--- title: "Series full converters: rectification and inversion" subject: "Power Electronics" chapter: 2 key: "semester_7/power_electronics:series-full-converters-rect-inversion" source: https://www.bctengineer.com/q/semester_7/power_electronics/series-full-converters-rect-inversion site: BCT Engineering --- # Series full converters: rectification and inversion Power Electronics · Chapter 2: Single phase AC to DC conversion - Asked in: 2081 Baishakh, 8 marks; 2080 Bhadra, 8 marks; 2074 Chaitra, 8 marks; 2074 Asoj, 8 marks (asked 4 times) - 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-rect-inversion - Whole chapter: https://www.bctengineer.com/md/semester_7/power_electronics/chapter-2 ## Question Explain the series connection of two single phase full converter with necessary circuit diagram and waveforms. How can these series connected circuits be operated in rectification mode and inversion mode? ## Short answer Two single-phase full converters fed from separate secondaries have dc outputs in series: Vo=(2Vm/pi)(cos alpha1+cos alpha2). Rectification: alpha1=0, vary alpha2, Vo=(2Vm/pi)(1+cos alpha2). Inversion: alpha1=180, vary alpha2, Vo=(2Vm/pi)(cos alpha2-1), giving better power factor and lower harmonics. ## Answer Two single-phase full converters are fed from two separate transformer secondaries and their d.c. outputs are connected **in series**. This is used for high-voltage d.c. output (for example traction) and, with sequence control, to improve the input power factor and reduce harmonics. #### Circuit ```text +------------------+ ~ ---+ S1 | Full converter 1|--+ + | )) | T1 T3 / T4 T2 | | Vo1 pri | +------------------+ | (one | +----+ trans-| +------------------+ | | load former| S2 | Full converter 2|--+ | (RLE) two | )) | T1'T3'/T4'T2' | | | sec.)| +------------------+ | | ~ ---+ +----+ - ``` Each secondary has rms voltage $V_s$ and peak $V_m$. The total output is $$ V_o = V_{o1} + V_{o2} = \frac{2V_m}{\pi}(\cos\alpha_1 + \cos\alpha_2) $$ The maximum output is $\frac{4V_m}{\pi}$ (both at $\alpha = 0$). #### Rectification mode (power flow a.c. to d.c.) Converter 1 is kept at $\alpha_1 = 0^\circ$ (output $+\frac{2V_m}{\pi}$) and converter 2 is phase-controlled, $\alpha_2$ from $180^\circ$ down to $0^\circ$: $$ V_o = \frac{2V_m}{\pi}(1 + \cos\alpha_2) $$ At $\alpha_2 = 180^\circ$ the two outputs cancel ($V_o = 0$); at $\alpha_2 = 0^\circ$, $V_o = \frac{4V_m}{\pi}$. So the output is varied smoothly from 0 to $\frac{4V_m}{\pi}$ with power flowing from the a.c. supply to the load. #### Inversion mode (power flow d.c. to a.c.) The load must contain a d.c. source $E$ of reversed polarity (for example a motor in regenerative braking). One converter is held at $\alpha_1 = 180^\circ$ (full negative voltage, $-\frac{2V_m}{\pi}$) and the other is controlled between $0^\circ$ and $180^\circ$: $$ V_o = \frac{2V_m}{\pi}(\cos\alpha_2 - 1) $$ so $V_o$ ranges from 0 to $-\frac{4V_m}{\pi}$; current direction stays the same, so power flows back to the a.c. supply. (In practice the angle is kept below about $165^\circ$ to allow a margin for commutation.) #### Waveforms (rectification, $\alpha_1 = 0$, $\alpha_2 = \alpha$, constant load current $I_o$) ```text vo1 /\ /\ /\ full-wave, alpha1 = 0 ----/--\/--\/--\-- vo2 _ _ _ full converter, alpha2, -----|-\-.|-\-.|-\ negative from pi to pi+a is1 ____ ____ ----| |____| square, in phase is2 ____ __ ------| |____| square, lags by alpha is (sum, on primary): ____ level 2Io from a to pi ----.___| |___. level 0 from 0 to a ``` - $v_o = v_{o1} + v_{o2}$: during $0$ to $\alpha$ of each half cycle $v_{o2}$ is negative and reduces $v_o$; from $\alpha$ to $\pi$ both add. - The primary current $i_s = i_{s1} + i_{s2}$ (equal turns) is a stepped wave: zero from $0$ to $\alpha$ and $2I_o$ from $\alpha$ to $\pi$. Its fundamental lags by only $\alpha/2$, so the displacement factor $\cos(\alpha/2)$ is better than $\cos\alpha$ of a single converter. #### Advantages - High output voltage with devices of half the voltage rating. - Better input power factor and less harmonic content than a single converter of the same rating, because only one converter is phase-controlled at a time (the other runs at $0^\circ$ or $180^\circ$). ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. For two series-connected single-phase full converters, the output voltage is: A. (2Vm/pi)(cos a1 + cos a2) B. (2Vm/pi) cos a1 cos a2 C. (Vm/pi)(cos a1 + cos a2) D. (4Vm/pi) cos(a1 + a2) Answer: A. Each full converter gives (2Vm/pi) cos alpha and the outputs add in series; the maximum is 4Vm/pi. (Key point: Output voltage expression) 2. In rectification mode, how is the output varied from 0 to 4Vm/pi? A. Both converters are varied together from 90 degrees B. Converter 2 is held at 0 and converter 1 is varied C. Converter 1 is held at alpha = 0 and converter 2 is varied from 180 degrees down to 0 D. Both are held at 180 degrees Answer: C. With alpha1 = 0, Vo = (2Vm/pi)(1 + cos a2), which is 0 at 180 degrees and 4Vm/pi at 0. (Key point: Rectification mode control) 3. What is needed for inversion mode, and how is it controlled? A. A resistive load; both converters at alpha = 0 B. A load with a reversed d.c. source E; one converter at 180 degrees and the other varied, giving Vo = (2Vm/pi)(cos a2 - 1) C. A load capacitor; converter 1 at 90 degrees D. An a.c. load; converter 2 at 90 degrees Answer: B. Power flows from d.c. to a.c. when the load has a reversed source such as a regenerating motor, with Vo ranging from 0 to -4Vm/pi. (Key point: Inversion mode requirement) 4. Why does this arrangement improve input power factor compared with a single converter? A. Both converters are phase-controlled at the same time B. The input current becomes sinusoidal C. The transformer provides reactive power D. Only one converter is phase-controlled at a time, so the displacement factor is about cos(alpha/2) Answer: D. The primary current fundamental lags by only alpha/2, giving a better displacement factor than cos alpha. 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