   ## Frequency Component of single-phase Half-Wave Rectifier Voltage and Current

The load current IL consist of a dc component IL(dc) and an ac component IL(ac). The Fourier series of the Half-Wave rectifier rectified current following through the load is found to be

#### iL = ILM(1/ π + sinωt/2 – 2 cos 2 ωt/3π – 2 cos 4 ωt /15 π + ……………)

As seen the Half-Wave rectifier current consists of a large number of ac components (which constitute the ripple) in addition to the dc component. The first term is ILM/2 which represents the dc component IL(dc). Sin ωt has a peak value of (ILM/2). It is called the fundamental or first harmonic component and it’s rms value is

#### IL1 = ILM/2√2

The third term represents the second harmonic component whose frequency is doubled that of the supply frequency. The rms value is

#### IL2 = peak value/√2 = 2ILM/3π× √2 = √2 ILM /3π.

The fourth terms represents the third harmonics component whose frequency is four times the supply frequency. It’s rms value is

#### 2 ILM/15π × √2 = √2× ILM / 15 π.

The rms values of other components can be similarly calculated, However, they found to be of continuously diminishing value.

The rectified output (or load ) current consist of

(i)                  dc component, IL(dc) = ILM

(ii)                ac component of rms value IL1, IL2 and IL3 etc. Their combined rms is given by

#### IL(ac) = √IL1­­­2+ IL2­2 + IL32 +…………

The rms (or effective) value of the total load current is given by

### = √IL(dc)2 + (IL12 + IL22 + IL32 + ……..)

Similarly, the Fourier series of the load voltage is given by

#### UL = VLM (1/π + sin ωt/2 – 2cos 2 ωt/3 π – 2 cos 4 ωt/15 π……..)

It also consist of

(i)                 a dc component, VL(dc) = VLM

(ii)               ac component of rms value VL1, VL2 and VL3 etc. which are given by

Again,

#### VL(ac) = √VL12 + VL22  +VL32 + ……

The rms value of entire load voltage is given by

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