**Single Phase Uncontrolled Halfwave Rectifier** is the simplest and probably the most widely used rectifier circuit albeit at relatively small power levels. The output voltage and current of this rectifier are strongly influenced by the type of the load. Rectification refers to the process of converting an ac voltage or current source to dc voltage and current. Rectifiers especially refer to power electronic converters where the electrical power flows from the ac side to the dc side.

Converter circuit may carry electrical power from the dc side to the ac side whereupon they are referred to as inverters. In this blog, the operation of Single Phase Uncontrolled Halfwave Rectifier with resistive, inductive, and capacitive loads will be discussed.

## 1.Single Phase Uncontrolled Halfwave Rectifier with Resistive load (R- Load)

Theis circuit diagram shows the Single Phase Uncontrolled Halfwave Rectifier with Resistive load (R- Load).

If the switch S is closed at t = 0, the diode D becomes forward biased in the interval 0 < ωt ≤ π. If the diode is assumed to be ideal.

For 0 < ωt ≤ π

v_0=v=\sqrt { 2 } V_i\sin { \omega t }

v_{D} = v_{i} – v_{0} = 0

Since the load is resistive

i_0=\frac { v_0 }{ R } =\frac { \sqrt { 2 } V_i }{ R } \sin { \omega t }

For ωt > π, v_{i} becomes negative and D becomes reverse biased. So in the interval π < ωt ≤ 2π

i_{i} = i_{0} = 0

v_{0} = i_{0}R =0

v_{D} = v_{i} – v_{0}= v_{i} =\sqrt { 2 } V_i\sin { \omega t }

**Average output voltage **

V_{0AV}=\frac { 1 }{ 2\pi } \int _{ 0 }^{ 2\pi }{ v_0d\omega t } =\frac { 1 }{ 2\pi } \int _{ 0 }^{ \pi }{ \sqrt { 2 } } V_i\sin { \omega td\omega t }=\frac { \sqrt { 2 } V_i }{ \pi }

**RMS voltage across the diode **

V_{DRMS}=\sqrt { \frac { 1 }{ 2\pi } \int _{ 0 }^{ \pi }{ 2V^2_i\sin^2 { \omega td\omega t } } }=\frac { V_i }{ \sqrt { 2 } }

High ripple content in the output voltage and current this rectifier is seldom used with a pure resistive load.

## 2. Single Phase Uncontrolled Halfwave Rectifier with R-L Load

The ripple factor of output current can be reduced to some extent by connecting an inductor in series with the load resistance as shown in Fig.

As in the previous case, the diode D is forward biased when the switch S is turned on. at ωt = 0. However, due to the load inductance i_{0} increases more slowly. Eventually, at ωt = π, v_{0} becomes zero again. However, i_{0} is still positive at this point.

Therefore, D continues to conduct beyond ωt = π while the negative supply voltage is supported by the inductor till its current becomes zero at ωt = β. Beyond this point, D becomes reverse biased. Both v_{0} and i_{0} remain zero till the beginning of the next cycle whereupon the same process repeats.

For 0 ≤ ωt ≤ β

v_{D} = 0

v_{i} = v_{0}

i_{i} = i_{0}

for β ≤ ωt ≤ 2π

v_{0}=0

i_{i} = i_{0}=0

v_{D} = v_{i} – v_{0}= v_{i}

**Average output voltage **

V_{OAV}=\frac { 1 }{ 2\pi } \int _{ 0 }^{ 2\pi }{ v_0d\omega t }=\frac { 1 }{ 2\pi } \int _{ 0 }^{ \beta }{ \sqrt { 2 } } V_i\sin { \omega td\omega t }=\frac { \sqrt { 2 } V_i }{ \pi } \left( \frac { 1-\cos { \beta } }{ 2 } \right)

**RMS output voltage **

V_{0RMS}=\sqrt { \frac { 1 }{ 2\pi } \int _{ 0 }^{ \beta }{ 2V^2_i\sin^2 { \omega td\omega t } } }=\frac { V_i }{ \sqrt { 2 } } \sqrt { \frac { 2\beta -\sin { 2\beta } }{ 2\pi } }

**Form factor of the voltage waveform** is

V_{OFF}=\frac { V_{0RMS} }{ V_{0AV} } =\pi \sqrt { \frac { 2\beta -\sin { 2\beta } }{ 2\pi \left( 1-\cos { \beta } \right)^2 } }

**The ripple factor**

However, the ripple factor of i_{0} decreases with increasing L. Therefore, in certain applications, where a smooth dc current is of prime importance this configuration of the rectifier is preferred.