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Three-phase full bridge rectifier with R load

Short answer

The six-diode bridge connects the load to the largest line voltage; each diode conducts 120° with commutation every 60°, giving a six-pulse output between 0.866VmL and VmL. Integrating one line-voltage segment gives Vdc=(3/π)VmL=1.35VL=1.654Vm, e.g. 540V for a 400V line supply.

Draw and explain the output waveforms of three phase full bridge rectifier with R load. Find the average value of output voltage.

A three-phase full bridge rectifier with six diodes (or six thyristors at α=0\alpha = 0) connects the load at every instant to the largest line voltage: the most positive phase through the upper group (D1, D3, D5) and the most negative phase through the lower group (D4, D6, D2).

        P (+) o-----+------+------+-----+
                    |      |      |     |
                   D1     D3     D5     |
                    |      |      |     R
 a o----------------+      |      |     |
 b o-----------------------+      |     |
 c o------------------------------+     |
                    |      |      |     |
                   D4     D6     D2     |
                    |      |      |     |
        N (-) o-----+------+------+-----+
 (a joins D1-D4, b joins D3-D6, c joins D5-D2)

Operation and output waveform (R load)

ωt\omega t (w.r.t. vav_a)Conductingvov_o
30∘30^\circ – 90∘90^\circD1, D6vabv_{ab}
90∘90^\circ – 150∘150^\circD1, D2vacv_{ac}
150∘150^\circ – 210∘210^\circD3, D2vbcv_{bc}
210∘210^\circ – 270∘270^\circD3, D4vbav_{ba}
270∘270^\circ – 330∘330^\circD5, D4vcav_{ca}
330∘330^\circ – 390∘390^\circD5, D6vcbv_{cb}
 line voltages (peak VmL), six per cycle:
 vo  vab  vac  vbc  vba  vca  vcb
    _/\__/\__/\__/\__/\__/\_    max VmL
                                 min 0.866 VmL
   30   90   150  210  270  330  390 (deg)
 io = vo / R (same shape)
  • Each diode conducts for 120∘120^\circ; a commutation occurs every 60∘60^\circ.
  • Output is a six-pulse wave (ripple frequency 6f=3006f = 300 Hz for 50 Hz), varying between 0.866VmL0.866V_{mL} and VmLV_{mL}; with R load the current has the same shape.
  • Each phase current is two 120∘120^\circ pulses (positive and negative) per cycle, so no dc flows in the supply.

Average output voltage

Taking one segment of vab=VmLsin⁡θv_{ab} = V_{mL}\sin\theta from π/3\pi/3 to 2π/32\pi/3 (period π/3\pi/3):

Vdc=3π∫π/32π/3VmLsin⁡θ dθ=3VmLπ[−cos⁡θ]π/32π/3=3VmLπ[12+12]=3VmLπ=32 VLπ=1.35 VL=1.654 Vm\begin{aligned} V_{dc} &= \frac{3}{\pi}\int_{\pi/3}^{2\pi/3}V_{mL}\sin\theta\,d\theta = \frac{3V_{mL}}{\pi}\left[-\cos\theta\right]_{\pi/3}^{2\pi/3} = \frac{3V_{mL}}{\pi}\left[\frac{1}{2} + \frac{1}{2}\right] \\ &= \frac{3V_{mL}}{\pi} = \frac{3\sqrt2\,V_L}{\pi} = 1.35\,V_L = 1.654\,V_m \end{aligned}

If the bridge uses thyristors fired at α\alpha (R load), Vdc=1.35VLcos⁡αV_{dc} = 1.35V_L\cos\alpha for α≤60∘\alpha \le 60^\circ and Vdc=1.35VL[1+cos⁡(α+60∘)]V_{dc} = 1.35V_L\left[1+\cos(\alpha+60^\circ)\right] for 60∘<α≤120∘60^\circ < \alpha \le 120^\circ.

Example: 400 V, 50 Hz supply gives Vdc=1.35×400=540V_{dc} = 1.35 \times 400 = 540 V, and Idc=540/RI_{dc} = 540/R.

✓ Answer checked by Claude Opus 5.5 (high effort)

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