Single phase AC to DC conversion - Asked in: 2072 Chaitra, 8 marks - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_7/power_electronics/half-wave-battery-charger-24v - Whole chapter: https://www.bctengineer.com/md/semester_7/power_electronics/chapter-2 ## Question For the circuit shown in figure below, the battery voltage is E = 12 V. The average charging current should be 10A. Calculate: i) The conduction angle of the diode ii) Power supplied to the battery iii) Average value of charging current iv) The rectifier efficiency [Figure: 24 V, 50 Hz ac source in series with R = 5 Ω, a diode and the battery E] ## Short answer With Vm=33.94 V and E=12 V, diode conducts from 20.7° to 159.3°, giving conduction angle 138.59°. Average charging current is 1.097 A, battery power 13.17 W, rectifier efficiency 40.86% for R=5 Ω; the stated 10 A is impossible with that R. ## Answer In a **half-wave battery charger**, the diode conducts only when the supply voltage is greater than the battery emf $E$, i.e. from $\alpha$ to $\pi-\alpha$ in the positive half cycle, where $V_m\sin\alpha = E$. The resistor limits the current. ```text R = 5 ohm D +----/\/\/\----->|----+ | | + (~) 24 V, 50 Hz === E = 12 V | | - +---------------------+ ``` ```text vs ___ / | \ --- E = 12 V level --/--+--\-----------/-- io /\ (current pulse from a to 180-a, ----/ \------ zero for the rest of the cycle) a 180-a 360 ``` *Note on data:* with $V_s = 24$ V and $R = 5\ \Omega$ the peak current is only $(33.94-12)/5 = 4.39$ A, so an average of 10 A is impossible. The quantities below are therefore calculated for the given $R = 5\ \Omega$; the resistance that would give 10 A is shown at the end. #### i) Conduction angle of the diode $$ \begin{aligned} V_m &= \sqrt2 \times 24 = 33.94\ \text{V} \\ \alpha &= \sin^{-1}\frac{12}{33.94} = 20.70^\circ \\ \theta_c &= 180^\circ - 2\alpha = 138.59^\circ\ (2.4189\ \text{rad}) \end{aligned} $$ The diode conducts from $20.70^\circ$ to $159.30^\circ$. #### iii) Average charging current $$ \begin{aligned} I_o &= \frac{1}{2\pi R}\left[2V_m\cos\alpha - E(\pi-2\alpha)\right] \\ &= \frac{1}{2\pi(5)}\left[63.498 - 29.026\right] = 1.097\ \text{A} \end{aligned} $$ #### ii) Power supplied to the battery $$ P_{batt} = E\,I_o = 12 \times 1.097 = 13.17\ \text{W} $$ #### iv) Rectifier efficiency RMS current: $$ \begin{aligned} I_{rms}^2 &= \frac{1}{2\pi R^2}\left[\left(\frac{V_m^2}{2}+E^2\right)(\pi-2\alpha) + \frac{V_m^2}{2}\sin2\alpha - 4V_mE\cos\alpha\right] \\ &= \frac{1741.58 + 380.99 - 1523.95}{157.08} = 3.8109 \\ I_{rms} &= 1.952\ \text{A}, \qquad P_R = I_{rms}^2R = 19.05\ \text{W} \end{aligned} $$ $$ \eta = \frac{P_{batt}}{P_{batt}+P_R} = \frac{13.17}{13.17 + 19.05} = 40.86\ \% $$ **Answer: conduction angle $= 138.59^\circ$; $I_o = 1.097$ A; $P_{batt} = 13.17$ W; $\eta = 40.86$ %.** If an average current of 10 A is really required, the conduction angle stays $138.59^\circ$, but $R$ must be reduced to $\frac{63.498 - 29.026}{2\pi \times 10} = 0.549\ \Omega$; then $P_{batt} = 12 \times 10 = 120$ W. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. For Vm = 33.94 V and E = 12 V, what is the diode conduction angle? A. 20.70 degrees B. 159.30 degrees C. 138.59 degrees D. 180 degrees Answer: C. alpha = asin(12/33.94) = 20.70 degrees, so theta = 180 - 2*20.70 = 138.59 degrees. (Key point: Conduction angle) 2. How does the half-wave charger average current expression differ from the bridge one? A. It uses 1/(2*pi*R) because current flows only once per cycle B. It uses 1/(pi*R) C. It does not depend on E D. It uses 2Vm instead of Vm Answer: A. Only one pulse occurs per cycle, so the average is taken over 2*pi, giving Io = 1.097 A. (Key point: Half-wave average current) 3. Why can an average of 10 A not be achieved with R = 5 ohm? A. The diode would overheat B. The battery voltage is too low C. The conduction angle is too large D. The peak current is only (33.94-12)/5 = 4.39 A Answer: D. The average cannot exceed the peak, so R would need to be reduced to about 0.549 ohm. (Key point: Why 10 A is impossible) 4. How is the rectifier efficiency obtained? A. P_batt/(P_batt + I_rms^2 R) = 40.86 % B. Vdc/Vm C. E/Vm D. P_R/P_batt Answer: A. P_batt = 13.17 W and P_R = 19.05 W, giving 13.17/32.22 = 40.86 %. (Key point: Efficiency) ## Related in Power Electronics - [Battery charger bridge 30 V: conduction, power](https://www.bctengineer.com/question.md?key=semester_7%2Fpower_electronics%3Abattery-bridge-30v-conduction-power) — Power Electronics, chapter 2, asked 2× - [Battery charger bridge 30 V: conduction period](https://www.bctengineer.com/question.md?key=semester_7%2Fpower_electronics%3Abattery-bridge-30v-conduction) — Power Electronics, chapter 2, asked 4× - [Battery charger bridge: average current, efficiency](https://www.bctengineer.com/question.md?key=semester_7%2Fpower_electronics%3Abattery-bridge-30v-avg-efficiency) — Power Electronics, chapter 2 - [Battery charger bridge 15 V: limit current to 0.5 A](https://www.bctengineer.com/question.md?key=semester_7%2Fpower_electronics%3Abattery-bridge-15v-extra-resistance) — Power Electronics, chapter 2 - [Battery charger bridge 15 V: currents, power, efficiency](https://www.bctengineer.com/question.md?key=semester_7%2Fpower_electronics%3Abattery-bridge-15v-efficiency) — Power Electronics, chapter 2, asked 3× - [Battery charger bridge 50 V: power and efficiency](https://www.bctengineer.com/question.md?key=semester_7%2Fpower_electronics%3Abattery-bridge-50v-efficiency) — Power Electronics, chapter 2 ## Related in other subjects - [Battery charger circuit and charging methods](https://www.bctengineer.com/question.md?key=semester_6%2Findustrial_power_distribution_and_illumination%3Abattery-charger-two-methods) — Industrial Power Distribution and Illumination, chapter 9 - [SCR analogy; half-wave rectifier for 1 A average](https://www.bctengineer.com/question.md?key=semester_4%2Fadvanced_electronics%3Ascr-analogy-hwr-120v) — Advanced Electronics, chapter 7 - [Diode circuit current, eta 1.74, 10 V](https://www.bctengineer.com/question.md?key=semester_3%2Felectronics_devices_and_circuit%3Adiode-current-eta-1p74-10v) — Electronics Devices and Circuit, chapter 1, asked 4× - [SCR half-wave rectifier numerical (200 V peak)](https://www.bctengineer.com/question.md?key=semester_4%2Fadvanced_electronics%3Ascr-hwr-1ma-200v) — Advanced Electronics, chapter 7