--- title: "Series full converters: average and rms output" subject: "Power Electronics" chapter: 2 key: "semester_7/power_electronics:series-full-converters-avg-rms" source: https://www.bctengineer.com/q/semester_7/power_electronics/series-full-converters-avg-rms site: BCT Engineering --- # Series full converters: average and rms output Power Electronics · Chapter 2: Single phase AC to DC conversion - Asked in: 2071 Shrawan, 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/series-full-converters-avg-rms - Whole chapter: https://www.bctengineer.com/md/semester_7/power_electronics/chapter-2 ## Question Describe the series operation of two single phase full converter to obtain high output voltage. Also find the average value and rms value of the output voltage waveform. ## Short answer Two full converters are connected in series to give high dc voltage and better power factor. With α1=0, average output is (2Vm/π)(1+cosα2), and RMS output is √2 Vm [(π−α2+0.5 sin2α2)/π]^1/2; maximum average is 4Vm/π. ## Answer **Series operation of two full converters** is used to obtain a high dc output voltage and a better input power factor. Two single-phase full converters are fed from two secondary windings of one transformer (each secondary $v_s = V_m\sin\omega t$), and their dc outputs are connected in series. ```text Sec 1 --> [Converter 1, a1] (+)--------+ (-) | | vo1 | (+) Load Sec 2 --> [Converter 2, a2] (Io) (-) vo2 | +---------------------+ Sec 1, Sec 2: two secondaries of one transformer vo = vo1 + vo2 ``` #### Operation - Output: $v_o = v_{o1} + v_{o2}$, each converter carrying the same load current $I_a$. - **Rectification:** converter 1 is held at $\alpha_1 = 0$ (full output, acts like a diode bridge); converter 2's $\alpha_2$ is varied from $0$ to $\pi$. Output goes from $4V_m/\pi$ down to $0$. - **Inversion:** converter 1 held at $\alpha_1 = \pi$, $\alpha_2$ varied from $0$ to $\pi$; output goes from $0$ to $-4V_m/\pi$. - Since only one converter is phase controlled at a time, the reactive power and harmonics are lower than in a single converter giving the same voltage. With $\alpha_1 = 0$, over one interval from $\alpha_2$ to $\pi + \alpha_2$: | Interval | $v_{o1}$ | $v_{o2}$ | $v_o$ | |---|---|---|---| | $\alpha_2$ – $\pi$ | $V_m\sin\omega t$ | $V_m\sin\omega t$ | $2V_m\sin\omega t$ | | $\pi$ – $\pi+\alpha_2$ | $-V_m\sin\omega t$ | $V_m\sin\omega t$ | $0$ | ```text vo ___ ___ | \ 2Vm | \ (2Vm sin wt from a2 ------+----\---------+----\--- to 180, zero from a2 180 180+a2 180 to 180+a2) ``` #### Average output voltage General case (continuous current): $$ V_{dc} = V_{dc1} + V_{dc2} = \frac{2V_m}{\pi}\left(\cos\alpha_1 + \cos\alpha_2\right) $$ With $\alpha_1 = 0$: $$ V_{dc} = \frac{1}{\pi}\int_{\alpha_2}^{\pi}2V_m\sin\omega t\,d(\omega t) = \frac{2V_m}{\pi}\left(1 + \cos\alpha_2\right) $$ Maximum value (both $\alpha = 0$): $V_{dm} = \frac{4V_m}{\pi}$. #### RMS output voltage With $\alpha_1 = 0$: $$ \begin{aligned} V_{rms}^2 &= \frac{1}{\pi}\int_{\alpha_2}^{\pi}4V_m^2\sin^2\omega t\,d(\omega t) = \frac{2V_m^2}{\pi}\left[\omega t - \frac{\sin2\omega t}{2}\right]_{\alpha_2}^{\pi} \\ &= \frac{2V_m^2}{\pi}\left(\pi - \alpha_2 + \frac{\sin2\alpha_2}{2}\right) \end{aligned} $$ $$ V_{rms} = \sqrt2\,V_m\left[\frac{1}{\pi}\left(\pi - \alpha_2 + \frac{\sin2\alpha_2}{2}\right)\right]^{1/2} $$ Special case $\alpha_1 = \alpha_2 = \alpha$: $v_o = 2v_{o1}$, so $V_{dc} = \frac{4V_m}{\pi}\cos\alpha$ and $V_{rms} = 2 \times \frac{V_m}{\sqrt2} = \sqrt2\,V_m$. #### Input power factor With $\alpha_1 = 0$, the fundamental input current is $I_{s1} = \frac{4\sqrt2 I_a}{\pi}\cos\frac{\alpha_2}{2}$ (referred to the primary, 1:1 ratio) lagging by $\alpha_2/2$, so the displacement factor is $\cos(\alpha_2/2)$, much better than $\cos\alpha$ of a single converter. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. How are the two converters combined for high voltage? A. Their dc outputs are added in series, each carrying the same load current B. Their ac inputs are in series C. Their outputs are in parallel D. One converter charges the other Answer: A. vo = vo1 + vo2 with each converter fed from its own transformer secondary. (Key point: Series connection) 2. How is the output varied in rectification mode? A. Both alpha values equal pi B. alpha1 = 0 and alpha2 varied from 0 to pi, so output falls from 4Vm/pi to 0 C. alpha1 varied and alpha2 = pi D. Both are fixed at 90 degrees Answer: B. Converter 1 acts as a diode bridge while converter 2 is controlled. (Key point: Rectification control) 3. With alpha1 = 0, what is Vdc? A. (2Vm/pi) cos(alpha2) B. (4Vm/pi) cos(alpha2) C. (2Vm/pi)(1 + cos(alpha2)) D. (Vm/pi)(1 + cos(alpha2)) Answer: C. vo = 2Vm sin(wt) from alpha2 to pi, so Vdc = (2Vm/pi)(1 + cos(alpha2)). (Key point: Average voltage) 4. Why is the input power factor better than for a single converter? A. Both converters are fully phase controlled B. Only one converter is phase controlled at a time, giving displacement factor cos(alpha2/2) C. The transformer ratio is higher D. The load is resistive Answer: B. Reactive power and harmonics are lower than a single converter giving the same output. 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