--- title: "Bridge with diode, α-dependent output and input current" subject: "Power Electronics" chapter: 2 key: "semester_7/power_electronics:bridge-d1-highly-inductive-fundamental" source: https://www.bctengineer.com/q/semester_7/power_electronics/bridge-d1-highly-inductive-fundamental site: BCT Engineering --- # Bridge with diode, α-dependent output and input current Power Electronics · Chapter 2: Single phase AC to DC conversion - Asked in: 2075 Chaitra, 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/bridge-d1-highly-inductive-fundamental - Whole chapter: https://www.bctengineer.com/md/semester_7/power_electronics/chapter-2 ## Question For a circuit shown in figure below, draw the wave forms of each diodes and thyristor, also derive the expression of average and RMS value of output voltage as the function of firing angle α. Assuming load current is constant at 10A, calculate the magnitude and phase of fundamental component of input AC current is. [Figure: single-phase bridge fed from Vs = 230 V, f = 50 Hz with input current is; top-left device labelled D1/T1, top-right T3, bottom T4 and T2; output to a highly inductive load carrying io] ## Short answer For the half-controlled bridge, Vm=325.27 V, average output Vdc=103.54(1+cosα) V, rms output Vrms=230√[(1/π)(π-α+sin2α/2)] V. With constant 10 A load, fundamental input current rms magnitude Is1=9.003cos(α/2) A, phase lagging the supply voltage by α/2. ## Answer The circuit is a **single-phase half-controlled bridge (semiconverter)**: two thyristors and two diodes. *Assumption:* the figure's labels are read as thyristors T1, T3 in the top (common-cathode) positions and diodes D4, D2 in the bottom positions; any valid half-controlled arrangement gives the same $v_o$ and $i_s$. ```text P (+) +-------+-------+ | | T1 T3 | | a o----+ +----o b | | D4 D2 | | +-------+-------+ N (-) T1, T3: thyristors (common cathode) D4, D2: diodes (common anode) Highly inductive load (io = 10 A) between P and N ``` #### Operation and device waveforms $V_m = \sqrt2 \times 230 = 325.27$ V, $I_o = 10$ A (constant). | Interval ($\omega t$) | Conducting | $v_o$ | $i_s$ | |---|---|---|---| | $\alpha$ – $\pi$ | T1, D2 | $v_s$ | $+10$ A | | $\pi$ – $\pi+\alpha$ | T1, D4 (freewheel) | $0$ | $0$ | | $\pi+\alpha$ – $2\pi$ | T3, D4 | $-v_s$ | $-10$ A | | $2\pi$ – $2\pi+\alpha$ | T3, D2 (freewheel) | $0$ | $0$ | - At $\pi$, $v_s$ reverses and D4 becomes forward biased; the load current freewheels through T1–D4, so $v_o$ never goes negative. - Each **thyristor** conducts for $180^\circ$ ($\alpha$ to $\pi+\alpha$, i.e. 10 A pulses of $180^\circ$). - Each **diode** also conducts for $180^\circ$, but from $0$ to $\pi$ (D2) and $\pi$ to $2\pi$ (D4), i.e. in step with the supply. ```text vo __ __ | \ | \ (vs pieces from a to 180, -----+---\-------+---\--- zero from 180 to 180+a) iT1 +-----------+ 10 A, a to 180+a iD2 +--------+ 10 A, 0 to 180 is +Io +-----+ +-- | | | 0 ------+ +--+ +-+ | | -Io +-----+ a 180 180+a 360 ``` #### Average and RMS output voltage $$ V_{dc} = \frac{1}{\pi}\int_{\alpha}^{\pi} V_m\sin\omega t\, d(\omega t) = \frac{V_m}{\pi}(1+\cos\alpha) = 103.54\,(1+\cos\alpha)\ \text{V} $$ $$ V_{rms} = \frac{V_m}{\sqrt2}\left[\frac{1}{\pi}\left(\pi-\alpha+\frac{\sin2\alpha}{2}\right)\right]^{1/2} = 230\left[\frac{1}{\pi}\left(\pi-\alpha+\frac{\sin2\alpha}{2}\right)\right]^{1/2}\ \text{V} $$ #### Fundamental component of input current $i_s$ is $+I_o$ from $\alpha$ to $\pi$ and $-I_o$ from $\pi+\alpha$ to $2\pi$ (pulse width $\pi-\alpha$, centred at $(\pi+\alpha)/2$). Fourier analysis gives: $$ \begin{aligned} a_1 &= \frac{2}{\pi}\int_{\alpha}^{\pi} I_o\cos\omega t\, d(\omega t) = -\frac{2I_o}{\pi}\sin\alpha \\ b_1 &= \frac{2}{\pi}\int_{\alpha}^{\pi} I_o\sin\omega t\, d(\omega t) = \frac{2I_o}{\pi}(1+\cos\alpha) \\ I_{s1,peak} &= \sqrt{a_1^2+b_1^2} = \frac{4I_o}{\pi}\cos\frac{\alpha}{2}, \qquad \phi_1 = -\frac{\alpha}{2} \end{aligned} $$ $$ \begin{aligned} I_{s1} &= \frac{2\sqrt2\,I_o}{\pi}\cos\frac{\alpha}{2} = \frac{2\sqrt2 \times 10}{\pi}\cos\frac{\alpha}{2} \\ &= 9.003\cos\frac{\alpha}{2}\ \text{A (rms)}, \quad \text{lagging } v_s \text{ by } \frac{\alpha}{2} \end{aligned} $$ Example: at $\alpha = 60^\circ$, $I_{s1} = 9.003 \times \cos30^\circ = 7.80$ A, lagging by $30^\circ$. **Answer: $I_{s1} = 9.003\cos(\alpha/2)$ A rms, phase $\alpha/2$ lagging.** The displacement angle is only half of that of a full converter, so the semiconverter has a better input power factor. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Why does vo never go negative in the half-controlled bridge? A. Thyristors turn off at pi B. A diode conducts with a thyristor during pi to pi + alpha (freewheeling) C. The load has a source D. The supply is clamped Answer: B. When vs reverses, the diode becomes forward biased and freewheels the load current, so vo = 0. (Key point: Freewheeling) 2. What is Vdc as a function of alpha? A. (2Vm/pi) cos alpha B. (Vm/pi)(1 + cos alpha) C. (Vm/2pi)(1 + cos alpha) D. (Vm/pi) cos alpha Answer: B. Output is vs from alpha to pi, so Vdc = (Vm/pi)(1 + cos alpha). (Key point: Average output voltage) 3. What is the phase of the fundamental input current? A. Lagging by alpha B. Lagging by alpha/2 C. Leading by alpha/2 D. In phase Answer: B. The pulse (alpha to pi) is centred at (pi + alpha)/2 so the fundamental lags by alpha/2. (Key point: Phase of the fundamental) 4. For Io = 10 A, what is Is1 (rms)? A. 9.003 cos(alpha/2) A B. 9.003 cos(alpha) A C. 12.73 A D. 10 A Answer: A. Is1 = (2sqrt(2)Io/pi) cos(alpha/2) = 9.003 cos(alpha/2) A. 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