Bridge Rectifier convers AC input into DC output. The rectification process is done by the junction diodes which are forward biased. When a junction diode is forward biased, it admits the current to flow through it. This application of junction diodes is mainly used in rectification. The Wheatstone Bridge system is used in Bridge rectifiers. Bridge rectifiers are also called Wave bridge rectifiers. It rectifies the waves also. Half-wave rectifiers and Full-wave rectifiers are examples of bridge rectifiers. There are many types of rectifiers but the Bridge rectifier is the more effective.
Bridge rectifiers effectively transforms the Alternating current into Direct current. A bridge rectifier circuit configuration consists of four or more diodes. Bridge rectifiers are another form of a full-wave rectifier.
The bridge rectifier circuit is shown below. The Wheatstone Bridge system is applied in the four diodes. $\mathrm{D_1,D_2,D_3}$,and $\mathrm{D_4}$ diodes are set in the Wheatstone Bridge configuration.
A and C ends are connected to the two ends of the secondary Transformer. The secondary ends of the transformer are denoted as $\mathrm{S_1}$ and $\mathrm{S_2}$. The AC signal is connected to the primary transformer. The positive half-cycle is conducted by the diodes $\mathrm{D_1}$ and $\mathrm{D_2}$. The negative half-cycle is conducted by the diodes $\mathrm{D_3}$ and $\mathrm{D_4}$. At the end of the two half-cycles, the load current flows through the circuit.
$$\mathrm{V_{p(out)}=V_{p(sec)}}$$
A Bridge Wave rectifier produces the output like a Bridge rectifier circuit. The wave rectifier also uses four separate diodes. It is a single-phase rectifier. Four separated diodes are linked with a closed-loop bridge circuit to get the required output. A special center-tapped transformer is not required for this Bridge wave rectifier. So, its cost and size needed are very less. It is the major advantage to use Bridge wave rectifiers. The AC signal is connected to the primary transformer. The positive half-cycle is regulated by the diodes $\mathrm{D_1}$ and $\mathrm{D_2}$. The negative half cycle is regulated by the diodes $\mathrm{D_3}$ and $\mathrm{D_4}$. A and C ends are connected to the two ends of the secondary Transformer. The secondary ends of the transformer are denoted as $\mathrm{S_1}$ and $\mathrm{S_2}$.
Bridge wave rectifiers have a configuration of the Bridge rectifier. $\mathrm{D_1, D_2, D_3}$, and $\mathrm{D_4}$ diodes are set in the Wheatstone Bridge configuration like a bridge rectifier. $\mathrm{R_L}$ is the load resistance. A and C ends are connected to the two ends of the secondary Transformer. The secondary ends of the transformer are denoted as $\mathrm{S_1}$ and $\mathrm{S_2}$. The AC signal is connected to the primary transformer. The positive half cycle is regulated by the diodes $\mathrm{D_1}$ and $\mathrm{D_2}$. The negative half cycle is regulated by the diodes. At the end of the two half-cycles, the load current flows through the circuit.
$\mathrm{D_1\:and\:D_2}$ diodes are forward biassed and current flows in the time of the positive half cycle of the input during this $\mathrm{D_3}$ and $\mathrm{D_4}$ are reverse biassed. Likewise, $\mathrm{D_3}$ and $\mathrm{D_4}$ diodes are forward biassed and current flow in the time of the negative half-cycle of the input during this $\mathrm{D_1}$ and $\mathrm{D_2}$ are reverse biassed. Therefore in the time of two input cycles current flows in the load resistance in a similar direction. So, the full-wave rectification process is done and the output of Direct Current is the wave developed across the load $\mathrm{R_L}$.
In the positive half cycle and negative half cycle, the load resistor is always linked in series with both diodes. The output voltage is calculated when the diode drops are carried into account.
$$\mathrm{V_{p(out)}=V_{p(sec)}-1.4V}$$
If $\mathrm{D_1}$ and $\mathrm{D_2}$ diodes are forward biassed and $\mathrm{D_3}$ and $\mathrm{D_4}$ are checked for reverse bias. $\mathrm{D_1}$ and $\mathrm{D_2}$ have avoidable onward resistance and they can be pulled to be connecting wires because the diodes are considered perfect. The peak inverse voltage of diodes $\mathrm{D_3}$ and $\mathrm{D_4}$ is similar to the peak secondary voltage. Secondary voltage is similar to the output voltage.
$$\mathrm{Peak\:inverse\:voltage=V_{p(out)}}$$
The peak inverse voltage value of the bridge diodes is lower than that needed for the center-tapped configuration.
The full-wave bridge rectifier has a similar equation of average value and output frequency as the same as the full-wave rectifier because it produces a full-wave output.
$$\mathrm{V_{dc}=\frac{2V_p}{\pi}}$$
$$\mathrm{f_{out}=2f_{in}}$$
The bridge rectifier has a soft Direct Current signal more than the output of the half-wave rectifier Direct Current.
Input AC signals are used fully in bridge rectifiers. In the time of the positive half cycle and negative half cycle, electric current is allowed. But in the Half wave rectifier input AC signal is used partially. The rest of the AC signal is stopped. So there is a huge loss in AC input signals.
During both positive and negative half-cycles, the electric current is enabled in the bridge rectifier. So, the output Direct current signal is nearly the same as the input Alternative Current signal.
There is a difficulty in forming the circuit configuration for bridge amplifiers.
The bridge rectifiers consist of four diodes in comparison to other rectifiers.
High power drops happen in bridge rectifiers. Because it consists of many diodes.
There is an excessive voltage drop in the bridge rectifier. Because in the time of both half-cycle diodes are connected in series with load resistors. But in other rectifiers, during both half-cycles, one diode is used. So, voltage drop is less .
The instrument used to rectify the current is called rectifiers. The rectification process is done by the junction diodes which are forward biased. When a junction diode is forward biased, it admits the current to flow through it. There are many types of rectifiers. A Bridge Wave rectifier produces the output like a Bridge rectifier circuit. The wave rectifier also uses four separate diodes. It is a single-phase rectifier. Four separated diodes are linked with a feedback-control circuit to get the required output. This rectifier is cost-effective and the size required are less space.
Q1. What is a Center-tapped full-wave rectifier?
Ans. It is a full-wave rectifier circuit configuration that has only two diodes. The step-down transformer is connected with these diodes. The primary part of the transformer is fed by the input Alternate Current signal.
Q2. Give the value of Ripple Factor of a full-wave rectifier?
Ans. The Ripple factor is 0.482 for a full-wave rectifier. This rectifier is more useful than a half-wave rectifier.
$$\mathrm{\gamma=0.482}$$
Q3. Describe the Rectifier efficiency of a rectifier?
Ans. Rectifier efficiency is defined as the proportion between the direct current output and the whole amount of alternate current signal provided to the rectifier.
$$\mathrm{\eta=\frac{dc\:output}{ac\:input}}$$
Q4. What is the RMS value?
Ans. The efficient value of a fluctuation of voltage or current is called the RMS value. (RMS - Root Mean Square).
$$\mathrm{RMS=\sqrt{\frac{1}{n}\displaystyle\sum\limits_i x_i^2}}$$
Q5. A half-wave rectifier has an input AC power of 100 watts. The applied power of output power is 50 watts. Calculate the rectifier efficiency.
Ans.
$$\mathrm{Rectifier\:Efficiency=\frac{dc\: output }{ac\:input}\times 100}$$
$$\mathrm{Rectifier\:Efficiency=\frac{50}{100}\times 100=50\%}$$
The rest of the 50 watts are converted into a sinusoidal wave because only 50 watts are given as input.
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