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Three-phase Bridge Rectifier Uncontrolled
Dec 28, 2017

Three-phase bridge rectifier uncontrolled[edit]

Disassembled automobile alternator, showing the six diodes that comprise a full-wave three-phase bridge rectifier.

For an uncontrolled three-phase bridge rectifier, six diodes are used, and the circuit again has a pulse number of six. For this reason, it is also commonly referred to as a six-pulse bridge. The B6 circuit can be seen simplified as a series connection of two three-pulse center circuits.

For low-power applications, double diodes in series, with the anode of the first diode connected to the cathode of the second, are manufactured as a single component for this purpose. Some commercially available double diodes have all four terminals available so the user can configure them for single-phase split supply use, half a bridge, or three-phase rectifier.

For higher-power applications, a single discrete device is usually used for each of the six arms of the bridge. For the very highest powers, each arm of the bridge may consist of tens or hundreds of separate devices in parallel (where very high current is needed, for example in aluminium smelting) or in series (where very high voltages are needed, for example in high-voltage direct current power transmission).

Controlled three-phase full-wave bridge rectifier circuit (B6C) using thyristors as the switching elements, ignoring supply inductance

The pulsating DC voltage results from the differences of the instantaneous positive and negative phase voltages {\displaystyle V_{\mathrm {LN} }}{\displaystyle V_{\mathrm {LN} }}, phase-shifted by 30°:

DC voltage profile of B6 three-phase full-wave rectifier.jpg

The ideal, no-load average output voltage {\displaystyle V_{\mathrm {av} }}{\displaystyle V_{\mathrm {av} }} of the B6 circuit results from the integral under the graph of a DC voltage pulse with the period duration of {\displaystyle {\frac {1}{3}}\pi }{\displaystyle {\frac {1}{3}}\pi } (from 60° to 120°) with the peak value {\displaystyle {\hat {v}}_{\mathrm {DC} }={\sqrt {3}}\cdot V_{\mathrm {peak} }}{\displaystyle {\hat {v}}_{\mathrm {DC} }={\sqrt {3}}\cdot V_{\mathrm {peak} }}:

  • {\displaystyle V_{\mathrm {dc} }=V_{\mathrm {av} }={\frac {1}{{\frac {1}{3}}\pi }}\int _{60^{\circ }}^{120^{\circ }}{\sqrt {3}}\cdot V_{\mathrm {peak} }\cdot \sin \varphi \cdot \mathrm {d} \varphi ={\frac {3\cdot {\sqrt {3}}\cdot V_{\mathrm {peak} }}{\pi }}\cdot \left(-\cos 120^{\circ }+\cos 60^{\circ }\right)={\frac {3\cdot {\sqrt {3}}\cdot V_{\mathrm {peak} }}{\pi }}\cdot {\Biggl [}-\left(-{\frac {1}{2}}\right)+{\frac {1}{2}}{\Biggl ]}={\frac {3\cdot {\sqrt {3}}\cdot V_{\mathrm {peak} }}{\pi }}}{\displaystyle V_{\mathrm {dc} }=V_{\mathrm {av} }={\frac {1}{{\frac {1}{3}}\pi }}\int _{60^{\circ }}^{120^{\circ }}{\sqrt {3}}\cdot V_{\mathrm {peak} }\cdot \sin \varphi \cdot \mathrm {d} \varphi ={\frac {3\cdot {\sqrt {3}}\cdot V_{\mathrm {peak} }}{\pi }}\cdot \left(-\cos 120^{\circ }+\cos 60^{\circ }\right)={\frac {3\cdot {\sqrt {3}}\cdot V_{\mathrm {peak} }}{\pi }}\cdot {\Biggl [}-\left(-{\frac {1}{2}}\right)+{\frac {1}{2}}{\Biggl ]}={\frac {3\cdot {\sqrt {3}}\cdot V_{\mathrm {peak} }}{\pi }}}

  • ⇒ {\displaystyle V_{\mathrm {dc} }=V_{\mathrm {av} }={\frac {3\cdot {\sqrt {3}}\cdot {\sqrt {2}}\cdot V_{\mathrm {LN} }}{\pi }}}{\displaystyle V_{\mathrm {dc} }=V_{\mathrm {av} }={\frac {3\cdot {\sqrt {3}}\cdot {\sqrt {2}}\cdot V_{\mathrm {LN} }}{\pi }}} ⇒ {\displaystyle V_{\mathrm {av} }={\frac {3\cdot {\sqrt {6}}\cdot V_{\mathrm {LN} }}{\pi }}}{\displaystyle V_{\mathrm {av} }={\frac {3\cdot {\sqrt {6}}\cdot V_{\mathrm {LN} }}{\pi }}} ≈ 2,34 ⋅ {\displaystyle V_{\mathrm {LN} }}{\displaystyle V_{\mathrm {LN} }}

3-phase AC input, half-wave and full-wave rectified DC output waveforms

If the three-phase bridge rectifier is operated symmetrically (as positive and negative supply voltage), the center point of the rectifier on the output side (or the so-called isolated reference potential) opposite the center point of the transformer (or the neutral conductor) has a potential difference in form of a triangular common-mode voltage. For this reason, the two centers must never be connected to each other, otherwise short-circuit currents would flow. The ground of the three-phase bridge rectifier in symmetrical operation is thus decoupled from the neutral conductor or the earth of the mains voltage. Powered by a transformer, earthing of the center point of the bridge is possible, provided that the secondary winding of the transformer is electrically isolated from the mains voltage and the star point of the secondary winding is not on earth. In this case, however, (negligible) leakage currents are flowing over the transformer windings.

The common-mode voltage is formed out of the respective average values of the differences between the positive and negative phase voltages, which form the pulsating DC voltage. The peak value of the delta voltage {\displaystyle {\hat {v}}_{\mathrm {common-mode} }}{\displaystyle {\hat {v}}_{\mathrm {common-mode} }} amounts ¼ of the peak value of the phase input voltage {\displaystyle V_{\mathrm {peak} }}{\displaystyle V_{\mathrm {peak} }} and is calculated with {\displaystyle V_{\mathrm {peak} }}{\displaystyle V_{\mathrm {peak} }} minus half of the DC voltage at 60° of the period:

  • {\displaystyle {\hat {v}}_{\mathrm {common-mode} }=V_{\mathrm {peak} }-{\frac {{\sqrt {3}}\cdot V_{\mathrm {peak} }\cdot \sin 60^{\circ }}{2}}=V_{\mathrm {peak} }\cdot {\Biggl (}1-{\frac {{\sqrt {3}}\cdot \sin 60^{\circ }}{2}}{\Biggl )}}{\displaystyle {\hat {v}}_{\mathrm {common-mode} }=V_{\mathrm {peak} }-{\frac {{\sqrt {3}}\cdot V_{\mathrm {peak} }\cdot \sin 60^{\circ }}{2}}=V_{\mathrm {peak} }\cdot {\Biggl (}1-{\frac {{\sqrt {3}}\cdot \sin 60^{\circ }}{2}}{\Biggl )}} = {\displaystyle V_{\mathrm {peak} }}{\displaystyle V_{\mathrm {peak} }} · 0,25

The RMS value of the common-mode voltage is calculated from the form factor for triangular oscillations:

  • {\displaystyle V_{\mathrm {common-mode} }={\frac {{\hat {v}}_{\mathrm {common-mode} }}{\sqrt {3}}}}{\displaystyle V_{\mathrm {common-mode} }={\frac {{\hat {v}}_{\mathrm {common-mode} }}{\sqrt {3}}}}

If the circuit is operated asymmetrically (as a simple supply voltage with just one positive pole), both the positive and negative poles (or the isolated reference potential) are pulsating opposite the center (or the ground) of the input voltage analogously to the positive and negative waveforms of the phase voltages. However, the differences in the phase voltages result in the six-pulse DC voltage (over the duration of a period). The strict separation of the transformer center from the negative pole (otherwise short-circuit currents will flow) or a possible grounding of the negative pole when powered by an isolating transformer apply correspondingly to the symmetrical operation.