--- title: "GTO full converter, extinction angle: RMS and input current" subject: "Power Electronics" chapter: 2 key: "semester_7/power_electronics:gto-extinction-angle-rms" source: https://www.bctengineer.com/q/semester_7/power_electronics/gto-extinction-angle-rms site: BCT Engineering --- # GTO full converter, extinction angle: RMS and input current Power Electronics · Chapter 2: Single phase AC to DC conversion - Asked in: 2073 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/gto-extinction-angle-rms - Whole chapter: https://www.bctengineer.com/md/semester_7/power_electronics/chapter-2 ## Question The average value of output voltage of single phase full converter with 4 GTO switches is controlled by extinction angle control method. The load current is constant and equal to 20Amp due to highly inductive load. For the extinction angle of 30°, draw the waveforms of load voltage, load current and input ac current. Also find RMS value of output voltage, magnitude and phase of fundamental component of the input ac current. [Figure: single-phase bridge of four GTO switches (S1, S3 top; S4, S2 bottom) fed from Vs = 230 V, 50 Hz; output Vo across a highly inductive load carrying Io] ## Short answer For extinction angle 30°, RMS output voltage is 226.66 V. The fundamental component of the input AC current is 17.39 A rms (24.60 A peak) and leads the supply voltage by 15°. ## Answer In **extinction angle control** the forced-commutated switches (GTOs) are turned ON at the zero crossing of the supply voltage and turned OFF $\beta$ before the next zero crossing ($\omega t = \pi-\beta$). The input current pulse then sits before the voltage peak, so its fundamental **leads** the voltage: the converter can supply reactive power and the power factor improves. Data: $V_s = 230$ V, $V_m = 325.27$ V, $I_o = 20$ A, $\beta = 30^\circ$. ```text P (+) +-------+-------+ | | S1 S3 | | a o----+ +----o b | | S4 S2 | | +-------+-------+ N (-) S1..S4: reverse-blocking GTOs (or IGBTs with series diodes); no anti-parallel diodes vs = 230 V, 50 Hz between a and b Highly inductive load (Io) between P and N ``` #### 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) | $0$ | $0$ | - From $\pi-\beta$ to $\pi$ the load current freewheels through the two switches of one leg, so $v_o = 0$ and $i_s = 0$. - The load current stays at 20 A because of the large inductance. #### Waveforms of $v_o$, $i_o$, $i_s$ ```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) ``` #### RMS value of output voltage $$ \begin{aligned} V_{rms} &= \left[\frac{1}{\pi}\int_0^{\pi-\beta}V_m^2\sin^2\omega t\,d(\omega t)\right]^{1/2} = \frac{V_m}{\sqrt2}\left[\frac{1}{\pi}\left(\pi-\beta+\frac{\sin2\beta}{2}\right)\right]^{1/2} \\ &= 230\left[\frac{1}{\pi}\left(2.618+0.433\right)\right]^{1/2} = 230 \times 0.9855 = 226.66\ \text{V} \end{aligned} $$ (Average value for reference: $V_{dc} = \frac{V_m}{\pi}(1+\cos\beta) = 193.20$ V.) #### Fundamental component of input current $$ \begin{aligned} a_1 &= \frac{1}{\pi}\left[\int_0^{\pi-\beta}I_o\cos\omega t\,d(\omega t) - \int_{\pi}^{2\pi-\beta}I_o\cos\omega t\,d(\omega t)\right] = \frac{2I_o}{\pi}\sin\beta = 6.366\ \text{A} \\ b_1 &= \frac{1}{\pi}\left[\int_0^{\pi-\beta}I_o\sin\omega t\,d(\omega t) - \int_{\pi}^{2\pi-\beta}I_o\sin\omega t\,d(\omega t)\right] = \frac{2I_o}{\pi}(1+\cos\beta) = 23.759\ \text{A} \\ I_{s1,peak} &= \sqrt{a_1^2+b_1^2} = \frac{4I_o}{\pi}\cos\frac{\beta}{2} = \frac{80}{\pi}(0.9659) = 24.60\ \text{A} \\ I_{s1} &= 24.60/\sqrt2 = 17.39\ \text{A (rms)} \\ \phi_1 &= \tan^{-1}\frac{a_1}{b_1} = \tan^{-1}\frac{\sin\beta}{1+\cos\beta} = \frac{\beta}{2} = 15^\circ\ \text{leading} \end{aligned} $$ **Answer: $V_{rms} = 226.66$ V; $I_{s1} = 17.39$ A rms (24.60 A peak), leading $v_s$ by $15^\circ$.** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. How are the GTOs operated in extinction angle control? A. Turned on at alpha and turned off at the zero crossing B. Turned on at the zero crossing and turned off at wt = pi - beta by forced commutation C. Turned on at the peak and off at the next peak D. Turned on and off by line commutation only Answer: B. The switches turn on at the zero crossing and are forced off beta before the next zero crossing. (Key point: Switching in extinction angle control) 2. What happens from pi - beta to pi? A. The load current freewheels through two switches of one leg, so vo = 0 and is = 0 B. The load current falls to zero C. The source supplies negative current D. vo equals -vs Answer: A. The inductive load current continues through the freewheel path, disconnecting the load from the supply. (Key point: Freewheeling interval) 3. How does the fundamental input current relate to vs for beta = 30 degrees? A. Lags by 30 degrees B. In phase C. Leads by 15 degrees D. Leads by 30 degrees Answer: C. phi1 = beta/2 = 15 degrees leading, since the current pulse is advanced. (Key point: Phase of the fundamental) 4. What is the rms fundamental input current for Io = 20 A and beta = 30 degrees? A. 17.39 A B. 24.60 A C. 18.0 A D. 12.73 A Answer: A. Peak = (4Io/pi) cos(beta/2) = 24.60 A, so rms = 24.60/sqrt(2) = 17.39 A. (Key point: Fundamental magnitude) ## Close variants set in other exams - [GTO full converter, extinction angle 30°](https://www.bctengineer.com/question.md?key=semester_7%2Fpower_electronics%3Agto-extinction-angle-avg-rms) — Power Electronics, chapter 2, asked 2× ## Related in Power Electronics - [Single-phase full converter, 10 A, α = 30°](https://www.bctengineer.com/question.md?key=semester_7%2Fpower_electronics%3Afull-converter-10a-30deg) — Power Electronics, chapter 2, asked 4× - [Full converter, 20 A, α = 30°, fundamental input current](https://www.bctengineer.com/question.md?key=semester_7%2Fpower_electronics%3Afull-converter-20a-30deg-fundamental) — Power Electronics, chapter 2, asked 2× - [Full converter, 25 A, α = 30°, fundamental input current](https://www.bctengineer.com/question.md?key=semester_7%2Fpower_electronics%3Afull-converter-25a-30deg) — Power Electronics, chapter 2, asked 4× - [Full converter, 25 A, α = 30°, input power factor](https://www.bctengineer.com/question.md?key=semester_7%2Fpower_electronics%3Afull-converter-25a-30deg-pf) — Power Electronics, chapter 2, asked 4× - [Extinction angle control: current leads by β/2](https://www.bctengineer.com/question.md?key=semester_7%2Fpower_electronics%3Aextinction-angle-fundamental-leads) — Power Electronics, chapter 2 - [Single-phase diode bridge with constant load current](https://www.bctengineer.com/question.md?key=semester_7%2Fpower_electronics%3Adiode-bridge-rl-load-numerical) — Power Electronics, chapter 2 ## Related in other subjects - [Single-phase fully controlled converter DC motor speed](https://www.bctengineer.com/question.md?key=semester_7%2Futilization_of_electrical_energy%3Afull-converter-dc-motor-speed) — Utilization of Electrical Energy, chapter 3 - [3-phase half controlled bridge: no-load speed, firing angle](https://www.bctengineer.com/question.md?key=semester_7%2Futilization_of_electrical_energy%3Ahalf-bridge-motor-no-load-speed) — Utilization of Electrical Energy, chapter 3, asked 3×