--- title: "GTO full converter, extinction angle 30°" subject: "Power Electronics" chapter: 2 key: "semester_7/power_electronics:gto-extinction-angle-avg-rms" source: https://www.bctengineer.com/q/semester_7/power_electronics/gto-extinction-angle-avg-rms site: BCT Engineering --- # GTO full converter, extinction angle 30° Power Electronics · Chapter 2: Single phase AC to DC conversion - Asked in: 2080 Baishakh, 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/gto-extinction-angle-avg-rms - Whole chapter: https://www.bctengineer.com/md/semester_7/power_electronics/chapter-2 ## Question Figure below shows a single-phase controlled rectifier with 4 nos of GTO switches. The average value of the output voltage is controlled by extinction angle control method. The load current is constant and equal to 20Amp due to highly inductive load. For an extinction angle of 30°, i) Draw the waveform of load voltage, load current and input ac current. ii) Average and RMS value of output voltage. iii) Calculate the magnitude and phase of fundamental component of the input ac current. [Figure: single-phase bridge of four GTO switches S1, S3 (top) and S4, S2 (bottom) fed from Vs = 230 V, 50 Hz with input current Is; output Vo across a highly inductive load carrying Io] ## Short answer Using extinction angle control with freewheeling, average output voltage Vdc = (Vm/π)(1+cosβ) = 193.20 V, RMS voltage Vrms = 226.66 V. The fundamental input current rms is 17.39 A, leading the supply voltage by 15°. ## Answer In **extinction angle control**, the switches are turned ON at the voltage zero crossing ($\omega t = 0$) and turned OFF at $\omega t = \pi - \beta$, where $\beta$ is the extinction angle. GTOs are needed because the turn-off occurs while the supply voltage is still positive (forced turn-off). The input current then **leads** the voltage, which improves the power factor. Data: $V_s = 230$ V, $V_m = \sqrt2 \times 230 = 325.27$ V, $I_o = 20$ A, $\beta = 30^\circ = \pi/6$. ```text P (+) +-------+-------+ | | S1 S3 | | a o----+ +----o b | | S4 S2 | | +-------+-------+ N (-) S1..S4: GTOs (or IGBTs) with anti-parallel diodes vs = 230 V, 50 Hz between a and b Highly inductive load (Io) between P and N ``` #### i) Waveforms and operation | Interval ($\omega t$) | ON switches | $v_o$ | $i_s$ | |---|---|---|---| | $0$ – $\pi-\beta$ | S1, S2 | $+v_s$ | $+I_o$ | | $\pi-\beta$ – $\pi$ | S1, S4 (freewheel) | $0$ | $0$ | | $\pi$ – $2\pi-\beta$ | S3, S4 | $-v_s$ | $-I_o$ | | $2\pi-\beta$ – $2\pi$ | S3, S2 (freewheel) | $0$ | $0$ | During the freewheeling intervals the load is shorted by two switches of one leg, so $v_o = 0$ and no current is drawn from the supply. ```text vo (load voltage, period 180 deg) _____ _____ / | / | / | / | -/-------+------/-------+----- 0 180-b 180 360-b 360 io ============================ Io (constant) is +Io +-------+ | | 0 -+ +---+ +--- | | -Io +---------+ 0 180-b 180 360-b 360 (b = beta) ``` #### ii) Average and RMS output voltage $$ \begin{aligned} V_{dc} &= \frac{1}{\pi}\int_0^{\pi-\beta} V_m\sin\omega t\, d(\omega t) = \frac{V_m}{\pi}(1+\cos\beta) \\ &= \frac{325.27}{\pi}(1 + 0.866) = 103.54 \times 1.866 = 193.20\ \text{V} \end{aligned} $$ $$ \begin{aligned} V_{rms} &= \left[\frac{1}{\pi}\int_0^{\pi-\beta} V_m^2\sin^2\omega t\, d(\omega t)\right]^{1/2} = \frac{V_m}{\sqrt2}\left[\frac{1}{\pi}\left(\pi-\beta+\frac{\sin2\beta}{2}\right)\right]^{1/2} \\ &= 230\left[\frac{1}{\pi}\left(2.618 + 0.433\right)\right]^{1/2} = 230 \times 0.9855 = 226.66\ \text{V} \end{aligned} $$ #### iii) Fundamental component of input current Fourier analysis of $i_s$ (amplitude $I_o$, width $\pi-\beta$, starting at $0$): $$ \begin{aligned} a_1 &= \frac{2I_o}{\pi}\sin\beta = \frac{40}{\pi}(0.5) = 6.366\ \text{A} \\ b_1 &= \frac{2I_o}{\pi}(1+\cos\beta) = \frac{40}{\pi}(1.866) = 23.759\ \text{A} \\ I_{s1,peak} &= \sqrt{a_1^2+b_1^2} = \frac{4I_o}{\pi}\cos\frac{\beta}{2} = 24.597\ \text{A} \\ I_{s1} &= \frac{24.597}{\sqrt2} = 17.39\ \text{A (rms)} \\ \phi_1 &= \tan^{-1}\frac{a_1}{b_1} = \frac{\beta}{2} = 15^\circ\ \text{(leading)} \end{aligned} $$ So $i_{s1} = 24.60\sin(\omega t + 15^\circ)$ A. **Answer: $V_{dc} = 193.20$ V, $V_{rms} = 226.66$ V, $I_{s1} = 17.39$ A (rms) leading $v_s$ by $15^\circ$.** (Displacement factor $= \cos 15^\circ = 0.966$ leading.) ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. How does extinction angle control operate? A. Switches turn ON at alpha and OFF at pi B. Switches turn ON at zero crossing and are forced OFF at pi - beta C. Switches turn on at the voltage peak only D. Switches stay ON all the time Answer: B. They turn ON at wt = 0 and OFF at pi - beta while vs is positive, needing GTOs. (Key point: Extinction angle control) 2. What is the effect on the input fundamental current phase? A. Lags by beta B. Lags by beta/2 C. Is in phase D. Leads vs by beta/2 Answer: D. The pulse spans 0 to pi - beta, so its centre is advanced and the fundamental leads by beta/2 = 15°. (Key point: Effect on input current) 3. What is Vdc for beta = 30° (Vm = 325.27 V)? A. (Vm/pi)(1 + cos beta) = 193.20 V B. (2Vm/pi) cos beta = 179.33 V C. (Vm/pi)(1 - cos beta) = 13.9 V D. 2Vm/pi = 207.07 V Answer: A. Integrating from 0 to pi - beta gives (Vm/pi)(1 + cos beta) = 193.20 V. (Key point: Average output voltage) 4. What is vo and is during pi - beta to pi? A. vo = -vs, is = -Io B. vo = vs, is = Io C. vo = 0, is = 0 because two switches of one leg short the load D. vo = 0, is = Io Answer: C. Load current freewheels through two switches, shorting the load and drawing no supply current. 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