# Full-wave Rectification(edit !

Dec 28, 2017 #### Full-wave rectification Full-wave rectifier, with vacuum tube having two anodes.

A full-wave bridge rectifier converts the whole of the input waveform to one of constant polarity (positive or negative) at its output. Mathematically, this corresponds to the absolute value function. Full-wave rectification converts both polarities of the input waveform to pulsating DC (direct current), and yields a higher average output voltage. Two diodes and a center tapped transformer, or four diodes in a bridge configuration and any AC source (including a transformer without center tap), are needed. Single semiconductor diodes, double diodes with common cathode or common anode, and four-diode bridges, are manufactured as single components. Graetz bridge rectifier: a full-wave rectifier using four diodes.

For single-phase AC, if the transformer is center-tapped, then two diodes back-to-back (cathode-to-cathode or anode-to-anode, depending upon output polarity required) can form a full-wave rectifier. Twice as many turns are required on the transformer secondary to obtain the same output voltage than for a bridge rectifier, but the power rating is unchanged. Full-wave rectifier using a center tap transformer and 2 diodes.

The average and RMS no-load output voltages of an ideal single-phase full-wave rectifier are:

• {\displaystyle {\begin{aligned}V_{\mathrm {dc} }=V_{\mathrm {av} }&={\frac {2\cdot V_{\mathrm {peak} }}{\pi }}\\[8pt]V_{\mathrm {rms} }&={\frac {V_{\mathrm {peak} }}{\sqrt {2}}}\end{aligned}}} Very common double-diode rectifier vacuum tubes contained a single common cathode and two anodes inside a single envelope, achieving full-wave rectification with positive output. The 5U4 and 5Y3 were popular examples of this configuration.

### Three-phase rectifiers

Single-phase rectifiers are commonly used for power supplies for domestic equipment. However, for most industrial and high-power applications, three-phase rectifier circuits are the norm. As with single-phase rectifiers, three-phase rectifiers can take the form of a half-wave circuit, a full-wave circuit using a center-tapped transformer, or a full-wave bridge circuit.

Thyristors are commonly used in place of diodes to create a circuit that can regulate the output voltage. Many devices that provide direct current actually generate three-phase AC. For example, an automobile alternator contains six diodes, which function as a full-wave rectifier for battery charging.

#### Three-phase, half-wave circuit Controlled three-phase half-wave rectifier circuit using thyristors as the switching elements, ignoring supply inductance

An uncontrolled three-phase, half-wave midpoint circuit requires three diodes, one connected to each phase. This is the simplest type of three-phase rectifier but suffers from relatively high harmonic distortion on both the AC and DC connections. This type of rectifier is said to have a pulse-number of three, since the output voltage on the DC side contains three distinct pulses per cycle of the grid frequency: The peak values {\displaystyle V_{\mathrm {peak} }} of this three-pulse DC voltage are calculated from the RMS value {\displaystyle V_{\mathrm {LN} }} of the input phase voltage (line to neutral voltage, 120 V in North America, 230 V within Europe at mains operation): {\displaystyle V_{\mathrm {peak} }={\sqrt {2}}\cdot V_{\mathrm {LN} }} . The average no-load output voltage {\displaystyle V_{\mathrm {av} }} results from the integral under the graph of a positive half-wave with the period duration of {\displaystyle {\frac {2}{3}}\pi } (from 30° to 150°):

• {\displaystyle V_{\mathrm {dc} }=V_{\mathrm {av} }={\frac {1}{{\frac {2}{3}}\pi }}\int _{30^{\circ }}^{150^{\circ }}V_{\mathrm {peak} }\cdot \sin \varphi \cdot \mathrm {d} \varphi ={\frac {3\cdot V_{\mathrm {peak} }}{2\pi }}\cdot \left(-\cos 150^{\circ }+\cos 30^{\circ }\right)={\frac {3\cdot V_{\mathrm {peak} }}{2\pi }}\cdot {\Biggl [}-\left(-{\frac {\sqrt {3}}{2}}\right)+{\frac {\sqrt {3}}{2}}{\Biggl ]}={\frac {3\cdot {\sqrt {3}}\cdot V_{\mathrm {peak} }}{2\pi }}} • ⇒ {\displaystyle V_{\mathrm {dc} }=V_{\mathrm {av} }={\frac {3\cdot {\sqrt {3}}\cdot {\sqrt {2}}\cdot V_{\mathrm {LN} }}{2\pi }}} ⇒ {\displaystyle V_{\mathrm {av} }={\frac {3\cdot {\sqrt {6}}\cdot V_{\mathrm {LN} }}{2\pi }}} ≈ 1,17 ⋅ {\displaystyle V_{\mathrm {LN} }} 