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**Application of junction diode as a Rectifier**

- Rectifier is a device which is used for converting alternating current or voltage into direct current or voltage
- A p-n junction diode can be used as a half-wave and full-wave rectifier
- The resistance of a p-n junction diode becomes low when forward biased and becomes high when reverse biased. This is the principle of the working of rectifier

**Half-wave Rectifier**

**Circuit Diagram**

- Transformer with primary and secondary coils
- Diode
- Load resistance R
_{L}

The AC voltage to be rectified is connected between the primary of the transformer

To one coil of the secondary, the p junction of the diode is connected

The output is measured across the load resistance R_{L}**Input and output**

**Working**

**➤ Case 1**

- During the positive half cycle of the input AC voltage, suppose P1 is negative and P2 is positive
- On account of inductance, S1 becomes positive and S2 becomes negative
- The p-n junction is forward biased and hence the resistance of the p-n junction diode becomes low
- Hence, current flows in the circuit and we get output across the load resistance R
_{L} - This is indicated in the graph above

**➤ Case 2**

- During the negative half cycle of the input AC voltage, suppose P1 is positive and P2 is negative
- On account of inductance, S1 becomes negative and S2 becomes positive
- The p-n junction is reverse biased and hence the resistance of the p-n junction diode becomes high
- Hence, no current flows in the circuit and we do not get any output across the load resistance R
_{L} - This is indicated in the graph above

The above process is repeated. Thus , we have current only in the positive half of the cycle. Hence, if is called as half-wave rectifier

The output signal is not continuous and available as bursts. Hence, this is not of much use.

**Full Wave Rectifier**

**➤ Circuit Diagram**

- The main difference between half and full wave rectifier in circuit, is the usage of two diodes – D1 and D2

**➤ Input and Output**

**Working**

**➤ Case 1**

- During the positive half of the input cycle of AC voltage, the junction diode D1 is forward biased as shown in the diagram above
- Hence, current flows in the above circuit as indicated
- The diode D2 is reverse biased and hence no current due to D2
- We get output when the same is measured across the load resistance R
_{L}due to the diode D1 alone

**➤ Case 2**

- The circuit diagram for the negative half of the input cycle of AC voltage:
- During the positive half of the input cycle of AC voltage, the junction diode D2 is forward biased as shown in the diagram above.
- Hence, current flows in the above circuit as indicated.
- The diode D1 is reverse biased and hence no current due to D1.
- We get output when the same is measured across the load resistance R
_{L}due to the diode D2 alone

We observe that one of the diode conducts and the flow of current across the load resistance is in the same direction Also, current flows during both cycles of the input AC voltage. However, the output though unidirectional has ripple contents. Ripple contents indicate both AC and DC components

We can get only the DC component by passing it through a filter circuit

The filter circuit consists of Resistance and Capacitance

**➤ Circuit Diagram**

- C has a high capacitance value and serves as a filter circuit
- R
_{L}is a load resistance

**➤ Working**

- The capacitance offers low impedance to AC component and offers infinite impedance to DC component
- Due to this the AC component is bypassed or filtered out
- The DC component produces a voltage drop across the load resistance which is almost DC voltage

**Problem: **In half-wave rectification, what is the output frequency if the input frequency is 50 Hz. What is the output frequency of full wave rectifier for the same input frequency?**Solution:** Half-wave rectifier –The output voltage is obtained is once in one cycle of input voltage, hence output ripple frequency after half-wave rectification = 50 Hz**Full-wave rectifier – **For one cycle of input voltage, we get output twice in the same direction. Hence, the output after full wave rectification = 100 Hz

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