Full Wave Rectifier | It’s Circuit, Formula, Important Factor and Efficiency

Content

Rectification & Rectifier

Types of Rectifier

Full Wave Rectifier

Full Wave Rectifier Working & Circuit

Full Wave Rectifier Formula

Full Wave Rectifier graph

Form Factor

Ripple Factor

Transformer Utilization Factor (TUF)

Full Wave Rectifier Efficiency

Difference between Full-wave Rectifier & Half Wave Rectifier

Rectification

Rectification is the electrical process to convert an alternating current (or voltage) to direct current (or voltage).

A rectifier is a device that has a low resistance to current in one direction and higher resistance in another order.

Types of Rectifier

Rectifiers can be classified into three types –

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

Full Wave Rectifier

Full-wave rectifiers are kind of rectifiers that converts ac to dc that is alternating current to direct current. This type of rectifier allows both halves of the ac input voltage to pass through the circuit. Two diodes are necessary to make a full-wave rectifier.

Full Wave Rectifier Working & Circuit

A full-wave rectifier is shown in the below circuit.

Full Wave Rectifier
Full Wave Rectifier using two diode.
Image Credit: WdwdFullwave.rectifier.enCC BY 3.0

There is a transformer T on the input side. The transformer T steps up or steps down the AC voltage supplied at the primary side. It is a center-tapped transformer. An ac input voltage of V = nVoSinwt is applied in the circuit. N is the turn ratio of the center-tapped transformer. Two diodes are connected to the course. Current flows through one diode for the first half of the cycle and flows through the other diode for the next half of the process. That is how a unidirectional current flow towards the load.

This is a modified and also an improved version of the half-wave rectifier. We use a center-tapped transformer. Each half of the transformer’s secondary has an equal number of turns; the voltage induced in each half of the secondary is equal in values and opposite in phase.

Now for any instance of the input half-cycle, point A has a positive voltage concerning O (center). The lower point B has an equal voltage but negative in magnitude (It is the center-tapped transformer). So, diode D1 conducts current, and diode D2 does not appear at this half of the cycle. For the next half of the process, diode D1 is at revere biased, and diode D2 is forward biased. So, diode D1 does not conduct the current while D2 does for this half of the input cycle. The load current is the sum of current from diode D1 and diode D2 from both the input voltage cycles.

Diodes D1 and D2 are identical, so the average value of load current for a full-wave rectifier circuit is double that of a half-wave rectifier.

As both halves of the cycle passed through the circuit, this is known as a Half wave rectifier. 

How a transformer works? Click to Know!

Full Wave Rectifier Formula & Equations

From the circuit,

Vi is the input voltage; Vb is the diode voltage, rd is the dynamic resistance, R is the load resistance, Vo is the output voltage.

Average O/p voltage:

Vo = VmSinωt; 0 ≤ ωt ≤ π

Vav = 1/π *  ∫ 0 Vo d(wt)

Or, Vav = 1/ π *  ∫ 0 VmSinwt d(wt)

Or, Vav = (Vm/π) [- Cosωt]0π

Or, Vav = (Vm / π) * [-(-1) – (-(1))]

Or, Vav = (Vm/ π) * 2

Or, Vav = 2Vm / π = 0.64 Vm

The average load current (Iav) comes as = 2* Im

The RMS (Root Means Square) Value of current:

Irms = [1/π * ∫ 0 I2  d(ωt)]1/2

I = ImSinωt; 0 ≤ ωt ≤ π

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

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

Now, Sin2ωt = ½ (1 – Cos2ωt)

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

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

The RMS voltage comes as – Vrms = Vm/√2.

The significance of the RMS value is that it is equivalent to DC Value.

Provided that RMS value is ≤ Peak Value

Peak Inverse Voltage (PIV):

Peak inverse voltage or PIV is an important parameter. It is defined as the maximum reverse bias voltage applied across the diode before entering the Zenner Region or Breakdown Region.

For a full-wave rectifier. Peak inverse voltage is given as PIV >= 2Vm

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.

Full Wave Rectifier graph

The following diagram represents the input and output graph.

Full wave rectifier graph, Image source – Basic Electronics

Form Factor

The form factor of a full-wave rectifier can be defined as RMS’s ratio (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) / (2*Vm/ π) = π/2√2=1.11

So, we can write, Vrms = 1.11 * Vav.

Ripple Factor

Ripple factor is given as the RMS (Root Means Square) Value of AC Component to the Average value of the output. The output current consists of both the AC and DC components. The ripple factor measures the percentage of AC components present in the rectified output.  The symbol represents the ripple factor – γ

Io = Iac + Idc

Or, Iac = Io – Idc

Or, Iac = [1/(2π) ∫ 0 (I-Idc)*(I-Idc) d(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

γFWR = 0.482

Transformer Utilization Factor

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

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

If Rf << RL, then,

TUF = 8 / π 2 = 0.812

The average transformer utilization factor comes as =

(0.574 + 0.812)/2 = 0.693

Increase in Transformer utilization factor suggests a better performance of the full wave rectifier.

Full Wave Rectifier Efficiency

Efficiency of Full 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 = 2*Im

So, η = (4Im2/ π2) / (Im2/2)

η = 8 / π2 * 100% = 81.2%

Efficiency of a ideal Full Wave Rectifier Circuit is = 81.2%

Specify Difference Between Half Wave and Full Wave Rectifier

Subject of ComparisonHalf Wave RectifierFull Wave Rectifier
No. of diodes usedOnly One diode is usedTwo diodes are used
Current flowCurrent flows in the circuit for only the positive half of the input cycle.Current flows in the circuit for all half of the input cycle.
Transformer RequiredAny step-down or step-up transformer,Centre Tapped transformers are the center required for full-wave rectifiers.
Peak Inverse VoltageFor a half-wave rectifier, peak inverse voltage is the maximum voltage across the transformer’s secondary winding.For the full-wave rectifier, each diode’s peak inverse voltage is twice the maximum voltage between the center tap and any other end of the transformer’s secondary winding.
Frequency of the load currentThe frequency of the load current in the half-wave rectifier is the same as the input frequency supply.The frequency of the load current is twice the input power supply.
Ripple FactorRipple factor is 1.21Ripple factor is 0.482
Transformer Utilization FactorsTransformer Utilization Factor is 0.287For a full-wave transformer, TUF is  = 0.693
EfficiencyThe efficiency of a half-wave rectifier is less than a full-wave rectifier and is  = 40.56%Efficiency is more than half-wave rectifier and is = 81.2%

Some Problems with Full Wave Rectifiers

1. A full-wave rectifier has a load of 1 kilo- ohm. The applied AC voltage is 220 V (RMS value). If the diodes’ internal resistances are neglected, what will be ripple voltage across the load resistance?

a. 0.542 V

b. 0.585 V

c. 0.919 V

d. 0.945 V

The ripple voltage is = γ * Vdc / 100

Vdc = 0.636 * Vrms * √2 = 0.636*220*√2 = 198 V.

The ripple factor of a full wave rectifier is 0.482

Hence the ripple voltage = 0.482*198/100 = 0.945 V

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

Peak inverse voltage is an important parameter defined as the maximum reverse bias voltage applied across the diode before entering the breakdown region. If the peak inverse voltage rating is less than the value, then breakdown may occur. The diode’s peak inverse voltage is twice the peak voltage = 2Vm -Vd for a full-wave rectifier. Vd is the diode cut-in voltage. Now for a silicon diode, the cut-in voltage  = 0.7 v. So, peak inverse voltage =2* 5 -0.7volts = 9.3 volts.

3. A input of 200Sin 100 πt volt is applied to a full-wave rectifier. What is the output ripple frequency?

V= VmSinωt

Here,  ω= 100

Frequency is given as – ω/2 = 100/2 = 50 Hz.

The output frequency of a center-tapped frequency is doubled the input frequency. Thus the output frequency = 50*2 = 100 Hz.

4. 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.

About Sudipta Roy

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