## Full wave bridge rectifier

A Full wave rectifier is a circuit arrangement which makes use of both half cycles of input alternating current (AC) and convert them to direct current (DC). In our tutorial on **Half wave rectifiers**, we have seen that a half wave rectifier makes use of only one half cycle of the input alternating current. Thus a full wave rectifier is much more efficient (double+) than a half wave rectifier. This process of converting both half cycles of the input supply (alternating current) to direct current (DC) is termed full wave rectification.

Full wave rectifier can be constructed in 2 ways. The first method makes use of a center tapped transformer and 2 diodes. This arrangement is known as **Center Tapped Full Wave Rectifier**. The second method uses a normal transformer with 4 diodes arranged as a bridge. This arrangement is known as a Bridge Rectifier.

### Full Wave Rectifier Theory

To understand full wave bridge **rectifier theory**** **perfectly, you need to learn half wave rectifier first. In the tutorial of half wave rectifier we have clearly explained the basic working of a rectifier. In addition we have also explained the **theory behind a pn junction** and the **characteristics of a pn junction diode**.

### Full Wave Rectifier Working & Operation

The working & operation of a full wave bridge rectifier is pretty simple. The circuit diagrams and wave forms we have given below will help you understand the operation of a bridge rectifier perfectly. In the circuit diagram, 4 diodes are arranged in the form of a bridge. The transformer secondary is connected to two diametrically opposite points of the bridge at points A & C. The load resistance R_{L} is connected to bridge through points B and D.

#### During the first half cycle

During first half cycle of the input voltage, the upper end of the transformer secondary winding is positive with respect to the lower end. Thus during the first half cycle diodes D1 and D_{3} are forward biased and current flows through arm AB, enters the load resistance R_{L}, and returns back flowing through arm DC. During this half of each input cycle, the diodes D_{2} and D_{4 }are reverse biased and current is not allowed to flow in arms AD and BC. The flow of current is indicated by solid arrows in the figure above. We have developed another diagram below to help you understand the current flow quickly. See the diagram below – the green arrows indicate beginning of current flow from source (transformer secondary) to the load resistance. The red arrows indicate return path of current from load resistance to the source, thus completing the circuit.

#### During the second half cycle

During second half cycle of the input voltage, the lower end of the transformer secondary winding is positive with respect to the upper end. Thus diodes D_{2} and D_{4} become forward biased and current flows through arm CB, enters the load resistance R_{L}, and returns back to the source flowing through arm DA. Flow of current has been shown by dotted arrows in the figure. Thus the direction of flow of current through the load resistance R_{L} remains the same during both half cycles of the input supply voltage. See the diagram below – the green arrows indicate beginning of current flow from source (transformer secondary) to the load resistance. The red arrows indicate return path of current from load resistance to the source, thus completing the circuit.

**Peak Inverse Voltage of a Full wave bridge rectifier:**

Lets analyse peak inverse voltage (PIV) of a full wave bridge rectifier using the circuit diagram. At any instant when the transformer secondary voltage attains positive peak value Vmax, diodes D1 and D3 will be forward biased (conducting) and the diodes D2 and D4 will be reverse biased (non conducting). If we consider ideal diodes in bridge, the forward biased diodes D1 and D3 will have zero resistance. This means voltage drop across the conducting diodes will be zero. This will result in the entire transformer secondary voltage being developed across load resistance RL.

Thus PIV of a bridge rectifier = Vmax (max of secondary voltage)

**Bridge Rectifier Circuit Analysis**

The only difference in the analysis between full wave and centre tap rectifier is that

- In a bridge rectifier circuit two diodes conduct during each half cycle and the forward resistance becomes double (2R
)._{F} - In a bridge rectifier circuit Vsmax is the maximum voltage across the transformer secondary winding whereas in a centre tap rectifier Vsmax represents that maximum voltage across each half of the secondary winding.

The different parameters are explained with equations below:

**Peak Current**

Instantaneous value of the voltage applied to the rectifier is given as

vs = Vsmax Sin wt

If the diode is assumed to have a forward resistance of R** _{F}** ohms and a reverse resistance equal to infinity, then current flowing through the load resistance is given as

i1 = Imax Sin wt and i2 = 0 for the first half cycle

and i1 = 0 and i2 = Imax Sin wt for second half cycle

The total current flowing through the load resistance R** _{L}, **being the sum of currents i1 and i2 is given as

i = i1 + i2 = Imax Sin wt for the whole cycle.

Where peak value of the current flowing through the load resistance R** _{L}** is given as

Imax = Vsmax/(2R** _{F}** + R

**)**

_{L}**2. Output Current**

Since the current is the same through the load resistance RL in the two halves of the ac cycle, magnitude od dc current Idc, which is equal to the average value of ac current, can be obtained by integrating the current i1 between 0 and pi or current i2 between pi and 2pi.

**3. DC Output Voltage**

Average or dc value of voltage across the load is given as

**4. Root Mean Square (RMS) Value of Current**

RMS or effective value of current flowing through the load resistance R** _{L } **is given as

**5. Root Mean Square (RMS) Value of Output Voltage**

RMS value of voltage across the load is given as

**6. Rectification Efficiency**

Power delivered to load,

**7. Ripple Factor**

Form factor of the rectified output voltage of a full wave rectifier is given as

So, ripple factor, γ = 1.11** ^{2} **– 1) = 0.482

**8. Regulation**

The dc output voltage is given as

**Merits and Demerits of Full-wave Rectifier Over Half-Wave Rectifier **

**Merits – **Lets talk about the advantages of full wave bridge rectifier over half wave version first. I can think about 4 specific merits at this point.

- Efficiency is double for a full wave bridge rectifier. The reason is that, a half wave rectifier makes use of only one half of the input signal. A bridge rectifier makes use of both halves and hence double efficiency
- The residual ac ripples (before filtering) is very low in the output of a bridge rectifier. The same ripple percentage is very high in half wave rectifier. A simple filter is enough to get a constant dc voltage from bridge rectifier.
- We know the efficiency of FW bridge is double than HW rectifier. This means higher output voltage, Higher transformer utilization factor (TUF) and higher output power.

**Demerits - ** Full-wave rectifier needs more circuit elements and is costlier.

### Merits and Demerits of Bridge Rectifier Over Center-Tap Rectifier.

A center tap rectifier is always difficult one to implement because of the special transformer involved. A center tapped transformer is costly as well. One key difference between center tap & bridge rectifier is in the number of diodes involved in construction. A center tap full wave rectifier needs only 2 diodes where as a bridge rectifier needs 4 diodes. But silicon diodes being cheaper than a center tap transformer, a bridge rectifier is much preferred solution in a DC power supply. Following are the advantages of bridge rectifier over a center tap rectifier.

- A bridge rectifier can be constructed with or without a transformer. If a transformer is involved, any ordinary step down/step up transformer will do the job. This luxury is not available in a center tap rectifier. Here the design of rectifier is dependent on the center tap transformer, which can not be replaced.
- Bridge rectifier is suited for high voltage applications. The reason is the high peak inverse voltage (PIV) of bridge rectifier, when compared to the PIV of a center tap rectifier.
- Transformer utilization factor (TUF) is higher for bridge rectifier.

**Demerits of Bridge rectifier over center tap rectifier **

### Uses of Full wave Bridge rectifier

### Full Wave Bridge Rectifier with Capacitor Filter

Output of full wave rectifier is not a constant DC voltage. You can observe from the output diagram that its a pulsating dc voltage with ac ripples. In real life applications, we need a power supply with smooth wave forms. In other words, we desire a DC power supply with constant output voltage. A constant output voltage from the DC power supply is very important as it directly impacts the reliability of the electronic device we connect to the power supply.

We can make the output of full wave rectifier smooth by using a filter (a capacitor filter or an inductor filter) across the diode. In some cases an resistor-capacitor coupled filter (RC) is also used. The circuit diagram below shows a half wave rectifier with capacitor filter.

#### Ripple factor in a bridge rectifier

**References **

**Principles of Electronics**.

**article**.

Give me value of diode in full wave bridge rectifier.

Dear sir!

How we calculate the V ripple and €^-t/RC

Rf =(vrms)”2/vdc-1

thank you explanation of these topic

Dear sir

I want to know that what will be the Output DC voltage if we give 220v AC.

THANKS FOR YOUR EXPLANATION ABOUT THIS TOPIC

Thanks a lot for the circuit and explaination, I’m a std 12th student and this information helped me a lot in making my school project. My teacher was very much impressed by this project and explanation. My course book didn’t explained that we need a capacitor and also that for diodes are better than two.

Thanks a lot for your help !!

amazing..well answerd

I am verymuch satisfied. Please inform me “what type of diode and transeformer is requred to form a bridge rectifire”

It depends upon the load voltage and current. So chose required Voltage/Current rating transformer and Diodes.

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