In half wave rectifier , rectifying element conducts only during
positive half cycle of input a.c. supply. The negative half cycles of
a.c. supply are eliminated from the output.

This rectifier circuit consists of resistive load, rectifying element,
i.e. p-n junction diode, and the source of a.c. voltage, all connected
in series. The circuit diagram is shown in the Fig. 1. Usually, the
rectifier circuits are operated from a.c. mains supply. To obtain the
desired d.c. voltage across the load, the a.c. voltage is applied to
rectifier circuit using suitable step-up or step-down transformer,
mostly a step-down one, with necessary turns ratio.

Fig. 1 Half wave rectifier |

The input voltage to the half-wave rectifier circuit shown in the Fig. 1
is a sinusoidal a.c. voltage, having a frequency which is the supply
frequency, 50 Hz.

The transformer decides the peak value of the secondary voltage. If N

where E

_{1}are the primary number of turns and N_{2}are the secondary number of turns and E_{pm}is the peak value of the primary voltage then,where E

_{sm}= The peak value of the secondary a.c. voltage.
As the return of E

e

ω = 2πf

f = Supply frequency

The average or dc value of alternating current is obtained by integration.

I

I

Where I

It is the product of average D.C. load current and the load resistance R

Note : Do not calculate E

_{sm}is sinusoidal the instantaneous value will be,e

_{s}= E_{sm}sinωtω = 2πf

f = Supply frequency

Let R

_{f}represents the forward resistance of the diode. Assume that, under reverse biased condition, the diode acts almost as an open circuit, conducting no current.__1.1 Operation of the Circuit__

During the positive half cycle of secondary a.c. voltage, terminal (A)
becomes positive with resoect to terminal (B). The diode is forward
biased and the current flows in the circuit in the clockwise direction,
as shown in the Fig. 2(a). The current will flow for almost full
positive half cycle. This current is also flowing through load
resistance R

_{L}hence denoted as I_{L}, the load current.
During negative half cycle when terminal (A) is negative with respect
to terminal (B), diode becomes reverse biased. Hence no current flows in
the circuit as shown in the Fig. 2(b). Thus the circuit current, which
is also the load current, is in the form of half sinusoidal pulses.

Fig. 2 |

The load voltage, being the product of load current and load
resistance, will also be in the form of half sinusoidal pulses. The
different waveforms are illustrated in Fig. 3.

Fig. 3 Load current and load voltage waveforms for half wave rectifier |

The d.c. output waveform is expected to be a straight line but the half
wave rectifier gives output in the form of positive sinusoidal pulses.
Hence the output is called pulsating d.c. It is discontinuous in nature.
Hence it is necessary to calculate the average value of load current
and average value of output voltage.

__1.2 Average DC Load Current ()__

For finding out the average value of an alternating waveform, we have
to determine the area under the curve over one complete cycle i.e. from 0
to 2π and then dividing it by the base i.e. 2π.

Mathematically, current waveform can be described as,I

_{L}= I_{m}sin t for 0 ≤ ωt ≤ πI

_{L}= 0 for π ≤ ωt ≤ 2πWhere I

_{m}= Peak value of load current
As no current flows during negative half cycle of a.c. input voltage,
i.e. between ωt = π to ωt = 2π, we change the limits of integration.

Applying Kirchhoff's voltage law we can write,

Where R

_{s}= Resistance of secondary winding of transformer. If R_{s}is not given it should be neglected while calculating I_{m}.__1.3 Average DC Load Voltage (E__

_{DC}

__)__

_{L}.
The winding resistance R

_{s}and forward diode resistance R_{f}are practically very small compared to R_{L}.
But as R

_{f}and R_{s}are small compared to R_{L}, (R_{f}+R_{s})/R_{L}is negligibly small compared to 1. So neglecting it we get,**Note**: When R_{f}and R_{s}are finite, calculate I_{m}, then I_{DC}and from that calculate as .Note : Do not calculate E

_{DC}as E_{sm}/π directly for finite R_{f}and R_{s}. Calculate I_{m1}then and then E_{DC}= I_{DC}R_{L}.__1.4 R.M.S. Value of Load Current (__I

_{RMS}

__)__

The R.M.S. means squaring, finding mean and then finding square root.
Hence R.M.S. value of load current can be obtained as,

**Note**: remember that this R.M.S. value is for half wave rectified waveform hence it is I

_{m}/2. For full sine wave it is I

_{m}/√2, which is derived later.

__1.5 R.M.S. Value of the Load Voltage__

The r.m.s. value of the load voltage is the r.m.s. value of the total
output voltage which includes d.c. output and the a.c. ripples. As the
load is resistive, the r.m.s. value of the load voltage is given by,

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