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Half-controlled bridge with R load, α = 45°

Short answer

Average output voltage V_dc = 176.75 V; output DC power P_dc = V_dc²/R = 3124 W. Input power factor = V_rms/V_s = 0.954 lagging.

Figure shows below a full wave controlled rectifier with resistive load. The load voltage is 230V, 50 Hz and it is operated at firing angle 45°. Draw the waveform of output voltage, output current and ac input current. Calculate average value of output voltage, output dc power and input power factor. [Figure: single-phase half-controlled bridge — thyristors T1, T2 on top and diodes D1, D2 on bottom, fed from Vs = 220V, 50 Hz, feeding R = 10 Ω]

The circuit is a single-phase half-controlled (semi-converter) bridge: thyristors T1, T2 on top and diodes D1, D2 at the bottom, feeding R=10 ΩR = 10\ \Omega. With a resistive load the output is a chopped sine wave from α\alpha to π\pi in each half cycle. Assumption: the supply is taken as 230 V, 50 Hz as stated in the text (the figure shows 220 V; results for 220 V are given at the end).

              P (+)
        +-------+-------+
       T1              T2
 a o----+               +----o b
       D1              D2
        +-------+-------+
              N (-)
 R = 10 ohm between P and N
 + half: T1 (a->P), D2 (N->b);  - half: T2, D1

Waveforms

IntervalConductingvov_oioi_oisi_s
00 – 45∘45^\circnone000
45∘45^\circ – 180∘180^\circT1, D2vsv_svs/Rv_s/R+vs/R+v_s/R
180∘180^\circ – 225∘225^\circnone000
225∘225^\circ – 360∘360^\circT2, D1−vs-v_s−vs/R-v_s/Rvs/Rv_s/R (negative)
 vo, io    __            __
          |  \          |  \
 ---------+---\---------+---\-----
 is        __
          |  \
 ---------+---\---------+---/-----
                        |__/
          45  180      225  360  (deg)

Average output voltage

Vm=2×230=325.27V_m = \sqrt2 \times 230 = 325.27 V, α=45∘\alpha = 45^\circ.

Vdc=Vmπ(1+cos⁡α)=325.27π(1.7071)=176.75 VIdc=VdcR=17.67 A\begin{aligned} V_{dc} &= \frac{V_m}{\pi}(1+\cos\alpha) = \frac{325.27}{\pi}(1.7071) = 176.75\ \text{V} \\ I_{dc} &= \frac{V_{dc}}{R} = 17.67\ \text{A} \end{aligned}

Output dc power

Pdc=VdcIdc=Vdc2R=176.75210=3123.97 WP_{dc} = V_{dc}I_{dc} = \frac{V_{dc}^2}{R} = \frac{176.75^2}{10} = 3123.97\ \text{W}

Input power factor

RMS output voltage:

Vrms=Vs[1π(π−α+sin⁡2α2)]1/2=230[2.3562+0.5π]1/2=230×0.9535=219.30 V\begin{aligned} V_{rms} &= V_s\left[\frac{1}{\pi}\left(\pi-\alpha+\frac{\sin2\alpha}{2}\right)\right]^{1/2} = 230\left[\frac{2.3562+0.5}{\pi}\right]^{1/2} \\ &= 230 \times 0.9535 = 219.30\ \text{V} \end{aligned}

With R load, isi_s has the same rms value as ioi_o: Is=Vrms/R=21.93I_s = V_{rms}/R = 21.93 A. All real input power is consumed in R:

Pac=Vrms2R=219.30210=4809.43 WPF=PacVsIs=4809.43230×21.93=VrmsVs=0.9535 (lagging)\begin{aligned} P_{ac} &= \frac{V_{rms}^2}{R} = \frac{219.30^2}{10} = 4809.43\ \text{W} \\ PF &= \frac{P_{ac}}{V_sI_s} = \frac{4809.43}{230 \times 21.93} = \frac{V_{rms}}{V_s} = 0.9535\ \text{(lagging)} \end{aligned}

Answer (230 V): Vdc=176.75V_{dc} = 176.75 V, Pdc=3124P_{dc} = 3124 W, input PF = 0.954 lagging.

With 220 V (figure value): Vdc=169.06V_{dc} = 169.06 V, Pdc=2858.2P_{dc} = 2858.2 W, PF = 0.954 (PF does not depend on the supply voltage).

✓ Answer checked by Claude Opus 5.5 (high effort)

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