**Half Wave Rectifier**

**Image By: RJB1, FRec var, CC BY-SA 4.0**

**Content**

**Rectification & Rectifier****Types of Rectifier****Half Wave Rectifier****Half Wave Rectifier Working & Circuit****Half Wave Rectifier Formula****Half Wave Rectifier graph****Form Factor****Ripple Factor****Transformer Utilization Factor (TUF)****Half Wave Rectifier Efficiency****Difference between Half wave Rectifiers & Full Wave Rectifiers**

**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.

- Half Wave Rectifiers
- Full Wave Rectifiers
- 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.

V_{input} – V_{diode} – I * r_{diode} – IR = 0

I = (V_{input} – V_{diode}) / (r_{diode} +R)

V_{ooutput} = IR

Or, V_{o} = (V_{i} – V_{b}) / (r_{d} +R) * R

Or, V_{o} = (R * V_{i})/ (R_{d} + R) – (R_{b} * V_{b}) / (R_{d }+ R)

V_{o }= V_{i }– V_{b}

Now, V_{o} = 0 for reverse bias condition.

**Average O/p voltage:**

V_{o} = V_{m}Sinωt; 0 ≤ ωt ≤ π

V_{o} = 0; π ≤ ωt ≤ 2*π

V_{av} = 1/(2π-0) *∫_{0}^{2π}Vo d(ωt)

Or, V_{av} = 1/(2π) * ∫_{0}^{2π}V_{m}Sinωt d(ωt)

Or, V_{av} = 1/(2π) * ∫_{0}^{π}V_{m}Sinωt d(ωt) + 1/(2π) * ∫^{π2π}V_{m}Sinωt d(ωt)

Or, V_{av} = (V_{m}/2π) [- Cosωt]_{0}^{2}^{π} + 0

Or, V_{av} = (V_{m} / 2π) * [-(-1) – (-(1))]

Or, V_{av} = (V_{m}/ 2 π) * 2

Or, V_{av} = V_{m} / π = 0.318 V_{m}

The calculated average load current (I_{av}) is = I_{m}/π

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

I_{rms} = [1/(2π) *∫ _{0} ^{2π} I^{2 }d(ωt)]^{1/2}

I = I_{m}Sinωt; 0 ≤ ωt ≤ π

I = 0; π ≤ ωt ≤ 2*π

Or, I_{rms} = [1/(2π) * ∫ _{0} ^{2π} I_{m}^{2 }Sin^{2}ωt d(ωt)]^{1/2}

Or, I_{rms} = [I_{m}^{2}/(2π) *∫ _{0} ^{2π} Sin^{2}ωt d(ωt)]^{1/2 }+ 0

Now, Sin^{2}ωt = ½ (1 – Cos2ωt)

Or, I_{rms} = [I_{m}^{2}/(2π) *∫ _{0} ^{2π} (1 – Cos2ωt)d(ωt)]^{1/2}

Or, I_{rms} = [I_{m}^{2}/4] ^{½ } Or, I_{rms} = I_{m}/2

The calculated RMS voltage is – V_{rms} = V_{m}/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 >= V_{m}

**For a half-wave rectifier. Peak inverse voltage is given as PIV >= V**_{m}

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

_{m}If, at any point, PIV<V_{m}, 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 – I_{dc} = 1/2π *∫_{0}^{2π}I_{load} d(ωt)

I_{load} is the instantaneous load current at time t, and is the source sinusoidal voltage’s angular frequency. A more excellent value of I_{dc} 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 = V _{rms} / V_{av}**

V_{rms} = V_{m}/2

V_{av} = V_{m} / π

Form Factor = (V_{m}/2) / (V_{m}/ π) = 1.57 > 1

So, we can write, V_{rms} = 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.

I_{o} = I_{ac} + I_{dc}

Or, I_{ac} = I_{o} – I_{dc}

I_{ac} = [1/(2π) * ∫_{0}^{2π}(I-Idc)^{2}d(ωt)]^{1/2}

Or, I_{ac} = [I_{rms}^{2} + I_{dc}^{2}– 2 I_{dc}^{2}]1/2

Or, I_{ac} = [I_{rms}^{2} – I_{dc}^{2}]1/2

So, Ripple factor,

**γ = I _{rms}^{2} – I_{dc}^{2} / I_{dc}^{2}**

or, γ = [(I_{rms}^{2} – I_{dc}^{2 }) – 1] 1/2

γ_{HWR} = 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 = P _{dc}/ P_{ac}(rated)**

Now, to find the Transformer Utilization Factor, we need the rated secondary voltage. Let us say that V_{s}. / √2. RMS current through the winding is I_{m}/2.

So, TUF = I_{dc}^{2} R_{L} / (V_{s}/ √2) * (I_{m} / 2)

TUF = (I_{m}/ π)^{2}R_{L} / ( I_{m}2 (R_{f} +R_{L})/(2r2) = 2√2/ π^{ 2} * (1 / (1 + R_{f}/R_{L}))

If R_{f} << R_{L}, 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 = P _{load} / P_{in} *100**

or, η = I_{dc}^{2} * R/ I_{rms}^{2} * R , as P = VI, & V= IR

Now, I_{rms} = I_{m}/2 and I_{dc} = I_{m}/π

So, η = (I_{m}^{2}/2) / (I_{m}^{2}/π)

Or, η = (I_{m}^{2}/4) / (I_{m}^{2}/π^{2})

η = 4 / π^{2 }* 100% = 40.56%

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

**Specify Difference between Half Wave and Full Wave Rectifier**

**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= V_{m}Sinωt

Here, V_{m} = 200

So the output voltage is V_{m} / π

So V_{o} = 200/ π volt

Or, V_{o} =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?**

V_{m} = 200 volt

The output DC power will be = V_{m}^{2} / (π^{2} *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.

Cover Photo By: Tumblr

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Soumyadeep GuinGreat explanation, solid content with proper diagram, tabulation and most important is it clarified all my doubts… Kudoos to the editor….