--- title: "Battery charger bridge 50 V: power and efficiency" subject: "Power Electronics" chapter: 2 key: "semester_7/power_electronics:battery-bridge-50v-efficiency" source: https://www.bctengineer.com/q/semester_7/power_electronics/battery-bridge-50v-efficiency site: BCT Engineering --- # Battery charger bridge 50 V: power and efficiency Power Electronics · Chapter 2: Single phase AC to DC conversion - Asked in: 2071 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/battery-bridge-50v-efficiency - Whole chapter: https://www.bctengineer.com/md/semester_7/power_electronics/chapter-2 ## Question In figure below shows a full wave rectifier circuit which is used to charge a 24 V battery through 5Ω resistor. Draw the necessary waveforms related with the operation of this circuit and hence determine: i) Average value of charging current ii) Power supplied to the battery iii) Gross output from rectifier iv) Efficiency of rectifier [Figure: single-phase diode bridge (four diodes) fed from 50 V, 50 Hz; output V0 feeds R = 5 Ω in series with a battery marked E = 20 V in the figure; charging current io] ## Short answer Diodes conduct when supply exceeds battery emf. With Vm = 70.71 V, α = sin⁻¹(24/70.71) = 19.84°. Average charging current Io = [2Vm cosα – E(π–2α)]/(πR) = 4.727 A. Power to battery = 24 × 4.727 = 113.45 W. Gross output = P_batt + I_rms²R = 288.15 W. Efficiency = 113.45/288.15 = 39.37%. ## Answer In a **battery-charging rectifier**, a diode conducts only when the instantaneous supply voltage exceeds the battery emf $E$. In each half cycle current flows from $\alpha$ to $\pi-\alpha$, where $V_m\sin\alpha = E$, and is limited by $R$: $$ i_o = \frac{V_m\sin\omega t - E}{R}, \qquad \alpha < \omega t < \pi - \alpha $$ ```text io R P o--->---/\/\/---+ | + | === Eb (battery) - | N o---------------+ (P, N = output of the diode bridge) ``` Data: $V_s = 50$ V, 50 Hz, $R = 5\ \Omega$, $E = 24$ V (as stated in the question; the figure shows 20 V, results for 20 V are given at the end). Diodes ideal. #### Waveforms ```text vs ___ / \ / ---/-----\-----------/--- \___/ vo ___ ___ vo = |vs| while a diode pair / | \ / | \ conducts, = E otherwise ==/==+==\===/==+==\== E io /\ /\ ----/ \------/ \------- a 180-a 180+a 360-a ``` - D1, D2 conduct from $\alpha$ to $\pi-\alpha$; D3, D4 from $\pi+\alpha$ to $2\pi-\alpha$. - Outside these intervals all diodes are off and $v_o = E$. $$ \begin{aligned} V_m &= \sqrt2 \times 50 = 70.71\ \text{V} \\ \alpha &= \sin^{-1}\frac{E}{V_m} = \sin^{-1}\frac{24}{70.71} = 19.84^\circ = 0.3463\ \text{rad} \end{aligned} $$ Conduction angle $= \pi - 2\alpha = 140.32^\circ$ per half cycle. #### i) Average charging current $$ \begin{aligned} I_o &= \frac{1}{\pi}\int_{\alpha}^{\pi-\alpha}\frac{V_m\sin\omega t - E}{R}\,d(\omega t) = \frac{1}{\pi R}\left[2V_m\cos\alpha - E(\pi-2\alpha)\right] \\ &= \frac{1}{\pi \times 5}\left[2(70.71)(0.9406) - 24(2.4490)\right] = \frac{133.026 - 58.776}{15.7080} = 4.727\ \text{A} \end{aligned} $$ #### ii) Power supplied to the battery and iii) gross output $$ \begin{aligned} I_{rms}^2 &= \frac{1}{\pi R^2}\left[\left(\frac{V_m^2}{2}+E^2\right)(\pi-2\alpha) + \frac{V_m^2}{2}\sin2\alpha - 4V_mE\cos\alpha\right] \\ &= \frac{7533.16 + 1596.32 - 6385.26}{78.540} = 34.940 \\ I_{rms} &= 5.911\ \text{A} \end{aligned} $$ $$ \begin{aligned} P_{batt} &= E\,I_o = 24 \times 4.727 = 113.45\ \text{W} \\ P_R &= I_{rms}^2R = 34.940 \times 5 = 174.70\ \text{W} \\ P_{gross} &= P_{batt} + P_R = 113.45 + 174.70 = 288.15\ \text{W} \end{aligned} $$ #### iv) Efficiency of the rectifier With ideal diodes all ac input power appears as gross output; the useful output is the power stored in the battery: $$ \eta = \frac{P_{batt}}{P_{gross}} = \frac{113.45}{288.15} = 39.37\ \% $$ **Answer: $I_o = 4.727$ A, $P_{batt} = 113.45$ W, gross output $= 288.15$ W, $\eta = 39.37$ %.** If $E = 20$ V (figure value): $\alpha = 16.43^\circ$, $I_o = 5.366$ A, $P_{batt} = 107.32$ W, gross $= 322.37$ W, $\eta = 33.29$ %. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. What is vo outside the conduction intervals? A. Zero B. Equal to the battery emf E C. Equal to |vs| D. Equal to -E Answer: B. With all diodes off, the battery sets the output terminal voltage at E. (Key point: Output voltage when diodes are off) 2. For Vm = 70.71 V and E = 24 V, what is alpha and the conduction angle? A. alpha = 19.84 degrees, conduction 140.32 degrees B. alpha = 19.84 degrees, conduction 19.84 degrees C. alpha = 24 degrees, conduction 132 degrees D. alpha = 70.7 degrees, conduction 38.6 degrees Answer: A. alpha = asin(24/70.71) = 19.84 degrees, conduction = 180 - 2*19.84 = 140.32 degrees. (Key point: Conduction angle) 3. What is the average charging current? A. 5.37 A B. 4.727 A C. 5.91 A D. 2.36 A Answer: B. Io = [133.026 - 58.776]/(5*pi) = 4.727 A; 5.91 A is the rms. (Key point: Average charging current) 4. How is the efficiency defined here? A. P_batt/P_gross = 113.45/288.15 = 39.37 % B. P_gross/P_batt C. P_R/P_gross D. Vdc/Vm Answer: A. Useful output is the power stored in the battery; the rest is lost in the resistor. 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