# Half Wave Rectifier | It’s Workings | 5+ Important Applications

Half Wave Rectifier

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## Rectification

Rectification is a way to change Ac to DC.

Rectifier: Rectifiers perform the rectification operation.

## Rectifier

Rectifier: Rectifiers perform the rectification operation.

## Types of Rectifier

Rectifiers are of three types based on their objectives. They are listed below.

1. Half Wave Rectifiers
2. Full Wave Rectifiers
3. Bridge Rectifiers

## Half Wave Rectifier

Half wave rectifiers perform the rectification operation in which the one half of AC voltage is allowed to pass through and another half is restricted. A single diode is enough to build a half wave rectifier.

## Half Wave Rectifier Working & Circuit

A half-wave transformer is shown in the below circuit-

A transformer T is placed at the input side. It helps to decrease or increase the input voltage as per the need.  Now, an input voltage is applied (should be an AC type). Let us say the voltage applied is V = nV0sinwt. Here ‘n’ represents the turn ratio of the transformer. Now, the current starts flowing through the diode because of the applied voltage. For the first half of the cycle, the diode is in forwarding bias. So, the current passes through the diode.

For the next half of the input cycle, the diode is in reverse discrimination. Thus, no current passes through the diode. The diagrammatic representation shows the output. Only one half of the cycle out of two comes in the output. That is why this circuitry is known as ‘Half Wave Rectifier’

## Half Wave Rectifier Formula & Equations

To obtain the HWR formulas and equations, observe the circuit carefully. Here Vinput is the input voltage and Vdiode represents the diode voltage. ‘Load R’ is the load resistance. Voutput represents the output voltage.

Vinput – Vdiode – I * rdiode – IR = 0

I = (Vinput – Vdiode) / (rdiode +R)

Vooutput = IR

Or, Vo = (Vi – Vb) / (rd +R) * R

Or, Vo = (R * Vi)/ (Rd + R) – (Rb * Vb) / (Rd + R)

Vo = Vi – Vb

Now, Vo = 0 for reverse bias condition.

### Average O/p voltage:

Vo = VmSinωt; 0 ≤ ωt ≤ π

Vo = 0; π ≤ ωt ≤ 2*π

Vav = 1/(2π-0) *∫0Vo d(ωt)

Or, Vav = 1/(2π) * ∫0VmSinωt d(ωt)

Or, Vav = 1/(2π) * ∫0πVmSinωt d(ωt) + 1/(2π) * ∫πVmSinωt d(ωt)

Or, Vav = (Vm/2π) [- Cosωt]02π + 0

Or, Vav = (Vm / 2π) * [-(-1) – (-(1))]

Or, Vav = (Vm/ 2 π) * 2

Or, Vav = Vm / π = 0.318 Vm

The calculated average load current (Iav) is = Im

### The RMS (Root Means Square) Value of current:

Irms = [1/(2π) *∫ 0 I2  d(ωt)]1/2

I = ImSinωt; 0 ≤ ωt ≤ π

I = 0; π ≤ ωt ≤ 2*π

Or, Irms = [1/(2π) * ∫ 0 Im2  Sin2ωt d(ωt)]1/2

Or, Irms = [Im2/(2π) *∫ 0 Sin2ωt d(ωt)]1/2 + 0

Now, Sin2ωt = ½ (1 – Cos2ωt)

Or, Irms = [Im2/(2π) *∫ 0 (1 – Cos2ωt)d(ωt)]1/2

Or, Irms = [Im2/4] ½   Or, Irms = Im/2

The calculated RMS voltage is – Vrms = Vm/2.

### Peak Inverse Voltage (PIV):

PIV or Peak Inverse Voltage is defined as the maximum voltage value which can be applied to the diode in the reverse bias condition. Voltage greater than PIV will lead to a Zenner breakdown of the diode. It is one of the crucial parameters of a diode.

PIV of a half-wave rectifier is: PIV >= Vm

### For a half-wave rectifier. Peak inverse voltage is given as PIV >= Vm

If, at any point, PIV<Vm, the diode will be damaged.

The load current of a rectifier circuit is fluctuating and unidirectional. The output is a periodic function of time. Using the Fourier theorem, it can be concluded that the load current has an average value superimposed on which are sinusoidal currents having harmonically related frequencies. The average of the dc amount of the load current is – Idc = 1/2π *∫0Iload d(ωt)

Iload is the instantaneous load current at time t, and is the source sinusoidal voltage’s angular frequency. A more excellent value of Idc implies better performance by the rectifier circuit.

## Half Wave Rectifier graph

The graphical representation below, shows the input as well as correspondent output for a half wave rectifier –

## Form Factor

The form factor of a half-wave rectifier is defined as the ratio of RMS (Root Means Square) Value of load voltage to the average value load Voltage.

Form Factor = Vrms / Vav

Vrms = Vm/2

Vav = Vm / π

Form Factor = (Vm/2) / (Vm/ π) = 1.57 > 1

So, we can write, Vrms = 1.57 * Vav.

## Ripple Factor

Ripple factor is defined as the ratio between the RMS of the AC voltage to the average output. The output of a half wave rectifier has both the AC part and DC part of current. The ripple factor helps us to determine the percentage of ripple factor present in the output.

Io = Iac + Idc

Or, Iac = Io – Idc

Iac = [1/(2π) * ∫0(I-Idc)2d(ωt)]1/2

Or, Iac = [Irms2 + Idc2– 2 Idc2]1/2

Or, Iac = [Irms2 – Idc2]1/2

So, Ripple factor,

γ = Irms2 – Idc2 / Idc2

or, γ = [(Irms2 – Idc2 ) – 1] 1/2

γHWR = 1.21

## Transformer Utilization Factor of Half Wave Rectifier

The transformer utilization factor is defined as the DC power ratio supplied to the load to the transformer’s AC power rating.

TUF = Pdc/ Pac(rated)

Now, to find the Transformer Utilization Factor, we need the rated secondary voltage. Let us say that Vs. / √2. RMS current through the winding is Im/2.

So, TUF = Idc2 RL / (Vs/ √2) * (Im / 2)

TUF = (Im/ π)2RL / ( Im2 (Rf +RL)/(2r2) = 2√2/ π 2 * (1 / (1 + Rf/RL))

If Rf << RL, then,

TUF = 2√2 / π 2 = 0.287 The TUF’s lower value suggests that the DC power delivered to a load in a half-wave rectifier is much less than the AC transformer rating.

## Half Wave Rectifier Efficiency

Efficiency of Half Wave Rectifier is defined as the ratio of the DC power available at the load to the input AC power. It is represented by the symbol – η

η = Pload / Pin *100

or, η = Idc2 * R/ Irms2 * R , as P = VI, & V= IR

Now, Irms = Im/2 and Idc = Im

So, η = (Im2/2) / (Im2/π)

Or, η = (Im2/4) / (Im22)

η = 4 / π2 * 100% = 40.56%

Efficiency of a ideal Half Wave Rectifier Circuit is = 40.56%

## Some Problems of Half Wave Rectifiers

### 1. If the input frequency is 60 Hz, then the ripple frequency of a half-wave rectifier circuit is equal to –

a. 40 Hz

b. 50 Hz

c. 60 Hz

d. 70 Hz

In the half-wave rectifier, the output load frequency is the same as the input frequency. So the output frequency is 60Hz.

### 2. If the peak voltage of a half-wave rectifier circuit is 5 V and the diode is silicon diode, what will be the peak inverse voltage on the diode?

PIV or Peak Inverse Voltage is defined as the maximum voltage value which can be applied to the diode in the reverse bias condition. Voltage greater than PIV will lead to a Zenner breakdown of the diode. It is one of the crucial parameters of a diode.

So, for a half-wave rectifier, the diode’s peak inverse voltage is equal to the peak voltage = Vm. So, peak inverse voltage = 5 volts.

### 3. A input of 200Sin 100 πt volt is applied to a half-wave rectifier. What is the average output voltage?

V= VmSinωt

Here, Vm = 200

So the output voltage is Vm / π

So Vo = 200/ π volt

Or, Vo =63.6619 Volt.

### 4. For a half-wave rectifier, the input voltage is 200Sin100 πt Volts. The load resistance is 10 kilo – ohm. What will be the DC power output of the half-wave rectifier?

Vm = 200 volt

The output DC power will be = Vm2 / (π2 *1000) = 200 *200 /(3.14 *3.14 *1000) = 4.05 watt

### 5. What is the main application of a rectifier? Which device does the opposite operation?

A rectifier transforms the AC voltage to the DC voltage. An oscillator converts a DC voltage to AC voltage.

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