--- title: "Single-phase diode bridge, average and fundamental" subject: "Power Electronics" chapter: 2 key: "semester_7/power_electronics:diode-bridge-220v-fundamental" source: https://www.bctengineer.com/q/semester_7/power_electronics/diode-bridge-220v-fundamental site: BCT Engineering --- # Single-phase diode bridge, average and fundamental Power Electronics · Chapter 2: Single phase AC to DC conversion - Asked in: 2075 Chaitra, 8 marks - Video: [Power Electronics | Lecture - 4A | Single-Phase Full-Wave Diode Rectifiers - Part-1](https://www.youtube.com/watch?v=Dz6SvQzfWIA) — Engineering Devotion - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_7/power_electronics/diode-bridge-220v-fundamental - Whole chapter: https://www.bctengineer.com/md/semester_7/power_electronics/chapter-2 ## Question Draw the circuit diagram and waveform of output voltage of a single phase full bridge diode rectifier. The input ac voltage is 220V, 50 Hz. Calculate average value and fundamental component of output voltage. ## Short answer Average output voltage Vdc=2Vm/π=198.07 V. The fundamental ripple component of the output is at 100 Hz, with peak value 132.05 V (93.37 V rms). ## Answer A **single-phase full bridge diode rectifier** uses four diodes; two diagonal diodes conduct in each half cycle so that the load always gets a positive voltage $v_o = |v_s|$. ```text P (+) +-------+-------+ | | D1 D3 | | a o----+ +----o b | | D4 D2 | | +-------+-------+ N (-) D1, D3: cathodes to P; D4, D2: anodes to N ``` #### Operation - Positive half cycle ($a$ positive): D1 and D2 conduct; $v_o = v_s$. - Negative half cycle ($b$ positive): D3 and D4 conduct; $v_o = -v_s$. - Each diode blocks a peak reverse voltage of $V_m$. ```text vs ___ / \ / ---/-----\-------/--- \_____/ vo ___ ___ / \ / \ ---/-----\-/-----\--- 0 180 360 (deg) vo = |Vm sin wt|, period 180 deg (100 Hz) ``` #### Average value $V_m = \sqrt2 \times 220 = 311.13$ V. $$ \begin{aligned} V_{dc} &= \frac{1}{\pi}\int_0^{\pi} V_m\sin\omega t\, d(\omega t) = \frac{2V_m}{\pi} \\ &= \frac{2 \times 311.13}{\pi} = 198.07\ \text{V} \end{aligned} $$ #### Fundamental (lowest-frequency ac) component of output The output has period $\pi$, so its Fourier series has only even harmonics of the supply frequency: $$ v_o(t) = \frac{2V_m}{\pi} - \frac{4V_m}{3\pi}\cos2\omega t - \frac{4V_m}{15\pi}\cos4\omega t - \frac{4V_m}{35\pi}\cos6\omega t - \cdots $$ General term: $\frac{4V_m}{\pi(n^2-1)}$ for $n = 2, 4, 6, \ldots$ The **fundamental ripple component** of the output is the $n = 2$ term at $2 \times 50 = 100$ Hz: $$ \begin{aligned} V_{2,peak} &= \frac{4V_m}{3\pi} = \frac{4 \times 311.13}{3\pi} = 132.05\ \text{V} \\ V_{2,rms} &= \frac{132.05}{\sqrt2} = 93.37\ \text{V} \end{aligned} $$ (The next component, at 200 Hz, has peak $\frac{4V_m}{15\pi} = 26.41$ V.) **Answer: $V_{dc} = 198.07$ V; fundamental output ripple = 132.05 V peak (93.37 V rms) at 100 Hz.** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. What is Vdc for a 220 V full-bridge diode rectifier? A. 155.6 V B. 311.13 V C. 198.07 V D. 220 V Answer: C. Vdc = 2Vm/pi = 2 x 311.13/pi = 198.07 V. (Key point: Average output) 2. What is the lowest ac frequency in the output (fundamental ripple) for a 50 Hz supply? A. 50 Hz B. 100 Hz C. 150 Hz D. 200 Hz Answer: B. The output period is pi, so only even harmonics appear and the lowest is 2 x 50 = 100 Hz. (Key point: Ripple frequency) 3. What is the peak of the fundamental ripple component? A. 4Vm/(3pi) = 132.05 V B. 2Vm/pi = 198.07 V C. 4Vm/(15pi) = 26.41 V D. Vm = 311.13 V Answer: A. The n = 2 term has amplitude 4Vm/(pi(n^2-1)) = 4Vm/(3pi). 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